vectors in three-dimensional space and solid analytic geometr

Vectors inthree-dimensional spaceand solid analytic geometr

19.1 Rt, THETHREE-DIMENSIONAL

NUMBER SPACE

L8.1.L Definition

18.1 tr, THE THREE-DIMENSIONAL NUMBER SPACE 811

In Chapter 1 we discussed the number line Rl (the one-dimensionalnumber space) and the number plane R2 (the two-dimensional numberspace). we identified the real numbers in Rl with points on a horizontalaxis and the real number pairs in R2 with points in a geometric plane. In ananalogous fashion, we now introduce the set of all ordered triples of realnumbers.

The set of all ordered triples of real numbers is called the three-dimensionalnumber space and is denoted by Rt. Each ordered triple (x, y, z) is called a

point in the three-dimensional number sPace.

To represent R3 in a geometric three-dimensional space we considerthe directed distances of a point from three mutually perpendicularplanes. The planes are formed by first considering three mutually perPe,n-dicular lineJwhich intersect at a point that we call the origin and denote by

the letter O. These lines, called the coordinate axes, are designated as the r

axis, the y axis,and the z axis. Usually the r axis and the y axis arc taken in

a horizontal plane, and the z axis is vertical. A positive direction is selectedon each axis. If the positive directions are chosen as in Fig. 18.1.1, the coor-

dinate system is called a right-handed system. This terminology follows

from the fact that if the right hand is placed so the thumb is pointed in thepositive direction of the r axis and the index finger is pointed in the posi-

iive direction of the y axis, then the middle finger is pointed in the posi-

tive direction of the z axis.If the middle finger is pointed in the negativedirection ofthe z axis, then the coordinate system is called lefthanded. A

left-handed system is shown in Fig. 78.1.2. In general, we use a right-handed system. The three axes determine three coordinate planes: tl]re xyplane containing the x and y axes, the xz plane containing the r and z ixes,and the yz plane containing the y and z axes.

F igu re 18 .1 .1 Figu re 18J .2

An ordered triple of real numbers (x, y, zpoint P in a geometric three-dimensional space.

) is associated with eachThe directed distance of P

812 VECTORS IN THREE-DIMENSIONAL SPACE AND SOLID ANALYTIC GEOMETRY

( G,o, o)

F igu re 18 .1 .3

from the yz plane is called the x coordinate, the directed distance of p fromthe xz plane is called the y coordinate, and the z coordinate is the directeddistance of P from the xy plane. These three coordinates are called the rec-tangular cartesian coordinates of the point, and there is a one-to-one corre-spondence (called a rectangular cartesian coordinate system) between allsuch ordered triples of real numbers and the points in a geometric three-dimensional space. Hence, we identify R3 with the geometric three-dimensional space, and we call an ordered triple (r, A, z) a point. Thepoint (3, 2,4) is shown in Fig. 18.1.3, and the point (4, -2,-S) is shownin Fig. 18.1,.4. The three coordinate planes divide the space into eightparts, called octants. The first octant is the one in which all three coordi-nates are positive.

zA

(4, 0, 0)

(4,

;*t l(o,0, -5)

F igu re 18 .1 .4

A line is parallel toon the line to the plane

O ILLUSTRATION 1.: A l ine

plane, and one parallel torespectively.

plane if and only if the distance from any pointthe same.

parallel to the yz plane, one parallel to the xzthe xy plane are shown in Fig. 18. 1.5a,b, and c,

a

is

at '

v(o, 2, o)

aa'

Fig u re 1 8. '1 .5

18.1 R3, THE THREE-DIMENSIONAL NUMBER SPACE 813

We consider all lines lying in a given plane as being parallel to theplane, in which case the distance from any point on the line to the planeis zero. The following theorem follows immediately.

18.1.2 Theorem ( i ) A line is parallel to the yz plane if and only if all points on the

line have equal x coordinates.A line is parallel to the xz plane if and only if all points on the

line have equal y coordinates.A line is parallel to the xy plane if and only if all points on the

line have equal z coordinates.

( i i )

( i i i )

In three-dimensional space, if a line is parallel to each of two inter-, secting planes, it is parallel to the line of intersection of the two planes.

Also, if a given line is parallel to a second line, then the given line is paral-lel to any plane containing the second line. Theorem 18.1.3 follows fromthese two facts from solid geometry and from Theorem 18.1.2.

18.1.3 Theorem (i) A line is parallel to the r axis if and only if all points on the linehave equal y coordinates and equal z coordinates.

(ii) A line is parallel to they axis if and only if all points on the linehave equal r coordinates and equal z coordinates.

(iii) A line is parallel to the z axis if and only if all points on the linehave equal r coordinates and equal y coordinates.

. rLLUsrRArroN 2: A line parallel to the x axis, a line parallel to the y axis,and a line parallel to the z axis are shown in Fig. 'l'8.1.6a, b, and c, re-sPectivelY. '

Figure 18.1.6

The formulas for finding the directed distance from one point toanother on a line parallel to a coordinate axis follow from the definition ofdirected distance given in Sec. 1.4 and are stated in the following theorem.

lII VECTORS IN THREE-DIMENSIONAL SPACE AND SOLID ANALYTIC GEOMETRY

18.1.4 Theorem (i) If A(xr,y, z) and B(xr,y,z) are two points on a line parallel tothe r axis, then the directed distance from Ato B, denoted by B,is given by

A B : x z - x 1

If C(x, At, z) and D(x, Az, z) are two points on a line parallel to

W V axis, then the directed distance from C to D, denoted byCD, is given by

CD : Uz- Ar

If E(x, y, z) and F (x, y, zr) are two points on a line parallel to thez axis, then the directed distance from E to F, denoted by EF, isgiven by

E F : z z - z r

o rLLUsrRAtroN 3: The directed distance PQ from the point P(2,-5,-4)to the point Q(2, -3, -4) is given by Theorem 1S.1.4(ii). We have

The following theorem gives a formula for finding the undirected dis-tance between any two points in three-dimensional space.

18.1.5 Theorem The undirected distance between the two points Pr(xr, yr, zr) andPr(xr, Az, z) is given by

(i i)

(i i i)

B (x r ,A r , z r lPt (xt , At, Zr)

pRooF: Construct a rectangular parallelepiped havingposite vertices and faces parallel to the coordinate planes

By the Pythagorean theorem we have

INF;P:ITEI'+ I7F,I'Because

lmP: lF,f l ,+ l$l ,we obtain, by substituting from (2) into (l) ,

Applying Theorem 18.1.4(i), (ii), and (iii) to the right side of (3), weobtain

ffpr-;z : (xr- xr), I (Ar- yr), * (zz- zr)z

So

and the theorem is proved.

Pr and P, as op-( see Fig. '/-,8.'J-,.7).

( 1 )

(2)

----),//

i-.. A(x2, Uz , zt)

,)-----/--, / Pz ( x2 , y r , z r )

F igure 18.1.7

I

EXAMPLE

distanceP (-3, 4,

l.: Findbetween-1) and

18.1 F ' , THE THREE-DIMENSIONAL NUMBER SPACE 815

the undirectedthe points

Q(2, 5, -4) .

soLUTroN: From Theorem 18.L.5, we have

:m-{gs

Note that the formula for the distance between two points in R3 isfor the distance be-merely an extension of the corresponding formula

tween two points in R2 given in Theorem 1-4.I- It is

the undirected distance between two points x2 and

l x r - r r l :@

The formulas for the coordinates of the midpoint of a line segment arederived by forming congruent triangles and proceeding in a manner anal-

ogous to the two-dimensional case. These formulas are given in Theorem

18.'/-..6, and the proof is left as an exercise (see Exercise 15).

18.1.6 Theorem The coordinates of the midpoint of the line segment having endpoints

Pr(xr,Ur, zr) and Pr(xr,Az,z2) are given bY

also noteworthy that

x1 in Rl is given by

18.1.7 Definition

18.L.8 Definition

The graph of an equation in RB is the set of all points (x, y, z) whose coor-dinates are numbers satisfying the equation.

The graph of an equation in RB is called a surface. One particular sur-face is the sphere, which is now defined.

A sphere is the set of all points in three-dimensional space equidistantfrom a fixed point. The fixed point is called the center of the sphere and the

measure of the constant distance is called t}:re tadius of the sphere.

pRooF: Let the point (h, k,l) be denoted by C (see Fig. 18.1.8). The pointP(x, y, z) is a point on the sphere if and only if

lcnl: t

or, equivalently,

@ : 7

Squaring on both sides of the above equation, we obtain the desiredresult. I

18.1.9 Theorem An equation of the sphere of radius r and centet at (h, k,l) is

(4)

(5 - 4) '

c(h, k,l)

P(*, v, z)


If the center of the sphere is at the origin, then L :k:l: 0, and so anequation of this sphere is

* + y ' l 22 : 12

If we expand the terms of Eq. (a) and regroup the terms, we have

* * y' * z2 - Zhx - 2ky - 2lz * (h2 + P + 12 - rz) : 0

This equation is of the form

x2 + y',+ 22 + Gx * Hy * lz + I - 0

where G,H,l, and/ are constants. Equation (5) is called the general formofan equation of a sphere, whereas Eq. (4) is called the center-radius form.Because every sphere has a center and a radius, its equation can be put inthe center-radius form and hence the general form.

It can be shown that any equation of the form (5) can be put in theform

(x - h) , + (y - k) , + (z - t ) , : K

where

h - - *C k - - tU l : - i I K -+ (Gr+H2+ tz -41 )

It is left as an exercise to show this (see Exercise 16).If K > 0, then Eq. (5) is of the form of Eq. (4), and so the graph of the

equation is a sphere having its center at (h, k,l) and radius l'R.lt K: 0,the graph of the equation is the point (ft, k, l). This is called apoint-sphere.If K < 0, the graph is the empty set because the sum of the squares ofthree real numbers is nonnegative. We state this result as a theorem.

18.1.10 Theorem The graph of any second-degree equation in r, y, and z, of the form

* + y , + 2 2 + G x * H y * t z * J : g

is either a sphere, a point-sphere, or the empty set.

(5)

x

F igu re 18 .1 .8

EXAMPLE 2: Draw a sketch of thegraph of the equation

x 2 + y ' + 2 2 - 6 x - 4 y + 2 2 - 2

(6)

solurroNt Regrouping terms and completing the squares, we have

f - 6x -t 9 -f y2 - 4y + 4 * z2 * 2z * l: 2 * 9 + 4 + l

( r - 3 ) ' + 0 - 2 )2 * (z * 1 )2 : 15

So the graph is a sphere having its center at (3,2,-1) and radius 4. Asketch of the graph is shown in Fig. 18.1.9.

F igu re 18 .1 .9

18.1 R', THE THREE-DIMENSIONAL NUMtsER SPACE 817

solurroN: The center of the sphere will be the midpoint of the line seg-

ment AB. Let this point be C(7,y,2).By Theorem 78.'l'.6' we get

- 9 - 5 ^ - - 4 + 6 - a = - 0 - 2 - - . .i : - : : z t : T : L z : - - : - L

So C is the point (2,1,-l). The radius of the sphere is given by

r -

Therefore, from Theorem 18.L.9, an equation of the sphere is

( x - 2 ) ' + Q - r ) ' , + ( z * r )2 : 75

or, equivalently,

x2 + y', + z2 - 4x - 2y * 2z - 69 : 0

Exercises 18.1

In Exercises 1 through 4, the given points A and B are opposite v,ertices of a rectangular parallelepiped, hawing its faces

p"r*"r to the coordin"t" ptunlr. In each problem 1a) drari a sketch of the figure, 1b) find the coordinates of the other sii

vertices, (c) find the length of the diagonal AB.

EXAMPT,n 3: Find an equation of

the sphere having the PointsA ( -5 ,5 , -2 ) and B(9 , -4 , 0 ) as

endpoints of a diameter.

1 . A ( 0 , 0 , 0 ) ; B ( 7 , 2 , 3 )

3 . A ( -1 , l , 2 ) ; B (2 , 3 , 5 )

2 . A ( I , 1 , l ) ; B (3 , 4 , 2 )

4. A(2, -1, -g); B(4, 0, -l)

distance between the points A and B, and (b) the midpoint of the line

7 . A ( 2 , - 4 , 1 ) ; B ( t , 2 , 3 1

9. A(-2, -L,5); B(5, 1 , -4)

5. The vertex opposite one comer of a room is 18 ft east, 15 ft south, and 12 ft up from th9 first comer. (a) Draw a sketch

of the figure, (b) determine the length of the diagonal joining two opposite vertices, (c) find the coordinates of all eight

vertices of the room.

In Exercises 6 throu gh g,find (a) the undirected

segment joining A and B.

6 . A ( 3 , 4 , 2 ) ; B ( 1 , 6 , 3 )

8 . A(4 , -3 , 2 ) ; B( -2 , 3 , -5 )

10. prove that the three points (7, -1,3) , (2, L,7) , and, (4,2,6) are the vertices of a right triangle, and find its area'

11. A line is drawn through the point (6,4,2) perpendicular to the yz plane. Find the coordinates of the points on this

line at a distance of 10 units from the point (0,4,0)'

12. Solve Exercise 11 if the line is drawn perpendicular to the ry plane.

13. Prove that the three points (-3,2,4), (6,1,2), and (-12,9,5) are collinear by using the distance fonnula'

14. Find the vertices of the triangle whose sides have midpoints at (3,2,3), (-1,1,5), and (0,3,4)'

15. Prove Theorem 18.1.6'

15. Show that any equat ion of the form **y"*z2lGx' lHy* lz+1:0 can be put in theform (x-h\"+ (y-k) '

* ( z - I ) ' : X .

In Exercises 17 through 21, determine the graph of the given equation.

17. x2 + y' + 22 -8r * 4y * 2z -4 : 0 L8. x2 + y', + 22 - 8y * 5z - 25 : 0 1 9 . x 2 + y ' + z , 2 - 6 2 * 9 : 0

20.

In

22.

23.

24.

25.

26.


x 2 + y ' + 2 2 - x - y - 3 2 * Z - - 0 2L. x2 + y' + zz - 6x * 2y - 4z *L9 : 0Exercises 22 through 24, find, an equation of the sphere satisfying the given conditions.

A diameter is the line segment having endpoints at (6,2,-5) and (-4,0,2).

It is concentric with the sphere having equation f * y, * z2 - 2y I 8;z - ) : g.

It contains the points (0,0,4)', (2,1,2),and (0, 2,6) and,has its center in the yz plane.

Prove analytically that the four diagonals joining opposite vertices of a rectangular parallelepiped bisect each other.I!!: Q'4, T{ S are four points in three-dimensional space and A, B, C, and D are the midpoints of pe, eR, RS, andSP, respectively, prove analytically that ABCD is a parailelogram.

x

Figu re 18.2.1

18'2 VECTO_RS IN The presentation of topics in solid analytic geometry is simplified by theTHREE-DIMENSIoNAL SPACE use of vectors in three-dimensional ,p".". irr" a"fir,itions and theolms

given in Secs. 17.1 and17.2 for vectors in the plane are easily extended.

18.2.1 Definition A aector in three-dimensional space is an ord.ered triple of real numbers(x, y, zl . The numbers tt, y, and z are called the componezlfs of the vector(x, y, zl.

We let Vs be the set of all ordered triples (x, y, zl for which x, y, and.zare real numbers. In this chapter, a vector is always in v3 unless otfrerwisestated.

Just as forvectors inv2, a vector in v, can be represented by a directedline segment. If A : (ar, a2, asl, then the directed line segmeni having itsinitial point at the origin and its terminal point at the polnt (ar, ar, af,) is:{td- the position representation of A. A directed line segment travinf itsinitial point at (x, y, z) and its terminal point at (x * ar, y * a2, z+ ar) isalso a representation of the vector A. See Fig. 1g.2.1.

The zero aector is the vector (0, 0,0) and is denoted by 0. Any point isa representation of the zero vector.

- The magnitude of a vector is the length of any of its representations. Ifthe vector A: (ar, ez, fls), the magnitude of A is denotei by lAl, and itfollows that

Wl: t/ar'177 t, ogThe direction of anonzero vector in v, is given by three angles, called

the direction angles of the vector.

The direction angles of a nonzero vector are the three angles that have thesmallest nonnegative radian measures a, B, andT measuied from the posi-tive x, A, and z axes, respectively, to the position representatio' o] thevector.

The radian measure of each direction angle of a vector is greater than

( r * a r , y * o r , z * a s )

(x, v, z)

18.2.2 Definition

18.2 VECTORS IN THREE.DIMENSIONAL SPACE 819

or equal to 0 and less than or equal to zr. The direction angles having radian

meaiu"es ot, F, andT of the vector A: (ar, az, asl are shown in Fig' 18'2'2'

In this figure the components of A are all positive numbers, and the direc-

tion angles of this .r"itot all have positive radian measure less than *2.From thl figure we see that triangle POR is a right triangle and

lml a1C O S C : _ _ : _

loPl lAlIt can be shown that the same formula holds if. Ln = q. < 7r. Similar

formulas can be found for cos B and cos 7, and we have

The three numbers cos a, cos B, and cos 7 are called the direction cosines of

vector A. The zero vector has no direction angles and hence no direction

cosines.

. rLLUsrRArroN 1: We find the magnitude and direction cosines of the

vector A : (3, 2, -6>.

lA l : m : !9T4+35: t /4s -7

From Eqs. (1) we get

cosa -+ cosF :? cosy : -+

If we are given the magnitude of a vector and its direction cosines, the

vector is uniqud determined because from (1) it follows that

n r : lA l cos a nz: lA l cos B ns: lA l cos 7

The three direction cosines of a vector are not independent of each

other, as we see by the following theorem'

1'8.2.3 Theorem If cos a, cos B, and cos 7 are the direction cosines of a vector, then

c o s z c * c o s 2 F * c o s z 7 : 1

pRooF: If A : (a1, az,a.), then the direction cosines of A are given by (1)

and we have

( 1 )

Figure 18.2.2

(2)

cos2 d+cosz P+cos2 y:f f i*#i . f f i

I- 1

VECTORS IN THREE-DIMENSIONAL SPACE AND SOLID ANALYTIC GEOMETRY

o rLLUsrRArroN 2: we verify Theorem 1.8.2.3 f.or the vector of Illustration 1.We have

'cos2 a+ cosz B+ aG i :

The vector A: (ar, ar, a"l is a unit vector if lAl : t, and from Eqs. (1)we see that the comPonents of a unit vector are its direction cosines.

The operations of addition, subtraction, and scalar multiplication ofvectors in Vs are given definitions analogous to the corresponding defini-tions for vectors in V2.

18.2.4 Def ini t ion I fA: (ar,nz,ar) andB:(br,br,br l , thenthesumof thesevectorsisgivenbv

A+ (a, * br , f lz * br , f ls * br)

EXAMPLE 1: Given A - (5, -2, 6>and B - (8, -5, -4), f ind A + B.

so lu r roN: A * B : (5 + 8 , ( -2 ) + ( -S) ,5 + ( -4 ) ) : (13 , -7 ,2>.

The geometric interpretation of the sum of two vectors in v3 is similarto that for vectors in Vr. See Fig. 18.2.3. If p is the point (x, y, z), A:(a1, a2, ar) and

"? tr a representation of A, then e is the point

(x * ar, A * az, z * ar).Let B : (br, br,b3) and let e? be a representationof B . Then R is the po in t ( r * (a1*br ) , A- f @r*br ) , z * (a r *b" ) ) .Therefore, p? is a representation of the vector A + B, and the parallel-ogram law holds.

x Figure 18.2.3

lf A: (ar, ar,ar), then the vector (-ar,-az,-ar) is defined to be thenega-tioe of. A, denoted by -A.

L8.2.5 Definition

18.2 VECTORS IN THREE-DIMENSIONAL SPACE 821

18.2.6 Definition The difierence of the two vectors A and B, denoted by A - B, is defined by

A - B : A * ( - B )

From Definitions 18.2.5 and 1'8.2.6 it follows that if A : (ar, ar, asl and

B: (b r ,b r ,b " l , then-B: ( -b r , -b2 , -b" l and

A - B : \a r - b1 , a2- b2 , asbs l

ExAMP rn 2: For the vectors A and

B of Example 1, find A - B.

SOLUTION:

A - B - ( 5 , 6> - (8,- (5 , -2 ,6> +: (-3 , 3, 10)

The difference of two vectors in Va is also interpreted geometrically as

it is in vz. see Fig. 1,8.2.4.4 representation of the vector A - B is obtained

by choosing representations of A and B having the same -initial point.

Tilen a ,"pi"r"-t t"tion of the vector A - B is the directed line segment

from the terminal point of the representation of B to the terminal point of

the representation of A.

o rLLUsrRAuoN 3: Givgl the points P(L,3,5) and Q(2,-L, a)' Fig- 19"2'5shorys p-i as well as O-P and O0. We see from the figure that V(PQ):

V(OQ) - v(OP). Hence,

v(t [) : (2, -1, 4> - <7,3, 5) : (L, -4, -L> '

-4>

4>5 ,

x

Figu re 18.2.4

P(L, 3, 5)

QQ, -L ,4 )

Lo__-i

Figu re 18.2.5

If c is a scalar and A is the vector (ar, fl2, as) ,

denoted by cA, is a vector and is given by

cA - c (a1 , f l2 , f lB ) - (car , c f l z , c f l s )

18.2.7 Definition then the product of c and A,


EXAMPrc 3: Given A -

(-4,7 , -2), f ind 3A and -5A.SOLUTION:

3A : 3(-4,7 , -2) - (-12, 2l , -5>

and-5A : (-S) (-4 ,7 , -2> - (20, -gS, 10)

Suppose thatcosines cos 0, cos(car, caz, c f ls) ; andcA, we have from

cos 01 cat: TrAJ

or/ equivalently,

cos cl

A - (ar, flz, azl is a nonzero vector having directionP, and cos y, and c is any nonzero scalar. Then cA :if cos o1, cos Fr, and cos yr are the direction cosines ofEqs. (1)

from which we get

cos a cos cos B cos

So if c > 0, it follows from Eqs. (3) that the direction cosines of vector cAare the same as the direction cosines of A. And if c < 0, the directioncosines of cA are the negatives of the direction cosines of A. Therefore, weconclude that if c is a nonzero scalar, then the vector cA is a vector whosemagnitude is lcl times the magnitude of A. If c > 0, cA has the same direc-tion as A, whereas if c < 0, the direction of cA is opposite that of A.

The operations of vector addition and scalar multlplication of anyvectors in Vr satisfy properties identical with those given in Theorem17.2.7. These are given in the following theorem, and the proofs are left asexercises (see Exercises 1 through 6).

18.2.8 Theorem If A,B, and C are any vectors in V3 and c and d are any scalars, then vectoraddition and scalar multiplication satisfy the following properties:

(i) A + B - B + A (commutative law)(ii) A + (B + c) : (A + B) + c (associative law)

(iii) There is a vector 0 in Vs for which

A+0:A (existence of addit ive identity)(iv) There is a vector -A in V, such that

A+( -A) -o (existence of negative)

(v) (cd)A - c(dl) (associative law)(vi) c(A + B) - cA * cB (distributive law)

(vii) (c * d)A: cA + dA (distributive law)(viii) 1(A) - A (existence of scalar multiplicative identity)

c a 1-FI IAT

ccos ot : IZI

cTr:1;1 cos 7 (3)

EXAMP rn 4: Express the vector of

Illustration L in terms of its mag-

nitude and direction cosines.

18,2 VECTORS IN THREE-DIMENSIONAL SPACE 823

realFrom Definition 17.2.2 and Theorem

'1.8.2.8 it follows that Vs is

vector space. The three unit vectors

i : ( 1 , 0 , 0 ) i : ( 0 , 1 , 0 ) 1 : ( 0 , 0 , 1 )

form a basis for the vector space V, because any vector (ar, az, at) can be

written in terms of them as follows:

(a r , a r , as l : a t (L ,0 , 0 ) + a r (o ,1 ,0 ) * c r (O, 0 , 1 )

Hence, if A : (ar, az, as), we also can write

A: ai* a" i * ask (4)

Because there are three elements in a basis, Vs is a three-dimensional

vector sPace.Substituting from Eqs. (2) into Eq. (4), we have

A: lAl cos ci + ll,l cos Bi + lAl cos 7k

or, equivalently,

A: lAl(cos ai * cos Bi * cos 7k) (s)

Equation (5) enables us to express any nonzero vector in terms of its

magnitude and direction cosines.

solurroN: In Illustration !, A: ( 3, 2, -6> , lAland cos y : -+,Hence, from Eq. (5) we have

A-7(? i++ i -+k)

cos a - ? , cos F :? ,

18.2.9 Theorem If the nonzero vector A: ai* a2i * ark, then the unit vector U having the

same direction as A is given bY

The proof of Theorem "l'8.2-9

17.2.3 for a vector in V2 and is leftis analogous to the Proof of Theorem

as an exercise (see Exercise 30).

EXAMPLE 5: Given the PointsR (2 , -1, 3) and S (3 , 4 ,6) , f ind the

unit vector having the same direc-

tion as V(ffi).

SOLUTION:

v( f f ) - <3, 4, 5> - (2, - r , 3) - (1, 5, 3>- i+5 i+3k

So-->

l v tns) l - f f i - \B

VECTORS IN THREE.DIMENSIONAL SPACE AND SOLID ANALYTIC GEOMETRY

Therefore, by Theorem L8.2.9 the desired unit vector is

u__1 . , 5 . , 3.F i r@i1155k

Exercises 1-8.21.. Prove Theorem 18.2.8(i). 2. prove Theorem 1S.2.S(ii).

3. Prove Theorem 18.2.8(iii), (iv), and (viii). 4. prove Theorem 19.2.g(v).

5. Prove Theorem 18.2.8(vi). 5. prove Theorem lg.2.g(vii).

I n E x e r c i s e s T t h r o u g h l S , l e t A : ( 1 . , 2 , 3 ) , 8 : ( 4 , - 3 , - 1 ) , C : ( - 5 , - g , S ) , D : < - 2 , 1 , 6 > .

7. Find A + 58

10. Find 48 + 6C - 2D

1 3 . F i n d C + 3 D - 8 A

16. F ind lA lc - lB lD

sum of their squares is L.

L9. P, (3 , -L , -4) ) Pz(7 , 2, 4)

21. Pr(4, -3, -L); Pr(-z, -4, -8)

8. Find 2A - C

1,1,. Find lTCl - lsDl14. Find 3A - 28 + C - tZD

9. Find 7C - 5D

12. Find l4Bl + l6cl - l2Dlls . F ind lA l ln l (c - D)

17. Find scalarc a and. b such that a(A + B) + b(C + D) : 0.

18. Find scalars a, b, artd c such that aA + bB + cC: D.

In Exercises L9 through 22, find. the direction cosines of the vector Vtfrl and check the answers by verifying that the

20. P, (1 , 3 , 5) ; Pr(z, -1 , 4)

22. P,( -2, 5 , 5) ; Pr(2, 4, ! )

23. Using the points P, and P, of Exercise 19, find the point e such trrat v(p?r) :3v(&a).

24. Using the points P, and P, of Exercise 20, find the point R such ttrat v(fr) : -2v(p?).

In Exercises 25 through 28, express the given vector in terms of its magnitude and direction cosines.E . - 6 i + 2 j + 3 k 2 5 . 2 i _ 2 i + k

27 . -2 i+ i -3k 28 .3 i+ 4 i_5k

29. lf the radian measure of each direction angle of a vector is the same, what is it?

30. Prove Theorem 18.2.9.

In Exercises 31 and 32, find the unit vector having the same direction as v1p,pr;.

31 . P {4 , -1 , -6 ) ;P , (5 ,7 , -2 ) 22 . pJ -8 , -5 ,2 ) ;p , ( -3 , - s ,4 )

33. Three vectorts in Vr are said to be independent if and. only if their position representations do not lie in a plane, andthree vectors Et, Ez, ar.d' E3 are said to form a basis for the vector space Vr if and only if any vector in V, can be writtenas a linear combination of E1, E2, and Eg. A theorem can be prorred which states thai three vectors form a basis for thevector sPace Vr if they are independent. Show that this theorem holds for the three vectors (1, O, O>, (1, 1, 0), and(1, 1, 1) by doing the following: (a) Verify that the vectors are independent by showing that their position represen-tations are not coplanar; (b) verify that the vectors form a basis by showing that any vector A can be written

18.3 THE DOT PRODUCT tN V3 825

r ( 1 , 0 , 0 ) * s ( 1 , 1 , 0 ) + t ( ' l . , l , l >

where r, s, and t are scalars. (c) If A : <6, -2,5), find the particular values of. r, s, and f so that Eq. (6) holds.

34. Refer to the first sentence of Exercise 33. A theorem can be proved which states that three vectors form a basis for thevector space V3 only if they are independent. Show that this theorem holds for the three vectors Ft: (1,0,7>,Fz:

<1,l,lr, and Fg : (2,l,2) by doing the following: (a) Verify that Fr, F2, and F3 are not independent by showing thattheir position representations are coplanar; (b) verify that the vectors do not form a basis by showing that every vec-tor in V3 cannot be written as a linear combination of Fr, Fz, and F3'

(6 )

1.8.3 THE DOTPRODUCT IN V3

L8.3.1 Definition

The definition of the dot product of two vectors in V3 is an extension of thedefinition for vectors in V2.

(br, br, brl , then the dot product of A and B,by

o rLLUSrRArroN 1 . : I f A : (4 ,2 , -6 ) l and B - ( -5 , 3 , -2 ) , then

A . [ - (4 ,2 , -6> . ( -5 ,3 , -2>- 4(-5) + 2(3) + (-6) (-2)

- - 20+6+12

- - nt

For the unit vectors i, j, and k, we have

i ' i : j ' i : k ' k : 1

and

i ' i : i ' k : j ' k : 0

Laws of dot multiplication that are given in Theorem 18.3'2 are thesame as those in Theorems 17.3.2 and17.3.3 f.or vectors in Vr. The proofsare left as exercises (see Exercises 1 through 4).

18.3.2 Theorem If A,B, and C are any vectors in V3 and c is a scalar, then

fl2,

Atiiiiri+

( i ) A ' B : B ' A( i i ) A ' ( B + C ) : A

( i i i ) c(A ' B) : (cA)( i v ) 0 ' A : 0( " ) A ' A : l A l '

(commutative law)' B + A' C (distributive law). B

Before giving a geometric representation of the dot product for vectorsin Vr, we do as we did with vectors in Vz. We define the angle between twovectors and then express the dot product in terms of the cosine of theradian measure of this angle.


18.3.3 Definition Let A and B be two nonzero vectors in V3 such that A is not a scalar mul-

tiple of B. If O? is the position representation of A and O? is the positionrepresentation of B, then the angle between the vectors A and B is defined

to be the angle of positive measure between O? and Oi ir,t""io, to the tri-angle POQ. If A: cB, where c is a scalar, then if c ) 0, the angle betweenthe vectors has radian measure 0, and if. c < 0, the angle between thevectors has radian measure n'.

Figure 18.3.1 shows the angle of radian measure 0 between the twovectors if A is not a scalar multiple of B.

18.3.4 Theorem If 0 is the radian measure of the angle between the two nonzero vectors Aand B in V3, then

x

F igu re 18 .3 .1

EXAMPLE 1: Given the vectorsA - 6 i - 3 i + 2k and B:2 i+ i

- 3k, find (a) the componentof B in the direction of A, (b) thevector proiection of B onto A; and(c) cos 0 if g is the radian measureof the angle between A and B.

The proof of Theorem17.3.5 for vectors in V2 and

If U is a unit vector in

( 1 )

1,8.3.4 is analogous to the proof of Theoremis left as an exercise (see Exercise 15).the direction of A, we have from (1)

- 7. Hence , a unit

the direction of A is

U . B : lUl l r l cos 6: lBl cos 0

As with vectors in Vr, lBl cos d is the scalar projection of B onto A and thecomponent of B in the direction of A. It follows from Theorem 18.3.4 thatthe dot product of two vectors A and B is the product of the magnitude,lAl, of one vector by the scalar projection, lBl cos 0, of the second vectoronto the first.

solurroN: lAl : :mvector in the direction of A is

u- + i -+ i +?kBecause U . f, - lBl cos 0, the component of B in

u ' t - S'_ 11 ;

+k) ' (2i+ i - 3k)

- 3_ T

The vector proiection of B onto A is therefore

tU: +8i - zti + abk

From Eq. (1)

c o s o = A ' B t I ' B=lAIlBT:lBT-

ExAMPtn 2: Find the distancefrom the point P (4,1,6) to theline through the points A (8 ,3 ,2)a n d B ( 2 , - 3 , 5 ) .

18.3 THE DOT PRODUCT tN V3

BecauseU . f , : t and ln l : : f f i

cos Q- 3 _ 3\m=ffi: gg

soLurroN: Figure 18.3.2 shows the point P and a sketch of the linethrough A and B. The point M is the foot of the perpendicular line from Pto the line through A and B. Let d units be the distance lF[1. fo find d wefind lffil and lZFl an{use the Pythagorean theore*. lIFl is the magni-tude of the vector V(AP).

v(A?): v(&) -v(A): (4 ,1 ,6> - <8,9 ,2>: (_4, _2, 4>

Hence,

lZPl :@TTR:f f i :5To find lMl we find the scalar projection of V(;F) onto V(A-i).

v(AT):v(ob)-v(o7): (2 , -1 ,5 ) - (8 ,3 ,2 ): (-6, -6 ' 3>

The scalar projection of v(a-F) onto V(Ai) is then

v(,4Ft . v(aEr _ (-4,-2, 4) . (-6,-6, g')

V(;;) | v36 + 36 +e

24+ 12 + 12\ET

_+1

Thus, r^r):

: f f i-6\ffi

:3 \ f f i

The definition of parallel vectors in V3 is analogous to Definition17.3.6for vectors inV2, and we state it formally.

18.3.5 Definition Two vectors in V3 are said to be parallel if. and only if one of the vectorsis a scalar multiple of the other.

B (2 , - 3 ,5 )

a P(4,L,6)

x

Figure 18.3.2

828

ExAMPln 3: Prove by usingvectors that the points A(4,9 , L) ,B ( -2, 5 ,3) , and C(6 ,3 , -2) arethe vertices of a right triangle.


The following theorem follows from Definition 18.3.5 and Theorem18.3.4. Its proof is left as an exercise (see Exercise 16).

18.3.6 Theorem Two nonzero vectors in Vs are parallel if and only if the radian measureof the angle between them is 0 or zr.

The following definition of orthogonal vectors in Vr corresponds toDefinition 17.3.7 f.or vectors in Vr.

18.3.7 Definition If A and B are two vectors in Vs, A and B are said to be orthogonal if ando n l v i f A . B : 0 .

solurroN: Triangle CAB is shown in Fig. 18.3.3. From the figure it looksas if the angle at A is the one that may be a right angle. We shall find

V(AB) and V(AC), and if the dot product of these two vectors is zero, theangle is a right angle.

-+ -+

v (AB) -v (oB) -v (oA)- ( -2 ,6 ,3 ) - (4 ,9 , ' ] . ,>: (_5, _3, 2>

v(A?) -v(oa) -v(o?)- (6 ,3 , -2 ) - <4 ,9 , ' j . ,>

: ( 2 , - 5 , - 3 )

v( f r ) . v( f r ) - F6,-g, 2> . <2,-6,-g) - -12Therefore; V(AB) and V(;E) are orthogonal, and so theangle CAB is a right angle.

+ L8 - 6 - 0

angle at A in tri-

Exercises 1-8.31. Prove Theorem 18.3.2(i). 2. Prove Theorem 18.3.2(ii).

3. Prove Theorem 18.3.2(iii). 4. Prove Theorem 18.3.2(iv) and (v).

I n E x e r c i s e s 5 t h r o u g h 1 4 , l e t A : ( - 4 , - 2 , 4 r , 8 : < 2 , 7 , - l t , C : < 5 , - 3 , 0 ) , a n d D : < 5 , 4 , - 3 > .

5 . F i n d A ' ( B + C )

8 . F i n d A ' D - B ' C

11. Find the cosine of the measure

L2. Find the cosine of the measure

13. Find (a) the component of C in

6 . F i n d A ' B + A ' C

9. F ind (B ' D)A - (D ' A)B

of the angle between A and B.

of the angle between C and D.

7. Find (A ' B) (C

10. Find (2A + 3B)

. D )

. ( 4 C - D )

B ( - 2 , 6 , 9 )

A(4, 9, L)

C(6, g, - z)

Figure 18.3.3

the direction of A and (b) the vector proiection of C onto A.

14.

15.

17.

18.

19.

20.

2L.

22.

18.4 PLANES

Find (a) the component of B in the direction of D and (b) the vector projection of B onto D.

Prove Theorem 18.3.4. 15. Prove Theorem 18.3.5.

Provebyusingvectorsthat thepoints (2,2,2) , (2,0,7) , (4,1, - l ) ,and (4,3,0) arethever t icesof arectangle.

Prove by using vectors that the points (2,2,2), (0,1,2), (-1,3,3), and (3,0, L) are the vertices of aparallelogram.

Find the distance from the point (2, -t,-4) to the line through the points (3,-2,2) and (-9,-5,6).

Find the distance from the point (3,2,1) to the line through the points (1,2,9) and (-3, -5,-3).

Find the area of the triangle having vertices at (-2, 3, l) , (l, 2, 3) ,,md (3, -l, 2).

If a force has the vector representation F : 5i - 3k, find the work done by the force in moving an object from the pointP J4, | ,3) along a straight line to the point Pr(-S , 6 , 2) . The magnitude of the force is measured in pounds and dis-tance is measured in feet. (nrxr: Review Sec. 17.3.)

23. A force is represented by the vector F, has a magnitude of 10 lb, and direction cosines of F are cos d : tV6 andcos B : +\/6. lf the force moves an object from the origin along a straight line to the point (7 , -4,2) , find. the workdone. Distance is measured in feet. (See hint for Exercise 22.)

24. l fAandBarenonzerovectors,provethat thevectorA-cBisor thogonal toBi fc :A 'B/ lB l 'z .

25. If A :l2iI 9i - 5k and B:4i + 3i - 5k, use the result of Exercise 24 to find the value of the scalar cso that the vec-tor B - cA is orthogond to A.

26. For the vectors of Exercise 25, use the result of Exercise 24 to find the value of the scalar d so that the vector A - dB

is orthogonal to B.

27. Prcve that if A and B are any vectors, then the vectors lBlA + lAlB and lBlA - lAlB are orthogonal.

28. Prove that if A and B are any nonzero vectors and C: lBlA + lAlB, then the angle between A and C has the samemeasure as the angle between B and C.

18.4 PLANES The graph of an equation in two variables, r and !, is a curve in the xy' plane. The simplest kind of curve in two-dimensional space is a straight

line, and the general equation of a straight line is of the form Ax * By* C:0, which is an equation of the first degree. In three-dimensionalspace, the graph of an equation in three variables, x, y, and z, is a surface.The simplest kind of surface is aplane, and we shall see that an equation ofa plane is an equation of the first degree in three variables.

1g.4.1 Definition If N is a given nonzero vector and Pe is a given point, then the set of allpoints P for which V(P'P) and N are orthogonal is defined to be a planethrough Ps having N as a normal aector.

Figure 18.4.1 shows a portion of a plane through the pointPo(ro, Ao, zo) and the representation of the normal vector N having itsinitid point at Ps.

In plane analytic geometry we c€rn obtain an equation of a line if weare given a point on the line and its direction (slope). In an analogousmanner, in solid analytic geometry an equation of a plane can be deter-


mined by knowing a point in the plane and the direction of a normalvector.

L8.4.2 Theorem If.Ps(xs, yo, zo) isapoint in a plane and a normalvector to the plane isN :(a, b, c) , then an equation of the plane is

pRooF: Refer to Fig. '1,8.4.'1,. Let P(x, !, z) be any point in

v(4?) is the vector havin g P;P as a representation, and so

V(P;P) - (x - xo, a - Uo, z- zo)

From Detinitions 1,8.4.1 and 18.3.7, it follows that

( 1 )

the plane.

V ( P g P ) . N : 0

Because N : (a, b, c>, from (2) and the above equation we obtain

a ( x - x s ) + a Q - A ) * c ( z - z o ) : 0

which is the desired equation.

(2)

+ y

(3)

IF igure 18 .4 .1

ExAMPLE L: Find an equation ofthe plane containing the point(2, 1, 3)and having 3i - 4i + kas a normal vector.

sot,urroN: Using (1) with(a, b, c l : (3, -4, ' ] - .>, we

3 (x -2 ) -4 ( v -1 )+

or/ equivalently,

3x -4y *z -5 :0

the point (xo, Uo,zo) -

have as an equation

( z - 3 ) : 0

(2,I , 3) and the vectorof the required plane

L8.4.3 Theorem If. a, b, and c are not all zero, the graph of an equation of the form

a x * b y * c z * d : 0 ( 4 )

is a plane and (a, b, cl is a normal vector to the plane.

PRooF: Suppose thatb * 0. Then the point (0,-dlb,0) is on the graph ofthe equation because its coordinates satisfy the equation. The given equa-tion can be written as

/ s \a(x - o) + b \a + t) + c(z - o) : 6

which from Theorem 18.4.2 is an equation of a plane through the point(0, -dlb,0) and for which (a, b , cl is a normal vector. This proves the the-orem if b # 0. A similar argument holds if b:0 and either a * 0 orc # 0 . r

Equations ( 1) and (4) are called cartesian equations of a plane. Equa-

18.4 PLANES 831

Figu re 18.4.2

tion (1) is analogous to the point-slope form of an equation of a line intwo dimensions. Equation (4) is the general first-degree equation in threevariables and it is called a linear equation.

A plane is determined by three noncollinear points, by a line and apoint not on the line, by two intersecting lines, or by two parallel lines. Todraw a sketch of a plane from its equation, it is convenient to find thepoints at which the plane intersects each of the coordinate axes. The rcoordinate of the point at which the plane intersects the r axis is called thex intercept of the plane ; the y coordinate of the point at which the plane in-tersects the y axis is called the y intercept of the plane; and the z intercept ofthe plane is the z coordinate of the point at which the plane intersects the zaxis.

. rLLUsrRArroN 1: We wish to draw a sketch of the plane having theequation

2 x - l 4 Y * 3 z : 8

By substituting zero for y and z, we obtain .x : 4; so the r intercept ofthe plane is 4. In a similar manner we obtain the y intercept and the zintercept, which are 2 and $, respectively. Plotting the points corre-sponding to these intercepts and connecting them with lines, we have thesketch of the plane shown in Fig. 18.4.2. Note that only a portion of theplane is shown in the figure. .

I rLLUSrRArrox 2: To draw a sketch of the plane having the equation

3 x * 2 Y - 6 2 : 0

we first notice that because the equation is satisfied when r, y, and zare all zero, the plane intersects each of the axes at the origin. If we setr: 0 in the given equation, we obtain y - 32: 0, which is a line in theyz plane; this is the line of intersection of the yz plane with the givenplane. Similarly, the line of intersection of the xz plane with the givenplane is obtained by setting A:0, and we get r-22:0. Drawing asketch of each of these two lines and drawing a line segment from a pointon one of the lines to a point on the other line, we obtain Fig. 18.4.3. .

In Illustration 2 the line in the yz plane and the line in the xz planeused to draw the sketch of the plane are called the traces of the given planein the yz plane and the xz plane, respectively. The equation r : 0 is anequation of the yz plane because the point (x, y , z) is in the yz plane if andonly if r : 0. Similarly, the equations y : 0 and z:0 are equations of thexz plane and the xy plane, respectively.

A plane parallel to the yz plane has an equation of the for:n x: k,where k is a constant. Figure 18.4.4 shows a sketch of the plane having theequation x: 3. A plane parallel to the xzplarre has an equation of the form

A : k, and a plane parallel to the xy plane has an equation of the form

Figure 18.4.3

Figu re 18.4.4


z: k. Figures 1.8.4.5 and 18.4.6 show sketchesequations y: -5 and z- 6, respectively.

of the planes having the

Figure 18.4.5 Figure 18.4.618'4'4 Definition The angle between two planesis defined to be the angle between the nonhal

vectors of the two planes.

EXAMPTB 2: Find the radian mea-sure of the angle between the twoplanes 5x- 2y * 5z- l2:0 and2 x * y - 7 2 * 1 1 : 0 .

solurroN: Let N1 be a normal vector to the first plane and N1 : Si-2i* 5k. Let N, be a normal vector to the second plane and N2 : 2i + i

- 7k.By Definition 78.4.4 the angle between the two planes is the angle

between N1 and N2, and so by Theorem 18.3.4 if d is the radian measure ofthis angle,

c o s o - N ' ' N '

:lN , l lN , l f f i f f i

Therefore,

o - &n

-27 15 4 2

18'4'5 Definition Two planes are parallel if and only if their normal vectors are parallel.

From Definitions 18.4.5 and 18.3.5,planes with equations

a r x * b r y + c f i + d t : 0

nzx + bzA * c2z * dr- 0

it follows that if we have two

Two planes areorthogonal.

18.4 PLANES

and normal vectors Nr : (4r, bt, ct) and Nz: (az, br, crl , respectively, thenthe two planes are parallel if and only if

N1 : kN2 where k is a constant

Figure 18.4.7 shows sketches of two parallel planes and representations ofsome of their normal vectors.

perpendicular if and only if their normal vectors are

From Definitions L8.4.6 and 18.3.7 it follows that two planes havingnormal vectors Nr and N, are perpendicular if and only if

N 1 ' N 2 : 0 Q )

soLUrIoN: Let Mbe the required plane and (a,b, cl be a normal vectorofM.Let Mr be the plane having equation te - y + z: 0.By Theorem 18.4.3 anormal vector of. Mr is <1, -l ,1). Because M and M, are perpendicular, itfollows from Eq. (7) that

( a , b , c l ' < 1 , - 7 , 1 ) : 0

or, equivalently,

a - b + c : 0

Let M2 be the planevector of Mz is (2 , l,

(a , b , c> ' (2 , L ,

or, equivalently,

2a *b -4c :0

L8.4.6 Definition

ExAMPLE 3: Find an equation ofthe plane pe{pendicular to each ofthe planes x- y + z - 0 and 2x* Y- 4z- 5 - 0 and containing thepoint (4, 0 , -2) .

(8)

having the equation 2x * y - 4z - 5: Q. A normal-$. Because M and M2 are PerPendiculat, we have

-4 ) :o


Solving Eqs. (8) and (9) simultaneously for b andb - 2a and c: a. Theref.ote, a normal vectot of M(4, 0 , -2) is a point in M, it follows from Theoremo f M i s

a(x - 4) + za(y - 0) * a(z + 2)

or, equivalently,

x *2Y*z -2 :0

Consider now the plane having the equation ax * by + d: 0 and thexy plane whose equation is z:0. Normal vectors to these planes are(a, b,0) and (0, O, 1.>, respect ively. Because (a, b,0) ' (0, 0, 1):0,the two planes are perpendicular. This means that a plane having anequation with no z term is perpendicular to the xy plane. Figure 18.4.8illustrates this. In a similar manner, we can conclude that a plane havingan equation with no r term is perpendicular to the yz plane (see Fig.1'8.4.9), and a plane having an equation with no y term is perpendicularto the xz plane (see Fig. 18.4.10).

An important application of the use of vectors is in finding the un-directed distance from a plane to a point. The following example illustratesthis.

c in terms of a, we getis (a, 2a, a>. Because

"1,8.4.2 that an equation

Figu re 1 8.4.9 F igu re 18 .4 .10

soLUrIoN: Let P be the point ('1,,4,5) and choose any point Q in theplane. For simplicity, choose the point Q as the point where the plane in-tersects the r axis, that is, the point (-5, 0, 0). The vector having QP as arepresentation is given by

v(a?) :6ir 4i + 6kA normal vector to the given plane is

N :2 i - i+zk

Figure 18.4.8

EXAMPI^.S 4: Find the distance

from the plane 2x - y + 2z * I0- 0 to the point (l , 4, 5) .

P(1 , 4 ,6 )

18.4 PLANES

The negative of N is also a normal vector to the given plane and

- N : - 2 i + i - 2 k

We are not certain which of the two vectors, N or-N, makes the smaller

angle with vector V(QIF). Let N' be the one of the two vectors N or -N

which makes an angle of radian measure 0 <Ln with V(Q]). In Fig.18.4.11 we show a portion of the given plane containing the point

Q(-5,0,0), the representation of the vector N' having its initial point

at Q, the point P(1., 4,6), the directed line segment Q?, and the pointR, which is the foot of the perpendicular from P to the plane' For sim-plicity, we did not include the coordinate axes in this figure. The dis-

tance lR?l is the required distance, which we call d. We see from Fig.18.4.11 that

d: lv(e=-P)l cos o (10)

Because g is the radian measute of the angle between N' and V(Q?),we have

- 5 r 0 , 0 )

( 1 1 )

Substituting from (1,1) into (10) and replacing lN'l by lNl, we obtain

lv(d l l (N' . v tOFl llNl lv(aF) l

Because d is an undirected distance, it is nonnegative; hence, we canreplace the numerator in the above expression by the absolute value of the

dot product of N and v(QF). Therefore,

Iw .v r& i l l (z i - i + 2k) . (5i + 4i + 6,k) | _ 20ffi

Exercises 78.4

In Exercises 1 through 4, find an equation of the plane containing the given point P and having the given vector N as a

normal vector.

L . P(3 , L , 2 ) ; N : (L , 2 , -3 )

3 . P (2 , t , - 1 ) ; N - - i + 3 i + 4k

In Exercises 5 and 6, find an equation of the

5 . ( 3 , 4 , L ) , ( L , 7 , L ) , ( - 1 , - 2 , 5 )

2 . P ( - ! , 8 , 3 ) ; N - ( - 7 , - t , l >

4 . P ( l , 0 , 0 ) ; N - i + k

plane containing the given three points.

6 . ( 0 , o , 2 ) , ( 2 , 4 , L ) , ( - 2 , 3 , 3 )

I


12, draw a sketch of the given plane and find two unit vectors which are normal to the plane.In Exercises'7 through

7 . 2 x - y + 2 2 - 5 : 0

1 0 . y + 2 2 - 4 : 0

8 . 4 x - a y - 2 2 - 9 : 0

1 . 1 . . 3 x * 2 2 - 5 : 0

9 . 4 x * 3 y - l 2 z - 0

L 2 . z : 5

In Exercises 13 through 17, find an equation of the plane satisfying the given conditions.

13. Perpendicular to the line through the points (2,2,-4) and, (7,-1,3) and containing the point (-5, 1,2).

14. Parallel to the plane 4x-2y t z- l:0 and containing the point (2,6,-l).

15. Perpendicular to the plane x*3y - z-7:0 and containing the points (2,0,s) and (0, 2,-l).

16. Perpendicularto each of the planes x-y + z:0 and 2x*y -42- 5:0 and containing the point (4,0,-2).

17. Perpendicular to the yz plane, containing the point (2,l,l\ , and making an angle of radian measure cos-r(i) with thep l a n e 2 x - y + 2 2 - 3 : 0 .

18. Find the cosine of the measure of the angle between the planes 2x-y-22- 5:0 and5x -2y -l3zt8:0.

19. Find the cosine of the measure of the angle between the planes 3x -l 4y: 0 and 4x - 7y I 4z - 6: 0.

20. Find the distance from the plane 2r + 2y - z - 6:0 to the point (2, 2, -4).

21. Find the distance from the plane 5r -l l7y * 2z - 30: 0 to the poinr (-2, 6, 3).22. Find the perpendicular distance between the parallel planes

k - 8 y - z * 9 : 0 a n d 4 x - B y - z - 6 : 0

23. Find the perpendicular distance between the parallel planes

4 y - 3 2 - 6 : 0 a n d 8 y - 6 2 - 2 2 : 0

24. Prove that the undirected distance from the plane ax -f by t cz * d:0 to the point (16, yo, zs) is given by

laxn-l byn* czs-t dl\/F 175+7

25. Prove that the perpendicular distance between the two parallel planes ax + by * cz * dr- 0 and ffic + by + cz + dz:0

26. If a' b, and. c are nonzero and are the r intercept, y intercept, and z intercept, respectively, of a plane, prove that anequation of the plane is

r , a z- * i * - : 1a o c

This is called the intercept form of an equation of a plane.

18'5 LINES IN R3 Let L be a line in R3 such that it contains a given point Po(ro, Ao, zo) and. isparallel to the representations of a given vectorR : (a, b, c). iigure 1g.5.1shows a sketch of line L and the position representation of vector R. Line Lis the set of points P (x, y ,z) such that v (p;F) is parallel to the vector R. so

( xo , Yo , zo )

(a, b, c)

P is on the line L

V(EB): fR

18.5 LINES IN R3

and only if there is a nonzero scalar t such that

(1 )

t b z - z s : t c

F igure 18 .5 .1

Figure 18.5.2

Because V(fl 'p) - (x- xo,A - Ao,z- zol we obtain from (1)

(x - xo , A - Uo, z - zo l : t (a , b , c l

from which it follows that

x - X s : t a Y - A o :

or, equivalently,

Q)

Letting the parameter t be any real number (i.e., f takes on all values

in the interval (-o, fm)), the point P may be any point on the line L.

Therefore, Eqs. (2) represent the line L, and we call these equationsparametric equations of the line'

. rLLUsrRArron 1: From Eqs. (2), parametric equations of the line that

is parallel to the representations of the vector R: (11, 8, 10) and that

contains the point (8,12,6) arc

r : 8 * 1 1 t y : L 2 + 8 t z : 6 * 1 0 t

Figure 18.5.2 shows a sketch of the line and the position representation

o f R . o

If none of the numbers A,b, or c is zero, we can eliminate f from Eqs.(2) and obtain

These equations are called symmetric equations of the line.

The vector R: (a, b, c) determines the direction of the line, and the

numbers a, b, and c are called direction numbers of the line. Any vector par-

allel to R has either the same or the opposite direction as R; hence, such avector can be used in place of R in the above discussion. Because the corn-ponents of any vectoiparallel to R are proportional to the components of

R, *e cur, conclude that any set of three numbers proportional to a, b, arrd

c also can serve as a set of direction numbers of the line. So a line has an

unlimited number of sets of direction numbers. We write a set of direction

numbers of a line in brackets as la,b, cf .

. rLLUsrRArroN 2: If. 12,3, -4] represents a set of direction numbers of

a line, other sets of direction numbers of the same line can be represented

as14,6 , -81 , [1 , t , -2 ] ,and l2 l1 ,U \ f29 , -41 \ /D l . o

. ILLusrRArroN 3: A set of direction numbers of the line of Illustration 1

(3)(11,8, 10)

Po (8, L2, 6


EXAMPLE 1: Find two sets of sym-metric equations of the linethrough the two points (-3, 2, 4)and (5 , 1 , 2 ) .

is [11,8, 10], and the line contains the point (8,12,6). Thus, from (3)wehave as symmetric equations of this line

sor.urroN: Let P1 be the point (-3, 2, 4) and P2 be the point (6,'1,,2).Then the required line is parallel to the representations of the vector

V (P.,P) , and so the components of this vector constitute a set of direction

numbers of the line. V(FE): \9,-L,-2>. Taking Po as the point(-3,2,4), we have from (3) the equations

x * 3 : y - 2 _ " - 49 - L - 2

Another set of symmetric equations of this line is obtained by taking P6 asthe point (6,'/.,,2), and we have

Equations (3) are equivalent to the system of three equations

b ( x - x s ) : a ( y - y ) c ( x - x s ) : a ( z - z ) c ( y - y ) : b ( z - z ) ( 4 )

Actually, the three equations in (a) are not independent because any oneof them can be derived from the other two. Each of the equations in (4) isan equation of a plane containing the line L represented by Eqs. (3). Anytwo of these planes have as their intersection the line L; hence, any two ofthe Eqs. (4) define the line. However, there is an unlimited number ofplanes which contain a given line and because any two of them will deter-mine the line, we conclude that there is an unlimited number of pairs ofequations which represent a line., If one of the numbers a, b, or c is zero, we do not use symmetric equa-

tions (3). However, suppose for example that b:0 and neither a nor c iszero. Then we can write as equations of the line

X - X o Z - Z o 1

a : T a n q A : U o

A line having symmetric equations (5) lies in the plane y: ys andhence is parallel to the xz plane. Figure 18.5.3 shows such a line.

x - 8 _ y - 1 2 _ z

- 61,1, g 10

x - 6 _ y - L9 - 1

2z-2

(5)

ExAMPtn 2: Given the two planes

x+3y -z -9 :0

If we solve the two given equations for x and y Ln terms of z,SOLUTION:

we obtain

y : - , >* - k

F igure 18 .5 .3

+2 y -?z+&

from which we get

18.5 LINES IN R3

(7)

for a and b in terms of- c we get

then has the set of direction

2 x - 3 Y * 4 2 * 3 : 0

For the line of intersection of thesetwo planes, find (a) a set of sym-metric equations and (b) a set ofparametric equations.

ExAMPrr 3: Find the direction

cosines of a vector whose rePre-

sentations are parallel to the line

of Example 2.

ExAMPLE 4; Find equations of theline through the Point (L, - t , !) ,perpendicular to the line

\ x : Z y - z

and parallel to the Plane

x * Y - z : 0

x -2- 1

or, equivalently,

x - 2 z - 0-3

which is a set of symmetric equations of the line. A set of parametric equa-

tions can be obtained by setting each of the above ratios equal to f, and we

have

x :2 :3 t y :&+2 t z :3 t

solurroN: From the symmetric equations of the line in Example 2, we see

that a set of direction numbers of the line is [-3,2,3]. Therefore, the vector(-g,2,3) is a vector whose representations are parallel to the line. The

direction cosines of this vector are as follows: cos a:-3llD'

cos F:21 \8, cos Y : 3l \8.

solurroN: Let la, b, cl be a set of direction numbers of the required line.

The equations 3r : 2y : z can be written as

x -0 z - 0

which are s;rmmetric equations of a line. A set of direction numbers of this

line is lt,t, tl. Becaus6 the required line is perpendicular to this line, it

follows that the vectors (a, b, cl and (*, *, 1) are orthogonal' So

(a ,b ,c> ' (+ ,+ , 1) :0

or, equivalently,

t a * *b+c :0

Anormal vector to the plane r * y - z:0 is <1,1,-tr' Because the

required line is parallel to thls plane, it is perpendicular to representations

of the normal vector. Hence, the vectors (a,b,cl and (L,1',-1) are

orthogonal, and so

( a , b , c > ' ( L , l , - 1 ) - 0

or, equivalently,

a + b - c - 0

Solving Eqs. (6) and (7) simultaneouslyfl : gc and b - -8c. The required line

y-+2

y -012

(5)


numbers l9c, -8c , cl and contains the point (1 ,metric equations of the line are

x - " 1 . _ y + I _ r - L% - 8 c c

or, equivalently,

x - l u * l- . t -

9 - 8z - L

1). Therefore, sym-

(8)

we have as

ExAMPrn 5: If I, is the linethrough A(L, 2,7) andB(-2,3, -4) and lr ts the l inethrough C (2, -'J,, 4) andD (5 , 7 , -3) , prove that I, and 12are skew lines (i.e., they do not liein one plane).

Exercises 18.5

In Exercises 1 throu gh 5, find parametric and

L. Through the two points (I ,2, t) and (5,

sor-urroN: To show that two lines do not lie in one plane we demonstratethat they do not intersect and are not parallel. Parametric equations of aline are

x - x s * t a y - y o + t b z - z s * t c

where.[a, b, cf is a set of direction numbers of the line and (16, A o, zo) is anypoint on the line. Because Vta7l: (-3, l, -!'!,>, a set of directionnumbers of It is l-3, 1, -111. Taking A as the point Po, we have as para-metric equations of h

x - l - 3 t y - 2 + t z - 7 - I L t

Because v(C-D) - (3,8, -7>, and Iz contains the point C,parametric equations of lz

x - 2 * 3 s y - - 1 * 8 s z - 4 - 7 s

Because the sets of direction numbers are not proportional, l, andl2are notparallel. For the lines to intersect, there have to be a value of f and a valueof s which give the same point (xr, Au zr) in both sets of Eqs. (8) and (9).Therefore, we equate the right sides of the respective equations and ob-tain

l - 3 t : 2 * 3 s

2 * t : - 1 * 8 s

7 - l l t : 4 - 7 s

Solving the first two equations simultaneously, we obtain s: f7 andt: -tl.This set of values does not satisfy the third equation; hence, thetwo lines do not intersect. Thus, l, andl, are skew lines.

symmetric equations for the line satisfying the given conditions.- L , l ) .

(e)

18.5 LINES IN R3

- 0 .

14 ,2 , "1 .1 and [ -3 , -2 , l f .

intersection.

841

2.

3.

4.

5.

5.

7.

23.

Through the point (5,3,2) with direction numbers 14,t, -1].

Through the point (4,-5,20) and perpendicular to the plane x*3y - 5z-8

Through the origin and pe{pendicular to the lines having direction numbers

Through the origin and perpendicular to the line i@ - L0) - *y - tz at their

Through the point (2,0,-4) and parallel to each of the planes 2x* y - z:0 and r * 3y * 5z:0.

Show that the lines

ry -a *=4 : ' =2 and2 - 5 3 -3- 2 5

are coincident.

8. Prove that the line r * t: -t(y - 6) : z lies in the plane 3x * y - z: 3.

The planes through a line which are perpendicular to the coordinate planes are called the projecting planes of the line. InExercises 9 through 72, find equations of the projecting planes of the given line and draw a sketch of the line.

9 . 3 x - 2 y I 5 z - 3 0 : 0 1 0 . x * y - 3 2 * ' 1 . : 0 1 7 . x - 2 y - 3 2 * 6 : 0 7 2 . 2 x - y - t z - 7 : 02 x * 3 y - 1 0 2 - 6 : 0 2 x - y - 3 2 * ' l ' 4 : 0 x I y * z - l : 0 4 x - y * 3 2 - 7 3 : 0

13. Find the cosine of the measure of the smallest angle between the two lines r:2yl4,z:-y *4, and x:y*7,2 y : y 1 2 .

14. Find an equation of the plane containing the point (6,2,4) and the line *(r - 1) : t@ + Z) : i(z- 3)-

In Exercises 15 and 16, find an equation of the plane containing the given intersecting lines.

x y - 2 z - 1) c - Z y + 3 z + 21 6 - - 4 : -L v ' 4 - L 3

L7. Show that the lines

Fx -Y -z :0L8x -2y -32* t

- ( 3 x + 2 y l z + 2 - 0a n d

t ) c - y + 2 2 - L : o

_ 0 a n d

x u - 2 z - " 1 ,r 5 . ; : T : 1

a n d 1.

[ x - 3 y + z + 3 : 0L 3 r - y - z + 5 : 0

19.

20.

21,.

22.

are parallel and find an equation of the plane determined by these lines.

18. F indequat ionsof thel inethroughthepoint (1, -1,1) ,perpendicular tothel ineSx:2y:z,and.paral le l to theplanex + y - z : 0 .

Find equations of thelinethroughthepoint (3,5,4),intersectingthezaxis,andparalleltotheplanex- 3y *52-6:0-

Find equations of ttre line through the origin, perpendicular to the line x : A .- 5 , z : 2y - S, u;iintersecting the line

! : 2 x t 1 ' z : x * 2 .

Find the perpendicular distance from the point (-1,3, -1) to the l ine x-22:7,Y:1.

Find the perpendicular distance from the origin to the line

x : - 2 - l t t y : 7 - i t z : 4 * t r t

Prove that the lines

!=:L:Jl : un4 ' -2:y+:ry5 - 2 - o r - J 2

are skew lines.

24. Find equations of the line through the point (3, -4, -5) which intersects each of the skew lines of Exercise 23.

) c - 3 y + 1 , 4 z - 8

L - L


Let A and B be two nonparallel vectors. Representations of these twovectors having the same initial point determine a plane as shown in Fig.18.5.1. We show that a vector whose representations are perpendicular tothis plane is given by the vector operation called the "cross product" of thetwo vectors A and B. The cross product is a vector operation for vectors inVr that we did not have for vectors in Vr. We first define this operation andthen consider its algebraic and geometric properties.

F igure 18 .6 .1 L8.6.L Definition If A : (ar., ar, a3) and B : (bt, br, b"l , then the cross producf of A and B,denoted by A x B, is given by

- atbs, f l tbz - arbt l

Because the cross product of two vectors is a vector, the cross productalso is called the aector product. The operation of obtaining the cross prod-uct is called aector multiplication.

. rLLUsrRArrou l.: If A : (2, 1,, -3, and B : (J,-1,4), then from Defini-tion 18.6.1 we have

A X B : (2, 1, -3) X <3, -1. , 4l

: ( (1) (4) - (-3) (-1), (-3) (3) - (2) (4) , (2) (-1) - (1) (3) ): <4 -3 , -9 -8 , -2 -3 ): (1 , _ lT, _Sl

: i - 17 i - 5k o

There is a mnenomic device for remembering formula (1) that makesuse of determinant notation. A second-order determinant is defined bvthe equation

la b ll ; alaa-a'

where a, b, and c are real numbers. For example,

I g G l ^ / F \| ; . l : 3 (5 )_ (G) (_z ) :zzl - z 5 l

Therefore, formula (1) can

( 1 )

AxB-lZ: f :) i-The right side of the above expression can be written symbolically as

18.6 CROSS PRODUCT 843

which is the notation for a third-order determinant. However, observethat the first row contains vectors and not real numbers as is customarywith determinant notation.

. rLLUsrRArioN 2: We use the mnemonic device employing determinantnotation to find the cross product of the vectors of Illustration 1.

- i - L7 i - 5k

18.5.2 Theorem If A is any vector Lrr Vy then

( i ) AxA :o( i i ) 0 x A :0

( i i i ) A x 0 :0

pRooF or (i): If A: (ar, ar, ar), then by Definition 18.6.1' we have

A X A: (a rnr - f l sa2 t asar - 41as , f ,1a2- Azar l

: ( 0 , 0 , 0 )

- 0

The proofs of (ii) and (iii) are left as exercises (see Exercise 13). I

By applying Definition 18.6.1 to pairs of unit vectors i, i, and k, we ob-tain the following:

i X i : i X i : k x k : 0

i x i : k i xk - - i k x i : i

j x i : - k k x l : - i i x k : - i

As an aid in remembering the above cross products, we first noticethat the cross product of any one of the unit vectors i, i, or k with itself isthe zero vector. The other six cross products can be obtained from Fig.18.6.2 by applying the following rule: The cross product of two consecutivevectors, in the clockwise direction,'is the next vector; and the cross productof two consecutive vectors, in the counterclockwise direction, is the nega-tive of the next vector.

It can be easily seen that cross multiplication of two vectors is notFigure 18.6.2


commutative because in particular i x I + iX i. However, i X;: Luttdi X i : - k ; a n d s o i x ; : - ( i x i ) . I t i s t r u e i n g e n e r a l t h a tA x B : - (B x A), which we state and prove as a theorem.

18.5.3 Theorem If A and B are any vectors in Vr,

pRooF: If A : \ar, az, a3) and B : (b,,,bz,bsl,then byDefinition 18.6.1 wehave

A X B : (azbs- asbz, f ,abr- arbr, nrb2- arb)

: -7(asb2- a2bs, atbs- a"b1, arb, - atbz)

: _ ( B x A ) I

Cross multiplication of vectors is not associative. This is shown by thefollowi.g example:

i x ( i x i ) - i x

( i x i )X i : ox

So

i x ( i x i l+Gx i )x i

Cross multiplication of vectors is distributive with reSpect to vectoraddition, as given by the following theorem.

If. A,B, and C are any vectors in 7r, then

Ax (B+C) :AxB+AxC

L8.5.5 Theorem

Theorem 18.5.4 can be proved by letting A: (ar, ar, a"), B:(bt, br, bs), and C: (cr, cz, cs) , and then showing that the components ofthe vector on the left side of (2) are the same as the components of thevector on the right side of (2). The details are left as an exercise (seeExercise 14).

If A and B are any two vectors in V, and c is a scalar, then

( i ) ( c A ) x B : A x ( c B ) ;( i i ) (cA) x B : c (A x B) .

The proof of Theorem 18.6.5 is left as exercises (see Exercises 15and 16).

Repeated applications of Theorems 18.6.4 and 18.6.5 enable us to com-pute the cross product of two vectors by using laws of algebra, providedwe do not change the order of the vectors in cross multiplication, which is

k : - i

i : 0

18.6.4 Theorem

(2)

18.6.6 Theorem


prohibited by Theorem 18.6.3. The following illustration demonstratesthis.

o rLLUerRArroN 3: We find the cross product of the vectors in Illustra-tion 1 by applying Theorems 18.6.4 and 18.6.5.

AxB: (2 i+ i=3k) x (3 i - i+4k ): 5 ( i x i ) -2 ( i x i ) +8( i x k ) +3( i x i ) - 1 ( i x i )

+4( i x k ) -e (k x i ) +3(k x i ) - 12(k x k ): 6(o) -2(k) + 8(- i ) + 3(-k) - 1(o)

+ 4(i) - e(i) + 3(-i) - 12(o): -2k -8 i -3k+4 i -9 i -g i

: i -L7i-Sk .

The method used in Illustration 3 gives a way of finding the crossproduct without having to remember formula (1) or to use determinantnotation. Actually all the steps shown in the solution need not be includedbecause the various cross products of the unit vectors can be obtained

immediately by using Fig.1'8.6'2 and the corresponding rule.

o rLLUSrRArrOw 4: We prove that if A and B are any two vectors in v3, then

lA x B l '? : lA l ' z lB l '? - (A ' n ; '

LetA: (a1, a2, a3) and B: \br,bz,bsl .Tl l ren

lA x Bl'z : (azbs- a"br)' * (agbr- arb")'* (nrbz- f,rbt)'

: azzbsz - Za2arbrbg * as2b22 * albrz - 2a$sb1bg

+ arzbrz * a12b22 - 2ararb1b, * a22b12

lAl'lBl' - (A' B\2 : (arz + a22 + ar') (b" + bzz + bsz)

- (arbr* a2b2* asbs)z

= arzbzz * arzb"2 * a22bt2 * ar2b"2 + aszbrz

+ a,zbzz - Zarasbrbs- 2arasb2bs- 2ata2b1b2

Comparing the two expressions, we conclude that

lA x Bl'z: lAl'zlBl'z - (A ' 3;' o

The formula proved in Illustration 4 is useful to us in proving the fol-

lowing theorem, irom which we can obtain a geometric interpretation of

the cross product.

If A and B are two vectors in v, and g is the radian measure of the angle

between A and B, then

ln x sl : lAl lBl sin o (3)

VECTORS IN THREE-DIMENSIONAL

Figure 18.6.3

EXAMrLE 1: Show that the quad-rilateral having vertices atP (1 , -2, 3) , Q(4, 3, -r) ,R(2, 2, 1,) , and S (5, 7,-3) is aparallelogram and find its area.

SPACE AND SOLID ANALYTIC GEOMETRY

PRooF: From Illustration 4 we have

lA x B l ' : lA l ' lB l ' - (A . B ) '

From Theorem '1,8.3.4,

if 0 is the radian measure of theand B, we have

A .B - lA l lB l cos0

Substituting from (5) into (4) we get

la x B l , : lA l , lB l , - lA l , lB l , cosz o: lA l r lB l t (1 - cosz 0)

So

lA x B l r : lA l r lB l2 s in , 0 (6 )

Because 0 < 0 < z, sin 0 > 0. Therefore, from Eq. (5), we get

In x n l : lA l lB l s in 0 r

We consider now a geometric interpretation of lA x Bl. Let p? be arepresentation of A and let p? f" a representation of B. Then the anglebetween the vectors A and B is the angle at P in triangle RPQ (see Fig.18.6.3). Let the radian measure of this angle be 0. Therefore, the number ofsquare units in the area of the parallelogram having pT ana FO ur adjacentsides is lAllnl sin 0 because the altitude of the parallelogram has iengthlBl sin 0 units and the length of the base is lAl units. Sofrom Eq. (g) itfollows that lA x Bl square units is the area of this parallelogram.

solurroN: Figure 18.6.4 shows the quadrilateral peSR.

v(P?) : (4- ' t ,s - ( -2) , ( - t ) -3) : (s ,s , -4 ;v(Ff t ) : (2- ' ! . ,2- ( -2) , t - 3) : ( t ,4 , -2>v(n-3) : (5- 2 ,7 -2 , -g- 1) : (3 , s , -41v(0-) : (5 - 4,7 - s, -g - (-1) ) : (t, 4, -2l

BecauseV(FQ) : V(n3) andV(P?) : V(Q*S), itfollows thatPQ isparal-lel to R3 and P? is parallel to Q3. Therefore, PQSR is a parallelogram.

Let A: v(pT) and B: V(P!), then

A x B : (i + 4i- 2k) x (3i + si -4k)

:3 ( i x i ) +5 ( i x i ) - 4 ( i x k ) +72 ( i x i ) +20 ( i x i )- t i ( i x k ) -5 (k x i ) - 10 (k x i ) + 8 (k x k )

: 3(0) + s(k) - 4(- i) + 12(-k) + 20(0) - 16(i)

(4)

angle between A

(s)

/

b ,R(2, z, !)

(4, 3, - L)

Figure 18.6.4 S(s,7, - g) - 5( i ) - 10(- i ) + 8(0)


2 i -7kHence,

lexBl :m- \@

The area of the parallelogram is therefore \@ square units.

The following theorem, which gives a method for determining if twovectors in V3 are parallel, follows from Theorem 18.6.5.

1g.6.7 Theorem If A and B aretwovectors inVg,AandB areparallelif and onlyif AxB:0'

pRooF: If either A or B is the zero vector, then from Theorem 1'8.6.2,

A x B:0. Because the zero vector is parallel to any vector, the theorem

holds.If neither A nor B is the zero vector, lAl + O and lBl # 0. Therefore,

from Eq. (3), lA x Bl :0 if and only if sin 0:0. Because lA x Bl : 0 if

and only if A X B : 0 and sin 0 : 0 (0' 0 = r) if and only if 0 : 0 ot r,we can conclude that

A x B : 0 i f a n d o n l y i f 0 : 0 o t r -

However, from Theorem 18.3.6, two nonzero vectors are parallel if and

only if the radian measure of the angle between the two vectors is 0 or zr.

Thus, the theorem follows. I

The product A '

vectors A, B, and C.A ' B i s a s c a l a r / a n dway.

(B x C) is called the triple scalar product of theActually, the Parentheses are not needed becausetherefore A ' B X C can be interpreted only in one

18.6.8 Theorem If A, B, and C are vectors in Vs, then

Theorem 18.6.8 can be proved by letting A: (at, az, ag),

B : (br, br, b"l , and C : (cr, c2, ca) and then by showing that the number

on the left side of (7) is the same as the number on the right. The details

are left as an exercise (see Exercise 17).

11 rLLUSrRArroN 5: We verify Theorem 18.6'8 if A: (1, -1, 21 , B:

(3 ,4 , 2> , and C - ( -5 ,1 , -4> .

B x c- (3 i+ 4 i - 2k) x ( -5 i + i -4k )- L6i + loi - 2(-i)- 3k - 12(-i) - zo(-k)

1,4 i+22i+23k

(7)


A. (B x C) : (L , -1 ,2 , . ( -L4 ,22 ,23>: -L4-22+ M:L0

Ax B - ( i - i + 2k) x (3 i + 4 i -2k): 4k -2 ( - i ) - 3 ( - k ) +2 i+6 i +8 ( - i )

: - 6 i+8 i +7k

(A x B ) . C : ( - 5 ,8 ,7> . ( - 5 ,1 , - 4>

:30+8 -28

:L0

This verifies the theorem for these three vectors.

EXAMPTu 2: Given the pointsP( -1 , -2 , -3 ) , Q ( -2 ,1 ,0 ) , andR(0, 5, L), find a unit vector

L8.5.9 Theorem If A and B are two vectors in 7r, then the vector A x B is orthogonal toboth A and B.

PRooF: From Theorem 18.6.8 we have

A . A X B _ A X A . B

From Theorem 18.5.2( i) , AxA:0.we have

Therefore, from the above equation

A ' A x B : 0 . 8 : 0

Because the dot product of A and A x B is zero, it follows from Definition18.3.7 that A and A x B are orthogonal.

We also have from Theorem 18.6.8 that

AxB.B :A .BxB

Again applytt g Theorem 1,8.6.2(i), we get B x B : 0, and so fromthe above equation we have

A x B . B : A . 0 : 0

Therefore, since the dot product of A X B and B is zero, Ax B and Bare orthogonal and the theorem is proved. I

From Theorem 18.5.9 we can conclude that if representations of thevectors A, B, and A X B have the same initial point, then the represen-tation of A X B is perpendicular to the plane formed by the representa-tions of A and B.

solurroN: Let A - V(p0) and g - v(p?). Then

A - ( - 2 - ( - 1 ) , 1 - ( - 2 ) , 0 - ( - s ) ) : ( - 1 , 3 , 3 >

whose representations are Per-pendicular to the plane through

the points P, Q,,and R.

ExAMPln 3: Find an equation of

the plane through the PointsP(1 , 3 , 2 ) ,Q(3 , -2 , 2 ) , andR ( 2 , L , 3 ) .

P A

Figure 18.6.5

18.6 CROSS PRODUCT

B - ( 0 - ( - 1 ) , 5 - ( - 2 ) , r - ( - 3 ) ) : ( L ,7 ,4>

The plane through P, Q, and'R is the plane formed by p0 and Pl,

which are, respectively, representations of vectors A and B. Therefore, any

representation of the vector A x B is perpendicular to this plane.

A x B: (- i + 3i + 3k) x ( i+7i + 4k) : -9 i+7i- 10k

The desired vector is a unit vector parallel to A X B. To find this unit

vector we apply Theorem 18.2.9 and divide A x B by lA x Bl, and we ob-

tain

A X B _ 9 _ r - 7 i _ _ 1 0 r -

l f f i I:-n5-r+ 1\/m-t- 16-x

solurroN: v(6) : -i * 3i + k and v(PR) : i- ?i + k. -A normal

vector to the required plane is the cross product Vtp?) x V(p?)' whichis

(-i+ 3i + k) x (i- 2i+ k) : si + 2i - k

So if Po : (1,3,2) and N: (5, 2,-ll, from Theorem18'4'2 we have as an

equation of the required Plane

5(r- 1) + z(V - 3) - (z- 2) : 0

or, equivalently,

5 x * 2 Y - z - 9 : 0

A geometric interpretation of the triple scalar product is obtained by

considering a parallelepiped having edges Pn,Frt.,and fi, and letting A:

V(P?), B:V(PR), and C:V(PS). See-Fig' 18'6 '5 ' The vectorA X B is

a normal vector to the plane of p0 and pA. fne vector-(A X B) is also a

normal vector to this plane. we are not certain which of the two vectors,

(AxB)or_(AxB) ,makesthesmal le rang lewi thC.LetNbetheoneofti\e two vectors (A x B) or - (A x B) that makes an angle of radian

measure 0 < tn with C. Then the representations of N and C having their

initial points at P are on the same side of the plane of F0 "",{

FR as shown

in Fig.'18.5.5. The area of the base of the paralletepiped i9 lA X Bl square

unitsl If h units is the length of the altitude of the parallelepiped, and if V

cubic units is the volume of the parallelepiped,

v - lAxB lh

Consider now the dot product N ' c. By Theorem "l'8.3.4,

lNl lg l cos 0. But h- lC l cos 0, and so N'C: lN lh. Because N

tA r b) ot -(A x B), it follows that lNl _ lR. x nl. Hence,

N .e : lAxB lh

(8)

N . C _is eitherwe have

(e)


Comparing Eqs. (S) and (9) we have

N . C : V

It follows that the measure of the volume of the parallelepiped is either(A x B) ' c or - (A x B) ' c; that is, the measure of the volume of theparallelepiped is the absolute value of the triple scalar product A X B . c.

ExAMPLE 4: Find the volume ofthe parallelepiped having verticesP ( 5 , 4 , 5 ) , Q ( 4 , 1 , 0 , 6 ) , R ( ' ! . , 9 , 7 ) ,and S(2 ,6 ,9 ) and edges &, ,P?,and u;

Figure 18.6.6

EXAMPLE 5: Find the distancebetween the two skew lines, /, and12, of Example 5 in Sec. 18.5.

N

v ( r t ) : -3 i+ i - l l k v(cB)-3 i+Bi - rk

solurroN: Figure 18.6.6 shows the parallelepiped. Let A: V(pQ) :( -1 , 6 ,1) , B: Vtp?) : ( -4 , 4 , 2) , and, C: V(p3) : ( -3 , 2 , 4) . Then

A x B : (-i + 6i + k) x (-4i + 4i +2k) : 8i- 2i+ 20kTherefore,

(A x B) . C: (8 , -2 ,20> . ( -9 ,2, 4 l : -24- 4 * g0:52

Thus, the volume is 52 cubic units.

solurroN: Because It and /, are skew lines, there are parallel planes p,and Pt containing the lines l, and 12, respectively. see Fig.1,g.6.7. Let dunits be the distance between planes p, and pr. th" distaice between /,andl, is also d units. A normar vector to the two planes is N: vtertix v(cD). Let u be a unit normar vector in the direction of N. Then

u: v(a-B) x v(c-i;lv(a-B) x v(6) |

Now we take two points, one in each plane (e.g.,Band C). Then the scalarprojection of V(CT) on N is V(CTI . U, and

d: lv(c?) - ul: Ittcll . v(AE) x v(cB) |

I lv(AB) x v(cD) | IPerforming the computations required, we have

s (2 ,6 ,9 )

R (1 , 9 ,7 )

Q@,TO, 6 )

P (5, 4, 5)

F igure 18.6.7

18.6 CROSS PRODUCT

k l-rrl: zr (3i - 2i - k)-7 1

851

Exercises 1'8.6

In Exercises 1. throu ghl2,let A - (1 ,2'3)' B - (4'-3'

L . F i n d A x B

3 . F i n d ( C x D ) ' ( E x F )

5. Verify Theorem 18.6.3 for vectors A and B.

6. Verify Theorem 18.6.4 for vectors A, B, and C.

7. Verify Theorem 18.6.5(i) for vectors A and B and c:3'

8. Verify Theorem 18.6.5(ii) for vectors A and B and c :3'

9. Verify Theorem 18.6.8 for vectors A, B, and c.

10. Find (A x B) x C and A x (B x C)'

L 1 . F i n d ( A + B ) x ( c - D ) a n d ( D - c ) x ( A + B )

12. Find lA x Bl lc x Dl.

L3. Prove Theorem 18.5.2(ii) and (iii)'

15. Prove Theorem 18.6'5(i)'


18. Given the two unit vectors

A -6 i+6 i -+k and B - -& i+3 i ++k

u - -? i -z i -k\n,Finally, vtdBl : -4i + 4i- 8k, and so

d-lv(c?)' t I | : # l-D- 8 + 8 | : h: + ffi

- 1 ) , C : ( - 5 , - 3 , 5 ) , p - ( - 2 , L , 5 ) , E - < 4 ' O ' - 7 > ' a n d F - < 0 ' 2 ' 1 ) '

2 . F i n d D x E

4 . F i n d ( C x E ) ' ( D x F )

and verify theY are equal.

14. Prove Theorem

16. Prove Theorem

L8.6.4.

18.5.5(i i ) .

If 0 is the radian measure of the angle between A and B, find sin 0 in two ways: (a)

(3) of this section); (b) by using the dot product and a trigonometric identity.

1g. Follow the instructions of Exercise 18 for the two unit vectofs:

- 1 1 1 1 R 1

a: u5 t -Vrt*+k and n:5f t t*5f t i+3vgk

by using the cross product (formula

zo. show that the quadrilateral having vertices at (1,1 ,3), (-2,'l ',-1), (-5 ,4,0), and (-8,4,-4) is a paral lelogram and

find its area.

21. show that the quadrilateral having vertices at (1,-2,3) , (4,3,

find its area.

-1) , (2,2,L), and (5,7,*3) is a paral lelogram and

8s2 VECTORS IN THREE-DIMENSIONALSPACE AND SOLID ANALYTIC GEOMETRY

+ -+22. Find the

23. Find the

24. Find the

26. Find a unit vector whose representations

tion of the vector i + 3i - Zk and FR i, u

28. Find the volume of the parallelepiped having( 1 , 3 , 4 ) , ( 3 , 5 , 3 ) , ( 2 , 1 , 6 ) , a n d ( 2 , 2 , 5 ) .

29. Find the volume of the parallelepiped peRS if2 i+ i

- k , and i - 2 i + k .

30. If A and B are any two vectors rnvs, prove that (A

In Exercises 31 and 32, use the cross product to find an31. ( -2 , 2 , 2 ) , ( -g , l , 6 ) , (3 , 4 , _L)

In Exercises 33 and i4, find the perpendicular distance

a ? x - " 1 , _ y - 2 _ z + 1 ^ - J x * 2 y + l z _ lr r ' 5 : 3 : 2 a n d i : T : = =

35. Let P, Q, and R be three noncollinearB, and C, respectively. Prove that theplane containing the points p , e, and

area of the parallelogram PQRS if V(pe) : 3i - 2i and V(pS) - 3i

area of the triangle having vertices at (0 , 2, z), (g , g, -2) , and (g ,area of the triangle having vertices at (4, 5 , 6) , (4, 4, 5) , and (3 , 5 ,

+ 4k.

12, 6) .

5 ) .

25' Let OP be the position representation of vector A , dQ b" the position representation of vector B, and o-i be the posi-tion representation of vector c. prove that the area of triangle pen is d (b - A) x (c - A) l.

-- - - ---- I

27 ' Given thepointsP(5,2, -D,Qe,4,-2) ,andR(11,1,4) .F indauni tvectorwhoserepresentat ionsareperpendicularto the plane through points p, e, and R.

are Perpendicular to the plane containi.g pQ urrd FR if P0 is a representa-representation of the vector 2i - i - k.

edges Pi, Fi, and E ir the points p, e, R, and s are, respectively,

the vectors v(P?), v(p?), and v(pB) are, respectively, i + 3i + zk,

- B ) x ( A + B ) : 2 ( A x B ) .

equation of the plane containing the given three points.

3 2 . ( a , b , 0 ) , ( a , 0 , c ) , ( 0 , b , c )

between the two given skew lines.

. A x t ' / . , y + Z z - lJ t . : : - : - and

x - l _ y - t _ z * t

5 3 2

points in R3 and dP' oa, and ol be the position representations of vectors A,representations of the vector A x B + B x c + c x A are perpendicular to theR.

I8.7 CYLINDERS ANDSURFACES OF REVOLUTION

18.7.L Definition

As mentioned previously, the graph of an equation in three variabres is asurface. A sudace is represented by an equation if the coordinates of everypoint on the surface satisfy the equatiorrand if every point whose coordi-nates satisfy the equation lies on the surface. we have already discussedtwo kinds of surfaces, a plane and a sphere. Another kind of surface that isfairly-simple is a cylinder. you are probably familiar with right-circularcyfinfels from previous experiet

"e. W" now consid", u -6r" general

rylindrical surface.,

A cylin.der is a surface that is generated by a line moving along a givenplane curve in such away that it always remains parallel to"a fixea tne notlying in the plane of the given curve. The moving line is called a generatorof the rylinder and the given plane curve is called a directix of the rylin-der. Any position of a generator is called a ruling of the cylinder.

we confine ourselves to cylinders having a directrix in a coordinate

18.7 CYLINDERS AND SURFACES OF REVOLUTION 853

plane and rulings perpendicular to that plane. If the rulings of a cylinder

are perpendicular to the plane of a directrix, the cylinder is said to be per-

pendicular to the plane.The familiar right-circular

circle in a plane pe{Pendicularcylinder is one for which a directrix is a

to the cylinder.

F igure 18 .7 .1

Figu re 18.7 .2

Figu re 18.7 .3

. rLLUSrRArroN 1: In Fig. 't8.7.'1,, we show a cylinder whose directrix is

the parabola y':8r in the xy plane and whose rulings are parallel to

the i axis. This cylinder is called a parabolic cylinder. An elliptic cylinder

is shown in Fig. 1'8.7.2; its directrix is the ellipsegxz*16y2:144 i4 the

ry plane and its rulings are parallel to the z axis. Figure 18.7.3 shows a

hyp-erbolic cylinder having as a directrix the hyperbola 25x2 - 4A2 :700

in the xy plane and rulings Parallel to the z axLS.

Let us consider the problem of finding an equation of a cylinder

having a directrix in a coordinate plane and rulings parallel to the coordi-

nate a-xis not in that plane. To be specific, we take the directrix in the ry

plane and the rulings paraltel to the z axis. Refer to Fig.1'8.7.4' Suppose

that an equation of the directrix in the xy plane is y : f (x). If the point

(xo,yo,0) in the xy plane satisfies this equatibn, any point (ra, Ao,.z)-iniftt"!-ai*""sionaispace, where z is any real number, will satisfy the'sdme

equation because z does not appear in the equatiqn. The points having

representations (ro, ao, z) all lie on the line parallel to the z axis through

ttre point (xo, yo,0). This line is a ruling of the cylinder' Hence, any point

whose x and y coordinates satiqf the equation A : f @) lies on t_he rylin-der. Conversely, if the point P(x, y, z) lies on the cylinder (see Fig'

18.7.5), then the point (r, y, 0) lies on the directrix of the cylinder in

the xy plane, and hence the r and y coordinates of P satisfy the equationy: f(x). Therefore, ity: f(x) is considered as an equation of a $raph in

Figu re 18.7 .4

(xo , yo , O)

F igure 18.7.5


three-dimensional space, the graph is a cylinder whose rulings are parallelto the z axis and which has as a directrix the curve y: f(x) in the planez:0. A similar discussion pertains when the directrix is in either of theother coordinate planes. The results are summarized in the followingtheorem.

18.7.2 Theorem In three-dimensional space, the graph of an equation in two of the threevariables x, y, and z is a cylinder whose rulings are parallel to the axis as-sociated with the missing variable and whose directrix is a curve in the

, plane associated with the two variables appearing in the equation.

. rLLUsrRAtroN 2: It follows from Theorem 18.7.2 that an equation of theparabolic cylinder of Fig. 18.7.1 is A" :8x, considered as an equation in RB.Similarly, equations of the elliptic cylinder of Fig. 18.7.2 and the hyper-bolic cylinder of Fig. 1.8.7.3 are, respectively , 9x? * I6y2 : 1,44 and. 25xz -4A2:100, both considered as equations in RB. o

A cross section of a surface in a plane is the set of all points of the sur-face which lie in the given.plane. If a plane is parallel to the plane of thedirectrix of a cylinder, the cross section of the cylinder is the iame as thedirectrix. For example, the cross section of the elliptic cylinder of Fig.18.7.2Ln any plane parallel to the xy plane is an ellipse.

EXAMPLE L: Draw a sketch of thegraph of each of the followirgequations: (a) y : In z;(b) z2 : )c3.

solurroN: (a) The graph is a cylinder whose directrix in the yz plane isthe curve a : ln z and whose rulings are parallel to the r axis. A sketch ofthe graph is shown in Fig. 1,8.7.6.

(b) The graph is a cylinder whose directrix is in the rz plane andwhose rulings are parallel to the y axis. An equation of the directrix is thecrrrve 22 : rf in the xz plane. A sketch of the graph is shown in Fig. 1,9.7 .7 .

Figu re 18.7.7Figure 18.7.6

18.7.3 Definition


If a plane cuwe is revolved about a fixed line lying in the plane of the

"..r*^", the sur{ace generated is called a surface of reoolution The fixed line

is called the axis of the surface of revolution, and the plane curve is called

the generating curae.

Figure 18.7.8 shows a surface of revolution whose generating cunre is

the curve c in the yz plane and whose axis is the z axis. A sphere is a par-

ticular example of a surface of revolution because a sphere can be gen-

erated by revolving a semicircle about a diameter.

o rLLUsrRArroN 3: Figure 18.7.9 shows a sphere which can be generated

by revolving the semicfucle y2*22:12, 22 0, about the y axis' An-

oih", "*"-ple

of a surface of revolution is a right-circular cylinder for

which the generating curve and the axis are parallel straight lines. If

the generating curve 1s the line z: k in the xz plane and the axis is the

r "*ir,

we obtain the right-circular cylinder shown in Fig' 1'8'7'10' '

we now find an equation of the surface generated by revolving about

the y axis the curve in the yzplane having the two-dimensional equation

z : f ( y ) ( 1 )

Refer to Fig. 1,8.7.1,1..LetP(x, y , z) be any point on the surface of revo-

lution. Through P, pass a plane perpendicular to the y axis. Denote the

point of interiection of this plane with the y axis by Q(0, y,0), and let

Po(o,a, z) be the point of intersection of the plane with the generating

"rr*". B".urrse the cross section of the surface with the plane through P is

a circle, P is on the surface if and only if

l@12- lOP,l 'Because l@l : \m and lOPol : zo, w€ obtain from (2)

Figure 18.7.8

x

Figure 18.7.9

(2)

x 2 + 2 2 : z o 2

The point Po is on the generating cun/e, and so

satisfy Eq. (1). Therefore, we have

zo : f Q )

(3)

its coordinates must

(4)

P(*, u, z)Po(0 , U,zo)

Q(o,y, o)

F igu re 18 .7 .10Y F igure 18 .7 .1 1


ExAMPLE 2: Find an equation ofthe surface of revolution gen-erated by revolving the parabolay' : 4x in the xy plane about the raxis. Draw a sketch of the graph ofthe surface.

From Eqs. (3) and (4), we conclude that the point p is on the surface of rev-olution if and onlv if

x2 + 22 : l f Q)1,

Equation (5) is the desired equation of the surface of revolution. Because(5) is equivalent to

_+t/ptry' : fe)we can obtain (5) by replacing z rn (1)

In a similar manner we can showhaving the two-dimensional equation

y : s (z )is revolved about the z axis, an equation of the surface of revolutiongenerated is obtained by replacing y in (6) by *I/FTT.Analogous remarks hold when a curve in any coordinate plane is revolvedabout either one of the coordinate axes in that plane. In summary, thegraphs of any of the following equations are surfaces of revolution havingthe indicated axis: ,'+ y' : lp(z)fz-z axis; f + z2 : lF(y)1"-y axis;y'+ z': [f(x)],-r axis. In each case, cross sections of the surface inplanes perpendicular to the axis are circles having centers on the axis.

soLUrIoN: In the equation of the parabola, we replace y by tWTzand obtain

y2 + z2: 4r

A sketch of the graph is shown in Fig. 18.7.1.2. Note that the same surfaceis generated if the parabola 22 :4r in the xz plane is revolved about the raxis.

(s)

that if the curve in the yz plane

(6)

r r r l r r ,

F igu re 18 .7 .12

The surface obtained in Example 2 is called aparaboloid of reaolution.Ifan ellipse is revolved about one of its axes, the surface obtained is calledan ellipsoid of reaolution. A hyperboloid of reaolution is obtained when ahyperbola is revolved about an axis.

EXAMPLE

surfac e x2

Draw a sketch of the22 - 4y ' :0 , i f y > 0.


solurroN: The given equation is of the formrP -f 22 : lF(y)lr,and so itsgraph is a surface of revolution having the y axis as axis. If we solve thegiven equation for y, we obtain

2Y:!\/7+VHence, the generating curve can be either the straight line 2y: r in the xyplane or the straight line2y : z in the yz plane. By drawing sketches of thetwo possible generating curves and using the fact that cross sections of thesurface in planes perpendicular to the y axis are circles having centers onthe y axis, we obtain the surface shown in Fig. 1,B..Z.lg (note that becausey > 0 we have only one nappe of the cone).

t. 4x2 * 9y' - 36

5. z : 2x2

F igu re 18 .7 .13

The surface obtained in Example 3 is called a right-circular cone.

Exercises 18.7

In Exercises 1 through 8, draw a sketch of the cylinder having the given equation.

2 . x : 2 - 2 2 : 4

6 . x 2 : A 3

In Exercises 9 througt:.L4, find an equation of the surface of revolutionthe indicated axis. Draw a sketch of the surface.

9. x2 : 4y tn the xy plane, about the y axis.

10. x2 : 4y in the xy plane, about the r axis.

11. x2 + 422: L6 in the xz plane, about the x axis.

12. x2 + 422 : L6 in the xz plane, about the z axis.

13. y : sin r in the .ry plane, about the x axis.

14. y2: d in the yz plane, about the z axis.

In Exercises 15 through 18, find a generating curve and the axis for the given surface of revolution. Draw a sketch of thesurface.

1 5 . x 2 + y ' - 2 2 : 4

19. The tractrix

1'6. y' + z2 - ezr 17. x2 + z' - lyl

x _ t - a t a n h j y _ - a s e c h la - a

from x - -a to x - 2a ts revolved about the x axis.

3 . y : l z l 4 . z - s i n y

7 . y : c o s h r 8 . z 2 - 4 y 2

generated by revolving the given plane curue about

Draw a sketch of the surface of revolution.

18. 4x2 * 9y' + 422 - 36

18.8 QUADRIC SURFACES

> y


The graph of a second-degree equation in three variables x, y, and z iscalled a quadric surface. These surfaces correspond to the conics in theplane.

The simplest types of quadric surfaces are the parabolic, elliptic, andhyperbolic cylinders, which were discussed in the preceding section.There are six other types of quadric surfaces, which we now consider. Wechoose the coordinate axes so the equations are in their simplest form. Inour discussion of each of these surfaces, we refer to the cross sections ofthe surfaces in planes parallel to the coordinate planes. These cross sec-tions help to visualize the surface.

The ellipsoid

- 1

where a, b, and c are positive (see Fig. 18.8.1).If in Eq. (1) we replace zby zero, we obtain the cross section of the

ellipsoid in the ry plhne, which is the ellipse *laz * yrlbr: 1.. To obtainthe cross sections of the surface with the planes z : k, we replace z by k inthe equation of the ellipsoid and get

If lkl < c, the cross section is an ellipse and the lengths of the semiaxesdecrease to zero as lkl increases to the value c. Iflkl: c, the intersection ofa planez : k with the ellipsoid is the single point (0,0, k). If lkl ) c, thereis no intersection. we may have a similar discussion if we consider crosssections formed by planes parallel to either of the other coord.inateplanes.

The numbers a, b, and c are the lengths of the semiaxes of the ellip-soid. If any two of these three numbers are equal, we have an ellipsoid-ofrevolution, which is also called a spheroid.If we have a spheroid and thethird number is greater than the two equal numbers, the spheroid is saidto be prolafe. A prolate spheroid is shaped like a foot6afl. An oblatespheroid is obtained if the third number is less than the two equalnumbers. If all three numbers a,b, and c in the equation of an ellipsoid areequal, the ellipsoid is a sphere.

The elliptic hyperboloid of one sheet

f u z z za+E 7 e)

\ (where a, b, and, c are positive (see Fig. 18.3.2).

The cross sections in the planes z: k are ellipses *laz * Azlbz : l+ kzle. when k: 0v the lengths of the semiaxes of tlie ellipse are smallest,

f * t - , - kza 2 b 2 ^ c 2

( 1 )

Figure 18.8.1

Figure 18.8.2


and these lengths increase as lkl increases. The cross sections in the planesx : k are hyperbolas y2 I b2 - zz I cL : L - kz.l n2. If I lcl < a, the transverse axisof the hyperbola is parallel to the y axis, and if lkl > o, the transverse axisis parallel to the z axis. If k:a, the hyperbola degenerates into twostraight lines: ylb - zlc: Oand ylb * zlc :0. In an analogous manner, thecross sections in the planes y : k are also hyperbolas. The axis of thishyperboloid is the z axis.

If a : b, the surface is a.hyperboloid of revolution for which the axis isthe line containing the conjugate axis.

The elliptic hyperboloid of two sheets

where a,b, and c are positive (see Fig. 18.8.3).

Replacing z by k in Eq' (3), we obtain x2la2 + A'lb'= Plcz - L' If

lkl < c, there is no intersection of the plane z = k with the surface; hence,

there are no points of the surface between the planes z: -c and z: c' lf

lkl: ,, the iirtersection of the. plane z: k wlth the surface is the single

point (0, 0, k). When lkl > t, ihe cross section of the surface in the plane

2: k is an ellipse and the lengths of the semiaxes of the ellipse increase as

lkl increases.The cross sections of the surface in the planes x: k are the hyperbolas

,z 1sz - y2lb2: l. * k2 | a2 whose transverse axes are parallel to the- z axis' In a

similaifashion, the cross sections in the planes A : k arc the hyperbolas

given by z2lc2 - f la2: 1 * Plbz forwhich the transverse axes are also par-

allel to the z axis.If. a : b, the surface is a hyperboloid of revolution in which the axis is

the line containing the transverse axis of the hyperbola'Each of the above three quadric surfaces is symmetric with respect to

each of the coordinate planes and symmetric with respect to the origin'

Their graphs are called ienttal quadrics and their center is at the origin. The

graph of any equation of the form

where a, b , and c ate Positiv€, is a central quadric.

soLUTIoN: The given equation can be written as

_ xz _y ' _ z ' _ 1- , r - w-7

- r (3)

Figu re 1 8.8.3

nls.rvrptE L: Dra* asketch of the

graph of the equation

4 x 2 - y ' + 100

+ t *Y ' * z ' - 1a 2 - b 2 - c 2

and name the surface. which is of the form of Eq. (2) with y and z interchanged. Hence, the sur-


face is an elliptic hyperboloid of one sheet whose axis is the y axis. Thecross sections in the planes y : k are the ellipses *125 + z2l4: | + k21700.The cioss sections in the planes x: k are the hyperbolas z2l4- y'1100:l-k2125, and the cross sections in the planes z:k ane thp hyperbolasfl25 - y21700: ! - kzl4. A sketch of the surface is shown in Fig. 18.8.4.

F igure 18 .8 .4

ExAMPLE 2: Draw a sketch of thegraph of the equation

4 x 2 - 2 5 y 2 - 2 2 : 1 0 0

and name the surface.

solurroN: The given equation can be written as

!--t-- , ' l_:1.25 4 100

which is of the form of Eq. (3) with r and z interchanged; thus, the surfaceis an elliptic hyperboloid of two sheets whose axis is the r axis. The crosssections in the planes x:k, where lkl >5, are the ellipses yzl4* 221100: k2125 - 1. The planes x = k, where lkl < 5, do not intersect thesurface. The cross sections in the planes A:k are' the hyperbolas fl2|- 221100:1* k214, and the cross sections in the planes z:k are thehyperbolas f 125 - U'14:1 + k,/100. The required sketch is shown in Fig.18.8.5.

Figure 18.8.5

Figure 18.8.6

F igure 18 .8 .7

ExAMPLE 3: Draw a sketch of the

graph of the equation

3y ' *1222 : "1 .6x



The following two quadrics are called noncentral quadrics.

The elliptic p araboloid

f uzz7* w: ; G)

where a and b are positive and c * 0. Figure 18.8.5 shows the surface ifc > 0 .

Substituting k for z in Eq. (4), we obtain *laz * y'lb' : k/c. Whenk: 0, this equation becomes *laz * y2lb2 -- 0, which represents a singlepoint, the origin. If. k + 0 and k and c have the same sign, the equation isthat of an ellipse. So we conclude that cross sections of the surface in theplanes z: k,where k and c have the same sign, are ellipses and the lengthsof the semiaxes increase as lkl increases. If k and c have opposite signs, theplanes z: k do not intersect the surface. The cross sections of the surfacewith the planes x: k and y : k arc parabolas' When c ) 0, the parabolas

open upward, as shown in Fig. 18.8'6; when c < 0, the parabolas opendownward.

lI a: b, the surface is a paraboloid of revolution.

The hyp erbolic p araboloid

where a and b are positive and c * 0. The surface is shown in Fig. 18.8.7

f o r c > 0 .The cross sections of the surface in the planes 2: ft, where k # 0, ate

hyperbolas having their transverse €xes parallel to the y axis if k and cv t

""e the same sign and parallel to the r axis if k and c have opposite signs.

The cross section of the surface in the plane z: 0 consists of two straight

lines through the origin. The cross sections in the planes x: k ate parab-

olas opening upward if c > 0 and opening downward if c < 0. The cross

sections in the pl".t"t y: k are parabolas opening downward if c ) 0 and

opening upward if c < 0.

solurroN: The given equation can be written as

which is of the form of Eq. (4) with r and z interchanged. Hence, thegraph of the equation is an elliptic paraboloid whose axis is the r axis. The

crois sections ir, the planes x: k ) 0 are the ellipses y2 lL6 -f zz | 4: kl3, and

the planes x: k 10 do not intersect the surface. The cross sections in the

planes y : k are the parabolas L2z2 :16x - 3k', and the cross sections in

u2 )c2 z- - - : -b 2 n 2 c

(s)


the planes z : k are the parabol as 3yz : L5x - Ue. A sketch of the ellipticparaboloid is shov,rn in Fig. 18.8.8.

Figure 18.8.8

EXAMPLE 4: Draw a sketch of thegraph of the equation

3y'- 1222 : l6x


z

Figure 18.8.9

solurroN: Writing the given equation as

tlz 22 x- - - : -1 6 4 3

we see it is of the form of Eq. (5) with x artd z interchanged. The surface istherefore a hyperbolic paraboloid. The cross sections in the planes x:k * 0 are the hyperbolas y'lL5 - z2l4: k/3. The cross section in the yzplane (r: 0) consists of the two lines y : h ffid y - -2z.In the planesz: k, the cross sections are the parabolas 3A, : L6x * l2t9; in the planesA : k, the clross sections are the parabolas l2z2 : 3k2 - t6x. Figure 1g.g.9shows a sketch of the hyperbolic paraboloid.

The elliptic cone

where a, b, arrd c are positive (see Fig. 18.8.10).The intersection of the plane z: 0 with the surface is a single point,

the origin. The cross sections of the surface in the planes z: k, *hetek * 0, are ellipses, and the lengths of the semiaxes increase as k increases.Cnr-ss se-ctions in the planes x: 0 and,y: 0 are pairs of intersecting lines.In the planes x: k arrd U: k, where k * 0, the cross sections a"e hyper-bolas.

5.#-$:o (6)

Figure 18.8.10


ExAMPLE 5: Draw a sketch of thegraph of the equation

4xz - yz * 2522 - 0


soLUrIoN: The given equation can be written as

x2

25

which is of the form of Eq. (6) with y and z interchanged. Therefore, thesurface is an elliptic cone having the y axis as its axis. The surface inter-sects the xz plane (y : 0) at the origin only. The intersection of the surfacewith the yz plane (r : 0) is the pair of intersecting lines U : +52, and theintersection with the xy plarre (z: 0) is the pair of intersecting linesy:-+zx. The cross sections in the planes A:k+ 0 are the ellipsesf 125 + z2l4:WlL00.In the planes x:k * 0 and z:k # 0, the cross sec-tions are, respectively, the hyperbolas y2lt00-z'14:l&125 and y21100- f IZS: k214. A sketch of the surface is shown in Fig. 18.8.11.

F igure 18 .8 .11

The general equation of the second degree in x, y, and z is of the form

af -t byz * czz * dxy * exz + fyz + gx * hy * iz * i : g

where a,b, . . . , j are constants. It can be shown that by translation androtation of the three-dimensional coordinate axes (the study of which is

beyond the scope of this book) this equation can be reduced to one ofthe following two forms:

A x z * B y ' , * C z z + l - 0

or

A* + Byz * lz:0 (8)

Graphs of the equations of the second degree will either be one of theabove six types of quadrics or else will degenerate into a cylinder, plane,line, point, or the empty set.

The nondegenerate cur:ves associated with equations of the form (7)are the central quadrics and the elliptic cone, whereas those associatedwith equations of the form (8) are the noncentral quadrics. Following areexamples of some degenerate cases:

f - y ' :0 ; two p lanes , x - Y :0 and r * Y =o

(7)


z2 :0; one plane, the ry plane

* * Y': 0; one line' the z axis

f * y'l z2:0; a single Point, the origin

f -t y' I z? * 1 : 0; the empty set

Exercises 1.8.8

In Exercises L through 12, draw a sketch of the graph of the given equation and name the surface.

t . 4 * + 9 y 2 l z 2 : 3 6 2 . 4 * - 9 y z - " 2 - * 3 . 4 * + 9 y 2 - * : 3 6 4 . 4 * - 9 y 2 * * : 3 6

s . f :Vz -2z 6 . f : y ,+2 , , . * * * :n , a , . f i+ { :+

,"2 ,2

9 . r e - * :gy 10 . f : 2y *42 17 . f +7622 :4y2 -76 12 .9y2 -4z2 t t 8 r : 0

13. Find the values of k for which the intersection of the plane x + W: 1 and the elliptic hyperboloid of two sheetsy'- * - z2 :1 is (a) an ellipse and (b) a hyperbola.

L4. Find the vertex and focus of the parabola which is the intersection of the plane y :2 with the hyperbolic paraboloidy ' _ x " _ 11 6 4 9

15. Find the area of the plane section formed by the intersection of the plane y : 3 with the solid bounded by the ellipsoid

f 7 f 2 2: * 1 * _ : 19 2 5 4

16. Show that the intersection of the hyperbolic phrabolo id yzlbz - * I a, : zl c and the plane z : bx I ay consists of twointersecting straight lines.

17' U1e the method of parallel plane sections to find the volume of the solid bounded by the ellipsoid rfl a2 + y17 * zzl cz : l.(The measure of the area of the region enclosed by the ellipse having semiaxes i and, b is nab.)

1'8.9 CURVES IN R3 We consider vector-valued functions in three-dimensional space.

18.9.1 Definition Let f u fzi and1" be three real-valued functions of a real variable f. Then forevery number f in the domain common to fr, fz, and fs there is a vector Rdefined by

R(r) : fJt)i+ f,(t)i+ fr(t)u (1)and R is called a aector-oalued function.

The graph of a vector-valued function in three-dimensional space isobtained analogously to the way we obtained the graph of a vector-valuedfunction in two dimensions in sec. 17.4. As f assumes all values in thedomain of R, the terminal point of the position representation of thevector R(f) traces a curve C, and this curve is called the graph of (l). Apoint on the curve C has the cartesian representation (r, a, z) , where

18.9 CURVES IN R3 865

x : f r ( t ) y : f r ( t ) z : f r ( t ) (2 )

Equations (2) are called parametric equations of C, whereas Eq. (L) iscalled a aector equation of. C. By eliminating f from Eqs. (2) we obtain twoequations in x, y , and z. These equations are called cartesian equations of. C.Each cartesian equation is an equation of a surface, and curve C is the in-tersection of the two surfaces. The equations of any two surfaces con-taining C may be taken as the cartesian equations defining C.

o rLLUsrRArroN 1: We draw a sketch of the curve having the vectorequation

R(f) : a cos fi * b sin fi + tk

Parametric equations of the given cuwe are

' x : a c o s t A : b s i n t z : t

To eliminate f from the first two equations, we write them as

f : a2 cosz t and A2: V sinz t

from which we get

x2

;: cosz t and

Adding corresponding members of these two equations, we obtain

f +a ' : 1a2 b2

Therefore, the cuwe lies entirely on the elliptical cylinder whose directrixis an ellipse in the xy plane and whose rulings are parallel to the z a><is.Table 18.9.1 gives sets of values of. x, y, and z fot specific values of f. Asketch of the curve is shown in Fig. 18.9.1.

Figure 18.9.1

Table 18.9 .1

t

0

TT

4

7T

23n4

7f

3rr2

v

0b\D

b

b\D0

- b

z

0l l

4

7T

23n47T

3n2

x

f l

a\D.

0

a- -t/z

-a

0

VECTORS IN THREE-DIMENSIONALSPACE AND SOLID ANALYTIC GEOMETBY

The curve of Illustration 1 is called a helix. If a : b , the helix is called acircular helix and. it lies on the right-circular cylinder f + yz: oz.

o rLLusrRATrow 2: The curve having the vector equation

R(f) - f i+ Pi+ f3k

is called a twisted cubic. Parametric equations of the twisted cubic are

x : t A : P z : F

Eliminating f from the first two of these equations yields A : *, which isa cylinder whose directrix in the xy plane is a parabola. The twisted cubiclies on this rylinder. Figure 18.9.2 shows a sketch of the cylinder and theportion of the twisted cubic from t:0 to t:2' o

Many of the definitions and theorems pertaining to vector-valuedfunctions in two dimensions can be extended to vector-valued functionsin three dimensions

If R(f) - f,(t)i + hgi * /3(f)k, then

lim R(f) : l im fi(t)i + lim hg)i + lim /s(f)kt-tr

, l- fi t-tr t-tr

if l im/r(f), l i lrl.f2Q), and lim/3(f) all exist.t-tr t-tr t-tr

The vector-valued function R is continuous at ft if and only if

(i) R(tl) exists;

(ii) ,lig

R(f) exists;

(ii i) l im R(t) : R(t,).t - h

18.9.4 Definition The derivative of the vector-valued function R is a vector-valued function,denoted by R' and defined by

R'(t):tt-M{f-E(1I

if this limit exists.

18.9.5 Theorem If R is the vector-valued function defined by

R(f) : f(t)i + h@i +/,(r)r.and R'(f) exists, then

R'(t) : fr ' (t) i + fr ' (t) i + fs'(t)kThe proof of Theorem 18.9.5 is left as an exercise (see Exercise 9).

18.9.2 Definition

18.9.3 Definition

Fig u re

z R(f + af) - R(f)

F igure 18.9.3


The geometric interpretation for the derivative of R is the same as that

for the derivative of a vector-valued function in R2. Figure 18.9.3 shows a

portion of the curye c, which is the graph of R. In the figure oF is the

position representation of R(f), O? is the position representation of

R(t + Af ), and so fQ is a representation of the vector R(f + At) - R(f )' As

Af approaches zero, the vector tR(t + af) - R(f)l/At has a representation

appioaching a directed line segment tangent to the curve C at P'

The definition of the unit tangent vector is analogous to Definition

17.8.'1, for vectors in the plane. So if T(t) denotes the unit tangent vector to

curve C having vector equation (1), then

. rLLusrRArroN 3: We find the unit tangent vector for the twisted cubic

of Illustration 2.

Because R(f) : ti+ t2i + fsk,

D'R(t) : i+ 2ti + 3t2k

and

lD'R(t)l: IGW49F

From Eq. (3), we have, then,

1T(t) : (i+ 2ti + 3f'zk)

Therefore, in particular,

(3)

r (1)- 1 i r 2 i r ?

\M r ' \m t '@k

Figure 18.9.4

Figure 18.9.4 shows the representation of T(1) at the point (1, 1, 1)' '

Theorems 17.5.6,77.5.7, and 17.5.8 regarding derivatives of sums and

products of two-dimensional vector-valued functions also hold for vectors

in three dimensions. The following theorem regarding the derivative of

the cross product of two vector-valued functions is similar to the colre-

sponding formula for the derivative of the product of real-valued func-

tions; however, it is important to maintain the correct cirder of the vector-

valued functions because the cross product is not commutative.

18.9.5 Theorem If R and Q are vector-valued functions, then

D' [R( t ) x Q( f ) ] : R( f ) x Q ' , ( t ) + R ' ( f ) x Q( f )

for all values of f for which R'(f) and Q'(t) exist'

The proof of Theorem 18.9.6 is left as an exercise (see Exercise 10).


ExAMPLE 1: Given the circularhelix R(f) - n cos ti* a sin fj+ tk, where a ) 0, find the lengthof arc from t - 0 to t - 2n.

We can define the length of an arc of a curve C in three-dimensionalspace in exactly the same way as we defined the length of an arc of a.curvein the plane (see Definition 77.6.1). If C is the curve having parametricequations (2), fr' , fr' , f"' are continuous on the closed interval fa, bf , andno two values of f give the same point (r, y, z) on C, then we can prove(as we did for the plane) a theorem similar to Theorem 17.6.3, whichstates that the length of arc, L units, of the curve C from the point(f'(n), fr(a), ft(a)) to the point (/,(b), f"(b), fs(b)) is determined by

If s is the measure of the length of arc of C from the fixed point(fr(t), fr(t), h! i) to the variable point (/ ,(f), fr(t), f"(t)) and s in-creases as f increases, then s is a function of fand is given by

du

As we showed in Sec. 17.6 for plane curves, we can show that if (1) is avector equation of C, then

D6 : lD 'R ( f ) l (5 )

units, given by (4), also can be determined byand the of atc, L

sol.urroN: DrR(t) : -a sin ti + a cos fi + k. So from (7) we obtain

dt

\m dt-zn\ff i

Thus, the length of arc is 2nTa4 "1, units.

The definitions of the curaature aector K(f) and the curoature K(t) at apoint P on a curye C in Rs are the same as for plane curyes given in Defini-non 17.9.1. Hence, if T(f) is the unit tangent vector to C at P and s is themeasure of the arc length from an arbitrarily chosen point on C to P, wheres increases as f increases, then

K(t) : D".1(t)

or, equivalently,

(4)

(5)

(7)

length

18.9 CURVES IN 83

and

K(r ) : IDJ( ' ) lor, equivalently,

K(r) : lP,I ! l1 | (e)l lD,R(f) l l

Taking the dot product of K(t) and T(f) and using (8), we get

(10)

( 1 1 )

(12)

Theorem 1,7.5.1,1, states that if a vector-valued function in a plane has aconstant magnitude, it is orthogonal to its derivative. This theorern arrd itsproof also hold for vectors in three dimensions. Therefore, becauselT(t) | : L, we can conclude from (10) that K(f) 'T(t) : 0. And so the cur-vature vector and the unit tangent vector of a curve at a point areorthogonal.

We define the unit normal aector as the unit vector having the samedirection as the curvature vector, provided that the curvature vector is notthe zero vector. So if N(t) denotes the unit normal vector to a curve C at apoint P, then if K(t) + 0,

From (L1) and the previous discussion, it follows that the unit normalvector and the unit tangent vector are orthogonal. Thus, the angle betweenthese two vectors has a radian measure of lzzr, and we have from Theorem'J,8.6.6,

l r ( r ) x N(r) | : l r ( r ) l lN(r) | s in #z' : (1) (1) (1) : 1

Therefore, the cross product of T(t) and N(f ) is a unit vector. By Theorem18.5.9 T(f) x N(t) is orthogonal to both T(f) and N(f); hence, the vectorB(f), defined by

B ( f ) - r ( f ) xN ( r )

F igure 18 .9 .5

EXAMP rn 2: Find the movin g tri-hedral and the curuature at anypoint of the circular helix of Ex-ample 1.

is a unit vector orthogonal to T(f ) and N(t) and is called the unit binormalaector to the curve C at P.

The three mutually orthogonal unit vectors,T(t), N(f), and B(f), of acurve C are called the moaing trihedral of C (see Fig. 18.9.5).

solurroN: A vector equation of the circular helix is

R(f ) - a cos t i * a s in t i+ tk

So DrR(f )get 1

T(f) : f f i (-a sin t i+

k and lD,R(r)l : \m. From (3) we

a cos fi + k)


So

D; r ( r ) : +?acos\/az + i

Applying (8), we obtain

sin fj )

K ( f ) (-a cos ti- a sin fj)

The cunrature, then, is given by

K(r ; - lK( f ) l

and so the curvature of the circular helix is constant. From (1r) we get

N(f) : -cos fi - sin fi

Applyrng (12), we have

1B(t) :

ffi (-a sin fi* a cos ti + k) x (-cos fi- sin fi)

1-- (sin t i- cos t i+ ak)\/a2 + 1

A thorough study of curves and surfaces by means of carculus formsthe subject of. diff er ential geometry. The use of the calculus of vectors furtherenhances this subject. The previous discussion has been but a short in-troduction.

we now consider briefly the motion of a particle along a curve inthree-dimensional space. If the parameter f in the vector e-quation (1)measures time, then the position at f of a particle moving along the curveC, having vector equation (t ), is the point p(f Jt) , fr(t), /r(r) ). fn e aeloc_ity aector, V(f), and the acceleration oector, A(f), are defined as in theplane. The vector R(f) is called the position aector, and,

V ( f ) : D ' R ( f )

and

A( f ) :D tY ( t ) : D fR( t )

The speed of the particle at t is the magnitude of theBy applying (5) we can write

l v ( t ) l : D t s

sol,urroN: A vector equation of the curye is

R ( f ) - 3 f i + t z i+& t ' k

Therefore,

V(f) : DrR(t) - 3i + 2ti + 2t2k

(13)

(14)

velocity vector.

EXAMPLE 3: A particle is movingalong the curve havin gparametricequations x: 3t, y : t2, andz : zrtt.Find the velocity andacceleration vectors and the speed

of the particle at t - L. Draw asketch of a portion of the cunre att - 1,, and draw representationsof the velocity and accelerationvectors there.


and

A(t; : DNG) :2i + 4tkAlso,

lv(r)l: \trTW+4So when t: !, Y: 3i * 2j + 2k, A:2i * 4k, and lv(l) | : \/F. Therequired sketch is shown in Fig. 18.9.6.

v(1)

Exercises 18.9

In Exercises 1 through 4, find the unit tangent vector for the curve having the given vector equation.

1 . R ( r ) : ( r + l ) i - r 3 i + ( 1 - 2 f ) k 2 . R ( t ) : s i n 2 f i l c o s 2 t i + 2 t } t ? k

3 . R ( f ) : e t cos t i - t e t s i n t i * e t k 4 , R ( f ) : t z i * ( t ++ f3 ) i+ ( t -+ t3 )k

In Exercises 5 through 8, find the length of arc of the curve fuom t' to t2.

5. The curve of Exercise 7; \ : -l; S": 2. 5. The curve of Exercise 2; tt : Q; 1, : 1.

7 . The curve of Exercise 3; tr: Q; 1r: 3. 8. The curve of Exercise 4; h: 0; tz: 'l'.

9. Prove Theorem 18.9.5. 10. Prove Theorem 18.9.5.

11. Wr i teavectorequat ionof thecurveof in tersect ionof thesur facesA:{and'z :x ! .

12. Prove that the unit tangent vector of the circular helix of Example 1 makes an angle of constant radian measure withthe unit vector k.

13. Find the moving trihedral and the curvature at the point where f : 1 on the tWisted cubic of Illustration 2.

14. Find the moving trihedral and the curvature at any point of the curve R(t) : cosh fi * sinh ti + tk.

In Exercises 15 through 18, find the moving trihedral and the curvature of the given curye at t: tt if they exist.

15. Thecurveof Exerc ise l ; t r : -1. 15. Thecurveof Exerc ise2; t t :9 .

17. The curve of Exercise 3; fr: 0. 18. The curve of Exercise 4; L:7.

In Exercises 19 through 22, aparticle is moving along the given curve. Find the velocity vector, the acceleration vector, and

the speed at t: ty Draw a sketch of a portion of the curve at t: tr and draw the velocity and acceleration vectors there.

19. The circular helix of Example L; tr: i tr. 20. x: t,y: $t2, z: iF; h:2.

2 1 . x : e 2 t , A : r - z t , 2 : t e z t ; t r : 1 . 2 2 . x : L l 2 ( t 2 + t ) , y : l n ( l * P ) , 2 : t a n - t t ; t r : 1 .

Figure 18.9.6


23. Prove that if the speed of a moving particle is constant, its acceleration vector is always orthogonal to its velocity vector.

24. Prove that for the twisted cubic of Illustration 2, iI t + 0, no two of the vectors R(f), V(f), and A(f) are orthogonal.

25. Prove that if R(t) : f ,(t)i + h@i + /r(t)k is a vector equation of curve C, and K(t) is the curvature of C, then

K(11 : lD,R(f ) x D,2R(f ) l

lD,R( f ) l '26. Use the formula of Exercise 25 to show that the curvature of the circular helix of Example I is al(az + l).

In Exercises 27 and28, find the curvature of the given curve at the indicated point.

2 7 . x : t , y : t " , 2 : f ; t h e o r i g i n . 2 8 . x : e ' , y : e - t , 2 : t ; t : 0 .

29. Prove that if R(f) : f'(t)i+ fr(t)j + /B(f)k is a vector equation of curve C, K(t) is the curvature of C at a point P, ands units is the arc length measured from an arbitrarily chosen point on C to P, then

D,R(t) . D"3R(f) : -[K(f)],

18.10 CYLINDRICAL AND Cylindrical and spherical coordinates are generalizations of polar coordi-SPHERICAL COORDINATES nates to three-dimensional space. The cylindrical coordinate represen-

tation of a point P is (r, 0, z), where r and 0 are the polar coordinates of theprojection of P on a polar plane and z is the directed distance from thispolar plane to P. See Fig. 18.10.1.

x

F igu re 18 .10 .1

EXAMPLE L: Draw a sketch of thegraph of each of the followingequations where c is a constant:( a ) r : c ; ( b ) 0 : c ; ( c ) z : c .

solurroN: (a) For a point P(r,0, z) on the graph of r: c, 0 and z canhave any values and r is a constant. The graph is a right-circular cylinderhaving radius lcl and thez axis as its axis. A sketch of the graph is shownin Fig. 18.10.2.

(b) For all points P(r, 0, z) on the graph oL 0: c, r andz can assumeany value while 0 remains constant. The graph is a plane through the zaxis. See Fig. 18.10.3 for a sketch of the graph.

(c) The graph of z: c is a plane parallel to the polar plane at adirected distance of c units from it. Figure 18.10.4 shows a sketch of thegraph.

(r ,0, z)

z

z

7v,/

P(*, a, z)(r, o, z)

Xx

18.10 CYLINDRICAL AND SPHERICAL COORDINATES

Figure 18 .10 .4

The name "cylindrical coordinates" comes from the fact that the graphof r : c is a right-circular cylinder as in Example 1(a). Cylindrical coordi-nates are often used in a physical problem when there is an axis of sym-metry.

Suppose that a cartesian-coordinate system and a cylindrical-coor-dinate system are placed so the ry plane is the polar plane of the cylin-drical-coordinate system and the positive side of the r axis is the polar axisas shown in Fig. 18.10.5. Then the point P has (r, y, z) and (r, 0, z) as twosets of coordinates which are related by the equations

solurroN: (a) Multiplying on both sides of the equation by r, we gett2: 5r sin 0. Because t2: X, * y, and r sin 0: y, we have X2 + yz: 6r.This equation can be written in the fonrt xz + (V - 3)': 9, which showsthat its graph is a right-circular cylinder whose cross section in the ryplane is the circle with its center at (0,3) and radius 3.

(b) Replacing'/ cos 0 by x and r sin 0 by y, we get the equation 3r*2y*62:0. Hence, the graph is a plane through the origin and has(3,2, 6> as a normal vector.

( 1 )

(2)F igure 18 .10 .5

ExAMPLEZ: Find an equation incartesian coordinates of the fol-lowing surfaces whose equationsare given in cylindrical coordin-ates and identify the surface:

( a ) r - 6 s i n 0 ;( b ) r ( 3 c o s 0 + 2 s i n 0 ) * 5 2 - 0 .

EI4 VECTORS IN THREE-DIMENSIONAL SPACE AND SOLID'ANALYTIC GEOMETRY

EXAMPLE 3: Find an equation incylindrical coordinates for each ofthe following surfaces whoseequations are given in cartesiancoordinates and identify the sur-face: (a) xz + !' : z;( b ) x 2 - A ' : 2 .

solurroN: (a) The equation is similar to Eq. ( ) of Sec. 18.8, and so thegraph is an elliptic paraboloid.lf. * * y2 is replaced by r', the equationbecomes f : z.

(b) Ttre equation is similar to Eq. (5) of Sec.18.8 with x and y in-terchanged. The graph is therefore a hyperbolig paraboloid having the zaxis as its axis. When we replace x by r cos 0 and y by t sin 0, the equationbecomes rz cosz 0 - rz sinz 0 : 4 andbecause cosz 0 - sinz 0: cos 20,wecan write this as z:12 cos20.

F igure 18 .10 .6

ExAMPrn 4: Draw a sketch of thegraph of each of the followirrgequations where c is a constant:( a ) p : c , a n d c > 0 ; ( b ) 0 _ c ;( c ) 6 - c , a n d 0

In a spherical-coordinate system there is a polar plane and an axis per-pendicular to the polar plane, with the origin of the z axis at the pole of thepolar plane. A point is located by three numbers, and the spherical-coor-dinate representation of a point P is (p, 0,S), where p : lOPl, I is theradian measure of the polar angle of the projection of P on the polar plane,and @ is the nonnegative radian measure of the smallest angle measuredfrom the positive side of the z axis to the line OP. See Fig. 18.10.6. The ori-gin has the spherical-coordinate representation (0,0,6), where 0 and $may have any values. If the point P (p, 0 , @) is not the origin, then p > 0and 0 < 6 < zr, where Q:0 if P is on the positive side of the z axis and6: r if P is on the negative side of the z axis.

soLUrIoN: (a) Every point P(p, 0, S) on the graph of p: c has the samevalue of p, 0 rnay be any number, and 0 = Q = zr.It follows that the graphis a sphere of radius c and has its center at the pole. Figure 18.1,0.2 shows asketch of the sphere.

(b) For any point P (p, 0, Q) on the graph of 0 : c, p may be any non-negative number, d may be any number in the closed interval 10, nf , and 0is constant. The graph is a half plane containing the z axis and is obtainedby rotating about the z axis through an angle of c radians that half of. the xzplane for which r > 0. Figure 18.10.8 shows sketches of the half planes for0 : i r , 0 : & r r , 0 : t r r , a n d g : - * 2 .

F igure 18 .10 .7

o: :27

F igu re 18 .10 .8

x

F i g u r e 1 8 . 1 0 . 1 0

ExAMPLE 5: Find an equation incartesian coordinates of the fol-lowing surfaces whose equations

18.10 CYLINDRICAL AND SPHERICAL COORDINATES 875

(c) The graph of 6: c contains all the points P(p,0, d) for which p isany nonnegative number, 0 is any number, and S is the constant c. Thegraph is half of a cone having its vertex at the origin and the z axis as itsaxis. Figure 18.10.9a and b each show a sketch of the half cone for 0 <c < *n and izr < c < zr, respectively.

(a)

F i gu re 18 .10 .9

I

+ "<c1r(b)

Because the graph of p: c is a sphere as seen in Example 4(a),wehave the name "spherical coordinates." In a physical problem when thereis a point that is a center of symmetry, spherical coordinates are oftenused.

By placing a spherical-coordinate system and a cartesian-coordinatesystem together as shown in Fig. 18.10.10, we obtain relationships be-tween the spherical coordinates and the cartesian coordinates of a pointP from

sin 0 z - laPlbecome

By squaring each of the equations in (3) and addingr w€ have

)c2 + y, + 22: p2 sin, + cos2 0 + p2 sin'0 sin2 0 + p2 cos'0

-- p2 sinz @ (cosz 0 + sin2 g) * p' cosz (0

: pz(sin2 Q + cosz 0)

: p 2

solurroN: (a) Because z: p cos {, the equation becomes z: 4. Hence,

the graph is a plane parallel to the xy plane and 4 units above it'(U) for spherical coordinates p>0 and sinS>0 (because 0<

4/' -* -

Because

(3)

0 < c < ;

P(*, a, z)(P ,0 ,6 )

c o s 0 y :

: p sin ,f and

876 VECTORS IN THREE.DIMENSIONAL SPACE AND SOLID ANALYTIC GEOMETRY

are given in spherical coordinatesand identify the surface:(a ) p cos Q: 4 ; (b ) p s in Q: 4 .

Q = rr); therefore, if we square on both sides of the given equation, we ob-tain the equivalent equation pz sinz 6:16, which in tum is equivalent to

P " ( l - c o s 2 { ) : 1 5or

Pz - P2 cos' 0: L6

Replacing p'by x2 + y' +

x 2 + y ' + 2 2 - 2 2 : L 5

or, equivalently,

x2 + A2 :15

z2 and p cos 6 by z, we get

Therefore, the graph is the right-circular cylinder having the z axis as itsaxis and radius 4.

sor,urroN: (a) A cartesian equation of the elliptic paraboloid of Example3(a) is *'+ y': z. Replacing xby p sin @ cos 0, y by p sin @ sin 0, and zby p cos Q, we get

p2 sin' Q cosz 0 * p'sinz ,f sin2 cos oor, equivalently,

p2 sin2 S(cosz d * sinz 0) : p cos S

which is equivalent to the two equations

P:0 and p sin2 S : cos S

The origin is the only point whose coordinates satisfy p : 0. Because theorigin (0, 0,En) lies on p sinz @: cos 6, we can disregard the equationp:0. Furthermore, sin { # 0 because there is no value of d for whichboth sin d and cos f are 0. Therefore, the equation p sin2 { : cos f can bewritten as p: cscz @ cos @, or, equivalently, p: csc @ cot {.

(b) A cartesian equation of the plane of Example 2(b) is 3x * 2y* 6z:0. By using Eqs. (3) this equation becomes

3p sin @ cos 0 *2p sin $ sin 0 * 6p cos g : g

ExAMPLE 6: Find an equation inspherical coordinates for: (a) theelliptic paraboloid of Example3(a); and (b) the plane of Ex-ample 2(b) .

Exercises 18.10

1. Find the cartesian coordinates of the point having the given cylindrical coordinates:

(a) (3, tn, 5) (b) (7 , 3n, -4)

2. Find a set of cylindrical coordinates of the point having the given cartesian coordinates:

( c ) ( L , 1 , ' 1 . )

( c ) ( 1 , 1 , ' 1 , )(a) (4, 4, -2) (b) (-3 \n, 3, 6)

18.10 CYLINDRICAL AND SPHERICAL COORDINATES 877

3. Find the cartesian coordinates of the point having the given spherical coordinates:

(a) (4, tn, tr) (b) (4, tn, *n) (c) $re, *n, *n)

4. Find a set of spherical coordinates of the point having the given cartesian coordinates:

(a) (1, -L, -rt) (b) (-1 , tE, 21 (c) (2, 2, 2)

5. Find a set of cylindrical coordinates of the point having the given spherical coordinates:

(a) (4,32r,*r) (b) (\/r, tu,n) (d Q|l, tu,*n)

5. Find a set of spherical coordinates of the point having the given rylindrical coordinates:

(a ) (3 , t n ,3 ) (b ) (3 ,1n ,2 ) ( c ) ( 2 , f u ' - 4 )

Exercises 7 through 10, find an equation in cylindrical coordinates of the given surface and identify the surface.

f * V ' 1 4 2 2 : 1 6 8 ' * - y 2 : 9

f - Y 2 : 3 2 ' l 0 ' * + A 2 : z z

Exercises 11 through 14, find an equation in spherical coordinates of the given surface and identify the surface.

In

7.

9 .

In

LL,

13.

x 2 + y ' , + 2 2 - 9 2 - 0

)c2 + a', :9

1 5 . ( a ) r - 4 ; ( b ) 0 - i n

L7. r2 cos 20 - zB

In Exercises 15 through 1g, find an equation in cartesian coordinates for the surface whose equation is given in cylindrical

coordinates. In Exercises 15 and 16, identify the surface'

L 6 . r : 3 c o s 0

L8. z2 sin3 0 : rg

L 2 . x 2 + A 2 : z '

1,4. x2 + U' : 2z

22 . p :6 s i n d s i n 0+ 3 cos t f

in cylindrical coordinates:

In Exercises 1g through 22, find, an equation in cartesian coordinates for the surface whose equation is given in spherical

coordinates. In Exercises 19 and 20, identify the surface'

19. (a) p: s; (b) e -- in; (c) Q: *r 20' p:9 sec @

2 I . P : 2 t a n 0

29. A curve C in R3 has the following Parametric equations

r : F J t ) 0 - F r ( t ) z - F ( f )

Use formula (4) of Sec. 18.9 and formulas (1) of this section to Prove that if L units is the length of arc of C from the

point where t: a to the point where f : b, then

lbr : l -661

J a

24. A curve C in Rr has the following parametric equations in spherical coordinates:

p : G , ( t ) o : GzQ) d : G ' ( f )

Use formula (4) of Sec. 18.9 and formulas (3) of this section to

point where t : a to the point where t : b, then

rbL- l f i ) z * p ' (D,Q)z d t

J a

prove that if L units is the length of arc of C from the

878 VECTORS IN THREE.DIMENSIONAL SPACE AND SOLID ANALYTIC GEOMETRY

25. (a) Show that parametric equations for the circular helix of Example

a 0 - t z - t

18.9 , are

(b) Use the formula of Exercise 23 to find the length of arc of the circular helix of part (a) from f : 0 to f : 2zr. Checkyour result with that of Example 1, Sec. 18.9.

26. A conical helix winds around a cone in a way similar to that in which a circular helix winds around a cylinder. Use theformula of Exercise 24 to find the length of arc from f : 0 to f :2r oI the conical helix having parametric equations

p : t 0 : t Q : t n

Reaiew Exercises (Chaqter 18)

1. Draw a sketch of the graph of x- 3 in Rt, R', and R3.

2. Draw a sketch of the set of points satisfying the simultaneous equations x: 6 and, y: 3 in Rz and R3.

In Exercises 3 through LL, describe in words the set of points in Rs satisfying the given equation or the given pair of equa-tions. Draw a sketch of the graph.

? f v :ov ' [ z - o

5 . x 2 + 2 2 : 4

9. x2 + A2 :92

In Exercises 12 through 17, letindicated vector or scalar.

1 2 . 3 A - z t s _ + C

1 5 . D . B x C

E - 5i - 2i, and find the

In Exercises 18 throrrgh 23,Insert the parentheses and

1 9 . B . A - C

2 1 , . A 8 . C

1 3 . 6 C + 4 D - E

1 6 . ( A x C ) - ( D x E )

there is only one way that a meaningfulfind the indicated vector or scalar if A -

1 , 9 . A + B . C

2 2 . A X B . C X A

( A . i ) i + ( A . i ) i + ( A . k ) k .

5 .

8.

1 1 .

i + 2 i - 2 k ,

T 4 . 2 8 . C + 3 D . E

17 . lA x B l lD x E lexpression can be obtained by inserting parentheses.(3 , - 2 , 4> , B : ( - 5 , 7 ,2 ) , and C - ( 4 , 6 , - 1 ) .

4 . { * _ ,l y : z

7. x - y

10. )c2 + A, : z,

A-- i + 3 i + 2k, B - 2 i+ i - 4k,

2 0 . A . B C

2 3 . A x B . A + B - C

24. If A is any vector, prove that A -

25. Find an equation of the sphere concentric with the sphere * * y, + zz + 4x +point (-4, 2,5).

2y - 6z * 10 : 0 and containing the

26. Find an equation of the surface of revolution generated by revolving the ellipse 9* * 422: 36 in the xz plane aboutthe r axis. Draw a sketch of the surface.

Determine the value of c so that the vectors 3i * ci - 3k and 5i - 4i * k are orthogonal.

Showtha t the rea re rep resen ta t i onso f the th reevec to rsA :5 i+ i -3k ,B : i *3 i -2k ,andC: -4 i+2 i * kwh ichform a triangle.

Find the distance from the origin to the plane through the point (-6,3,-2) and having 5i - 3j + 4k as a normal vector.Find an equation of the plane containing the points (1,7 , -3) and (3, 1, 2) and, which does not intersect the r axis.Find an equation of the plane through the three points (-1,2,l) , (1,4, 0), and (1,-1,3) by two methods: (a) usingthe cross product; (b) without using the cross product.

29.

30.

3r.

REVIEW EXERCISES

Find two unit vectors orthogonal to i - 3i * 4k and whose representations are parallel to the yz plane.

Find the distance from the point P(4, 6,-4) to the line through the two points A(2,2,1) and B(4,3,-1').

Find the distance from the plane 9r - 2y 'f5z * 44: 0 to the point (-3, 2, 0).

I f 0 is the radianmeasure of theanglebetween thevectors A:2 i+ i+ kandB:4i -3 i * 5kf indcos 0 in two ways:

(a) by using the dot producf (b) by using the cross product and a trigonometric identity.

36. Prove that the lines

x - l a + 2 z - 2 . x - 2 Y - S z - 5_ _ _ : 2 : 2 a n d - n - : 3 : - - -

are skew lines and find the distance between them'

32. Find symmetric and parametric equations of the line through the origin and perpendicular to each of the lines of

Exercise 36.

Find the area of the parallelogram two of whose sides are the position representations of the vectors2i - 3k and 5i * 4k'

Find the volume of the parallelepiped having vertices at (1, 3, O) , (2, -1, 3) , (-2,2' -1), and (-1,l' 2) '

Find the length of the arc of the curve

l : f c o s t y : t s i n t z : l

from f: 0 to t: i tr.

A particle is moving along the curve of Exercise 40: Find the velocity vector, the acceleration vector, and the speed at

t j-

tn. Draw a sketch of iportion of the curve at t: Ln and draw the representations of the velocity and acceleration

vectors there.

r12. Find the unit tangent vector and the curvature at any point on the curve having the vector equation

R(t ) : e t i * e- t i I2 tk

43. Find an equation in cylindrical coordinates of the graph of each of the equations: (a) (r * !)2 + t : z; (b') 25* r +yz : 1gg'

44. Find an equation in spherical coordinates of the graph of each of the equations: (a) r' i Y'+ 422:4; (b) 4f - 4y2

l 9 z 2 : 3 6 '

415. If R, e, and W arethree vector-valued functions whose derivatives with respect to f exist, prove that

D , [ R ( r ) . e ( r ) x w ( f ) ] : D , R ( f ) . Q ( f ) x w ( t ) + R ( t ) ' D r Q ( t ) x w ( t ) + R ( t ) ' Q ( t ) x D ' w ( t )

879

32.

33.

34.

35.

38.

39.

40.

4r.

Differential calculusof functionsof several variables

I9.I FUNCTIONSTHAN ONE

OF MOREVARIABLE

19.1 FUNCTIONS OF MORE THAN ONE VARIABLE

In this section we extend the concept of a function to functions of n vari-

ables, and in succeeding sections we extend to functions of n variables the

concepts of the timit of a function, continuity of a function, and derioa-

tive of a function. A thorough treatment of these topics belongs to a course

in advanced calculus. In this book we confine most of our discussion of

functions of more than one variable to those of two and three variables;

however, we make the definitions for functions of n variables and then

show the applications of these definitions to functions of two and three

variables. W" utto show that when each of these definitions is applied to a

function of one variable we have the definition previous$ given.

To extend the concept of a function to functions of any number of vari'

ables, we must first consider points in n-dimensional number space. Justas we denoted a point in Rl by a real numb et x, apoint in R2 by an ordered

pair of real numterc (x, y), and a point in R3 by an ordered triple of real

numbers (x, y, z), we rePresent a point in n-dimensional number sPace'

R", by an'oriered n-taple of real numbers customarily denoted by P :

( r r , * r , . . . , rn ) . Inpar . t i cu la r , i f .n :7 ,we le tP :x i i f n :2 'P : (x 'y ) ;

if. n: g, P : (x, y, z)t if. n: 6, P : (xr., x2, xg, xa, x5, x6).

The set of all ordered z-tuples of real numbers is called the n'dimensional

number space andis denoted by R'. Each ordered n-tuple (1,r, xr, . ' , xo)

is called a point in the n-dimensional number space'

Afunctionofnaariablesisasetoforderedpairsoftheform(P-.al)inwhich,ro t"o distinct ordered pairs have the same fiist element. P is a point in

n-dimensional numbe, -rpu."

and w is a real number' The set of all

p"rriUi" values of P is called the domain of the function, and the set

ff ail possible values of w is called the rnnge of the function'

From this definition, we see that the domain of a function of. n vari-

ables is a set of points in R" and that the range is a set of feal numbers or,

equivalently, " ""t

of points in Rl' When rt: L' we have a function of one

variable;thus,thedomainisasetofpointsinRlor 'equivalent ly 'asetofreal numbers, and the range is a set

-of real numbers' Hence' we see that

Definition .1..7.L is d specia-i case of Definition 19.1..2.lf. n:2, wehave a

function of two variables, and the domain is a set of points in R3 or, equiv-

;iliit, a set of ordered pairs of real numbets (x' y)' The range is a set of

real numbers.

. rLLUsrRArroN 1: Let the function f of two variables r and y be the set

of all ordered pairs of the form (P, z) such that

z: {zs777l

The domain of I is the set of all ordered pairs (x' y) for which 25 - * -

y, > o.This is tire set of all points in_the xy plane on the circle I * Az :25

and in the interior region bounded by the circle'

1r9.L.1 Definition

19.1.2 Definition

DIFFERENTIAL CALCULUS OF FUNCTIONS OF SEVERAL VARIABLES

Figure 1 9.1 .1

Because z: ry'F=@Tfr, we see that 0 < z < 5; therefore, therange of / is the set of all real numbers in the closed interval [0, 5]. InFig. 19.1.1 we have a sketch showing as a shaded region in R2 the set ofpoints in the domain of /. o

. rLLUsrRArron 2: The function g of two variables r and y is the set of allordered pairs of the form (P, z) such that

\/F t 7 -,5

vThe domain of g consists of all ordered pairs (r, y) for which * + yz - 25and y # 0. This is the set of points, not on the r axis, which are either onthe circle * * y, :25 or in the exterior region bounded by the circle.Figure 19.7.2 is a sketch showing as a shaded region in R2 the iet of pointsin the domain of g. o

Figure 19.1 .2 Figure 19.1 .3

o rllusrRArror.r 3: The function F of two variables x au';rd, y is the set of allordered pairs of the form (P, z) such that

z :yG+f , -2slf y:0, then z -- 0 regardless of the value of r. Howeveg iI y # 0, thenf * y'- 25 must be nonnegative in order for z to be defined] Therefore,the domain of F consists of all ordered pairs (r, y) for which either y = gor rP * y' - E > 0. This is the set of all points on the circle rp * yr-:25,all points in the exterior region bounded by the circle, and all poitrts onthe r axis for which -5 . r < 5 In Fig. 19.1.3 we have a sketclishowingas a shaded region in R2 the set of points in the domain of F. o

o rr,r,usrRArror.r 4: The function G of two variables r and y is the set of all


ordered pairs of the form (P, z) such that

f (9, -4) -

f (-2, r) :

f (u, 3a) :

: f f i -g: f f i -2 \E_f f i

vz -

Figure 19.1 .4

ffilf y:0, then z:0 provided that * * y'- 25 + 0. lf. y # 0, then 12 *y' - 25 must be positive in order f.or z to be defined. Hence, the domain ofG consists of all ordered pairs (x, y) for which x' * y' - 25 > 0 and those

' for which 3r : 0 and x # +-5. These are all the points in the exterior regionbounded by the circle 12 * y' :25 and the points on the r axis for which-5 ( r < 5. Figure 19.1.4 is a sketch showing as a shaded region in R2the set of points in the domain of G. o

If f is a function of n variables, then according to Definition 19.'1"2'

/ i sase to f o rderedpa i rso f the form (P,w) ,whereP: (x r ,x2 , . - . , xn)is a point in R" and ar is a real number. We denote the particular value of ar,which corresponds to a point P, by the symbol f (P) or f (xr, h, . . , xo).

In particular, 7f. n:2 and we let P : (x, A) , we can denote the function

value by either /(P) or f(x,y). Similarly, if. n:3 and P: (*, y, z), we

denote lhe function value by either /(P) or f(x, y, z). Note that if n: l,

P : xi hence, if f is afunction of one variable, f (P) : /(r). Therefore, thisnotation is consistent with our notation for function values of one variable.

A function f of. n vaiables can be defined by the equation

w : f ( \ , x 2 , . . , x n )

The variabl ES X1, Xz, . . ., xn ate called the indep endent a ariables, arrd w

is called tl:.e dependent aariable.

o ILLUsrRArron 5: Let / be the function of Illustration 1; that is,

f(x' Y) : \/8=7-

EXAMPLE 1: The domain of a

function I is the set of all ordered

triples of real numbers (x, Y, z)

such that

g (x , y , z ) Sxz * yz'

Find (a) S(1 , 4, -2);

(b) S(2 a, -b, 3c); (c) 8(x', y', z');(d) Sg, z,-x).

SOLUTION:

( a ) 8 ( 1 , 4 , - 2 ) : 1 2 - 5 ( 1 ) ( - 2 ) + 4 ( - Z ) ' : 1 ' + 1 0 + 1 6 : 2 7

(b) B(2a, 3c) - (2a)2 - 5(2a) (3c) + (-b) (3c)'

: 4a2 - 30ac - 9bc2

(c) g@r, Ar, zr) : (x')2 - 5(x')(z') + (y')(z')2: rc4 - 5x222 * y'zn

(d) SQ, z, -)c) : yz - Sy(-r) * z(-r)2 : y' + Sxy * xzz


19.1.3 Definition If / is a function of a single variable and g is a function of two variables,then the composite function f ' I is the function of two variables defined by

(f ' s)(x, y) : fQ@' y))and the domain of f " I is the set of all points (x, y) in the domain of gsuch that g@, y) is in the domain of /.

EXAMPTr 2: Given f (t) - ln fand g@,y ) - x2 * y , f i nd h (x , y )if h - f o S, and find the domainof. h.

19.1.4 Definition

EXAMPLE 3: Given F(r) - sin-l rand G(x, A, z) - ,find the function F o Q and itsdomain.

SOLUTION]

h(x, y) : ( f " i l@, y) : f k@' y)):11* + y): ln(* I y)

The domain of g is the set of all points in R2, and the domain of / is(0, +.;. Therefore, the domain of h is the set of all points (r, y) for whichf * y > 0 .

Definition 19.1.3 can be extended to a composite function of n vai-ables as follows.

If / is a function of a single variable andg is a functionof.n variables,then the composite function f " g is the function of. n vartables defined by

U " i l ( x r , x r , . . . , x ) : f ( g @ r , x 2 , . . , , x n ) )

and the domain of f " g is the set of all points (x1, x2, . . .,r,) in thedomain of g such thatg(rr, x2, . . . , x,\ is in the domain of f.

SOLUTION:(F . c) (x, y, z) :ii:##_r,: sin-l tp+f+Za

The domain of G is the set of all points (x, y, z) in Rs such that I * y, *zz - 4 > 0, and the domain of F is [-1, 1]. So the domain of F. G isthe set of a l l po ints (x , A, z) in R3 such that 0 <x2*yr*22-4 < l or ,equivalently, 4 < x2 * y, * z2 < 5.

- A.polynomial function of two variables r and y is a function / suchthat f (x, y) is the sum of terms of the form cxoy^, where c is a real num-ber and n and rfl are nonnegative integers. The degree of the porynomialfunction is determined by the largest sum of the exponents of x nd yappearing in any one terrn. Hence, the function / defined by

f (x, y) : 6ruy' - Sry" r 7*y - 2f + y

is a polynomial function of degree 5.

19.1.5 Definition

Figure 19 .1 .5

EXAMPLE 4; Draw a sketch ofthe graph of the function / havingfunction values f (x, A) : x' + y'.


A rational function of two variables is a function h such that h(x, y) :

f(x,y)lg@, y), where f and,g aretwopolynomialfunctions. Forexample,the function / defined by

, x2ll2l \ x , y ) : f f iE

^ ' !

is a rational function.The graph of a function / of a single variable consists of the set of

points (x, y) in R2 for which y : f (x). Similarly, the graph of a functionof two variables is a set of points in R3.

If / is a function of two variables, then the graph of f is the set of all points.(x, y, z) in Rs for which (x, y) is a point in the domain of / and z : f(x, Y).

Hence, the graph of a function f of. two variables is a surface whichis the set of all points in three-dimensional space whose cartesian coordi-nates are given by the ordered triples of real numbers (x, y , z). Becausethe domain of. f is a set of points in the xy plane, and because for eachordered pafu (x, y) in the domain of / there corresponds a unique valueof. z, no line perpendicular to the xy plane can intersect the graph of / inmore than one point.

o rLLUsrRArroN 6: The function of Illustration 1 is the function f which isthe set of all ordered pairs of the form (P, z) such that

z:VFE=TSo the graph of / is the hemisphere on and above the xy plane having aradius of 5 and its center at the origin. A sketch of the graph of this hemi-sphere is shown in Fig. 19.1.5. o

soLUrIoN: The graph of / is the surface having the equation z: f * y'.

The trace of the surface in the xy plane is found by using the equationz = 0 simultaneously with the equation of the surface. We obtain fI y' :0, which is the origin. The traces in the xz and yz planes are foundby using the equations y:0 and x:0, respectively, with the equationz: * * yz.We obtain the parabolas z: f andz: A2. The cross section ofthe surface in a plane z:k,Parallel to the xy plane, is a circle with its

center on the z axis and radius Vt. wittr this information we have the

required sketch shown in Fig. 19.1..6.

F igure 19 .1

886 DIFFERENTIAL CALCULUS OF FUNCTIONS OF SEVERAL VARIABLES

Figu re 19.1 ,7

EXAMPLE 5: Let f be the func-tion for which f(x, y) - 8 - x2- 2y. Draw a sketch of the graphof f and a contour map of fshowing the level curyes of f atL0, g, 6, 4, 2, 0, -2, -4, -6,

and -8.

There is another useful method of representing a function of two vari-ables geometrically. It is a method similar to that of representing a three-dimensional landscape by a two-dimensional topographical map. Supposethat the surface z: f (x, y) is intersected by the plane z: k, and the curveof intersection is projected onto the xy plane. This projected curve hasf (x, y): k as an equation, and the curve is called the leael curae (ot con-tour curce) of the function / at k. Each point on the level curve correspondsto the unique point on the surface which is k units above it if k is positive,or k units below it if k is negative. By considering different values for theconstant k, we obtain a set of level curves. This set of curves is called a con-tour map. The set of all possible values of k is the range of the function /,and each level curve, f (x, V) : k, in the contour map consists of the points(x, y) in the domain of / having equal function values of k. For example,for the function f of Example 4, the level curves are circles with the centerat the origin. The particular level curyes for z:7,2,9, 4,5, and 6 are shownin Fig. 19.1,.7.

A contour map shows the variation of z with x and y. The level curvesare usually shown for values of z at constant intervals, and the values of z

solurroN: A sketch of the graph of f is shown in Fig. 19.1.9. This is thesurface z:8- f -2y. The trace in the xy plane is obtained by settingz : 0, which gives the parabola x2 : -2(V - 4). Setting y : 0 and r : 0, w6obtain the traces in the xz and yz plarres, which are, respectively, theparabola rf : - (z - 8) and the line 2y * z: 8. The cross section of the sur-face made by the plane z : k is a parabola having its vertex on the line 2y* z : 8 in the yz plane and opening to the left. The cross sections for z : g,6, 4,2, -2, -4, -6, and -8 are shown in the figure.

The level curves of. f arc the parabolas f : -Z(y - 4++k). The con-

z : 2z - - 3

z - - 4

z : 5

z : 6

19.1 FUNCTIONS OF MORE THAN ONE VARIABLE 887

tour map of / with sketches of the required level curves is shown in Fig.1,9.1,.9.

x8

1.0

F igure 19 .1 .9

Figure 19.1 .8

Extending the notion of the graph of a function to a function of nvari-

ables, we have the following definition.

L9.1.6 Definition If / is a function of n variables, then the graph of. f is the set of all points

(xr, xr, . . . , xn, ar) in Ro+l for which (xr, xr, , xn) is a point in thedomain of. f and w : f(xr, xz,' . ., xn).

Analogous to level curves for a function of two variables is a similar

situation for functions of three variables. If / is a function whose domain is

a set of points in R3, then if k is a number in the range of /, the graph of the

equation /(r , y , z) : k is a surface. This surface is called the leoel surf ace of

1 it tc. Every sur{ace in three-dirnensional space can be considered as alevel surface of some function of three variables. For example, if the func-tion g is defined by the equation g(x, y, z) : f * y' - z, then the surface

shown in Fig. 19.1,.6 is the level surface of g at 0. Similarly, the surfacehaving equation z- f -y'+ 5:0 is the level surface of g at 5'

Exercises 19.L

l. Let thefunctionf of twovariables xandybethesetof allorderedpairsof the form(P,z) suchthatz: (r* y)l(x- V)'

F ind: (a) f (4,$; (b) f ( f ,y ' ) ; (c) l f (x ,V) l ' ; (d) f ( -x ,y) - f (x , -V) ; G) the domain of . f ; ( f ) therange of f 'Dtawa

sketch showing as a shaded region in R2 the set of points not in the domain of /.

2 . Le t t he func t i ongo f t h reeva r i ab les , x , y ,andz ,be these to f a l l o rde redpa i r so f t he fo rm (P ,w) such tha ta r :


w . F i n d : ( a ) s ( l ' , - t , _ t ) ; ( b ) s ? a , 2 b , * c ) ; ( c ) g ( y ,( f) lgQ, U,z) l ' - ISQ + 2, y * 2, z) l ' . Draw a sketch showing as

-x, -y); (d) the domain of g; (e) the range of g;a shaded solid in RB the set of points in the do-

InR2

3 .

main of g.

Exercises 3 through L1, find the domain and range of the function / and draw a sketch showing as a shaded region inthe set of points in the domain of./.

f (x ,a) : f f i 4 . f ( * , A ) : x \ f r - x 2 - Y z

7 . f ( x , t t \ : U ', x - y

10. f (x, A) : sin-l (x + Y)

the domain and range of the function f .

5 . f ( x ' a ) : ffi

13. f (x, y, z): sin-r x * cos-l y + tan-r z

the domain and range of the function / and draw a sketch of the graph.

5. f(x,a):rffis . f ( x , D :1lv l

In Exercises 12 through 14, find

12. f (x, y, z) - (x * y) tF.

1,4. f (x, y, z) - lxlsur'

In Exercises L5 through 20, frnd

15. f (x, A) : 4x2 * 9yz

1 8 . f ( * , A ) :

8 . f (x , a) : )c + Yxy

ll. f (x, y) : ln (xy - I)

17. f (x, A) : 16 - x ' - A'

20. f (r ,n : [7 : : . * vf . 0 i f . x : y

In Exercises 21 through 26, draw a sketch of a contour map of the function / showing the level curves of / at the givennumbers.

21. The function of Exercise 15 at 36, 25, 16,9, 4, l, and,0. 22. T]ne function of Exercise 16 at 5, 4, g,2,1, and 0.23. The function of Exercise 17 aL 8, 4, 0, -4, and -8. 24. The function of Exercise 18 at 10, g, 5, 5, and 0.25. The tunction / for which f(r, V) : t(* + y2) at g, 6, 4, 2, and 0.

26 . The func t i on / f o rwh ich f ( x , y ) : ( r - 3 ) l (V+2 ) a t4 ,2 ,1 , i , + ,0 , -+ , -+ ; , - t , - 2 ,and . -4 .

In Exercises 27 and 28, a function / and a function g are defined. Find, h(x, y) it h: f . g, and also find the domain of Iz.2 7 . f ( t ) : s i n - r t ; S ( x , y ) - @ 28. f (t): tan-r t; S(x,y) - tff iz2e. Givenl (x , ! ) :x-y ,sG): t / l ,n?) :s2.F ind(a)(8. f ) (s , r ) ; (b) f (h(s) ,5(q) ;G) f@(x) ,h(y) ) ; (d)S(( r , . f ) (x ,y) ) ;

(e) (s . h) ( f (x ,v) ) .30. Givenf(x ,y) :x ly ' ,sk) : * ,h(x) :16. r ind(a)(h" f ) (z , r ) ; (b) f (s(z) ,hG)) ; (c) f (s6/ i ) ,h( r , ) ) ; (d)h((S " f ) (x , i l \ ;(e) ( / t . i l ( f@,v)) .

The electric potential at a point (x, y) of the ry plane is V volts and V: +t {i=7=7. Draw the equipotential curvesfor V : 75, 12, 8, 4, l , t, and *.

The temperature at a point (x,y) of a flat metal plate is f degrees and f : 4* *2y2. Draw the isothermals for t:12,8, 4, 1, and,0.

For the production of a certain commodity, if r is the number of machines used and y is the number of man-hours,the number of units of the commodity produced. is f (x, y) and f (x, y): 5ry. Such a function / is called a productionfunction and the level curves oI f are called constant product curyes. Draw the constant product curves foi this func-tion / at 30,24, 18,12,5, and0.

3L.

32.

33.

19.2 LIMITS OF FUNCTIONS OF MORE THAN ONE VARIABLE 889

In Exercises 34 and 35, draw sketches of the level surfaces of the function f at the given numbers.

i l . f ( * , y ,2 ) : * * y ' I z2 a t9 ,4 ,7 , and0 . 35 . f ( x , ! , 2 ) : f * y " - 4z a t8 ,4 ,0 , -4 , and -8 .

I9.2 LIMITS OF FUNCTIONSOF MORE THANONE VARIABLE

19.2.1 Definition

19.2.2 Definition

19.2.3 Definition

In Rr the distance between two points is the absolute value of the differ-ence of two real numbers. That is, lx - al is the distance between thepoints r and a. In R2 the distance between the two points P(x, y) andPo(xo, y6) is given av ffi. In R3 the distance betweenthe two points P(x, y, z) and Po(xo, !o, zo) is given by

In R" we define the distance between two points analogously as follows.

l f .P (x r , x2 t . . . , xn ) andA(a1 ,a2 , . . , an ) a re twopo in t s i nR ' , t hen

the distance between P and A, denoted by llp - All, is given by

If in Rl we take P : x and A: a, (L) becomes

llr- nll: {G=W: lx- al Q)

If in R? we take P : (x, y) and A: (xo, A), $) becomes

l l (x ,y ) - (xo ,yo) l l : f f i (3 )

And, if in Rs we take P : (x, A, z) and A: (xr, !0, zo), (1) becomes

ll(x, y, z) - (xo, Ao, zr)ll: (4)

llP - All is read as "the distance between P and A-" lt is a nonnegativenumber.

lf A ]s a point in R" and r is a positive number, then the open ball B(A; r)

is defined to be the set of all points P in R' such that llP - All < t'

If A is a point in R, and r is a positive number, then the closed ball BIA; rl

is defined to be the set of all points P in Rn such that lp - a1; = '1'

To illustrate these definitions, we show what they mean in R', R',and R3. First of all, if a is a point in Rl, then the open ballB(a; r) is the setof all points r in Rr such that

l x - a l < r ( 5 )

The set of all points r satisfying (5) is the set of all points in the open

interval (a-r , n- l r) ; so the open bal l B(a;r) in R'(see Fig. 19.2.1) is

simply an open interval having its midpoint at a and its endpoints at

( 1 )

f , - r a

oPen ballB(a; r) in Rl

Figu re 19.2.1

a * r


a - r a a * r

closed ball Bla; rl in Rl

Figure 19,2.2

Figure

closed ball B [(ro ,Uo); r ] in R2

Figure 19,2.4

a - r and, a * r. The closed ball Bla; rl in Rr (Fig.19.2.2) is the closed in-terval [a - r, a * rf .

If (ro, ys) is a point in R2, then the open ball B( (xo, yr); r) is the setof all points (x, y) in R' such that

ll(*' y) (xr, yo) llFrom (3) we see that (5) is equivalent to

) '

So lhe open ball n(@p, yi; r) in R2 (Fig. tg.Z.g) consists of all pointsin the interior region bounded by the circle having its center at (xs, ys)and radius r. An open ball in R2 is sometimes called an open a*i*. rn6closed ball, or closed disk, B[(re, yo); r] in R, (Fig. 19.2.4) is the set of allpoints. in the open ball B ( (ro, yo) ; r) and on the circle having its center at(xr, yi and radius r.

If (xo, !0, zo) is a point in R3, thenthe set of all points (x, y , z) in Rs such

ll(x, y, z) (xo,Ao, zo)ll < t

(5)

the open ball B((xo, Ao, zo); r) isthat

(7)

open ball B((r o, Ao, zo); r) in R 3

F igure 19 .2 ,5

closed ball B[(r o,ao, zo); r] in R3

F igure 19 .2 .6

Let / be a function of. n vaiables which is defined on some open ballB(A; r),,except possibly at the point A itself. Then the limit o1 f @) as eapproaches A is L, written as

l i m f ( P ) - LP - A (8)

19.2.4 Definition


if for any € ) 0, however small, there exists a 6 > 0 such that

l f (P) - Ll

If f is a function of one variable and if in the above definition we

take A: a in Rr and P: r, then (8) becomes

f'T /t') : r

and (9) becomes

l f ( x ) - L l < " w h e n e v e r 0 < l t - r l < 6

So we see that the definition (2.1.1) of the limit of a function of one vari-

able is a special case of Definition 19'2'4'

we now state the definition of the limit of a function of two variables.

It is the special case of Definition 7g.z.4,where A is the point (re, ys) and

P is the Point (x' Y).

19.2.5 Definition Let f be a function of two variables which is defined on some open disk

B((n, Yr); r), except possibly at the point (16' y6) itself' Then

,".,1t-T".r", f (x' Y) : L

i f fo ranye)0 ,howeversmal l , thereex is tsa6>0suchtha t

lf (x,Y) - Ll < € whenever 0

(e)

(10)

In words, Definition 19.2.5 states that the function values f(x, v)

approach a limit L as the point (r, y) approaches the point (rq' yq) if the

a'bsolute value of the diffeience between f(x, V) and I, can bemade arbi-

tt*ify small by taking the point (!,y) sufficiently close to (xo' yo) but

""i "lf"J," (io, yJ. frote that in Definition 19.2.5 nothing is said about

the fulnction valul at the point (xo,yJ; that is, it is not necessary that the

function be defined at (xo, /o) in order for ,,,rli1.,*, f(x,y) to exist'

A geometric interpretation of Definition 19.2.5 is illustrated in Fig'

1g.2.7.1he portion above the open disk B ( (ro, Ao\;r) of the surface having

ug";,io" ,': f(*, y) is shown] W" t"" that f(x, U). on the z axis will lie

between f - 2 and L + e whenever the point (r, y) in the ry plane is in

tfr" op"r, disk 8((16, /o); S). Another way of stating this.is tnatf(x'A)

on th" z axis can be."ttti.t"d toliebetweenL-e and L* e byrestrict-

iig ifr" point (r, y) in the xy planeto be in the open disk B((ro' /o);6)'

. rLLUsrRArroN 1: We apply Definition l'9'2'5 to prove that

lim (2x * 3Y) = ll(c,rr)-(r,s)

We wish to show that for any € > 0 there exists a 6 > 0 such that

(xo , yo , f (xo , vo) )

( x o , U o , o )

Figu re 19.2.7


l (2x+31) -111 <e whenever O< VG=I I+ fy -g)z<6

Applying the triangle inequality, we get

l2x+3y - 111: l2x-2 t3y -e l =z lx - 1 l + s lV -31

Because

lx- 1l - t/@ffiQ$ and ly-gl - I/G=TTQ=Twe cnn conclude that

2 lx -11+} lv -31 <26+36whenever

o < V{FT)ITIY= a)z < aSo if we take 6 : *e, we have

l2x+3y -111 =z lx -L l+3 lV-31 <56 :e

whenever

o < V(I= r)zTly= s)z < aThis proves ,hu,

,",llT,r, (2x + 3y) : tt. o

ExAMPTE 1.: Prove that

lim (3x' * v) : 5(r ,a ) - ( r ,z ) '

\ r '

by "pplying

Definition 79.2.5.

sor,urroN: Wewish to show that foranye ) 0 thereexistsa6 > 0suchthat

l(3* +y) -51 < e whenever O < V(7=lzTly-Z)t < aApplyirg the triangle inequality, w€ get

l3* + y -51: lzxz -t *y -2]t = slx- l l l r+ 1l + lV -zl . (11)If we require. the 6, for which we are looking, to be less than or equal to1., then lr- 1l < 6 < 7 andly - 2l <D < 1 whenever

o< t /GVTQ-2 )2<6<1

Furthermore, whenever lr-11 ( L, then -l 1x-1< 1, and so 1()c +'/., < 3. Hence,

3 lx - r l l r+ 1 l + l y -2 | , < 3 . 6 .3 + 6 - 106whenever

0 < (x - 1 )

(12)

+ Q -z)',

0 we take E - m in (L ,So if forand (12)

any €

l3* '*y -51 < 100 < € whenever

This proves that r;,lift,zt

(Jx, * y) : 5.

#e), then we have from (11)

0

Figure 19 .2 .8

19.2.5 Definition

19.2.7 Definition

( * , , + 1 )I

( m - T , n * T ) | ( * + T , n * I )

( m ' ' - 1 )


We now introduce the concept of an "accumulation point," which weneed in order to continue the discussion of limits of functions of twovariables.

A point Po is said to be an accumulation point of a set S of points in R' ifevery open ball B(P6; r) contains infinitely many points of S.

o rLLUsrRArron 2: If S is the set of all points in R2 on the positive side ofthe r axis, the origin will be an accumulation point of S because no matterhow small we take the value of r, every open disk having its center at theorigin and radius r will contain infinitely'many points of S. This is anexample of a set having an accumulation point for which the accumula-tion point is not a point of the set. Any point of this set S also will be anaccumulation point of S. .

I rLLUSrRArror 3: If S is the set of all points in R2 for which the cartesiancoordinates are positive integers, then this set has no accumulation point.This can be seen by considering the point (m, n), where m and n are posi-tive integers. Then an open disk having its center at (m, n) and radius lessthan 1 will contain no points of S other than (m, n); therefore, Definition19.2.6 will not be satisfied.(see Fig. 19.2.8). r

a function of two variables as a point

where (x, il is restricted to a specificWe now consider the limit of

(x, y) approaches a point (xo, Ui,set of points.

Let f b" a function defined on a set of points S in R2, and letan accumulation point of S. Then the limit of f(x, V) as (x, y)(xr, U ) in S is L, written as

l i m f ( x , v ) - L(r,a)-( l ,o,ud

(p in s)

if for any €

l f (x ,y) - L l < € whenever 0

and (x, y) is in S.

( xo , y ) beapproaches

(13)

A special case of (13) occurs when S is a set of points on a curve con-taining (xo, yo).In such cases the limit in (1.3) becomes the limit of afunction of one variable. For example, consider,,,llTr,, f(x' a). Then if

S, is the set of all points on the positive side of the r axis, we have

l im f (x , y) : l im f (x ,0)(r,r,)-(oo) r-o'

(P inSr)


If 52 is the set of all points on the negative side of the y axis,

,",li,To ,orf (x' v) : lt-T f

(o' Y)(p in sz)

If Ss is the set of all points on the x axis,

l im , f (x, A) : l im f (x, 0)(r,a)-(o,o) r-o

(p in ss)

If 54 is the set of all points on the parabola A : x2,

,,,1,1T, ,orf (*'a) : lg f (x'x')@ in s+)

79.2.8 Theorem Suppose that the function f is defined for allhaving its center at (xo, A) , except possibly at

r",rliTo ,uorf (*'Y) - L

Then if S is any set of points in R2 havin g (xo, yo)

l im f (x ,y)(r,g)-(co,go)

exists and always has the value L.

l i m f ( x , A ) : L(t,u)-(to,uo)(p in s)

and L does not depend on the set S through which(xo, Ao).This proves the theorem.

PRooF: Because li+ . f (x, y) : L, then by Definition 19.2.5, for anyU,i-bo,ur) '

-

e ) 0 there exists a 6 > 0 such that

l f (x,y)-Ll < e whenever 0 ( l l (x,y) - (xo,/o) l l < S

The above will be true if we further restrict (x, y) by the reguire-ment that (x, y) be in a set S, where S is any set of points having (io, yo)as an accumulation point. Therefore, by Definition 1.9.2.7,

points on an open disk(xo, y) itself, and

as an accumulation Point,

(x, y) is approachingI

The following is an immediate consequence of Theorem 19.2.g.

19.2.9 Theorem If the function / has different limits as (x,y) approaches (ls, yo) throughtwo distinct sets of points having (xr, yJ as an accumulation point, thett

t"''li["'" f(x'Y) does not exist'

PRooF: Let 5r and Sz be two distinct sets of points in Rz havin g (xr, yo)

19.2 LIMITS OF FUNCTIONSOF MORE THAN ONE VARIABLE 895

ExAMPtn 2: Given

f ( x ,a ) : xY' x 2 * y '

find lim f (x, y) if( r ,a ) - (o 'o ) -

as an accumulation point and let

lim f (*, a) : Lt and(r,u)-(ro,uo)

(p in sr)

sol.urroN: Let S, be the set of all

l im f ( x , y ) - l im f ( x ,0 ) :(r,u)-(o,o) r -o

(p in sr)

l im f (x, y) : Lz(r,d-( to,uo)

(p in sz)

points on the r axis. Then

r . 0t;9ffi-o

Now assume that lim f (x,y) exists. Then by Theorem 19.2'8 Lt must(r,y)-(ro,yo)

equal L", but by hypothesis Lr * L2, and so we have a contradiction.Therefore'

,r,rl1T, ,rorf (x'y) does not exist' I

it exists.Let 52 be the set of all

l im f ( x , y ) :(r,g) - (o'o)

@ in sz)

Because

(p in sr)

points on the line y - r. Then

l im f (x ,y ) + l i * f Q,y )(r,s)*(o,o) (r 'u)*(o'o)

(p in sz)

it fotlows from Theorem 1,9.2.9 that lim f (x , y) does not exist.( r ,u ) - (o ,o ) -

In the solution of Example 2, instead of taking s1 to be the set of all

the points on the x axis, we could just as well have restricted the points

of S, to be on the positive side of the r axis because the origin is an accu-

mulation point of this set.

solurroN: Let 51 be the set of all points on the x axis. Then

,,,11%,or f (*' Y) : t"T ',h - o

EXAMPLE 3: Given

f(x,Y):ffifind lim f (x,y) if it exists-

(r 'g)-(o,o)

(p in sr)

Let Sz be the set of all points on any line through the

any point (x, y) in Sz , A : mx.

l im f (x, a) : l im :{ ' } :^ r ' 3mx : o

(r,s)-(o,o) '--o x' + m'x': tJT

Jt, ntz: u

(p in sz)

that for

Even though we obtain the same limit of 0,if. (x, y) approaches (0,0)

through a set of points on any line through the origin, we..cannot con-

cludelhat *,ll%., f(*, y) exists and is zero (see Example4)' However'

let us attempt to prove an",,,,l lTo,orfG,V):0' FromDefinition 19'2'5'


EXAMPTr' 4: Given

if we can show that for any e ) 0 there exists a E > 0 such that

l%l .€ whenever o< tW<Dl x ' * y ' l

\ \ ' Y r

then we have proved that lim f (x, A) : g.

Because rp - xz * yr"fi'ilj''! W, w€ have

| 3* 'y ^1 :g l lv l 3(x '+U\W:_g Wlry,l:m

So if 6 - *e, we can conclude that

l+ l .€ whenever o< \W<El f+v l

which is (1a). Hence, we have proved that.r,ll% ,orf(r, U) : g.

solurroN: Let Sr be the set of all points on either the r axis or the y axis.So if (r, y) is in Sr, rA:0. Therefore,

,",}}1r, f(x' Y) : o(P tnSr)

Let Se be the set of all points on any line through the origin; so if. (x, y)is a point in 52, y : mx.'We have, then,

,,,11T.,, f(x, y): L,T #: fi ffi : o(P in s2)

Let Ss be the set of all points on the parabola y: f. Then

l i ry . f k , A) : l im =x : :e : l im* : *x4

f (*,y\ : : '! ,' f * y ,

find ,,,liT, ,orf

(x'v) if it exists'

EXAMPLE

; ; x 4 + x 4 ; * o 2 - 2

* , l i q _ f (x ,y )(r,s)-(o,o)'(p in sr)

it follols tha, ,,,1111 , rf

(x,U) does not exist.

soLUrIoN: Let Sr be the set of all points on the y axrs. Then

l i m , f ( x , A ) : l i m 0 : 0

(t,a)-(o,o)'(p in ss)

Because

,,,llT, ,r,f (x'u)(p in ss)

(a,g)-(o,o)(p in sr)

t f . x + A

if x - 0

I - O

f ( x ,a ) :

find ,",11% ,orf

(x,u) if it exists.Let s2 be the set of all points on any line through the origin except pointson the y axis; that is, it (x, y) is a poinf in 52, y: kt, where r * 0. Ttren

19.2 LIMITS OF FUNCTIONS OF MORE THAN ONE VARIABLE

, liq, ,f (x, A) :lim (r * kx) rir, 1(a,g)-(o,o) ' 1^-o X(P in ^gz)

To find the above limit we make use of the fact that lim (r * kr) : g.

Because lim sin(Ur) does not exist, we cannot uppry tri"-lrreorem aboutc-0

the limit of a product. However, because LtT

(r * kx) :0, it follows that

for any e ) 0 there exists a 6 > 0 such that

l x *kx l<e whenever 0< l r l <6

The 6 is, in fact, elll + kl. ButI r l I r ll(r + kr) sin:l : lx + kxl lsin jl = lr * kxl . Ll x l l x l

Hence, for any e ) 0 there exists a 6 > 0 such that

| 1 ll ( x+kx )s in j l <e whenever 0< l r l <6l 4 l

Thus,

. l i - . f ( x , A ) : 0(r,g)-(o,o) -

(p in sz)

Let 53 be the set of all points (x, y) for which U:k*, where n is anypositive integer and, x * 0. Then by an argument similar to that used forproving (15) it follows that

,",5T0," f(x,il: lt1 (t *kxn) sin]:o(P in,ca)

We now attempt to find a E > 0 for any € > 0 such that

lf (x, v) - ol (15)

two cases:

( ls)

which will prove

r : 0 a n d x + 0 .

Case L : I f x - 0 , l f ( * , V ) - 0 l = l 0 - 0 l which is less than e for any

that lim f (x, A) :0 We distinguish(c,a)-(o'o)

V) sin(Ur) l .

6 > 0 .

Cqse 2: l f x

Il (x + v )I

+ 0 , l f ( * ,y ) - 0 l : l ( r +

sin*l : V+vt 1""11


:zvf + y2Then

| 1 l

l@ + v) sin jl < z .*e whenever o < lTf < tel ' x l

So take 6: te.Therefore, in both cases we have found a E > 0 for any e ) 0 such

that (16) holds, and this proves that lim f(x, y):0.- (c,g)-(o,o)' -

The limit theorems of Chapter 2 and their proofs, with minor modi-fications, apply to functions of several variables. We use these theoremswithout restating them and their proofs.

o rLLUsrRArrow 4: By applying the limit theorems on sums and products,we have

l im (r '* 2*y - y ' + 2): (-2) ' + 2(-2) ' (1,) - (1) ' * 2:7 oG,il-Gz,r)

Analogous to Theorem 2.6.5 fot functions of a single variable is thefollowing theorem regarding the limit of a composite function of twovariables.

l9.2.l0 Theorem If g is a function of two variables and !i+ .g@, y):b, and f is a

tunction of a single variable continuous "gi?;tfi:"rl. l i+ . ( f " d@,y) : f (b)(;rl l)-(co,aol

or, equivalently,

,",,IiT".,u fQ@, D) :/(",,1iT".,", s@,v))

The proof of this theorem is similar to the proof of Theorem 2.6.5and is left as an exercise (see Exercise 20).

ExAMPtn 6: Use Theorem19.2.10 to find lim ln (xy - 1).

(t ,u)-(2,1)

soLUrIoN: Let g be the function such that g@, y) - ry - 1, and / be thefunction such that f(t):ln t.

l im (w - L\ : ' tG,i-(z,i

-

and because / is continuous at 1, we use Theorem 19.2.10 and get

",ll%,,, rn(xv - t) : t"(,,.11T.,, (v - t))

- l n L

- 0

Exercises 19.2

In Exercises L through 6,establish the

1. lim (3x - 4V) - 1(c,u)-(g,z)

4. lim (2x' - yz) : -L(c,u)-(z.g)

19,2 LIMITS OF FUNCTIONS OF MORE THAN ONE VARIABLE

limit by findi.g a 6 > 0 for any e ) 0 so that Definition 1,9.2.5 holds.

2. lim (5x - 3y) - -z(x,u)-(2,4\

5 . l i - ( x ' * 2 x - y ) : 4(t,g)-(2,4)

3. l im (x' * A') :2(r ,g ) - (1 ,1 )

5' ,r,jt-T,-r,

(x' + yz - 4x * 2y) : -+

f (x,y) does not exist.In Exercises 7 through 12, prove that for the given function f ,,r,llTo,o,

8 . f (x ,a) :J7. f (x, A) :.?- V1x ' + y '

L0 . f ( x , A ) :xa*3* 'y ' *Zxyg

(x' * y')'11. f (x, A) : , =t'n *(x' + yn)t

x ' + y '

In Exercises L3 through 15, prove that,",llTo, orf @,U) exists.

13. f (x,a): f f i u. f (x,y)( , . \ L L

1 5 . f ( x , a ) : l ( r + y ) s i n 2 s r n ' = r f x + 0 a n d y + 0 r c . f @ , y )

L0 i f ei ther x:0 or y - 0

In Exercises 17 through 19, evaluate the given limit by the use of limit

17. lim (3x' * xy - 2y')(c,u)-(2$)

20. Prove Theorem 19.2.10

18. lim v+m(c,a)-(-2,E)

In Exercises 2L through 23, show the application of Theorem 19.2.L0

u21. lim tan-l 1

(c,y)-(2,2'1 X

In Exercises 24 through 29, determine if the indicated limit exists.

i f x + 0

r f . x - 0

lg. I im e* * eu

(a,s)-(o,o) COS X * Sin y

to find the indicated limit.

l r23. lim t

(x,u\-(t,2, \'l g* - +V

: * t * Y t

x '+ y '

[f i" (xy): l x \ ' .

lvtheorems.

22. lim [5r + ty'\(t,s)-(-2,g)

24. lim !'a' ,(c,s)*(o,o> f * Yn

f-*v- +28. f (x,a):wtv

zs. f (x, a) :{ t t i" i* '

L0

sin(r + y)lim

(c,u)-(2,-2) X + y22. tim

*: + Y=(c,s)-(o,o) X' + y225. lim !'a' =

-l

i f x + 0 1-h l im f ( x , y )if x - 0J (e's)-(o'o)

t f x + L a n d a * 0 I' l I l i m f ( x , y )

if either x:0 or y : o) (c'a)'(o'o)

26.

. Lsln -x

. 1sln -

x

30. (a) Give a definition, similar to Definition 19.2.5, of. the limit of a function of three variablesproaches a point (xo, yo, zo). (b) Give a definition, similar to Definition 19.2.7, of the limit of aables as a point (x, y, z) approaches a point (xo, yo, zs) in a specific set of points S in R3.

31. (a) State and prlove a theorem similar to Theorem 19.2.8 f.or a function / of three variables. (b)orem similar to Theorem 19.2.9 for a function f of three variables.

as a point (x, y, z) ap-function of three vari-

State and prove a the-

9OO DIFFERENTIAL CALCULUS

In Exercises 32 through 35, usenot exist.

g2. f (x, y, z) - f * Yx, '* z2x2'/

f + Yn * zn

u . f (x ,A , z ) : ,O. * ! "' ' f + Y '+ zn

In Exercises 36 and 37, use the

96. f (x, A, z) : *r!l;r-( ,,

; ) : { e * y * z ) r i r , 1 r i r , 1 i f x + o a n d , y + o

L0 if either x :0 or y : 0

19.3 CONTINUITY OF We define continuity of a function of. n variables at a point in RzFUNCTIONS OF MORETHAN ONE VARIABLE

OF FUNCTIONS OF SEVERAL VARIABLES

the definitions and theorems of Exercises 30 and 31 to prove anu,,",r,ltTo,o ,orf

@,A, z) does

* + y 2 - 2 233. f(x,v,"):FTFr,

g5. f(x,y,z):Fffidefinition in Exercise 30(a) to prove that

,,.r.1l1o.o.o, f @, y, z) exists.

19.3.1 Definition Suppose that / is a function of n variables and A is a point in R". Ttren /is said to be continuous at the point A if and only if the following threeconditions are satisfied:

(i) f(A) exists;(ii) lim /(P) exists;

P.A

( i i i ) l im f(P): f(A).

If one or more of these three conditions fails to hold at the point A, then/ is said to be discontinuous at A.

Definition 2.5.1 of continuity of a function of one variable at a num-ber a is a special case of Definition 19.3.1.

If / is a function of two variables, A is the point (re, ys), and. P is apoint (r, y), then Definition 19.3.1 becomes the following.

19.3.2 Definition The function f of. two variables r and y is said to be continuous at thepoint (rs, ye) if and only if the following three conditions are satisfied:

(1) f(xo, yo) exists;(ii) lim f(x, y) exists;

(t,il-(so,ud -

(iii) _ !i+ . f(x, y) : f (xo, y).(a, i - (h,sd '

soLUrIoN: We check the three conditions of Definitionpoint (0, 0).

(i) f (0, 0) : Q. Therefore, condition (i) holds.+ (0, 0)- (0, 0)

19.3.2 at the

determine If f is continuous at(0 , 0 ) .

19.3 CONTINUITY OF FUNCTIONS OF MORE THAN ONE VARIABLE 901

(tt) ,,.11%,, f(r, v):,",llTo,o, ffi:0, which was proved in Ex-ample 3, Sec. 19.2.

(ttt),".ll1, rf(x' u) : /(0, o).

So we conclude that / is continuous at (0, 0).

solurroN: Checkingfollowing.

the conditions of Definition "1.9.3.2, we have the

(i) f (0, 0) - 0 and so condition (i) holds.( i i ) When ( x , y ) + (0 , 0 ) , f ( x ,U ) : xy l ( x '+ y ' ) . I n

'!.9.2, we showed that ,r,llT, ,rr*Al@'

+ y') does

lim f (x,y) does not exist. Therefore, condition(t,a)-(o,o)

We conclude that / is discontinuous at (0, 0).

EXAMPLE 2: LCt

be defined by

f *Yf ( * ,a) : |T f

LoDetermine tf. f ts(0 , 0 ) .

the function f

i f (x, y) + (0, 0)

i f ( x , y ) - ( 0 , 0 )

continuous at

Example 2, Sec.not exist and so

(ii) fails to hold.

If a function f of. two variables is discontinuous at the point (ro, yo)but tim f(x, y) exists, then / is said to have a remoaable disconti-

(r ,u)-(co,ui ' -

nuity at (xs, !) because if f is redefined at (xo, /o) so that f(xo, y):

. li* f (x, y), then / becomes continuous at (xs, y). It the disconti-(t,il-(co,uo)

nuity is not removable, it is called art essential discontinuity.

o rLLUsrRArron 1: (a) If g(r, y):3x'yl(f * yz), then g is discontinuousat the origin because g(0, 0) is not defined. Howevet, in Example 3, Sec.19.2, we showed an"a

,,,11T0, or3*yl(xz * y'):0. Therefore, the discon-

tinuity is removable if g(0, 0) is defined to be 0. (Refer to Example 1.)(b) Let h(x, y):xyl(xz *y2). Then ft is discontinuous at the origin

because h(0, 0) is not defined. In Example 2, Sec. 79.2, we showed thatlim xyl(x2 * y) does not exist. Therefore, the discontinuity is essential.

k,u)-(0,0)

(Refer to Example 2.) o

The theorems about continuity for functions of a single variable canbe extended to functions of two variables.

19.3.3 Theorem If f andg are two functions which are continuous at the point (16, yo), then

(i) / + g is continuous at (xo, Y);(ii) / - g is continuous at (xo, Y);

(iir) /S is continuous at (ro, yo);(iv) fl7 is continuous at (ro, /o), provided that g(xo, yr) * 0.

The proof of this theorem is analogous to the proof of the correspond-ing theorem (2.6.1) for functions of one variable, and hence it is omitted.


19.3.4 Theorem A polynomial function of two variables is continuous at every point in R2.

PRooF: Every polynomial function is the sum of products of the func-tions defined by f (x, y) : x, g@, y) : y, and h(x, y): c, where c is areal number. Because f , g, and ft are continuous at every point in R2, thetheorem follows by repeated applications of Theorem 19.3.3, parts (i)and (iii). I

19.3.5 Theorem A rational function of two variables is continuous at every point in itsdomain.

pRooF: A rational function is the quotient of two polynomial functionsf and g which are continuous at every point in R2 by Theorem 1,9.9.4.If.(xo, y) is any point in the domain of flg, then g(ro, y) # 0; so by The-orem 19.3.3(iv) f lS is continuous there. I

EXAMPTT 3: Let the functi on fbe defined by

solurroN: The function / is defined at all points in R2. Therefore, con_dition (i) of Definition 19.3.2 holds for every point (rs, y6).

Consider the points (xr, yi if. xsz I Ao2 + 1.If. xoz + yoz ( 1, then

,",,f1T,,u f k' y): (",,liTo.,o) (f + y') : xo2 * yo': f (xr, yo)

lf xoz * yo2 ) 1, then

,,.,11T".,u f@' y) :,".rli2",,u 0 : 0 : f(xo' yo)

Thus, / is continuous at all points (xo, y) for which xoz + yo2 + L.To determine the continaity of. f atpoints (re, yo) for which xoz + yoz :

L, we determttu tt,",rrti2" ,uorf

(*, y) exists and equals 1.

Let 51 be the set of all points (r, y) such that f * y, = 1. Then

f (x, y) : [ r ' * y ' i f x 2 * y , < L

L 0 r f x z * y ,

Discuss the continuity of f .Whatis the region of continuity of f?

lim f (x,(r,u)-( to,uo) '

(p in sr)

lim f (x,(t ,a)-(ro,go) '

(p in sz)

Because

lim f (x,(t ,u\-(co,ao) '

(p in sr)

Let s2 be the set of all points (x, y) such that x, + y,

a) :,r,rfi2o,ro, Q' + y') - xoz * yo' - 1@ in sr)

A ) : l i m 0 : 0(t,a)-(ro,go)

(p in sz)

y ) * . l i r y , f @,y)(r,a)-( to,uo) '

@ in sz)

we conclude that li+ . f (x, y) does not exist. Hence, / is discontinu-@,i-bo,ud

ous at-all points (xo,yi fo-rwhic! xo2 + yo2: L. Theregionof continuityof/consists of all points in the ry plane exiept those or, tl.r" circle rp * y, ! t.

I

19.3.6 Definition

19.3.7 Theorem

4

3

2

- 5 - 4 - 3 - Z - L L

L 2 3 4 5- 1

- 2

- 3

- 4

Figure 19.3.1

ExAMPrn 4: Given that

f(x,u):ffidiscuss the continuity of fanddraw a sketch showing as a

shaded region in R the region of

continuity of f .

19.3 CONTINUITY OF FUNCTIONS OF MORE THAN ONE VARIABLE 903

The function f of. n variables is said to be continuous on an open ball if. it

is continuous at every point of the open ball.

As an illustration of the above definition, the function of Example 3

is continuous on every open disk that does not contain a point of the

circle f * y ' :7.The following theorem states that a continuous function of a con-

tinuous function is continuous. It is analogous to Theorem 2'6'5'

suppose that / is a function of a single variable and g is a function of two

.r""ilbles. Suppose further that g is continuous at (ro, ye) and f is con-

tinuous at g(xo, yo). Then the composite function.f .8 is continuous at(xo , y l -

The proof of this theorem, which makes use of Theorem 19.2.70' is

similar to tt "

proof of Theorem 2.6.6 and is left as an exetcise (gee Ex-

ercise 7).

o rLLUsrRATroN 2: Let

h(x, Y) : ln (rY - 1)

we discuss the continuity of.h.If g is the function defined by 8@,y):

x xa - 7, gis continuous at all points in R2. The natural logarithmic function

is continuous on its entire domain, which is the set of all positive num-

bers. So if / is the function defined by f(t):.ln t,.f is-continuous for all

f>0 .Thenthe func t ionh is thecompos i te func t ion f "gand 'byTheo-rcmlg .3 .T , iscont inuousata l lpo in ts (x ,y ) inR2forwh ichxy-L)0or 'equivalently , xy > 1.. The shaded region of Fig. 19.3.1 is the region of

continuity of h' '

soLUrIoN: This is the same function as the function G of lllustration 4 in

sec. 19.1. we saw there that the domain of this function is the set of all

points in the exterior region bounded by the circle rc *y',:25 and the

points on the r axis for which -5 < x < 5'The function / is the quotient of the functions g and h for which

g@, y) : y and h(x, V): \/FW=E The tunction g is a polynomial

irnction and therefore is continuous everywhere. It follows from The-

orem 19.3.7 that h is continuous at all points in xP for which * * y2 > 25.

Therefore, by Theorem 19.3.3(iv) we conclude that / is continuous at all

points in the exterior region bounded by the circle f * y' :25'

Now consider the points on the x axis for which -5 < r < 5, that is,

the points (a, 0) wherl -5 I a < 5. If Sr is the set of points on the line

x: a, (a, 0) is an accumulation point of Sr, but

* .}T,r, f(x 'Y):tu#x(P tn Sr)

gO4 DIFFERENTIAL CALCULUS OF FUNCTIONS OF SEVERAL VARIABLES

Figure 19.3.2

which does not exist because yl{ffi is not defined if lyl =\E-. Therefore, / is discontinuous it *re points on the r axis forwhich -5 ( r < 5. The shaded region of Fig. 19.3.2 is the region of con-tinuity of /.

(nrNr: See Example 4, Sec. 19.2.)

(x, y) + (0, 0)

(x ,y ) : (0 , o )

if (x, y) + (0, 0)

if (x, y) : (0, 0)

Exercises 19.3

In Exercises 1 through 6, discuss the continuity of /.

t+ ir @,y) + (o,o)1. f (r, V) : ]{Ff lnrNr: See Exercise 13, Exercises 19.2)

t0 i f (x,y) : (0,0)

( x ' y .2 . f (x ,a) : | f f i

i r (x 'Y) * (o ' o)f.0 if (x, y) : (0, 0)

3. f (x ,a) : t# i f @,v) + (o, o)L0 if (x, y) : (0, 0)

( r y5. f(x,a) ={lffiT ir (x'Y) + (o' o)

L0 i f (x ,y) : (0 , 0)


In Exercises 8 through 17, determine the region of continuity of I and draw a sketch showing as a shaded regron in R2 theregion of continuity of /.

L1..

L3.

f ( x , a ) : Wf (x, A) : r ln (xV')

L4. f (x, A) : sin-l (xy)

19.4 PARTIAL DERIVATIVES

15. f (x, A) : tan-r L + sec-r (xV)v

(* - y' rsin(x * u)l - i f . x * y 1 j . t _ " . , _ l # i f x + y + O1 6 . f ( x , D : l x - y - " - ' r t z . f ( x , y ) : t , x - t y

L x - y i f x : y U i f x * y : g

In Exercises 18 through 27, prove that the function is discontinuous at the origin. Then determine if the discontinuity isremovable or essential. If the discontinuity is removable, define /(0, 0) so that the discontinuity is removed.

905

( 1 )

t o y r spoint

(2)

( Sxyz

25 . f ( x , y , z ) : ]WL0

( x , y , z ) # ( 0 , 0 , 0 )

(x , y , z ) : (0 , 0 , 0 )

1 9 . f ( x , A ) : ( x + y ) s i n lv

21 . f ( x ,A) :Yx u * y n

I xz -v ' '25. f (x, A,z) -

f r ' * y ' * z2L0

( x , y , z ) + ( 0 , 0 , 0 )

( x , y , z ) - ( 0 , 0 , 0 )

22. (a) Give a definition of continuity at a point for a function of three variables, similar to Definition 19.3.2. (b) State the-orems for functions of three variables similar to Theorems 19.3.3 and 79.3.7. (c) Define a polynomial function of threevariables and a rational function of three variables.

In Exercises 23 through 25, use the definitions and theorems of Exercise 22 to discuss the continuity of the given function.

23. f (x, A, z) : W24. f(x, y, z) : ln(36 - 4x2 - y2 - 9z')

if

if

if

if

I9.4 PARTIAL DERIVATIVES

19.4.1 Definition

We now discuss differentiation of real-valued functions of n variables.The discussion is reduced to the one-dimensional case by treating a func-tion of n variables as a function of one variable at a time and holding the

others fixed. This leads to the concept of a "partial derivative." We firstdefine the partial derivative of a function of two variables.

Let / be a function of two variables, r and y. The partial deriaatitse of fwith respect to x is that function, denoted by Dtf , such that its functionvalue at any point (x, y) in the domain of / is given by

if this limit exists. Similarly, the partial deriaatiae of f u)ith respectthat function, denoted by Drf , such that its function value at any(x, y) in the domain of / is given by

The process of finding a partial derivative is called partial dif-ferentiation.

f ind Drf (x,y) and Drf (x,y) byapplying Definitio n'!.9.4.1.

FUNCTIONS OF SEVERAL VARIABLES

Drf is read as "D sub t of f ," and this denotes the partial-derivativefunction. Drf (*, y) is read as '? sub L of. f of. x and yi' and this denotesthe partial-derivative function value at the point (x, y). Other notationsfor the partial-derivative functionDrf are fr, f ,, and O/dr. Other notationsfor the partial-derivative function value D/(r, A) arc fr(x, A), f ,(x, y),and |f(x, y)lax. Similarly, other notations for Drf are fz, fa, and Aflay;other notations for Dd@, y) are fr(x, y), fu(x, y), and \f(x, V)laV.Itz : f (x, A) , we can write 0zl 0x for D rf (r, y) . A partial derivative cannotbe thought of as a ratio of 0z md Er because neither of these symbols hasa separate meaning. The notation dyldx can be regarded as the quotientof two differentials when y is a function of the single variable r, but thereis not a similar interpretation for 0zl0x.

SOLUTION:

Drf(r,/) : l im f@ + ax' Y.) - f(x' Y)- r/ \.- ' " '

: ;; 4* * d^r1|'- 2(x + ax)y * y, - (g* - 2ry-!n-An-o Ar

: l im 3 f *6x Ax+3(Ax )2 -2 ry :2y Lx+y2 -3 f+z ry -yz

Ar-o Ax

: ^tlll

5x Lx + 3(4t!)2 - 2u a'x

: l im (Gx+ j Lx-2y)Ao-0

:6x - 2A

Drf (x,/) : lim f @' v + AY) - f(x' v)aa-o Ly

: tim 3* - Zx(y + AV) + (y * Ay), - (gf - 2ry * yr)au-o Ly

_ r. ̂ 3* - 2xy - 2x A,! * y2'* 2y Ay + (Ay)z - 3* I 2xy - yz

Au-o Ly

: lim -2x LY + 2Y LY + (LY)z

aa-o Ly' : lim (-2r + 2y + Ay)

Aa-O

:-2x * 2A

(3)

If (xr, y) is a particular point in the domain of /, then

19.4 PARTIAL DERIVATIVES

if this limit exists, and


o rLLUSrRArroN L: We apply formula (3) to find Drf(3,-2) for the function

/ of Example 1.

D,f (3,-2):ll* At + ̂ x'-fI- f@'-2\

:,:: ,

: lim 27 + LB Ar * 3(Ax)'? + L2 + 4 Lx * 4- 43Ar*o Ax

: l im (18+3 Lx t4 )A.r-0

: 2 2 o

Altemate formulas to (3) and (4) for Drf (xo, ys) and Ozf (n,/o) are

given by


o rLLUSrRArrow 2: We apply formula (5) to find Drf(3, -2) for the function

f of Example 1.

D,f(3,-2):n^t@AFA" ' t

x - 3

(4)

(s)

(5)

and


ExAMPrn 2: Given f(x, A) : 3x3- 4x'y * 3xyz a Tx - By, findDr f (x ,y ) and Dr f (x ,y ) .

o rLLUsrRArrON 3: In Example 1, we showed that

Drf (x, V) :6x - 2y

Therefore,

Dt f (9 , -2 ) :18 + 4: 2 2

This result agrees with those of Illustrations 1 and 2. o

To distinguish derivatives of functions of more than one variabrefrom derivatives of functions of one variable, we call the latter derivativesordinary derioatiaes.

comparing Definition 19.4.1 with the definition of an ordinary deriv-ative (3.3.1), we see that Drf (x, y) is the ordinary derivative of f if f isconsidered as a function of one variable x (i.e., y is held constant), andDrf (r, y) is the ordinary derivative of. f if/ is considered as a function of

soLUTroN: Considering f as a function of r and holding y constant,we have

Dtf (x, A) :9* - *ry + 3y2 + 7

Considering / as a function of y and holding r constant, we have

Drf (x, A) : -4* * 5ry - 8

solurroN: (a) If y # 0, from (5) we have

/ , (0, y)- l im f@' Y) - f (o ' Y)r - 0

+ (0, o): (0 , 0)

(a) f '(0, (b) f ,(x, 0).

19.4 PARTIAL DERIVATIVES 909

If y: 0, we have

/ ' (0 ,0) : l - iT f : tTT-0

Because /, (0 , y) - -y

/t (O , y) - -y for all y.(b) lf x * 0, from

i f y + 0 and / ' (0 , 0) :0 ,

(5) we have

we can conclude that

f r (x, o) : tJT '

- limA ' 0

I f x: 0, we have

fr(O, o) - l im f (0 ,y ) - f (0 ,0 ) _

Because f ,(x,f r ( x , 0 ) - x f o r

y -0

i f x# 0 andfr(O, 0) :0, we can conclude that

Geometric interpretations of the partial derivatives of a function oftwo variables €ue similar to those of a function of one variable. The graphof a function f of two variables is a surface having equation z:f(x,U).If y is held constant (say, y : yo),then z : f (*,yo) is the equation of thetrace of this surface in the plane y: yo. The curve can be represented bythe two equations

a : a o a n d z - f ( x , y )

because the curve is the intersection of these two surfaces.Then Dtf(rq, /o) is the slope of the tangent line to the curve given

by Eqs. (7) at the point P6(16, Ao, f (xo, yo) ) in the plane A : Ao.In an analo-gous fashion, Drf (xs, ye) represents the slope of the tangent line to thecurve having equations

x : x o a n d z : f ( x , y )

at the point Po in the plane r:xe. Figure L9.4.La and b shows the por-tions of the curves and the tangent lines.

: limU - 0

x3

x:2

limU - 0

o ) : r ,all x.

(a) x

(b) x

(7)

Figure 19,4.1


EXAMPIE 4: Find the slope ofthe tangent line to the curyeof intersection of the surfacez -+@wi th thep lane y :2a t t hepo in t(2 ,2 , \E) .

EXAMPTn 5: According to theideat gas law for a confined gars,if P pounds per square unit isthe pressure, V cubic units is thevolume, and T degrees is thetemperature, w€ have the formula

PV: kT

where k is a constant of propor-tionality. Suppose that the vol-ume of gas in a certain containeris 100 in.3 and the temperature is9 0 o , a n d k : 8 .

(a) Find the instantaneousrate of change of P per unitchange in T if V remains fixed at100. (b) Use the result of part (a)to approximate the change in thepressure if the temperature isincreased to 92". (c) Find the in-stantaneous rate of change of Vper unit change in P if T remainsfixed at 90. (d) Suppose that thetemperature is held constant. Usethe result of part (c) to find theapproximate change in the vol-

The required slope is the value of 0zl0x at the point

-x

we obtain

SOLUTION:

(2, 2, \E).

0z

(8)

A partial derivative can be interpreted as a rate of change. Actually,every derivative is a measure of a rate of change. lf. f is a r,ri.tio'of thetwo variables r and y, the partial derivative oi1*iit respect to r at thepoint P1(ro, /o) gives the instantaneous rate of change, at ps, of f(x, y)per unit change in r (r alone varies and y is held fixed at ye). similariy,the partial derivative of / with respect to'y atpo gives the instantaneousrate of change, at Po, of f(x,y) per unit change in y.

soLUrIoN: Substituting V:100, T:90, andk:8 in Eq. (g), we obtainP : 7 .2. (a) Solving Eq. (8) for P when k : 8, we get

Therefore,

a P 8- : -a T v

When T:90 and V: l0o,lPl0T:0.08, which is the answer required.(b) From the result of part (a) when T is increased by z 1ind, v re-

mains {ixed) an approximate increase in p is 2(0.09) :0.16. We conclude,then, that if the temperature is increased from 90" to 92", the increase inthe pressure is approximately 0.1,6 lblin.z

(c) Solving Eq. (8) for V when k: 8,

Therefore,

av 8T_ :AP P2

W h e n T - 9 0 a n d P -

t25

ume necess ary to produce thesame change in the pressure asobtained in part (b).

EXAMPtn 6: Given

f (x, y, z) -- x'U * Yz' * zg

verify that xf t(x, Y , z)* y f r (x ,y , z ) t , f t (x ,Y , z ) :

3 f ( * , y , 2 ) .

function, denoted by D rcf , such that its function value at any

the domain of f ,t given by

. . , X n ) :

( x 1 , X 2 , . .

19.4 PARTIAL DERIVATIVES 911

which is the instantaneous rate of change of V per unit change in P whenT: 90 and P :7.2 if T remains fixed at 90.

(d) If P is to be increased by 0.L6 and T is held fixed, then from theresult of part (c) the change in V should be approximately (0.16) (-435) :-#. Hence, the volume should be decreased by approximately # in.3 ifthe pressure is to be increased from 7.2lblin.z to7.36lblin.'z.

We now extend the concept of partial derivative to functions of n

variables.

19.4.2 Definition Let P(xt , x2, . x) be a point in R', and let f be a function of the n* 2 t . . o i * n l l v ' \ - 1

I / L t r r | ' L I I r r V s r a Y r v L

/ I

x1r x2 t . . . , xn. Then the partial derivative of f with respect tovariablesxk is thatpoint P in

Duf@t, x2,

limA'ro-6

(xr ,


In particular, If f is a function of the three variables r,

the paftLal derivatives of f are given by

x+Lx , y , z ) - f ( x , y , z )

y, and z, then

f(Dt f (x ,U, z) : l im Ar

and

if these limits exist.

soLUTroN: Holding y and z constant, we get

f t(x, v, z) :2xY

Hotding x artd z constant, we obtain

f r ( x ,Y , z ) : f +z 'Holding x and y constant, we get

f"(x, y, z) :2Yz * 3z'

Therefore,

x f {x ,y ,z) +yf r (x ,y ,z) + z f r (x ,A,z) :x(2x1) +A(* * z2) * z(2yz*3zz)

Lxrc


:' 2x2A * r'y * yz' * 2yz2 * 3zg- 3 ( x ' y * y z r * z t )

: 3f (x, y, z)

Exercises 19.4

In Exercises L through 6, apply Definition '!.g.4.'!, to find each of the partial

1 . f (x , A) :6x * 3y -7 ; Dr f (x ,y ) Z . f (x , A) :4xz - 3xy ; Dr f (x ,y )

4. f (x , A) : ryz - Sy * G;Drf (x ,y) 5. f (x,a): f f i ' ;Drf(x,y)

In Exercises 7 through 10, apply Definition 1,9.4.2

7. f (x , y ,z) : fy - 3xy, * 2yz; Dr f (x ,y ,z)

to find each of the partial derivatives.

8. f (x, y, z) : x2 * 4y, * 9zz; Drf (x, y, z)9. f(x, !, z, r, t) - xyr * yzt * yrt * zrt; Dnf@, !, z, r, t)

70. f (r, s, t, u, o, 7u) : 3rzst I spa - 2tup2 - tow * 3uutz; Dsf e, s, t, u, o, w)LL. Given f (x, y) - x! - 9y2. Find Drf (2,1) by (a) apglyrng formula (3); (b) apptnng formula (S); (c) applyrng formula (t)and then replacing x and, y by 2 and 1, respectively.12' For the tunction in Exercise 11, find Drf (2,11by tul applyrr,g formula (l); (b) applying formula (5); (c) apprying for-mula (2) and then replacing x and, y by-2'and't,i"#"tii"iy. "

*

llrtil::'.Hl?i[::?l# j1fi,1i,'*fted partial derivatives bv holdins alt but one or the variables constant and appry-

13. f(x, A) :4yt + ff ir; Drf (x,y)

15. f @, Q) : sin 3d cos 2g; Dzf @, O)

1 7 . z - s a t r : "

0 zt" v' uv

derivatives.

3. f (x, A) : 3xy * 6x - Ar; Drf (x, y)

6. f (x, a):#, D,f (x,y)

L4. f (x ,A) :& ,=@i;Dr f (x ,Y)

1,6. f (r, 0) : r2 cos 0 - 2r tan 0; Drf (r, 0)

19. u - (x' + y, + rr1-t,r. 4dz

2A. u - tan-r (xyzut), #

22. f(x, y, z): stu sinh 2z- eru cosh2z; fr(x, A, z)

26. f (x, A) : fu e"*t dtJ r

28. Given rp - *y + y,z I zzx.Ver i fy U + @, du) - ," y A x - W n E : ( r + y * z ) ,

L8. r : e-o cos( 0 + e); !a0

21.. f (x, A, z) : 4xyz * ln(2xyz); fs(x, y, z)

29. f (x, y, z) : suuz * tan- t!, fr(x, y, z)

24- f ( r ,0 ,0) :4rz s in 0 * se ' cos 0 s in r f -2 cos @; f r ( r ,0 , o)

In Exercises

25. f(x, A) :

27. Given u -s in | * h ' - . v"nfy t f f +, #: o.

25 and 25, f ind fr(x,y) and fu(x,y).fu

J , ln s in t d t

It+ u'29. Given f (x, u) :1ffi if (x' Y) + (o' o)

L0 rf (x, y) : (0, 0)

Find (a\ f'(0, 0); (b) fr(O, 0).

19.5 DIFFERENTIABILITY AND THE TOTAL DIFFERENTIAL

, \ l - ' t - l i f (x,y) + (0, o)30. Given f(x, A) :1 , + y ^r \"

L0 if (x, y) _ (0, 0)

Find (a) f '(0,y) i f y * 0; (b) /,(0, 0).

913

31. For the tunction of Exercise 30 find (a) fr(x,0) if r + 0; (b) fr(O,0).

32. Find the slope of the tangent line to the curve of intersection of the surface 36f - 9yz + 4z2 * 36: 0 with the plane

x:1 at the point (t, t/tZ, -3). Interpret this slope as a partial derivative.

33. Find the slope of the tangentline to the curve of intersection of the surface z: * * y2 with the planey: l atthepoint(2, 1,5). Draw a sketch. Interpret this slope as a partial derivative.

34. Find equations of the tangent line to the curve of intersection of the surface * * y'* z2:9 with the plane y: 2 at

the point (1,2,2) .

35. The temperature at any point (r, y) of a flat plate is T degrees and 7 : S+ - W - 4y2. lf. distance is measured in feet,

find the rate of change o1 the temperature with respect to the distance moved along the plate in the directions of the

positive r and y axes, respectively, at the point (3, 1).

36. Use the ideal gas law for a confined gas (see Example 5) to show that

v-_ . {. e!_: -raT dP dV

g7.lf.V dollars is the present value of an ordinary annuity of equal payments of $100 Per year for f years at an interest

rate of 100i percent Per Year, then

11 - (1+ t ) - ' lV :100 l -t i l

(a) Find the instantaneous rate of change of V per unit change in i if f remains fixed at 8' (b) Use the result of part (a)

io find the approximate change in the lresent .raloe if the interest rate changes from 6vo to 7vo and the time remains

fixed at 8 years. (c) Find the instantaneous rate of change oIV per unit change in f if I remains fixed at 0'06'' (d) Use the

result of part (c) to find the approximate change in tlie present value if the time is decreased from 8 to T.years and

the interest rate remains faed at 6Vo.

19.5 DIFFERENTIABILITY AND In Sec. 3.6 in the proof of the chain rule we showed that if I is a differen-

THE TOTAL DIFFERENTIAL tiable function Of the single variable r and y : f (x), then the increment

Ay of the dependent variable can be expressed as

Ay: f' (r) Ar * q Lx

where 4 is a function of Ar and q -+ 0 as Ax -+ 0'From the above it follows that if the function / is differentiable at

ro, the increment of. f at ro, denoted by Lf (xs) ' is given by

A/ (xo) - f ' ( xo) Ar* nLx

where ligo r:

O.

For functions of two or more variables we use an equation colTe-

sponding to Eq. (1) to define differentiability of. a function. And from the

almition we determine criteria for a function to be differentiable at a

914I


poinl we give the details for a function of two variables and begin bydefining the increment of such a function.

19.5.1 Def ini t ion I f / isafunct ionof twovariablesxand ! , thentheincrementof fat thepoint (ro, ye), denoted by Lf(xs, AJ, is given by

Figure 19.5.1 illustrates Eq. (2) for a function which is continuouson an open disk containing the points (xo, yo) and (16 * Lx,ys+ Ay).The figure shows a portion of the surface ,J f(*, y). Af(xo, iJ : ffn,lvhere^Q is the point (16 * M,yo!.Ly, f(xo, 'y))-ana'i isihe point( xs * Lx ,Ao* NA, f@o+ Ax ,yo+Ay) ) . -

z: f(x,y) \ (ro * ax,uo * Lv,f@o * ax,ys + av))

Figure 19.5.1

o rLLUsrBArrow 1: For the function / defined by

f(x, y) :3x - xazwe find the increment of / at any point (xn, ys).

Lf (xo, Ai : f (xr* M, yo+ Ail - f Go, yo):3(ro * Ar) - (ro * Lx)(yo+ Ay), _ (3ro _ xoAoz): 316 * 3 Ar - xeiloz _ Aoz Lx _ 2*oyo Ay _ 2yo Ax AV

- xo(Ay), - Ax(Ay)z _ 3xs * x6ysz: 3 Ax - Ao2 Ax - Zxoyo Ly - 2ys Ax Ay _ xs(Ay), _ ArlAyy, .

19'5'2 Definition rf f is afunction of two variables x and y and the increment of f at (xo, yo)can be written as

Af (xo, y) : Drf(xo,Ui Ax * Drf (xs, A) Ly * e1 Ar + ez Ly. (3)

(2)

+ Ay)

o)

z ( xo , yo , f ( xo ,VJ )

19.5 DIFFERENTIABILITY AND THE TOTAL DIFFERENTIAL 915

where e1 and e2 are functions of Ar and Ay such that e1 -+ 0 and €z + 0 as(M, Ly) - (0,0), then/is said tobe differentiable at (xs,ys).

. rLLUsrRArrou 2: We use Definition 19.5.2 to Prove that the function of

Illustration 1 is differentiable at all points in R2. We must show that for

all points (xo, yo) in R2 we can find an e1 and an e2 such that

Af (xo, y) - Dtf (xo,U) Lx - D"f (xo, y) Ly: e1 Ar I e2 Ay

and e1-+ 0 and €z + 0 as (Ar, Ay) -+ (0, 0).Because f(x, y):3x- rA2,

Drf (xo, y) :3 - voz and D"f (xo, ao) : -2xoyo

Using these values and the value of Lf (xo' yd from Illustration L, we have

Lf (xo,yo)-Dtf (xo,Uo) Lx-Drf (xo,yi AA:-xo(N)'-2ys Lx LV- Lx(LV)'

The right side of the above equation can be written in the following ways:

l-2yo Ay - (AV)'l Ar * (-ro Ay) LV

or(-2y, Ly) Ar * (-Ar Ly - xo LY) LV

or

[-(Ay)'] Lx* (-2ysAr-re Ly) LV

or0 ' A x * f - Z y o A r - L x L y

So we have four possible pairs

€1 : -2yo Ly - (Lv)' and

or

- xs Lyl Ly

of values for €1 efld €21

€2: -xo LA

G2: -Lx Ly - xo LUand€r : -2yo Ly

or

€1 : - (Ly)'

or

€ r : 0

For each paut, we see that

l im €r :0 and(Lr,au)-(o,o)

and €z: -2Yo Ax - xo LA

and €z: -2Yo Lx - Lx LY - xo LU

l i m € z : 0(ar,Lu)-(o,o)

It should be noted that it is only necessary to find one pair of values

for e1 and e2. o

L9.5.3 Theorem If a function f of two variables is differentiable at a point, it is continuous

at that Point.


ExAMpLE 1.: Given

pRooF: If / is differentiable at the point (ro, a), it follows from Defini-tion 19.5.2 that

f(xo+ Ax, yo+ Ay) - f (xo, y i : Drf(xo, y) Lx * Drf(xo, y) Ly* e r A r i - e 2 L y

where €r + 0 and e2 + 0 as (Lx, Ly) -t (0,0). Therefore,

f (xo+ Ax, Ao+ Ay) : f (xr, ys) * Dlf @s,U) Ax * Drf (xo, y) Ay

Taking the limit on both sides of the above as

l im. , f (xo* Ax , Ao* Ay) : f (xo ,Ao)(Ar ,ag) - (o ,o ) '

* e r A r * e z L A

(Lx, Ly) - (0, 0), we obtain

If we let xo* Ax:, ?n!yoi- AA:y,, '(Ax, Ay) -- (0,0)" is equivalentto "(x, y) t (xo, yo)." Thus, we have from (4)

,,,,1\T,*rf(x' y) : f(x" yo)

which proves that / is continuous at (xs, ys). r

Theorem 19.5.3 states that for a function of two variables difierenti-ability implies continuity. However, the mere existence of the partial de-riv.ativr D'! -ya Drf at a point does not imply differentiability at thatpoint. The following example illustrates this.

(4)

f ( x , y )

SOLUTION:

Dr f (0 ,0) - t im fk '0) - f - (0 '0) - t im 0 - 0-

0

D,f (0, a'r): ; t *' t, l,l'' g: ; &],u ' 0 Y - 0 u - 0 Y

\prove that Dl(0, 0) and Dd@, 0)exist but that f is not differen-tiable at (0, 0).

i f (x , y) # (0, 0)

i f (x , y) : (0, 0)

Therefore, both Drl(0, 0) and Drf(0,0) exist.In Example 2 of Sec. L9.2 we showed that for this function

lim f(x, a)(.r,g)-(o,o)' -

does not exist; hence, / is not continuous at (0,0). Because/is not con_tinuous at (0, 0), it follows from Theorem 19.5.3 that / is not differenti-able there.

... P"f9f stating a theorem that gives conditions for which a tunctionwill be differentiable at a point, we Jonsider a theorem needed in its proof.It is the mean:value t*reorem for a function of a single variaule appli"d to

"function of two variables.

19.5.4 Theorem

( t r , a o , f G r , Y o ) )


Let / be a function of two variables defined for all r in the closed interval

la, bl and all y in the closed interval lc, d).(i) If.D1f@, /o) exists for some yoinlc, d] and for all xinfa, b], then

there is a number f, in the open interval (a,b) such that

f (b ,y ) - f (a ,AJ : (b - a )D ' . f ( t ' ,A ) (s

(ii) lt Drf (xo, y) exists for some xo in la, bl and for all y in lc , dl , then

there is a number f, in the open intewal (c, d) such that

f (xo, d) - f (xo, c) : (d - c)Drf (xo, t r)

Before proving this theorem, we interpret it geometrically. For part (i)

refer to fig. tS.S.i, which shows the portion of the surface z:f(x,y)

above the rectangular region in the xy plane bounded by the lines x: a,

x: b,y -- c, and y : d: The plane y : yo intersects the surface in the curve

represlnted by the two equations y: /o and z: f(x, y). The.slope of the

line through the points A(a, yo, f@, y)) and B(b, Ao, f(b, /o)) is

f (b, v) - f(a' Yo)b - a

Theorem 19.5.4(i) states that there is some point (tt, Ao, fGr, yJ ) on the

curve between the points A and B where the tangent line is parallel to the

secant line through A and B; that is, there is some number (1in (a, b) such

that D/(f1, yJ: lf (b, y) - f(a, y)ll(b - a), and this is il lustrated in

the figure, for which Drf (€t, /o) < 0.Figure 19.5.3 illustrates part (ii) of Theorem 19.5.4. The plane x: xo

interseits the surface ,: f (x, y) in the curve rePresented by the two equa-

tions r:xo ;lfld z:f(x,y).The slope of the line through the points

- f (xo, c)

(5 )

D(*o , d , f ( ro , d ) )

z : f ( x , Y )

(ro, tr, f(xo ' tr))

f (o ,y ) - f (b ,vo)

z : f ( x , Y

B(b, Ao , f (b, yo ))

Figu rc 19.5.2 Figure 19.5.3


C(xo, c, f (xo, c)) and D (xs, d, f (xo, d) ) is [ f (xo, d) - f (xo, c)] tG- c), and

Theorem 19.5.4(ii) states that there is some point (xo, tr, f(A, €")) onthe curve between the points c and D where the tangent tine is parallel tothe-secant line through c and D; that is, there is som-e number {, in 1c, d,1such that Drf (xo, €r) : lf (xo, d) - f (xo, c)llG - c).

PRooF oF THEoREM 19.5.4(i): Let g be the function of one variable rdefined by

g ( r ) : f ( x , y o )

Then

8'(r) : Drf(x, yo)

Because-Daf (x, y) exists for arl r in fa, bl, itfoflows thatg'(r) exists for alrx in fa, bf , and therefore g is coniinuous on [a, bl. so u]' ir," mean-varuetheorem (4.7.2) for ordinary derivatives there exiits a number (, in (a, b)such that

or, equivalently,

Drf (€r,Ao) - f (b ' Yo) - f (o ' Yo)a - a

from which we obtain

f(b, yr) - f(a, Ao) : (b - a)Dl(€r, Ao)The proof of part (ii) is similar to the proof of part

an exercise (see Exercise 11).

SOLUTION:

(2, 5) such

f (s' 4)

8 ' ( f ' )

(i) and is left asI

Equation (5) can be written in the form

f(xo + h, Ui - fko, A) : yDJ(€r, Ur)where f1 is befween xe and xs * h and h is either positive or negative (seeExercise 12).

Equation (6) can be written in the form

f (xo, yo+ I() - f(xo, yr) : lDrf (xo, €r) (8)

Yh"r" f, is between yo and ao * 11and k is either positive or negative (seeExercise L3).

(7)

EXAMPIr 2: Given By Theorem "l'9-5.4, there is a number f1 in the open intervalthat

- f (2, 4) : (s - z)Dl(€,, 4)f (x, y)

find a19.5.4

Theorema n d y : 4 .

fr reguired byif x is in 12, 5l

So


F r 6 , ^ 2 4c - 5 : J ( 3 + f J

9 7 25-re-|F

(3+ t ) ' : 40

+ l f (xo * Ar ,A) - f (xo,

Because Drf and Drf exist on B(Po; r) and (ro * Lx, Ao*B (Po; r) , rt follows from (8) that

f (xo* A ,x , Ao* Ay) - f (xo* Lx , y ) : (AV)Dr f (x0 + Lx , t r )

where fr is between lo and Ao * Ly.From (7) it follows that

f (xo* Lx , ys ) - f (xo ,Ao) - (Ar )D ' f (€ ' , Ao)

where fr is betwe€n .rs and xo * A,x.Substituting from (10) and (11) in (9), w€ obtain

Lf (xo,Ao) - (Ly)Drf ( ro * Lx, t ) + (Ar)D' f (€r ,Uo)

919

a) l (e)

Ly) is in

(10)

Therefore,

3 + f r : +2\m

But because 2

€r:z\rc - 3sign and obtain

The following theorem states that a function having continuous par-tial derivatives at a point is necessarily differentiable at the point.

19.5.5 Theorem Let/be a function of two variables x andy. Suppose thatDtf NrdDrf existon an open disk B (Po; r), where P6 is the point (16, y6). Then if Dtf andD2fare continuous at Ps, f is differentiable at Pq.

pRooFr Choose the point (xs* Lx, yo* Ly) so that it is in B(Po; r).Then

Lf(xo, y) : f(x, * Lx, ys + LV) - f(xo, yo)

Subtracting and adding f (xo + Lx, lJ to the right side of the above equa-tion, we get

Lf (xo, y) : lf (xoi- Lx, ys+ LV) - f (xo* A,x, yr)f

( 1 1 )

(12)


Because (ro * Lx,Ao* Ly) is in B(Po;r) ,€zis between yoandAo* Ly,and Drf is continuous at P0, it follows that

l im.- ,Drf (xo * Lx, (r) : Drf (xo, Uo)(Ar,Au)-(o,o)

and, because f, is between xs and xo * Lxit follows that

l im. ,Dr f (h ,A) : Dr f (xo ,Ao)(Ar,As)- (o,o)

-'

If we let

er: Drf(tv y) - D,,f (xo, yr) 1rs)it follows from Eq. (14) that

l im €r :0ta',lir[to,ot

-' - (16)

and if we let

e2: Dd(xs * Ax, {r) - Drf (xr, yJ Gz)it follows from Eq. (13) that

, . l i + . - . . G 2 : 0 ( 1 8 )(AJ',au)-(o,o)

Substituting from Eqs. (15) and (12) into (12), we get

Lf (n, yd : LylDd@o, yi * erl * AxlDi@s, yo) * erl

from which we obtain

Lf (xo, yi : Drf (xo, Ao) Lx * D2f ks, y) Ly * e1 Ar * e2 Ay (19)

From Eqs. (15), (18), and (19) we see that Definition 19.5.2 holds; so/ is differentiable at (xo, Ai. I

A function satisfying the hypothesis of Theorem 19.5.5 is said to becontinuously diferentiable at the point po.

(13)

and Drf rs continuous at Po,

(14)

ExAMpu 3: Given

f x2u2 .t| : - r i r ( x , y ) *

f ( x , A ) : 1 r ' * y ' - -t 0 i f ( x , y ) :

use Theorem 1.9.5.5 to provef is differentiable at (0, 0).

solurroN: To find Drf we consider two cases: (x,( x , y ) * (0 ,0 ) . I f ( x , y ) : ( 0 ,0 ) , w€ have

f ( x , O ) - f ( 0 , 0 ) , . 0 - 0 AD l ( 0 , 0 ) - l i m : t T ; - 0

(0, 0) and

(0, 0)

(0, 0)

that lf. (x, y) * (0, 0) , f (x, y) : *yrl(* * yr). To find Drf (x, y) we can usethe theorem for the ordinary derivative of a quotient and consider y asa constant. We have

Drf(*, r, -2rY'z(xz !!'z) :-?x(fY2)

(* * y')'


2xyn: 1x4 11'

The function Dtf is therefore defined by

921

(20)

y) + (0, 0)

y) - (0, o)

function Drf defined by

, t , ry:^, ir (x,y) + (o, o)Drf (x, y) : ] (x2 - y t

L0 1f. (x, y) : 10, 0)

Both D/ and Drf exist on every open disk having its center at theorigin. It remains to show that Dtf and D2f are continuous at (0, 0).

Because Drf(O,0) :0, Drf will be continuous at (0,0) if

l im D ' f (x , U) :O(e,s)-(oo)

Therefore, we must show that for any e ) 0 there exists a 6 > 0 such that

| 2*Yn irDr f (x , A) : l ( * ' * y ' ) '

r^

Lo if

In the same manner we obtain

| ' r * o l 4 I . A - ^ / -

| ^nJ

l(* *'r1'l ' e whenever o

l , ?* !n " r r l - , l , lyn - , =nWfW)^ - 2\W

lry| @'* y') ' : (x'+ Y') 'Therefore,

| 7 r t14 |I

tnl

l(x'+fuI "u whenever o

(x,

(x ,

the

So if we take 6 : !e, we have (20). Hence, Dt;f is continuous at (0,0)' In

the same way we show that D2f is continuous at (0' 0). It follows from

Theorem 19.5.5 that f is differentiable at (0,0).

If we refer back to Eq. (3), the expression involving the first two

terms on the right side, Dgf(ro , A) Lx * D2f (xs, yr) NA, is called the prin-

cipal part of. Lf (xs, yq) or ttre "total differential" of the function/ at (n, y).

We make this as a formal definition'

19.5.5 Definition lf f isafunctionof twovariables xandy,and/isdifferentiableat(x,y),ttren the total differential of f is the function df having function values

given by(2r)

y, Lx, and Ly.IfNote that df is a function of the four variables x,


z: f(x, A), we sometimes write dz in place oI df(x, y, Lx, Ay), and thenEq. (21) is written as

(22)

If in particular f (x, y) : x, then z : x, D 1f (x, y) : 1, and D2f (x, y) : 0,and so Eq. Q2) gives dz: Ar. Because z: /, we have for this functiondx: Ax.In a similar fashion, if we take/( x,U) : y, then z: U,Drf (x,y) :0, and Drf (x,A) : l, and so Eq. Q2) gives dz: Ay. Becausez: y, we havefor this function dy : Ay. Hence, we define the differentials of the inde-pendent variables as dx: Ar and dy : Ly. Then Eq. e2) can be writtenas

ExAMPrn 4: A closed metal canin the shape of a right-circular

rylinder is to have an insideheight of 5 in., €ln inside radiusof 2 in., and a thickness of 0.1. in.If the cost of the metal to be usedis L0 cents per in.t, find by differ-entials the approximate cost ofthe metal to be used in themanufacture of the can.

dz- Dr f (x ,y ) dx * Dr f (x ,y ) dy

and at the point (xo, Uo) , we have

dz - Drf (xo, A) dx * Drf (xo, U) dy

In Eq. (3), lett ing Lz: Lf (xo,U), dx: Lx, and dy: Ly, we have

Az : Drf (xo, Ao) dx + Ddjco, Ao) dy + e1 dx * e2 dy

dv: ff a, +{, anFrom Eq. (27) it follows that

av _ ^_-.t- _,_ r av q,; :2nrh and

f f i : Trrz

(23)

(24)

(25)

Comparing Eqs. @a) and (25), we see that when dx (i.e., Ax) and dy (i.e.,Ay) are close to zero, and because then e, and eralso will be close to zero,we can conclude that dz is an approximation to Az. Because dz is ofteneasier to calculate than az, we make use of the fact that ilz - Az incertainsituations. Before showing this in an example, we write Eq. (23) with thenotation 0zl0x and 0zlOy insteadoLDlf(d, y) and Drf(x, yl, respectively:

(26)

solurroN: The formula for the volume of a right-circular cylinder, wherethe volume is V in.8, the radius is r in., and the height is h in., is

V - nrzh

The exact volume of metal in the can is the difference between the vol-umes of two right-circular rylinders for which r : 2.'!., h: 6.2, and, r : 2,h: 5, respectively.

AV would give us the exact volume of metal, but because we onlywant an approximate value, we find dV instead. Using (26), we have

(zz7

(28)

t


Substituting these values into Eq. (28) gives

dV :Znrh dr * nrz dh

Because r - 2 , h :6 , d r - 0 .L , and dh :0 .2 , w€ have

dV :2n (2 ) (5 ) (0 .1 ) * n (2 ) ' ( 0 .2 )

- 3.2Tr

Hence, AV : 3.2rr, and so there is approximately 3'2rr in'8 of metal in the

can. Because the cost of the metal is 10 cents per in.3 and 1.0 ' 3.2tt:32tr :

100.53, the approximate cost of the metal to be used in the manufacture

of the can is $1.

we conclude this section by extending the concepts of differenti-

abilitv and the total differential to a function of n variables.

19.5.7 Definition If l is a function of the n variables x1'1 1z" " ' xn' antd P is the point

(xi, xr, . . . 'r'), then the increment of f at P is given by

Af(P) : f(x'+ Lx1, ir* Lx2, . . - ' ?n* Lx) - f(P) (29)

L9.5.g Definition If f is a functio_n of the n variables x1, x2,

f at the Point P can be written as

Lf (P) : D'f (P) Ar, + Drf (P) Lx2 * '

* er Lxt

w h e r e € 1 + 0 r € z - ) 0 , . . ' , € n - - + Q ' a ; s

( A r r , L x 2 , . . . , L x r ) + ( 0 1 0 , "

then f ," said to be differentiable at P '

. . , xn, and the increment of

. . +D"f(P) Axn

* e z L x 2 * " ' + e n L x n ( 3 0 )

, 0 )

Analogously to Theorem 19.5.5, it can be proved that sufficient con-

ditions foia functio n f of. n variables to be difflrentiab-le at a point P are

that Df , Drf , . . . , bnf all exist on ar-r open ball B(P; r) and that D/'

Dzf, .-. . ,-'Dnf are all continuous at P. As was the case for functions

of" two variabies, it follows that for functions of n vatiables differen-

tiability implies continuity. Howevet, the existence of the partial deriv-

atives Drf, Drf, . . ., Dof ata point does not imply differentiability of

the function at the Point.

1.9.5.9 Definition If f isafunctionof thenvariables xy,x2,. . . 'xnarrd/isdifferentiable

at,P, then the total difi er ential of / is the function df havtng fu nction. values

given bY

d f ( P , L x r , L x 2 , " " A r o ) : D l ( P ) L x 1 1 - D 2 f @ ) M z * " ' + D J ( P ) A r " ( 3 1 )

92/T DIFFERENTIAL CALCULUS OF FUNCTIONS OF SEVERAL VARIABLES

Letting w: f (xr, x2, . . . , xn), defining dx,dxn: Lx, and using the notation 6w | 0x; insteadEq. (31) as

: Lxt, dxr: L,x2, .of D1f(P), w€ can

(32)

V in.3 be the volume of a box whose dimensions are trwe have the formula

EXAMPTn 5: The dimensions ofa box are measured to be 1o ln.,12 in., and 15 in., and the mea-surements are correct to 0.02 in.Find approximately the greatesterror if the volume of the box iscalculated from the given mea-surements. Also find the approxi-mate percent error.

solurroN: Lettingin., y in., and z in.,

xyz

The exact value of theas an approximationables, we have

av, ,avdV :+ - dx t+dx dy

and so

error would be found fromto LV. Using Eq. (32) for

da * {a ,v d z

LV; however, we use dVthree independent vari-

d V : y z d x * x z d y + x y d z (33)

From the given information lArl < 0.02,lAyl < 0.02, and lArl = 0.02. Tofind the greatest error in the volume we take the greatest error in themeasurements of the three dimensions. So taking dx:0.02, dy:0.02,dz: 0.02, and r : 10, ! : 12, z:1.5, we have from Eq. (33)

6y: (12) (1s) (0.02) + (10) (15) (0.02) + (10) (12) (0.02)- 9

so, AV - 9, and therefore the greatest possible error in the calculation ofthe volume from the given measurements is approximately 9 ins.

The relative error is found by dividing the error by the actual value.Hence, the relative error in computing the volume from the given mea-surements is LVIV;= ilVlV:r*o:zfo:0.005. So the approximateper-cent error is 0.57o.

Exercises 19.5

1'. If f(x, A) : 3x2 * 2xy - A', Ar - 0.03,o f f a t ( 1 , 4 ) .

and Ay : -0.02, find (a) the incremen t of f ar ( 1 , 4) and (b) the total differential

2. It f(x, y) : xye"u, Ax: -0.1, and Ay : 0.2, find (a) the increment oI f at (2,-4) and (b) the(2,-4) .

3. If f(1, y, z): xy *ln(yz), Ax:0.A2, Ay:0.04, and &:-0.03, f ind (a) the increment of /total differential of f at (4, 7, S).

total differential of f at

at (4, I , 5) and (b) the


4. If f (x, y, z) - x'y * Zxyz - z.3, Lx: 0;0L , Ay: 0.03, andthe total differential of f at (-3 , 0, 2) .

find (a) the increment of f ut (-3 , 0, 2) and (b)

In Exercises 5 through 8, prove that / is differentiable at all points in its domain by doing each of the following: (a) FindLf (xo, y) for the given function; (b) find an e1 and an €2 so that Eq. (3) holds; (c) show that the et and the e2 found in part(b) both approach zero as (Lx, Ay) --, (0,0).

5. f (x, U) : x'Y - 2xY f (x ' y)

f (x , y)

6. - 2x2 i 3y'

vx

Prove that Dl[,l) and, D2f(1,1) exist, but / is not differentiable at (1, 1).

. l3!I- i r (x,d + @,0\10. Given f (x, y) : 1f + Vn

-' \-" r '

l0 it (x, y\ : (0, 0)

Prove that DlQ, 0) and D2f (0,0) exist, but / is not differentiable at (0, 0).

11. Prove Theorem 19.5.4(ii).

12. Show that Eq. (5) may be written in the form (7) where {1 is between 16 and xs I h.

13. Show that Eq. (6) may be written in the form (8) where {2 is between yn and lo * k.

In Exercises 14 through 77, use Theorem19.5.4 to find either a f1 or a f2, whichever applies.

7. f (x, ,) : i 8'

s. Given f (x, a) : { ; . Y - 2

; l : } ' r ; i ; t . ,

14. f(*, a) : x2 + 3xY - Y';r is in [1, 3] ; Y - 4

16. f(x, A) :h, y LS in l-2, 2f; x : 4

, \ | ?:Y:i ir (x, y) + (0, o)L8. Given f (x, A) : 1*' + y' \ '- '

L0 r f (x, y) : (0, 0)

This function is continuous at (0, 0) (see Example

Drf (0,0) exist but Dtf and Drf are not continuous

?ve-n :rs. Given f (x, a) : I #

if (x' Y) + (o' o)

L0 t f (x , y) _ (0, 0)

15. f (x, y) : xs - y'; x is in 12,61; y - 3

tr. f (x, y) : ' f f i ; y Ls Ln lo,4l; x - Z

3, Sec. 19.2, and Illustration L, Sec. L9.3). Prove that Drf (0,0) and

at (0, 0).

21,. f (x, y, z) : )cy - xz * z2

Prove that f is differentiable at (0, 0) by using Theorem 19.5.5.

In Exercises 20 and 21, prove that / is differentiable at all points in R3 by doing each of the following: (a) Find, Af (xo) ys, zn);(b) find drr G11 G2, and e3, so that Eq. (30) holds; (c) show that the e1, e2, and e3 found in (b) all approach zero as (Ax, Ly, Lz)approaches (0, 0, 0).

20. f (x, A, z) :2x22 - 3yz'

| * "22. Given f(x,y,z) :Vi fu

i I (x 'v 'z) + (0 '0 '0)

L0 it (x, y, z) : (0, 0, 0)

(a) Show that D1l(0, 0, O) , D2f @ , 0, 0) , and D"f (0 , 0, 0) exist; (b) make use of the fact that differentiability impliescontinuity to prove that / is not differentiable at (0, 0, 0).

24.

23.


( xttzzGiven/( r , ! ,2) : l7T7I7

i f (x 'v 'z) * (0 ,0,0)

|.O if (x, y, z): (0, 0, 0)Prove that / is differentiable at (0, 0, 0).

Use the total differential to find approximately the greatest error in calculating the area of a rightlengths of the legs if they are measured to be 6 in. aid 8 in., respectively, with i possible error of 0.1surement. Also find the approximate percent error.25. Find approximately, by using the total differential, the greatest error in calculating the length of the hypotenuse of theright triangle from the measurements of Exercise 24. Alio find the approximate percent error.26' If the ideal gas law (see Example 5, sec. 19.4) is used to find P when T and, v are given, but there is an error of 0.37oin measuring T and an error of 0.8vo in measuring V, find approximately the greatlst percent ertor in p.

27. The specific gravity s of an object is given by the formulaA

S : -- A - W

where A is the number of pounds in the weight of the object in air and W is the number of pounds in the weight of theobject in water. If the weight of an obiect inair is read is 20 lb with a possible error of 0.0i lb and its weighiin wateris read as 12 lb with a possible enor of 0.02.1b, find approximately the ligest possible ertor in calculating ifrom thesemeasurements. Also find the largest possible relative error.

28' A wooden box is to be made of lumber that is 3 in. thick. The inside length is to be 5 ft, the inside width is to be 3 ft,the inside depth is to be 4 ft, and the box is to have no top. use the total differential to find the approximate amountof lumber to be used in the box.

29' A- company has contracted to manufacture 10,000 closed wooden crates having dimensions g ft, 4 ft,and 5 ft. The costof the wood to be used is 5c per square foot. If the machines that are used to cut the pieces of wood have a possibleeror of 0'05 ft in each dimension,-find approximately, by using the total differential, the greatest possible error inthe estimate of the cost of the wood.

In Exercises 30 through 33, we show that a function may be differentiable at a point even though it is not continuously dif-ferentiable there' Hence, the conditions of Theorem 19.5.5 are sufficient but not necessary for differentiability. The func-tion / in these exercises is defined bv

triangle from thein. for each mea-

f ( x , y ). 1sin

u76, if (x, Y) + (0, o)

if (x, y) : (0, 0)

CHAIN RULE In Sec. 3.6 we had the following chain rule (Theorem 3.d.1) for functions ofa single variable: rf y is afunction of a, defined by y : f (u) , and,Dzl exists;and z is a function of x, definedby u: g(x), anaoru exists; their y is afunction of x, and D"y exists and ii given by

Dry : D;y Dru

30.

32.

33.

Find A/(0, 0). 31. Find Drf(x, y) and Drf (x, y).Prove that / is differentiable at (0, 0) by using Definition 19.5.2 and the results of Exercises 30 and 31.Prove that D1f and.Drf arc not continuous at (0, 0).

19.6 THE

19.6 THE CHAIN RULE

or, equivalently,

da dy du"/dx du dx

We now consider the chain rule for a function of two variables, whereeach of these variables is also a function of two variables.

lf. u is a differentiable function of x and y, defined by u:f(x,y)' andx: F(r, s) , U : G(r, s), and \xlEr, \xl0s, 0yl0r, and Eyl0s all exist, then u isa function of r and s and

pRooF: We prove (2). The proof of (3) is similar.If s is held fixed and r is changed by an amount Ar, then r is changed

by an amount Ar and y is changed by an amount Ay. So we have

Lx: F(r * Ar, s) - F(r, s) (4)

and

Ly: G( r * L r , s ) - G( r , s ) (5 )

Because / is differentiable,

Lf (x, y) : Drf (x, y) A'x * Drf (x, v) Ay * e1 Ar * e2 Ly (6)

where er and e2 both approach zero as (Lx, Ly) approaches (0, 0). Further-more, we require that e1 :0 and €z:0 when Ax: Ly:0' We make this

requirement so that e1 and er, which are functions of Ax and Ly, will becontinuous at (Ax, Ly\: (0,0).

If in (6) we replace Lf (x, y) by Lu,Drf (x, V) by \ul6x, andDzf G, U)by \ulay and divide on both sides by Lr (Lr * O) , we obtain

Lu 0u Lx 0u La A'x , Ly- : - - T € r - - : - T C r . -

Lr 0x Lr 0Y Lr 'Lr - Lr

Taking the limit on both sides of the above as Ar approaches zero, we get

H # :#rin ̂ +,.H Hql # + (1iq.,) li$ # * ,1i*1.,) li$ fl vtBecause u is a function of r and y and both r and y are functions of r and s,

a is a function of r and s. Because s is held fixed and r is changed by an

( 1 )

19.6.1 TheoremThe Chain RuIe


amount Lr, we have

limA,r-O

Also,

limAr- 0

L u _Lr

r :_ u(r * Ar , s) - u( r , i l _ O,'m or

Lx _ r :_ F( r * L r , s ) - F ( r ,s ) 0xLr: ll$

: a,

(8)

(e)

and

l im ry:hmAr-o Lf Ar-o

0u

ay(10)

0r

Because 0xl0r and 0yl6r exist, F and G are each continuous withrespect to the variable r. (rorr: The existence of the partial derivatives ofa function does not imply continuity with respect to au of the variablessimultaneously, as we saw in the preceding section, but as with functionsof a single variable it does imply continuity of the function with respect toeach variable separately.) Hence, we have from (4)

tF( r + L,r, s) - F(r, s) ]

s) - F(r , s)- 0

and from (5)

lim AY: lim lc(r + Lr, s) - G(r, s)fAr-0 Ar-0

: G(r, s) - G(r, s)- 0

Therefore, as Ar approaches zero, both Ar and Ay approach zero. Andbecause both e1 and e2 approach zero as (Lx, Ay) approaches (0,0), wecan conclude that

li-^ €r : 0 and lim €z: 0Ar-0 Ar -0

Now it is possible that for certain values of Lr, Lx: Ly:0. Because werequired in such a case that er:€2:0, the limits in (11,) are still zero.Substituting from (8), (9), (10), and (1.1) into (Z), we obtain

which proves (2),

l im Ar: l imAr-0 Ar -0

: F(r ,

(11 )

#:(,H(#) .(,H(y)EXAMPLE ].: GiVCN

u - l n@ 0x_ _ p 8

a V

dray

19.6 THE CHAIN RULE 929

x : T€r, and y : f o-r, find 6ul0r

and oulos.0xE : " r

From (2) we get

0 u xAr :W

From (3) we get

0uds

As mentioned earlier the symbols 0ul0r,0ul0s, Eul\x,0ul0y' and so

forth must not be considered as fractions. The symbols 0u, 0x,, and so on

have no meaning by themselves. For functions of one variable, the chain

rule, given by Uq. (t), is easily remembered by thinking of an ordinary

derivative as- the quotient of two differentials, but there is no similar

interpretation for partial derivatives.Another trouLlesome notational problem arises when considering

u as a function of x and y and then as a function of r and s.lf. u: f (x, y) ,

x: F(r, s), and y: G(r, s), then u: f (F(r, s), G(r, s)) . [Note that i t is

incorrect to write u : f (r, s).1

o rLLUsTRArroN 1: In Example 1,

u: f(x, y) : lnty'p +V

x: F(r, s) : re"

A : G( r , s ) : re - "

and so

L t - fG?,s ) , G( r , s ) ) : ln f f i

f f ( r , s ) : 6 t f f i + u ' f

If we let f(E(r, s), G(t, s)):h(r, s), then Eqs. (2) and (3) can be

written resPectivelY as

hr(r , s) : f , (x, y)Er(r , s) + fr(x, Y)Glt , s)

and

hr(r , s) : f r(x, A)F"(t , s) + 1r@, Y)Gr(r ' s)

In the statement of Theorem 19.6.1 the independent variables are

r antd. s, and, u is the dependent variable. The variables r and y can be

:alled the intermediate variables. We now extend the chain rule to z inter-

mediate variables and m independent variables.

19.6.2 Theorem Suppose that u is a differentiable function of the n variables.rr, x2t . . . ,The General Chain Rule xn,' ind each of these variables is in turn a function of the m variables

x , ^\ U / .. _c\ r(xes - ye-r): f f i ( rer)+ff i (-r t-r)-W


Ar, Uz, . . . , A*. Suppose furtherA x i l A y j ( i : 1 , 2 , . . . , n ; j - 1 , 2 , .

that each of the partial derivatives, m) exists. Then u is a function

)+

,) *

ffi(,H(,HW)#,:mGH.(#:d z \

)xr/

:

#--(,Hffi).(,H(,H.(.9

EXAMPLE Given

u -xy+xz *yz

X: r, A - r cos t, and Z: r stn t,

find 0ul0r and 0ul0t.

The proof is an extension of the proof of Theorem 19.6.1..Note that in the general chain rule, there are as many terms on the

right side of each equation as there are intermediate varia-bles.

solurroN: By applying the chain rule, we obtain

#:e;e).w)(,*).(Hff): (y * z)(1) + (x I z) (cos f) * (r * y) (s in f): y * z * r cos t * z cos f * r sin t i- y sin t: / cos f * r s in f * r cos f * (r s in t)(cos t) * rs inf * (rcos f)(s in t):2r(cos f * sin t) + r(2 sin f cos f): 2r(cos f * sin t) * r sin 2t

#:(rJ(#J .ffi)(y).ffie,): (y * z) (O) + (x *-z) (-r sin r) + (s * y) (r cos f): (r* r s in f)(-r s in t) + (r* r cos f)(r cos f):-r2 sin f - rz sinz t * 12 cos t I rz cosz t: r2(cos f - sin t) * rz(cosz f - sin2 f): rz(cos I - sin t) + 12 cos 2t

Now suppose that z is a differentiable function of the two variablesr and y, and both r and y are differentiabre functions of the single vari-able f. Then z is a function of the single variable t, and so instead of thepartial derivative of z with respect to f, we have the ordinary derivative

19.6 THE CHAIN RULE

of. u given by

(12)

EXAMPLE Given

u -x2+2xy *y '

)c: f cos t, and y : t sin t,

find duldt by two methods:(a) using the chain rule; (b) ex-

pressing u in terms of f before

differentiating.

EXAMPLE 4: lt f is a differ-entiable function and a and bare constants, prove that z:

f Gax'- *ayt) satisfies the Partialdifferential equation

soLUTroN: Let u - tbxz -

the given equation. By the

Y:+Y:f ' (u)(bx)0x du 0x J' \-

Therefore,

*ay'. We wish to show that z - f (u) satisfieschain rule we get

and * :+Y: f ' (u) ( - ny ' )!*a'lE ay du ay

we call duldt given by Eq. (12) the total deiaatiae of u with respect

to f. If u is a differentiable function of the n variables x1, x2, . . ., xn andeach ri is a differentiable function of the single variable t, then z is a func-tion of f and the totdl derivative of a with resPect to f is given by

solurroN: (a) 6uldx: 2x * 2y, \ul0y - 2x *dyldt: sin t + f cos f. So from (12) we have

dxldt: cos t- t sin

# : (2x *2y ) ( cos t - f s i n f ) + (2x *2y ) ( s in t+ t cos

- 2 ( r * i l ( c o s t - f s i n t + s i n t + f c o s f )

: / ( t c o s t + f s i n f ) ( c o s t - f s i n t + s i n t + f c o s f )

:2 t (cos2 t - f s in f cos t+ s in f cos t+ f cos2 t+ s in

- t s i n2 t+ s in2 t+

- 2 t [ 1 + 2 s i n f c o s t + f ( c o s 2 t - s i n 2 t ) ]

:2t(1 + sin 2t + t cos 2t)

(b) u : (f cos t) ' + 2(t cos t) (t sin f ) + (t sin t)2

- sz cos2 t + t'(2 sin f cos f ) + t2 sinz f

: t2 + tz stnZt

4:2t + 2t stn 2t + Ztz cos 2t

t )

f cos

f sin

t

f cos f)

dt

' 02 * bx* : ayz l f ' (u ) (bx) l * bx l f ' (u ) ( -ay ' ) l :0a'Y' ax dy

which is what we wished to Prove.

gilI DIFFERENTIAL CALCULUS OF FUNCTIONS OF SEVERAL VARIABLES

EXAMPLE 5: Use the ideal gaslaw (see Example 5, Sec. "1,9.4)

with k - 10 to find the rate atwhich the temperature is chang-ing at the instant when the vol-ume of the gas is L20 in.3 and thegas is under a pressure of 8 lb/in.2if the volume is increasing at therate of 2 in.t/sec and the pressureis decreasin g at the rate of0.1. lb/in.2 per sec.

soLUTroN: Let t -

7Tr

T

D -

T f -

the number of seconds in the time that has elapsedsince the volume of the gas started to increase;the number of degrees in the temperature at t sec;the number of pounds per square inch in the pressureat t sec;the number of cubic inches in the volume of the gasat t sec.

dP ldt: -0.1, and dvldt - 2.

PV: L0T and s F^ PV

o r : 1 o

At the given instant, P : 8, V - !20,Using the chain rule, we obtain

dT aTdP .aTdv- : -d t aP i f t ' a vd t

: ydP _L u70 itt ' t0 dt

:1#(_0.1) + fo(Z): -1 .2 * 1 .6:0 .4

Therefore the temperature is increasing at the rate of 0.4 degree persecond at the given instant.

Exercises 19.6

In Exercises 1 through 4, find-the indicated partial derivative by two methods: (a) Use the chain rule; (b) make the substi-tutions for r and y beforc differentiating.

1. u- xz - y'; x: 3r - s; y : r * 2s;#ry,

3. u - sutc' x:2r cos t; A : 4r sin t, !, U0r' at

In Exercises 5 through 10, find the indicated partial

5. Lt : sin-l(3r * il; x: rler; y : sin rs; #, X

7. u -cosh X, * : 3r2s; y : 5se'; #r#

9. u- xz I y' + 22; x - rsin @ cos 0; y :r sin @ sin

2. u - 3x2 * )cy - 2y, + 3x- y; x: 2r- 3s; y : r * s;#, X

4. u- x2 + y, ; x : cosh r cos t ; A: s inh r s in t ;U=Ouor' at

derivative by using the chain rttle.

5. u - xe-o; )c - tan-r (rst) ; y : ln(3rs

8. u: ,cy + xz * Az; x : rs ; y : rz - s2;

o; z: r cos ' ott ou oua; *; ao,M

10. u - x'yz; )c -!; V : res; z - re-'r; ou. ou0r ' ds

LL. u: ye' + )cea; x: cos f; y - sin f

13. u- W; x: tan t ; y - cos f ; z- s tn t ;0 < t < Ln

In Exercises L5 through 18, find the total derivative duldt by using the chaindifferentiating.

19.6 THE CHAIN RULE 933

In Exercises L1 through 14, find the total derivative duldtby two methods: (a) Use the chain

tions for x and y or fior x, y, and z before differentiating'(b) make the substitu-

: l n x y + y ' ; x : e t ) y :

t * e r: - . t ; x : 3 s i n t ; A :y - e '

do not expres s u as a function of t before

1s. u -tan-r e), x - rn t; a - et

L7 . u -H, x : ln f ; y : t "1y t

In Exercises 1"9 through 22, assume that thefind 0zl0x and 0zlAy.

19. 3x2 + y'+ z2 - 3xy * 4xz - 15 - 0 20. z-- (x'* yz) sin xz 21.

Zg. If f is a differentiable function of the variable u,let u: bx - ay and

a(azlAx) + b(azlay) : 0, where a and b ate constants.

24. If. f is a differentiable function of two variables u arrd. o,let u: x - y and a : ! - r and prove that z: f (x - y ' y - ')

satisfies the equation 0zl6x * Azl0y - 0.

ZS. If f rs a differentiable function of x and y andu- f (x,A), x- r cos 0, and

0u 0u ^ 6usin0c o s t - - -

d x 0 r - - - - A 0 r

0 u 0 u ^ d u c o s 0a " : a r

s r n 0 + a e r

26. | f fandgaredi f ferent iablefunct ionsof xand yandu: f (x , i l anda: g(x,V) ,suchthat dulax: |o ldyandaul |y :-dvl}x, then if x: r cos 0 and'y: r sin 0, show that

du 'l,. 6v 0o | 0u

ar :V N and ar : -7 ao

27. Supposefis a differentiable function ofr and y and.u: f(x,y). Then ifx: cosh 7) cosTo andy: sinh o sin at, exPress

6ul0p and duldw in terms of 0ul0x and 6ul6y.

28 . Suppose / i sad i f f e ren t i ab le func t i ono f x , y ,andzandu : f ( x ,V ,z ) .Then l f x : r s i n@cos0 ,

,:, "o"

{, express dul0r, 6ul0Q, arrd aulae in terms of. |ul6x, \ul0y, and |ul0z'r sin d sin 0, and

29. At a given instant, the length of one leg of a right triangle is 10 ft and it is increasing at the rate of 1 ftlmin and the

lengti of the other leg of tie right triangle is 12 ft and it is decreasing at the rate of 2 ftlmin. Find the rate of change

of the measure of the acute angle opposite the leg of length 12 ft at the given instant'

30. A vertical wall rnakes an angle of radian measure Sa with the ground. A ladder of length 20 ft is leaning against the

wall and its top is sliding doiyn tne wall at the rate of 3 ftlsec. How fast is the area of the triangle formed by the ladder,

the wall, and ihe g"orttJ changing when the ladder makes an angle of ta'radians with the ground?

31. A quantity of gas obeys the ideal gas law (see Example 5, Sec. 19.4) with k: 12, and the gas is in a container which

is being heated at a raie of 3o per second. If at the instant when the temperature is 300o, the pressure is 5 lb/in'2 and is

decreasing at the rate of 0.1 lb/in.'z per second, find the rate of change of the volume at that instant.

e-t

l n f

12. u

1,4. u

1 , 6 . u - x y + x z + f cos f sin

18. u- ln(xz * y '+ t2); x - f s in t ; U: cos f

given equation defin es z as a function of x and y. Dlfferentiate implicitly to

ye*o" cos 3xz - 5 22. zeaz I Zxe*' - 4e'o : 3

prove that z: f (b* - ay) satisfies the equation

r sin 0, show that

9:II DIFFERENTIAL CALCULUS OF FUNCTIONS OF SEVERAL VARIABLES

ZZ. l'tater is flowing into a tank in the form of a right-circular cylinder at the rate of *z fd/min. The tank is stretching insuch a way that even though it remains cylindrical, its radius is increasing at the rate of 0.002 ftlmin. How fast is thesurface of the water rising when the radius is 2 ft and the volume of water in the tank is 20z- fp?

19'7 HIGHER-OR_DER If f is a function of two variables, then in general Drf and.D4f are alsoPARTIAL DERIVATIVES functions of two variables. And if the pa*iil derivatives of these func-

tions exist, they are called second partial derivatives of. f. In contrast,Dtf and D"f are called first partial derivatives of f. There are four secondpartial derivatives of a function of two variables. If f is a function of thetwo variables r and !, the notations

Dr(Dl) Drrf f ,, f *oArf

0y 6x

all denote the second partial derivative of /, which we obtain by firstpartial-differe_ntiating / with respect to r and then partial-differentiatingthe result with respect to y. This second partial derivative is defined bi

f t ( x , U ) : l i mAu-O

rLy

if this limit exists. The notations

Dr(Drf ) Drrf fn f ,,Arf0xz

all denote the second partial derivative of /, which is obtained by partial-differentiating twice with respect to r. we have the definition

r (x * Ax, y) - fr(x, y)fu@,y):lntr e)if this limit exists. we define the other two second partial derivatives inan analogous way and obtain

( 1 )

frr(x, a) : l im

and

(3)

(4)f ,r(x, a) : lim f'@ ' Y * LY) - fr(x ' Y)As-0 ay

if these limits exist.The definitions of higher-order partial derivatives are similar. Again

we have various notations for a specific derivative. For example,

Drrrf frr" f atf tl-rcu av ax ax ay$

all stand for the third partial derivative of /, which is obtained by partial-differentiating twice with respect to r and then once with respect to y.

Note that in thefrom left to right;

19,7 HIGHER-ORDER PARTIAL DERIVATIVES 935

subscript notation, the order of partial differentiation isin the notati on 03f l0y 6x 0r, the order is from right to left.

EXAMPLE Given

e* sin

SOLUTION:

Drf (x, Y)

So (a) Dr,;f(r ,A) : e" siny - Ll*; and (b) Drrf (x, y) : s" cos y. (c) To find

Asflax Ay2, we partial-differentiate twice with respect to y and then oncewith respect to x. This gives us

a'f6x 0y2

Higher-order partial derivatives of a function of n variables have

definitions which are analogous to the definitions of higher-order partial

derivatives of a function of two variables. lf f is a function of n variables,

there may be n2 second partial derivatives of f at a particular point. That

is, for a function of three variables, if all the second-order partial deriva-

tives exist, there are nine of them: /rr, f ,r, fr", frr, frr, fr", f"r, 1"r, and flr.-

er stn V +*. , 1

s rn u t -'/ )cy

. Lslnu - -,

t' y'

f (x, u) : y+ lnxy

(b) D 'rf (x,find: (a) D'f (x,(c) atf li.x oyz.

EXAMPLE Given

f (x, A, z) - sin (xy * 2z)

f i n d D r t f ( i c , y , z ) .

EXAMPLE Given

A , z ) - y c o s ( x y * 2 2 )

y , z) - -2y sin (xy * 2z)

cos sin y

SOLUTION:

D'f (x,

D"f (x,

f ( x ,a ) : fY -find: ( a) Dt , f (x, U) ;

y cosh xy

( b) Dr , , f ( * ,y) .

Dtr r f (x ,y , z) sin (xy * 2z) - 2xy cos (xY * 2z)

SOLUTION:

(a) Drf (x,y) - 3x2U - Y2 srnh xY

Drrf (x,Y) :3x2 - 2Y sinh xY - xY' cosh xY

(b) Drf (x,Y) : x3 - cosh )cY - xY sr'.h xY

Dr,, f (x,y):3x2 - y sinh xy - y sinh xy - xy'cosh xy

:3)c2 - 2y sinh xy - xy' cosh ry

We see from the above results that

"mixed" partial derivatives Dtrf (x, y)

this particular functioh, when finding

respect to x and then !, the order of

condition holds for many functions'

shows that it is not alwaYs true.

for the function of ExamPle 3 the

and Drrf(x, V) are equal. So for

the second partial derivative with

differentiation is immaterial. This

However, the following examPle

93G DIFFERENTIAL CALCULUS OF FUNCTIONS OF SEVERAL VARIABLES

ExAMPLE 4: Let f be the func-tion defined by

t . \ x 2 - 7 r zf (x ,a) : l (xv) f f i i r (x ,Y)

L o if (x, y)

Find frr(0, 0) and frr(0, 0).

(0, 0)

(0, 0)

solurroN: In Example 3, Sec . 19.4, we showed that for this function

f '(0, Y) - -y for all y

and

fr(x,O) : r for al l x

From formula (1) we obtain

ft (0, o) : limAg-0

19.7.1 Theorem

But from (5), f ,(0, Ly) Ay and f ,(0, 0) and so we have

frr(0,0) : l im -O{,, O:

l im (-1) : -1Au-o ay Au-o

From formula (3), we get

frr(o, o) : lim f'(o * Lx ' o) - fr(o ' o)A,x

From (5), fr(Lx,0) : Ax and fr(0,0) :0. Therefore,

f r , , (o , o) : l im #: hm L: lAal-0 AI Atr_Q

For the function in Example 4 we see that the mixed partial deriva-tives/tr(r, y) and frr(x, y) are not equal at (0,0). A set of'conditions forthi+ ft(xo, yo):fu(xo, ye) is given by Theorem 1g.7.'!., which follows.The function of Example 4 does not satiify the hypothesis of this theorembecause both f' and fn are discontittuou" at (0, 0). It is left as an exerciseto show this (see Exercise 20).

*qp::g that / is a funcrion of rwo variables x and y defined on an opendisk B((16, yJ; r) and f*, fu, f,r, ^d fu, also are aefinea on B. Further-more, suppose that f ,u and fu, are continuous on B. Then

f ,u(h, uo) : fo,(xo, yo1

pRooF: consider a square hgifg its center at (xs, yo) and the length ofits side 2lhl such that 0 < frlhl < r. Then ail the ioi"t, in the interior9f the square and on the sides of the square are irithe open disk B (seeF ig . . l9 .7 . l_ ) . So the po in ts (x r *h ,yo*h) , (x r *h ,yo) , ind 1xo ,yo ih1are in B. Let A be defined by

A : f ( x o * h , y o + h ) - f @ o + h , y ) - f ( x o , V o + h ) + f ( x o , y o ) ( z )

Consider the function G defined by

c(r) - f(x, Ao * h) - f(x, yr) (8)Figure

19.7 HIGHER.ORDER PARTIAI- DERIVATIVES

Then

G(x+h ) - f ( x+h ,Ao+h ) - f ( x+h ,A r )

So (7) can be written as

A - G ( r o + h ) - G ( r o )

From (8) we obtain

G' ( x) : f"(x, Uo + h) - f"(x, yo)

Now, because fr(x, Ao*h) and fr(x, Ao) are defined on B, G'(r)exists if r is in the closed interval having endpoints at re and xn*h.Hence, G is continuous if r is in this closed interval. By the mean-valuetheorem (4.7.2) there is a number c, between xq and xs1' h such that

937

(10)

G ( r o + h ) - G ( r o ) : h G ' ( c ' )

Substituting from (11) into (9), we get

A - hG ' ( c t )

From (12) and (10) we obtain

A - h l f " (c r , Uo * h ) - f , (c t , U) f

Now Lf I is the function defined bY

s@) - f " ( c r ,Y )we can write (L3) as

A - hls@o + h) - s(yo) l

From (L4) we obtain

8 ' ( Y ) - f " o ( h , a )

( 1 1 )

(r2)

(13)

(14)

( ls)

(15)

Because f,u(cr, y) is defined on B,g'(y) exists if y is in the closedinterval having endpoints at yq and AoI h; hence, g is continuous if y isin this closed interval. Therefore, by the mean-value theorem there is anumber d1 between Ao and Ao * h such that

g\o + h) - sg) : h7'(d') (J7)

Substituting from (1,7) into (15), we get A:h2{ (dr); so from (16) it fol-lows that

L: hzf,o(ct, dt) (18)

for some point (c1, d1) in the open disk B. We define a function { by

QQ) : f (xo+ h ,y ) - f (xo , V) (1e)

a n d s o Q @ + h ) : f ( x o * h , y + h \ - f ( x o , A + h ) . T h e r e f o r c , Q ) c a n b e

written as

a - 6uo+ h) - fuo) (20)


From (19) we get

Q '@) : fo ( ro * h ,Y) - fo7o,Y)

{'exists if y is in the closed interval having yo and Ao*h as end.-points because by hypothesis each term on the right side of (21) existson B. Therefore, Q is continuous on this closed interval. so by the mean-value theorem there is a number d2 between ye and ao * h such that

QQo+ h) - AQJ = hO'@,)

From (20), (2L), and (22) it follows that

A - hlfr(ro * h, dr) - fr(xo, dr)f

(30 )

on both

(2r)

(22)

(23)Define the function 1 by

x(r) : fo@, k) e4)and write (23) as

A: h ly(x6+ h) - x( ro) l (2s)From (24) we get

x'(x) : fo,@, d") e6)and by the mean-value theorem we conclude that there is a number c2between xo and xs * h such that

x( ro + h) - X@): hx ' (cr ) e7)From (25), (26), and e7) we obtain

A: h2fnr(c2, dr) (2S)Equating the right sides of (1g) and (2g), we get

hzf,n(cr, ilr) : hrfur(cz, ilz)

and because h # 0, we can divide by hr, which gives

fru(cr, d): fnr(cr, dr) eg)where (cr, dr) and (c2, d) arc in B.

Because c, and c, are each between ro and xs * h, we can twite c, :xs*e1h, where 0 ( e, < L, and cz:xo*e2h,wheie0 ( e, < 1. Similarly,because both i\ and d2,are between youia!o*h,*".1n iit" d,r:yo! ,"h, where 0 ( es <_-1a. and ilr: Ao * enh,- where 0 ( ea ( 1.. Makingthese substitutions in (29) gives

f ,o(xo I e1h, Uo I esh) : f o*(xo * €zh, Uo * e+h)

Because fry and far are continuous on B, upon taking the limitsides of (30) as h approaches zero, we obtain

f ro (xo ,U) : f o * ( xo , yo ) I

EXAMPLE 5: Given that u -

f ( x , y ) , x - F ( r , s ) , a n d Y -

G (r , s) , and assuming that

f ,o: f o*, Prove bY using the chainrule that

62u-6or : f , * (x ,Y) [ r '? ' s ) f '

* 2 f *o(x, y)F, ( r , s)G,( r , s)

* f oo(x , y ) [G, ( r , s ) l '

+ f "(x , y) F,,(r , s)

* f o(x, U)G,,(r , s)

Exercises 19,7

In Exercises L through 8, do each of the following:

L . f (x , y )

19.7 HIGHER-ORDER PARTIAL DERIVATIVES 939

As a result of the above theorem, if the function / of two varidbleshas continuous partial derivatives on some open disk, then the orderof partial differentiation can be changed without affecting the result;that is,

Drrrf : Dt rf : Drttf

Drrrrf : Dt rrf : Drrrrf : Drr,rrf : D"t tf : Dr"rtf

and so forth. In particular, assuming that all of the partial derivativesare continuous on some open disk, we can Prove that Drttf : Dtt"f bYapplying Theorem 19.7.7 rcpeatedly. Doing this, we have

Drrrf : Dr(Drrf) : Dr(Drrf) : DlD2(D1f)l: DzIDJDrf)]

: D r ( D t f ) : D t t f

sol-urroN: From the chain rule (fheorem 19.6.1\ we have

#: f ,(*, y)F,(r, s) + fu@, y)G,(r, s)

Taking the partial derivative again with respect to r, and using the for-mula for the derivative of a product and the chain rule, we obtain

t2n '

f f i : l f,,(x,y)F,(r, s) * f*o(x, y)G,(r, s)lF,(r, s)'f f,(x, y)F,,(r, s)

* lf o,(x, y) F,(r, s) + f oo(x, y) G,(r, s)fG,(r, s) + f uQ, y) G,,(r, s)

Multiplying and combining terms, and using the fact that f ,u(x, A):f ,,(x, y) , we get

t#: f , ,(r, y)lF,(r, s)l ' + 2f ,o(x, y)F,(r, s)G'(r, s)

I fou(x, y) lG,( r , s) l ' * f , (x , y)F*( r , s) + fo@, y)G, , ( r , s)

which is what we wished to Prove.

2 ' f (x ,

(a) Find Dr,. f (x, i l ; b) f ind Drrf (x,y); (c) show that Drrf (x,A) : Dr ' , f (x,y)-

a) :2xB - 3x'y * xy' 3. f (x, U) : e2r svt Y- t -Ly x 2

4 . f ( x , y ) - t - r t a * l n

7. f (x , A) :4r s inh y

In Exercises 9 through

Y - t r t / * . , \ - ( u 2 t U

x 5. f (*, v) -- (x' * Y') tan-r 1

+ 3y cosh r 8. f(x, A) : tr cos y - ye*

14, find the indicated partial derivatives.

6. f (x, A) : "rn-t !


9. f (r, y, z) : ye, I zev t ez; (a) f ,"(x, y, ); 1ay yo,Q, y, z) 10. g\x, y, z)

l l. f(r,s) :2rtsl r2* -5rs3; (a) frrr(r,s); @) frrr(r,s) 12. f(u, o) :

73. g(r, s, t) : ln(r, * 4s2 - SP); (a) grrr(r, s, t); b) gp2e, s, t)

14. f(x, A, z) : tan-t(3xyz); (u) fu"k, y, z); (b) frrr(x, y, z)

In Exercises 15 through 18, show that u(x, y) satisfies the equation

: sin (xyz); (a) gza(x, y , z); (b) gr, (x, y , z)

ln cos (u - a); (a) f uuo(u, u); (b) f ,uo(u, a)

17. u(x, A) : tan- l

R2.

16. u(x, A) : er sin y + ea cos x

#,.#-owhich is known as Laplace's equatidn in

15. u(x, A) : ln(rz * y ')

18. u(x, A) : tan-l ;l! ,,

( 2xu2r . f (x ,a) : ] f f i i r (x 'Y) + (0 ,

L0 i f (x, y) - (0,

2s. f (x, a) : { t ' tu" - 'X- Y2 tan- 'L

1.0

19. Laplace's equation in R3 is

u" *ry+92-oa- r - t v r -Mshow that u(x, y, z) : (x, * y, * zr)-u2 sTlisfies this equation.

20' For the function of Example 1 tlo* that f nis discontinuous at (0, 0) and hence that the hypothesis of Theorem 19.7.1is not satisfied if (xs, go) : (0, 0).

In Fxercises 21 through 23, find frr(O,0) and fzJ},0), if they exist.

o)o)

if (x, y) + (0, 0)

i f (x , y) : (0, 0)

if x + 0 and y + 0

if ei ther x :0 or y : 024. Given that u --

f (x,y), F(t), and y : G(t) , and assumin g that f ,u: f ax, ptove by using the chain rule thats2,,

f f i : f * t r ' oLF ' ( t ) l ' t 2 f * ( x , y )F ' ( t )G ' ( t ) * f * ( x , y ) [G ' ( t ) ] r t f , ( x , y )F , , ( t )+ t ' " k , y )G , , ( t )25. Given that u- f (x , U) , x - F(r ,s) , and

26. Given Lt- ea cos x, x:2t, a - tz. Find dzuldtz in threeformul a of Exercis e 24; (c) by using the chain rule.

27' Given u:3xy - 4y', x:2se', ! : re-". Find, }z_urdr2 in three ways: (a) by first expressing zusing the formula of Example 5; (c) by using the chain rule.

28. For u, x-, and, y as given in Exercise 27, hnd A2ulAs 0r in three ways: (a) by first expressing zusing the formula of Exercise 25; (c) by using the chain rule.

G(r, s), and assumi.g that f*o: fo* prove by using the chain rule that

y ) lF , ( r ,s )G, ( r , s ) + F , ( r , s )G, ( r , s ) f

* foo?, y)G,(r , s)G,(r , s) * f r (x , y)F, , ( r , s) * fu@, y)G,"( r , s)ways: (a) by first expressing u in terms of t; (b) by using the

in terms of r and s; (b) by

in terms of r and s; (b) by

29. Given Lt - 9x2 * 4y,, x - r cos 0, y: r sin 0. Find dzulyrz in three(b) by using the formula of Example 5; (c) by using the chain rule.

ways: (a) by first expressing u in terms of r and, 0;

30' For u' x' and' y as given in Exercise 29, find,02ul0tr in three ways: (a) by first expressing z in terms of r and 0; (b) byusing the formula of Example 5; (c) by using the chain rule.

REVIEW EXERCISES 941

31. For Lt , x, and y as given in Exerciseusing the formula of Exercise 25; (c)

92. I f u : f (x ,y) and a - g(x, y) , then

0 u _ 0 a a n d

0 r - _ A !0x Ay 0x AY

are called rhe Cauchy-Riemann equations.If / and g and their first and second partial derivatives are continuous, Prove

that if z and u satisfy the Cauchy-Riemann equations, they also satisfy Laplace's equation (see Exercises 15 through 18)'

33. The one-dimensional heat-conduction partial differential equation is

0u , , aru- : e -

At " 6xz

Show that if / is a function of r satisfying the equation

,flt

ftrt+ x'11x1:oand g is a function of f satisfying the equation dgldt * k'zI'Dg(f ) : 0, then 11 v: f (x)g(t), the partial diffetential equa-

tion is satisfied. k and )t are constants.

34. The partial differential equation for a vibrating string is

azu , ozuat,:

o' ar'

Showthat if f isatunctionof xsatisfyingtheequationfrftdf+fT(r):Oand{isafunctionof f satisfyingtheequa-

iion Bgldp t az\zg(t): 0, then lf. u: f(i)g(t), the partial differential equation is satisfied. a and )t are constants'

35. prove that if / and g are two arbitrary functions of a real variable having continuous second derivatives andu: f (x + at)

t g@ - at), theni satisfies the partial differential equation of the vibrating string given in Exercise 34' (ruNr: Let

a: xi at andTo: x- at; lhenals a function of o and w, anda andw are in tum functions of x and l')

35. prove that if / is a function of two variables and all the partial derivatives of. f up to the fourth order are continuous on

some open disk, then

Dtrrrf : Drt tf

Reaiew Exercises, ChaPter 19'

In Exercises I through 5, find the indicated partial derivatives'

& - t t

7. f (r, D :ff; D,f (x, y) , D"f (x' Y) , D'"f (x, Y)

2 . F ( x , A , z ) : 2 x y ' 4 3 y 2 ' - S x z s ; D j ( x , y , z ) , D s f ( x , y , z ) , D p f Q ' y , z ) r3 . g G , f ) : s i n ( s P ) * t e s ; D 1 g @ , t ) , D 2 g @ , t ) , D z , l G , t ) 4 ' h ( x ' y ) : t a n - 1

f o ; o ' h ( x ' y ) ' D z h ( x ' y ) ' D t h ( x ' v )

s. f (u, u, w) : t#t Di@, t t , w), Dpt '@, a' u) ' D' , i (u ' t t ' w)

6 . f ( u , 7 ) , 7 D ) : w c o s 2 t t * 3 o s i n u - 2 u a t a n w ; D z f ( u ' t t ' w ) ' D i @ ' a ' w ) ' D ' " ' f \ u ' a ' w )

In Exercises 7 and' 8 find, 6ul 6t and aulas by two methods'

7 . u : y l n ( r 2 + y ' ) , x : 2 s - l 3 t , y - - 3 t - 2 s 8 ' u : e 2 ' + o c o s ( Z y - x ) ' x : 2 s 2 - t ' ' y : s 2 * 2 t 2

29,by

the

find 62ul 0r d0 in three ways: (a) by first expressing utn terms of r and 0;\b)by

using the chain rule.

equations


9 . f ru : xy l f , r : 4cos t , andy :3s in f , f i nd theva lueo f t he to ta lde r i va t i ve du ld ta t t : * rnby twomethods : (a )Donot express u in terms of f before differentiating; (b) express z in terms of f before differentiating.

l$. If f (x, !, z) :3ry'- 5xz2 - 2ryz, Lx:0.02, Ay: -0.01, and Az : -0.02, find (a) the increment of. f at (-1,3,2) and

(b) the total differential of f at (-7 , 3, 2) .

In Exercises 11 through 14, find the domain and range of the function / and draw a sketch showing as a shaded region inR2 the set of points in the domain of /.

rr. f(x, i l : \ /@ V 72. f(x,y):sin- '{r=7=T13. f (x, y) :161;r{p=fi ra. fQ, y) : flrn + [{1,111

In Exercises 15 through 17, hnd. the domain and range of the function /.

xv xu l xz t r t l z x7 5 . f ( x , ! , 2 ) : 2 . 1 6 . f ( x , ! , 2 ) : - - r

' - - - r - 7 7 . f ( x , ! , 2 ) : - -Yz xyz lyl - lzl

InExerc ises18through20,establ ishthel imi tbyf indinga6>0foranye)0sothatDef in i t ion lg.2.5holds.

1 8 . . 1 i + . ( x - S y ) : z t 1 9 . ! i + ( 3 f - 4 y , ) : - e 2 0 . l i m ( f - y r * 2 x - 4 y ) : t 0(t,r)-(4,-1) (t,t)-Q,-i l

- (c.s)-(e,r)

In Exercises 21 and 22, determine if the indicated limit exists.

2r. lim :I- 22. lim f,- v',

t",yt<ool **y' (c,y)-(o,o) f *yn

In Exbrcises 23 through 27, discuss the continuity ofl.

( f'us23. f(x, il :|1ri ,, if (x' y) + (0' 0)

(nwr: see Exercise 21.)1.0 i f (x ,y) : (0,0)

( f - u at-----=- if (x,u) + (0,0)

2a. f@, a) : lf + yt (HrNr: See Exercise 22.)t0 it (x, y) : (0, 0)

25, f (x, A) : ' :* +.Y:xz - 4Yz

29 . f (x , A) : '? - IYv'

cos2 tnx * cos2 tryIn Exercises 28 and 29, ptove that the function / is differentiable at all points in its domain by showing that Definition19.5.2 holds.

28. f (x, A) :3xy'- 4x2 * y'

30. Suppose c is the radian measure of an acute angle of a right triangle and sin a is determined,by alc,where a in. is thelength of the side opposite the angle and c in. is the length of the hypotenuse. If by measurement a is found to be 3.52and c is found to be 7.14, and there is a possible error of 0.01 in eich, find the possible error in the computation ofsin c from these measurements.

31. A painting contractor charges l2A pet square foot for painting the four walls and ceiling of a room. If the dimensionsof the ceiling are measured to be 12 ft and 15 ft, the height of the room is measured to be10 ft, and these measurementsare correct to 0-05 ft, find approximately, by using the total differential, the greatest error in estimating the cost of thejob from these measurements.

32. At a given instant, the length of one side of a rectangle is 6 ft and it is increasing at the rate of 1 ftlsec and the length

REVIEW EXERCISES 94II

of another side of the rectangle is 10 ft and it is decreasing at the rate of 2 fitlsec. Find the rate of change of the area of

the rectangle at the given instant.

gg. If f is a differentiable function of the variable z, let u: * + yz and, prove that z: ry + f(f + y') satisfies the equation

dz 6zu ? - x ? : 1 t 2 - f- i|x dy

34. Verify rhat u(x, y) : (sinh r) (sin y) satisfies Laplace's equation in R2:

02u *02u _ n0 * ' a Y z

3.5. Verify that' nffi

e(ntnzktllzYu\x, t) : srn ,

satisfies the one-dimensional heat-Conduction partial differential equation:

du , , dzu- : P -

at '"

a*

35. Verify that u(x, t) : A cos(kat) sin(kr), where A and k are arbitrary constantS, satisfies the partial differential equa-

tion for a vibrating string:

d2u ^ dzuaP:

a" a*

97. lf f is a differentiable function of x and y and,u: f(x,V),I: / cos 0, andy: r sin 0, show that

(#)' .i,G*;': (*J', .(#)',38. Given

, l_t!_ if (x, d + (0, o)f ( x ' V ) : l # + Y z

- - \ - - ' r ' \ - '

[0 if. (x, y) : (0, 0)

Prove thatD,/(0,0) andDd@,0) existbutthat/isnotdifferentiableat (0,0). (nnrr:SeeExample 4,9ec.19.2, and Ex-

ercise 2 in Exercises 19.3.)

39..Given

t_ fa2z2

f (x ,v ,O: t ro f f i i r (x 'Y 'z) + (o 'o 'o)

[o if. (r ' Y ' z) : (0, 0, 0)

Prove that / is differentiable at (0, 0, 0).

t(). Let / be the function defined by

I eatr"u

f (x , i l : l e -u"T f t i f x *o

L 0 i f r : 0

Prove that f is discontinuous at the origin'

41. For the function of Exercise 40, prove that Dl(0,0) and Drf(0'0) both exist.

Directional derivatives, gra-dients, applications of partialderivatives, and line integrals

20j DIRECTIONAL DERIVATIVES AND GRADIENTS

20.L DIRECTIONAL We now generalize the definition of a partial derivative to obtain'theDERMTIVES AND rate of change of a function with respect to any direction. This leads to

GRADIENTS the concept of a "directional derivative."Let / be a function of the two variables x and y and let P(r, y) be a

point in the xy plane. Suppose that U is the unit vector making an angleof radian measure 0 with the positive side of the r axis. Then

LI: cos 0i * sin 0i

Figure 20.1.1 shows the representation of U having its initial point atthe point P(x, y).

20.1.1 Definition

Figure 20.1.1

Let f b" a function of two variables r and y. If. U

* sin ei, then the directional deriaatiae of f in the

by Duf , is given by

is the unit vector cos 0idirection of U, denoted


The directional derivative gives the rate of change of the functionvalues f(x, y) with respect to the direction of the unit vector U' This isillustrated in Fig. 20.L.2. The equation of the surface S in the figure isz: f(x, A). Po(xo, Ao, zo) is a point on the surface, and the points R(rs, ys, 0)and Q(ro * h cos 0, yo * h sin 0,0) are points in the xy plane. The planethrough R and Q, parallel to the z a>tis, makes an angle of 0 radians withthe positive direction on the r axis. This plane intersects the surface S inthe curve C. The directional derivative Dol, evaluated at Pe, is the slope ofthe tangent line to the curve C at Po in the plane of R, Q, and Pe.

If U : i, then cos 0 : 1 and sin 0: 0, and we have from Definition20'1"1'

.. f (x + h, y) - f (x, y)D'f(r, /) : liq'- hh-o

P(x, y)

z : f ( x ' Y )

+ h sin d,0)

Figure 20.1.2

DIRECTIONAL DERIVATIVES, GRADIENTS, APPLICATIONS OF PARTIAL DERIVATIVES, AND LINE INTEGRALS

\

which is the partial derivative of / with respect to r.If U : i, then cos 0: 0 and sin 0 : l,and we have

f (x ,V+h) - f ( x ,y )Dtf G, u) : limtr \ ' . J , - ; - ;

h

which is the partial derivative of / with respect to y.So we see that f , and /, are special cases of the directional derivative

in the directions of the unit vectors i and i, respectively.

o rLLUsrRArrow 1: We apply Definition 20.1.1 to find Dul if

f (x, Y) :3* - y2 * 4x

and U is the unit vector in the direction *zr. Then LJ : cos *zri * sin f7i :iltfgi+ *i. So from Definition 20.1.1 we have

4,f @,/) : lim f@ + +\/-gh' Y ! +h) - f (x' v)h-o h

_ r_ 3(x + +\/3h)2 - (y + +h)2 + 4(x + +\/-gh) - (g* - y2 + 4x): [ m + h

- - ) ^ r * * r { r r r * *Zn" -y , - r r , - * t , *n * r t / - ! , -u l , * r , -n

: rrm 3\/3hx + Zh'z- UY - +h2 + 2\fgh

h-o n

: lim @{3x + *h - V - +h + 2\/g)lr-0

: 3 {3x -y+2 f i .we now proceed to obtain a formula that will enable us to calculate

a directional derivative in a way that is shorter than if we used the defi-nition. We define g as the function of the single variable t, keeping x, y,and d fixed such that

g ( t ) : f ( x + t c o s d , y * t s i n ? ) ( 1 )

and let U: cos 0i * sin 0i. Then by the definition of an ordinary deriva-tive we have

8,(0) : , .* /(x+ (0 + h) cos 0, y * (0 + f t ) qin d) - / (r*0 cos 0, y +0 sin d)

" ' , r r - . h o a c A t t - . i t r c i - A \

h

g,(0) : rr^, '- ' h cos 0' Y'*-h sin 0) - f(x, Y)

h-o h

Because the right side of the above is Dof (x, y), it follows that

8'(o) : Duf@, y) e)We now find g'(t) by applying the chain rule to the right side of

20.1 DIRECTIONAL DERIVATIVES AND GRADIENTS 947

d(r * f cos 0)at

f c o s o , y + t s i n g ) o ( Y t t s r n o )

at

cos 0

* f r ( r * f c o s 0 , Y + t s i n 0 ) s i n 0

Therefore,

g'(0) : f "(x, y) cos 0 * f oQ, Y) sin 0 (3)

From (2) and (3) the following theorem is obtained'

20.1.2 Theorem If f is a differentiable function of r and u, and LI: cos 0i * sin 0i, then

o rLLUSrRArrow 2: For the function / and the unit vector U of Illustration 1,

we find Duf by Theorem 20.1.2'geca;e fir, y):3f - y2 * 4x, f ,(x, y): 6x * 4 ul1 ["\*'

y) - -2y'

Because U: cos tri * sin tzi, we have from Theorem20"l"2

Duf (x, 11 : (6x + 4)+\/g + (-2y)l: gtfgx + Z{3 - y

which agrees with the result in Illustration 1' '

The directional derivative can be written as the dot product of two

vectors. Because

f,(x, y)cos 0 * fu@, y) s in g: (cos 0i + sin 0i) ' l f "(x ' y) i+ f"(x ' y) i l

it follows from Theorem 20.1,.2 that

(1), which gives

g ' ( t ) : f r ( x + t c o s 0 , y + t s i n 0 )

* f'(r *

: f r ( r * f c o s 0 , y + t s i n 0 )

Duf k ,A) : (cos 0 i+ s in 0 i ) ' l f * (x ,y) i + f r (x , y) i l

The second vector on the right side of Eq' (4) is a very important one'

and it is called the "gradient" oithe function/. The symbolthat we use for

,t" gt"af""t of f is"if, where V is an inverted capital delta and is read

"del" sombtimes the abbreviation grad f is used'

20.1.3 Definition If / is a function of two variables x and y and f* ffid fu exist' then the

grad ien to f f ,denotedbyVf ( read: "de l f " ) ' i sde f inedby

Using Definition 20.!.3, Eq. (4) can be written as

Du f @,y ) - U , .Y f ( x ,Y )

(4)

(5)


ExAMPLE 1: If

f (x,a):#.n;find the gradient of f at the point(4, 3) . Also find the rate ofchange of f (x, y) in the direc-tion *n at (4, 3) .

Figure 20.1 .3

Therefore, any directional derivative of a differentiable function can beobtained by dot-multiplying the gradient by a unit vector in the desireddirection.

solurroN: Because f"(x, y): $r and f ,(x, y) : &y, we have

9f (x, y) : txi * &yjTherefore,

Vf (4 ,3) : t i+ ? iThe rate_ of change of f(x, y) in the direction |rr at (4,3) is Dvf(4,3),where U is the unit vector

1 . , 1v. '*v.l

This is found by dot-multiplying Vf (4, g) by U. We have, then,t 1 1 . \ / 1 , . . 2 . \ 7 -Duf(4,3): ( - i *# t / 6 t+i i ) : ; \n

If a is the radian measure of the angle befween the two vectors u andVf, then

U . Yf : lullv/l cos aFrom Eqr. (5) and (d) it follows that

Duf : lullv/l cos dWe see from Eq. (7) that Duf will be a maximum when cos a : 1., that is,when U is in the direction of Vf; and in this case, Duf : lV/|. Hence, thegradient of a function is in the direction in which thelunciion has its max_imum rateof change. In particurar, on a two-dimensional topographicalmap- of a landscape where z units is the elevation at a point 1i, yy anaz: f(x, y), the direction in which the rate of change of z is the gr€atest isgiven byYf (x' y); that is,yf (x, y) points in the direction of steepest ascent.This accounts for the name "gradient" (the grade is steepest ir, tr," direc-tion of the gradient).

' rLLUsrRArron 3: In Figure 20.1.3 we have a contour map showing thelevel curves of the function of Example '1. at'!.,2, and.3. The level curves are9U]ns9s. The figure also shows thJrepresentation of.Vf(4,3) having itsinitial point at (4, 3). .

(6)

(7)

EXAMPTN 2:

f (x, y)

Given

- 2 x 2 - y 2 * J x - y

sot,urroN: f*(x, U) : 4x * 3 and fo(x, A) : -2y- L.

Yf (*, A) : @x + 3)i + (-2y - l) i

v f(4, 3)

So,

find the maximumat the point where

Y : - 2 -

value of Dvfx - l a n d

20.1 DIRECTIONAL DERIVATIVES AND GRADIENTS

Therefore,

v f ( \ , -2 ) -7 i+3 i

So the maximum value of Du f ut the point (L,-2) is

lv / ( r , -2) l : \M: \68-

solurroN: (a) We wish to findDuT(x, y), where

U : cos *zri * sin *ni : Li + +\/3i

Because T(x, y) : f * y', T r(x, y) : 2x, and To(x, y) : 2y - Hence,

VT(x, Y) : T,(x, Y) i + Tok, Y) i :2xi + 2Yi

Therefore,

DvT(x, / ) : g 'YT(x, Y): (t i+ +{' i l .Qrt+2vi): x * $y

Hence,

DuT(3, 4) : 3 + +tE = 3 * 4(1.732) : 9'93

(-3, 1) is a maximum in the direction making an angle of radian measure

zr - tan-l * with the positive side of the r axis.

we extend the definition of a directional derivative to a function of

three variables. In three-dimensional space the direction of a vector is

determined byits direction cosines. So we let cos d/ cos F, and cos 7 be

the direction cosines of the unit vector U; therefore, IJ: cos oi * cos Bi

* cos 7k.

Suppose that / is a function of three variables x, y, atrd z' lf U is the unit.tr""io, cos ai i cos Bi * cos 7k, then the directional derivative of / in the

direction of U, denoted by Duf , is given by*+u*ul+$**u'.;i*+i


The directional

ExAMPLE 3: The temperature at

any point (x, y) of a rectangular

plate lying in the xy plane is deter-

mined by T(x, y) - xz * Y'.(a) Find the rate of change of the

temperature at the Point (3, 4) in

the direction making an angle of

radian measute +n with the Posi-tive r direction; (b) find the

direction for which the rate of

change of the temPerature at the

point (-3 , "l') is a maximtlm.

20.1.4 Definition

derivative of a function of three variables gives the


rate of change of the function values f (x, y, z) with respect to distance inthree-dimensional space measured in the direction of the unit vector U.

The following theorem, which gives a method for calculating a direc-tional derivative for a function of three variables, is proved in a mannersimilar to the proof of Theorem 20.7.2.

20.1.5 Theorem If. f is a differentiable function of x, y , and z and

u: cos ai * cos Bi * cos 7kthen

' (8 )

ExAMPrn 4: Given f (x, A, z) :

3x2 * xy - 2y' - yz * 22, find therate of change of f (x, y, z) at(I , -2, -L) in the direction ofthe vector 2i - 2i - k.

solurroN: The unit vector in the direction of 2i - 2i - k is given by

U:3 i -g i -+kAlso,

f(x, U, z) :3* * rY - 2Y2 - Yz * z2

So from (8)

Duf@, y, z) : &(6x + y) - 3@ - 4y - z) - tFy + 2z)Therefore, the rate of change of f(x,y,z) at (!,-2,-1) in the directionof U is given by

D"f(r, -2, -r) - 3@) - 3(10) - +(o)

20.1.6 Definition If / is a function of three variables x, y, and z and the first partial deriva-lives f ,, fu, and f , exist, then the gradient of /, denoted,byvf , is defined by

Just as for functions of two variables, it follows from Theorem 20.1.5and Definition 20.1.6 that if IJ : cos ai * cos Bi * cos 7k, then

Du f Q , y , z ) - u 'Y f ( * , y , 2 ) (e)

Also, the directional derivative is a maximum when U is in the directionof the gradient, and the maximum directional derivative is the magnitudeof the gradient.

Applications of the gradient occur in physics in problems in heatconduction and electricify. Suppose that w:f(x, y, z). A level surfaceof this function f at k is given by

f(x, y, z) : k (10)

If w is the number of degrees in the temperature at point (x, y, z), then

ExAMPLE 5: If V volts is theelectric potential at anY Point(x, y , z) in three-dimensionalrpu . .and v -L | f f i ,find: (a) the rate of change of V

at the point (2, 2, -L) in the di-

rection of the vector 2i - 3i + 6k;

and (b) the direction of the

greatest rate of change of- V at

(2 ,2 , - 1 ) .( x ' * y ' + 2 2 ) 3 t 2 ( x ' * y ' + 2 2 ) s t 2

20.1 DIRECTIONAL DERIVATIVES AND GRADIENTS 951

all points on the surface of Eq. (10) have the same temperature of k de-grees, and the sur{ace is called an isothermal surface. lf w is the number

of volts in the electric potential at point (x, y , z) , then all points on thesurface are at the same potential, and the surface is called an equipoten-tial surface. The gradient vector at a point gives the direction of greatest

rate of change of w. so if the level surface of Eq. (10) is an isothermal sur-

face,Yf(x, y, z) gives the direction of the greatest rate of change of tem-

peratuie at (x , y , z) . If. Eq. (10) is an equation of an equipotential surface,

ihen V/(r, y, z) gives the direction of the greatest rate of change of poten-

tial at (x, y , z).

solurroN: Let f (x, y , z) -'J,l\m2 + 22.(a) A unit vector in the direction of 2i - 3i * 6k is

u-? i -+ i ++kwe wish to f ind D"f (2,2, -1).

V f ( x , y , z ) : f * ( x , y , z ) i + f , ( x , A , z ) i + f " ( x , U , z ) kk-x

The direction cosines of this

rection of the greatest rate ofvector are -?, -3, and *, which

change of V at (2, 2, -1).

Then we have

D"f (2,2, -1) : lJ ' Yf (2, 2, -L)

:g;,Jru?hi-+j++k):T€g

- 0.042

Therefore, at (2, 2,-1) the potential is increasing at the rate of 0.042 volt

per unit change in the distance measured in the direction of u.

(b )V f ( z ,2 , -L ) : - * i -+ i+#keun i t vec to r i n thed i rec t i ono fvf (2 ,2, -1) is

y-\7, ?, -:),, --f,T i - &i + +k : -1i- Ai + +k-lvf(2,2,-7)l-

+'give the di-

Exercises 20.1

In Exercises 1 through 4, tind the directional derivative of

by using either pefinition zo.I.L or Definition 20.I.4, and

Theorem 20.1.5, whichever one applies'

the given function in the direction of the given unit vector U

then verify your result by upplying either Theorem 20.L.2 or

952 DIRECTIONAL DERIVATIVES, GBADIENTS, APPLICATIONS OF PARTIAL DERIVATIVES, AND LINE INTEGRALS

l. f(r, V) :21' + 5y\ V: cos *zri * sin *zj1

2. gG, v) : r, + yr;u : gi- * i

3. h(x' A ' z) :3* -f y2 - 4z2i U: cos izri * cos {zi * cos 3zk' " 4 . f ( x ,V , z ) : 6 * -2xy tVz ;U :+ i++ j++k

In Exercises 5 through 1.0, find the value of the directional derivative at the particular point Po for the given function inthe direction of U.

5. g(*, y\ : y, tan2 r; U : -i{Si * f i; Po: (*n, 2)

5. f (r, !) : xezu; U: +i + t!' i; Po: (2, 0)

7 . h ( x , y , z ) : c o s ( r y ) * s i n ( y z ) ; U : - + i + A j + 3 k ; f o : e , 0 , - 3 )

s . f ( r , a, z) : tn( f * y2 + 22);u:+ F+t-+k; po: 0,s,2)9. f (r, y) : s-et cos 3y; U: cos(-#z)i f sin(-# n)j; Po: t#n, 0)

r0. g(x, a, z) :cos 2r cos 3y sinh 4z; rt :+ t - # t *$ u,

". : (tn, o, 0)

In Exercises 11 through 14, a function /, a point P, and a unit vector U are given. Find (a) the gradient oI f atP,and (b) therate of change of the function value in the direction of U at P.

11.. f (x, U) : x2 - 4y ; P : (-2, 2) ; V : cos *zri * sin *zj

12. f (x , y) : ez"u; P: (2,1) ; U: € i - g i

13. f (x, !, z) : y' + zz - 4xz; P : (-2, l, 3) ; U : +i - +i + +k

M. f ( x , ! , 2 ) : 2 f * t yz * xzz ; P : (1 ,1 ,1 ) ; U : + \ / - 21 i -+ \nk

1.5. Draw a contour map showing the level curves of the function of .Exercise 11 at 8, 4,0,-4, and -8. Also show the rep-resentation otVf (-2,2) having its initial point at (-2,2).

L6' Draw a contour map showing the level curyes of the function of Exercise 12 at eB , ea,7, e-a, and e-s. Also show the rep-resentation of Yf(z,1) having its initial point at (2,1).

In Exercises 17 through 20,IindDvf at the given point P for which U is a unit vector in the direction of f!. atso atpfindDvf , if. U is a unit vector for which Dul is a maximum.

1 7 . f ( x , A ) : e " t a n - l y ; P ( 0 , 1 ) , Q ( 3 , 5 ) 1 8 . f ( x , ! ) : e " c o s y * e u s i n x ; p ( t , O ) , e ( - 3 , 3 )

7 9 . f ( x , A , z ) : x - 2 y * 2 2 ; P ( 3 , 1 , - 2 ) , Q 0 0 , 7 , 4 ) 2 0 . f ( x , ! , 2 ) : f - t y r - 4 x z ; p ( 3 , 1 , - 2 ) , e e 6 , g , 4 )

21. Find the direction from the point (1, 3) for which the value of / does not change if. f (x, y) : e2o tan-l(yl3x).

22. The density is p slugs/fP at any point (x, y) of a rectangular plate in the ry plane and p : U \/F + it-+ j. (a) Find therate of change of the density at the point (3,2) in the direction of the unit vector cos Szri * sin |r,j. (b) Find the direc-tion and magnitude of the greatest rate of change of p at (3,2).

23. The electric potential is V volts at any point (x, y) in the ry plane and V: e-2" cos 2y. Distance is measured in feet.(a) Find the rate of change of the potential at the point (0, *z) in the direction of tlie unit vector cos *zi + sin *zi.(b) Find the direction and magnitude of the greatest rate of change of V at (0, *z).

24. The temperature is T degrees at any point (x, y, z) in three-dimensional space and T: 601(* t y, + zz -t 3). Distanceis measured in inches. (a) Find the rate of change of the temperature at the point (3,-2,2) in the direction of the vec-tor -2i + 3i - 6k. (b) Find the direction and magnitude of the greatest rate of change of T at (g , -2, 2) .

20,2 TANGENT PLANES AND NORMALS TO SURFACES 953

25. An equation of the surface of a mountain is z : 1200 - 3f - 2y2, where distance is measured in feet, the r axis pointsto the east, and the y axis points to the north. A mountain climber is at the point corresponding to (-10,5,850). (a)What is the direction of steepest ascent? (b) If the climber moves in the east direction, is he ascending or descending,and what is his rate? (c) If the climber moves in the southwest direction, is he ascending or descending, and what ishis rate? (d) In what direction is he traveling a level path?

20.2 TANGENT PLANES ANDNORMALS TO SURFACES

Let S be the surface having the equation

F ( x , y , z ) - 0

and suppose that P6(16, !o, zo) is a point on S. Then F(xo, yo, zo) : 0' Sup-pose further that C is a curve on 5 through P6 and a set of parametric equa-tions of C is

( 1 )

x - f ( t ) y : 8 ( f ) z - h ( t )

Let the value of the Parameter t at the point Ps beo f C i s

(2)

fo. A vector equation

(3)

upon substituting from (2)

R( f ) - f ( f ) i + s ( f ) i + h ( t )k

Because curue C isin (1) ,

F ( f ( t ) , g G ) , h ( t ) )

on surface 5, we have,

- 0 (4)

Let G (f ) : F (f (t), 8(t), h(t)) . If F,, F a, and F" are continuous and not

all zero at Po, and If f , (t) , g' (to) , and h'(fe) exist, then the total derivative

of F with respect to f at Ps is given by

G'(fo; : F,(xo, ao, zo)f ' (to) + Fr(ro , !s, zs)g'(to) + F,(ro, ys, z)h' (tr)

which also can be written as

G'(fo) : VF(xo, !o, zo) ' DrR(to)

Because G'1f;:0 for all f under consideration (because of (4)),

G'1t01 :0; so it follows from the above that

YF(xs, Ao, zo) ' D1R(fs) : 0 (5)

From sec. 18.9, we know that DrR(fo) has the same direction as the

unit tangent vector to curve C at P6. Therefore, from (5) we can conclude

that the lradient vector of F at P6 is orthogonal to the unit tangent vector of

".r"ry ".i*" C on S through the point Ps. We are led, then, to the following

definition.

20.2.L Definition A vector which is orthogond to the unit tangent vector of every curve C

through a point Pe on a surface S is called a normal aector to S at Ps.

From this definition and the preceding discussion we have the follow-

ing theorem.

DIRECTIONAL DERIVATIVES, GRADIENTS,'APPLICATIONS OF PARTIAL DERIVATIVES, AND LINE INTEGRALS

20.2.2 Theorem

20.2.3 Definition

, zo)

Figure 20.2.1

EXAMPLE 1: Find an equation ofthe tangent plane to the ellipticparaboloid 4x2 * y, - l,Gz: 0 atthe point (2, 4, 2).

20.2.4 Definition

EXAMPLE 2: Find symmetricequations of the norrnal line tothe surface of Example L at( 2 , 4 , 2 ) .

The denominators innormal vector to S atnormal line at a pointthere.

SOLUTION:

equations

x - 2

If an equation of a surface 5 is F(r, y, z) : 0, and F a, F u,and F, are continu-ous and not all zero at the point Po(xo, yr, zs) on S, then a normal vector to5 at Po is VF(ro, Ao, zo).

We now can define the "tangent plane" to a surface at a point.

If an equation of a surface S is F(r, y, z) :0, then the tangent plane of S ata point Po(ro, !o, zi is the plane through Pe having as a normal vectorVF(xo, Ao, zo).

An equation of the tangent plane of the above definition is

Fr(xo, Uo, zo) Q - ro) * Fu(xo, !0, z) (U - A) * Fr(xo, !o, zo) k - zo) : 0 (6)

Refer to Fig. 20.2.1, which shows the tangent plane to the surface S atPo and the representation of the gradient vector having its initial pointat Po.

A vector equation of the tangent plane given by (5) is

VF(ro,Ao, zo) . l (x- xs) i + (y - A) i + e- zo)kl - 0

solurroN: Let F(x, y, z) .: 4xz + y2 - 16z. Then VF(r, A, z) : gxi * 2yi-76k, and so VF(2, 4, 2):tli+Si - 15k. From 1Z; ii toUows that inequation of the tangent plane is

1 .6 ( x -2 ) +8 (y -4 ) - 76 (z - 2 ) : 0

2 x * y - 2 2 - 4 : 0

The normal line to a surface s at a point po on s is the rine through po havingas a set of direction numbers the components of

"ny t ot-al vector to s

at Ps.

, If an equation of a surface S is. F(r, y, z) :0, symmetric equations ofthe normal line to S at Ps(xs, ys, zs) are

(7)

(8)

(8 )Po)on

are components of YF(xs, Ao, zo), which is athus, (8) follows from Definition 20.2.4. The

a surface is pe{pendicular to the tangent plane

Because VF (2,4,2) :16i + 8i - '!,6k, it follows that symmetricof the required normal line are

y -4 _ r -21 - 2

20.2 TANGENT PLANES AND NORMALS TO SURFACES

20.2.5 Definition The tangent line to a curve C at a point Po is the line through Ps having asa set of direction numbers the components of the unit tangent vector toC at Po.

From Definitions 20.2.3 and 20.2.5 we see that all the tangent linesat the point Po to the curves lying on a given surface lie in the tangentplane to the surface at Pe. Refer to Fig. 20.2.2, showing sketches of a sur-

?ace and the tangent plane at Po. Some of the curves through Ps and their

tangent lines are also sketched in the figure.Consider a curve C which is the intersection of two surfaces having

equations

Figu re 20.2.2

ExAMPLE 3: Find symmetric

equations of the tangent line to

the curye of intersection of the

surfaces 3r2 * Zy', * z2 - 49 and

x2 + y2 - 222 - L0 at the point

( 3 , - 3 , 2 ) .

and

F ( x , y , z ) - 0

G ( x , y , z ) - 0

(e)

(10)

(9) is given by

N1 : VF(x6, !o, zo) : Fr(rco, Uo, zo)i * Fo(xs, Ao, z)i I Fr(xo, Ao, zo)k

and a normal vector at Ps to the surface having Eq. (10) is given by

N2: VG(ro , Uo, zo) : G,(xo, Ao, zo) i* Go(xs, Ao, z) i* G"(xs' Ao' zo)k

of direction numbers of the tangent line. From this set of direction num-

bers and the coordinates of Pe we can obtain symmetric equations of the

required tangent line. This is illustrated in the following example.

solurroN: Let F(x, y, z) :3x2 * 2y2 * z2 - 49

and

G(x, Y, z) : f f- Y' - 2zz - L0

Then VF(r, y, z) : 5xi + ayj * 2zk and VG (r, y, z) : 2xi + Zyi' 4zk' So

Nr : VF(3 , 2) :18i - rzi + 4k:2(9i- 6i + 2k)

Nz : VG (3, -3, 2) - 6i - 6i- 8k - 2(3i - 3i - 4k)

N, X Nz: 4(9 i - 6 i + 2k) x (3 i - 3 i - 4k)


:4(30i + 42i-9k)

: 12(10i + 14i :3k)

Therefore, a set of direction numbers of the required tangent line is110,'J-.4, -3]. Symmetric equations of the line are, t'h"rr,

x -3 a *3 z -210 14 -3

If fwo surfaces have a common tangent plane at a point, the twosurfaces are said lo be tangenf at that point.

Exercises 20.2In Exercises 1 through 72, frnd an equation of the tangent plane and equationsthe indicated point.

of the normal line to the given surface at

3. x2 + y2 - 3z:2; ( -2, -4, 6)

6. z: e3" sin Jy; (0, trr, l)

9. xrtz + ytt' + zrtz : 4; (4, I , l.)

1 . x2 + y, + z2 : lT; (2, -2,9)

4. x2 * yz - z2 : 6; (3, -L, 2)

7. x2 : L2y; (6, g, g)

13. x2 + y2 - z- 8, x - y2 * z2 - -2; (2, -2, 0)

15. y : x2, A : L6 - z2; (4, L6, 0)

20.3 EXTREMA OFFUNCTIONS OF

TWO VARIABLES

2. 4x2 * y' + 222 : 26; (1 , -2, g)

5. y : e0 cos z; (1, e, 0)

8. z: 76rtz -t, yttr; (1, L, 2)L0 . z f - xy ' - yzz :18 ; (0 , -2 ,3 ) l l . x2 t s l yz ta l zz1s :14 ; ( -g ,27 ,1 ) 12 . xuz+z l t z :g ; ( 8 ,2 ,9 )In Exercises 13 through 18, if the two given surfaces intersect in a curve, find equations of the tangent line to the curve ofintersection at the given point; if the two given surfaces are tangent at the given point, prove it.

1 7 . x 2 * 2 2 + 4 y : 0 , * * y " + 2 2 - 6 2 I 7 : 0 ; ( 0 , - 1 , 2 ) 1 8 . f + y " + z r : g , y z : 4 ; ( 0 , 2 , 2 )19' Prove that every normal line to the sphere * + y" * 22: a2 passes through the center of the sphere.

L4. xz + y2 - 2z *1 - 0, x2 * y , - zz- 0; (0, 1, ! )

L5. x: 2 + cos rTyz, y- 1 * sin nxz; (9, l, Z)

An important application of the derivatives of a function of a single vari-able is-in the study of extreme values of a function, which leads to a varietyof problems involving maximum and minimum. we discussed this thor_oughly in chapters 4 and 5, where we proved theorems involving the firstand second derivatives, which enableilus to determine relative maximumand minimum values of a function of one variable. In extending the theoryto functions of two variables, we see that it is similar to the oire-variablecase; however, more complications arise. I

- -

The function / of trvo variables is said to have an absolute maximum aalueon a disk B_in the xy plane if there is some point (re, /o) in B such thatf (xo, y) > f (x, y) for all points (x, y) inA. h such

" .*", f (xs, ys) is the

absolute maximum value of f on B.

20.3.1 Definition

20.3.2 Definition

20.3.3 TheoremThe Extreme-V alue Theorem

for Functions of Two Variables

20.3.4 Definition

20.3.5 Definition

20.3.6 Theorem

20.3 EXTREMA OF FUNCTIONS OF TWO VARIABLES 957

The function / of two variables is said to have an absolute minimum aalue ona disk B in the ry plane if there is some point (x6,90) in B such that

f (xo, Ai - f (x, y) f.ot all points (x, y) in B. In such a case, f (xs, ys) is theabsolute minimum value of f on B.

Let B be a closed disk in t]|te xy plane, and let f be a function of two vari-

ables which is continuous on B. Then there is at least one point in B where

f has an absolute maximum value and at least one point in B where/has an

absolute minimum value.

The proof of Theorem 20.3.3 is omitted because it is beyond the

scope of this book.

The function / of trvo variables is said to have arelatiue maximum aalue at

the point (xs, !) if there exists, an open disk B((xo, y); r) such that

f(x, y) < f (xo, y) fot all (r, y) in the open disk'

The function / of two variables is said to have a relatiae minimum oalue

at the point (ro, yo) if there exists an open disk B((xo, yr); r) such that

f(x, y) > f (xo, V) for all (x, y) in the open disk'

lf f (x, y) exists at all points in some open disk B((xo, yr); r) and if f has

a relative extremum it (xo, yo), then if f,(xo, ys) and fu(xn, ys) exist,

f,(xo, Yr) : fo(xo, U) : o

pRooF:Weprovethat i f /hasarelat ivemaximumvalueat(xo'yiandif

1,t o, yr) exists, then f,(xo, /o) :0. By the definition of a partial deriva-

tive,

Because / has a relative maximum value at (xo, ao), by Definition 20.3.4 it

follows that

f (xo * Ar ,AJ - f Qo,Uo) < o

whenever A,x is sufficientlY smallapproaches zeto from the right, Ar

Ax

Hence , by Theorem 4.3.4, if f *(xo,Ao) exists, f *@o, Ui < 0'

Simila rly, tf L,x aPProaches zero from the left, Ar < 0; so

f (xo * Lx, yo) - f (xo,y) > o

Lx

so that (xo * L,x, Y) is in B. If L'x> 0; therefore/


20.3.7 Definition

Therefore, by Theorem 4.3.5, if fr(xo, Ad exists, f ,(xo, Ui > 0. We con-clude, then, that because fr(xo, yi exists, both inequalittes, fr(xs, yo) = 0and f ,(xo, Ao) > 0, must hold. Consequently, f ,(xo, Ad:0.

The proof that f u(xs, Ad: 0, if. f u@n, ye) exists and/ has a relative max-imum value at (xo, yo), is analogous and is left as an exercise (see Exercise13). The proof of the theorem when /(rs, ys) is a relative minimum valueis also left as an exercise (see Exercise 14). I

A point (xo, A) for which both f ,(xo, /o) : 0 and f n(xs, U) :0 is called acritical point.

Theorem 20.3.5 states that a necessary condition for a function of twovariables to have a relative extremum at a point, where its first partialderivatives exist, is that this point be a critical point. It is possible fo"

"function of two variables to have a relative extremum at a point at whichthe partial derivatives do not exist, but we do not consider this situation inthis book. Furthermore, the vanishing of the first partial derivatives of afunction of two variables is not a sufficient condition for the function tohave a relative extremum at the point. such a situation occurs at a pointcalled a saddle point.

. rLLUsrRArroN 1: A simple example of a function which has a saddlepoint is the one defined by

f(x, y) : y'- *For this function we see that f,(x, a):-2x and fu@, !):2y. Bothf*(0'0) and fr(0,0) equal zero. A skeich of the grapiiof tfre function is$own in Fig. 20.3.1, and we see that it is saddle srrapea at points close tothe onsin. It is apparent that this function / does not-satisfy either Defini-tion 20.3.4 or 20.3.5 when (x6, yo) : (0, 0). .

we have a second-derivative test which gives conditions that guaran-tee a function to have a relative extremum at i point where the first partialderivatives vanish. However, sometimes it is possible to determine rela_tive extrema of a function by Definitions 20.3.+ and 20.3.5, as shown inthe following illustration.

o rLLUsrRArrow 2: Let / be the function defined by

f ( x , y ) : 6 r - 4 y - * - z y z

We determine if f has any relative extrema.Because / and its first partial derivatives exist at all (x, y) in R2, The_

orem 20.3.6 is applicable. Differentiating, we get

f , (x ,y ) :6 -2x and fu@, y ) : -4 - 4y

Setting fr(x,y) andfu(x, y) equal to zero, we get r:3 and y:-1. The

Figure 20.3.1

20.3 EXTREMA OF FUNCTIONS OF TWO VARIABLES 959

graph (see Fig. 20.3.2 f.or a sketch) of the equation

z : 6 x - 4 y - f - 2 y '

is a paraboloid having a vertical axis, with vertex at (3,-L,1.L) and oPen-ing downward. We can conclude that f(x,y) <f(9, -1) for all (x,y) *(3, -1); hence, by Detinition 20.3.4, f(3, -1) - LL is afunction value. It follows from Definition 20.3.'l' thatmaxinrum function value of f on R2.

relative maximum1,1, is the absolute

20.3.8 TheoremS e cond-D eria atia e T est

Figure 20.3.2

The basic test for determining relative maxima and minima for func-

tions of two variables is the second-derinative test, which is given in the

next theorem.

Let f be a function of two variables such that f and its fils!- 1d second-

order partial derivatives are continuous on some oPen disk B((4, b); r\'

Suppoie further that f ,(a, b) : f u(a, b) : 0. Then

(i) / has a relative minimum value at (a' b) if

f , , (a ,b ) fuu@,b) - f 'n '@,b) >0 and f , , (a ' b ) >0

(ii) f has a relative maximum value at (a ' b) if

f , , (a ,b ) fuu@,b) - f * ' (a ,b ) > 0 and f * , (a 'b ) <0

(1ii) f(n, b) is not a relative extremum if

f , , (a ,b ) f ,u@,b) - f * ' (a ' b ) < 0


(iv) We can make no conclusion if

f , "(a, b)f oo@, b\ - f * ' (a, b) : 0

pRooF or (i): For simplicity of notation, let us define

QQ' y) : f,,(x, a)foo@, Y) - f ,o'k' Y)

We are given S(a, b) > 0 and f ,,(a, b) > 0, and we wish to prove that

f(a, b) is a relative minimum function value. Because fr, f ,n, and, fuo arecontinuous onB((a,b);r), it follows that f is also continuous on B. Hence,there exists an open disk B' ((a, b) ; r' ), where r' = r,such that Q@, y) > 0and f",(x, A) > g for every point (r, y) in B'. Let h and k be constants,not both zero, such that the point (a + h, b + k) is in B'. Then the twoequations

x : a * h t a n d y : b + k t 0 < t < 1

define all the points onand all these points arefined by

F ( t ) : f ( a+h t , b+

By Taylor's formula (formula (9), Sec. 15.5), we have

F( t ) : F(0) + F ' (O) f *# n

where f is between 0 and f. lf t - 1 in

F( l ) : F(0) + F ' (0) * tF" ( f )

where 0 < f < L. Because F(0) - f(a,follows from Eq. (3) that

f(a + h, b + k) : f(a, b) + F'(0) +w h e r e 0 < f < L .

To find F'(t) and F" (t) from (1),

F'(t) : hf *(a * ht, b + kt) * kfo@

the line segment from (a, b) to (a * h, b + k) ,in B'. Let F be the function of one variable de-

kt) (1)

Eq. (2), we get

(3)

b) and F(1) - f (a + h, b + k), i t

+F" (€) (4)

we use the chain mle and obtain

+ ht, b + kt) (s)and

F" (t) : h2f ,, + hkf u" + hkf ,a + Pf uu

where each second partial derivative is evaluated at (a * ht , b + kf ) . FromTheorem lg.7.l,itfollows thatf,u(x,A) : fu,(x,y'| for all (x,y) in B'. So

F" (t) : hzfr, + zhkfw * k'fou (6)

where each second partial derivative is evaluated at (a * ht,b + kt). Sub-stituting 0 for f in Eq. (5) and { for t in Eq. (5), we get

F ' ( 0 ) : h f , ( a , b ) * k f u @ , b ) : 0 ( z )

20.3 EXTREMA OF FUNCTIONS OF TWO VARIABLES ,961

ExAMPLE L: Given that / is the

function defined bY

f ( x ,A ) : 2 f *Y ' - x2 -2Y

determine the relative extrema of

f it there are arry.

and

F ' (€ ) : h ' f * ,+2hk f *u*k ' f ru (8 )

where each second partial derivative is evaluated at (a*ht, b+kf)'where 0 < f < 1. Substituting from (7) and (8) into (4), we obtain

f (a *h ,b+k) - f (a , to ) :$ (hz f , ,+2hk f "u*k ' f * ) (e )

The terms in parentheses on the right side of Eq. (9) can be written as

hzf ,, + 2hkf ,u * tsf uo : f ,, lr,' + 2hk * * (o'#)' - (r'#)' . "'#l

So from (9) we have

f (a * h, b + k) - f (,, D :+lV .tr)' +rJTjld wl (10)

Because f ,,f ,o - f ,o2 evaluated at (a + ht, b * kf) equals

6@+ht ,b+ k f ) > o

it follows that the expression in brackets on the right side 9f Eq. (10) is

positive. Furthermori, b""uttt" fr*(a*ht, b+k0 > 0, it follows from^eq.

( fO) thatf(a+h,b +k) - f (a,b) ) 0. Hence,wehave provedthat

f ( a * h , b + k ) > f ( a , b )

(iv) is included to cover all possible cases. I

SOLUTION:

f , (x , y) :8xB - 2x f u@, Y) :2Y - 2

Sett ing fr(x,y):0, we get x:-t, x:0, and x:* ' Sett ing

fok, y):6, *" git y:1. Therefore , f , and f n bolh vanish at the Points(-+, L) , (0, 1), and (t, 1).

To apply the second-derivative test, we find the second partial deriva-

tives of / and get

f , ,(x, Y) :24f - Z foo(r, Y) :2 f,u(x' Y) :0

f , , ( - i , 1 ) : 4 > 0

and

f . ,?E , l ) foo? t , t ) - f *o 'e+ ,1 ) :4 ' 2 - o :8 > o

Hence , by Theorem z}.g.g(i) , f has a relative minimum value at r)

DIRECTIONAL DERIVATIVES, GRADIENTS, APPLICATIONS OF PABTIAL DERIVATIVES, AND LINE INTEGRALS

f ,,(0, 1)fou(0, L) - f ,u' (0, 1) : (-2) (2) - 0 : -4 ( 0

and so by Theorem 20.3.8(iiil, / does not have a relative extremum at(0, 1).

f ""(* ' l)

and

f ,*(t, L)foo(L, t) - f,o'G, t1

Therefore, by Theorem 20.3.8(i)Hence, w€ conclude that / has athe points (-i, t) and (t, L7.

>0

ExAMP:.;. 2. Determine the rela-tive dimensions of a rectangularbox, without a top and having aspecific voluril€, if the leastamount of material is to be usedin its manufacture.

- 4 ' 2 - 0 : 8 > 0

/ has a relative minimum value at (+, 1).relative minimum value of - I at each of

solurroN: Let )c - the number ofbox;

y - the number ofbox;

z - the number ofS - the number of

box;V - the number of

(V is constant).x, y, and z are in the interval (0, +o;. Hence, the absolute minimum valueof S will be among the relative minimum values of S. We have the equa-tions

S : x y * 2 x z * 2 y z a n d V : r y z

Solving the second equation fot zin terms of the variables r and y and theconstant V , we get z : V lry . And substituting this into the first equationgives us

S : x 1 t * ! + !- y x

Differentiating, we get

as 2v ds 2vE:y- f ay : r -T

A2S 4Voy' y'

Setting 6Sl0x:0 and 0Sl0y:0, and solving simultaneously, we get

x2Y - 2V:0

ry2 - 2V:0

from which it follows that r : 9fr , and y : W. For these values of r

units in the length of the base of the

units in the width of the base of the

units in the depth of the box;square units in the surface area of the

cubic units in the volume of the box

azs 4v azsM:7 ff i_l

20.3 EXTREMA OF FUNCTIONS OF TWO VARIABLES

and A, we have

azs_ 4v _4v : )>ndxz (W)3

- 2v

- ' / v

a2S0x2

From Theorem 20.3.8(i), it follows that s has a relative minimum value

and hence an absoluie minimum value when x:flTv nd y:{N'

From these values of r and Y we get

etr-

We therefore conclude that the box should have a square base and a

depth which is one-half that of the length of a side of the base'

Our discussion of the extrema of functions of two variables can be ex-

tended to functions of three or more variables. The definitions of relative

extrema and critical point are easily made. For example, if. f is a function

of the three variablis r, y, alrrd z, arrd f ,(xo, Yr, zo) : fo(xo, yo' zo) :

fr(xo, yo,zo):0, then (rs, ys, zo) is a critical point of /' Such a point is'oUtui""a

by solving simultaneously three equations in-three unknowns.

For a functio n of. n variables, the cltical points are found by setting all the

n first partial derivatives equal to zero and solving simultaneously the n

equati6ns in n unknowns. The extension of Theorem 20.3'8 to functions

of tht"u or more variables is given in advanced calculus texts.

In the solution of Example 2 we minimized the function having func-

tion values ry * 2xz * 2yz, iubject to the condition that r, A, ad:rd z satr;sfy

the equatioi ryr:V. Cbmpare this with Example 1, in w-hich we found

the relative extrema of the filnction f for which/(', y) :2f * y2 - *- 2V'

These are essentially two different kinds of problems because in the first

case we had an additional condition, called a constraint (or side condition).

Such a problem is called one in constrained extterna, whereas that of the sec-

ond type is called a problem in free extrema'Ihe solution of txample 2 involved obtaining a function of the two

variables x and y by replacing z in the first equation by-its value from fhe

second equation. Another method that can be used to solve this example is

due to ]oseph Lagrange, and it is known as the method of. Lagrange multi-

pliers. The- theory Uifrina this method involves theorems known as

implicit function theorems, which are studied in advanced calculus.

He^nce, a proof is not given here. The procedure is outlined and illustrated

by examples." Supiose that we wish to find the critical points of a function /of the

three variables x, y , and z, subiect to the constraint g (r, y ' z) : 0' We in-

a2S_ ( a r s \ r __4V L_ l_3>o@-w)

:@'798ff-r

964 DIRECTIONAL DEBIVATIVES, GRADIENTS, APPLICATIONS OF PARTIAL DERIVATIVES, AND LINE INTEGRALS

EXAMPT,N 3:

the methodmultipliers.

Solve Example 2 byof Lagrange

troduce a new variable, usually denoted by )t, and form the auxiliaryfunction F for which

F(x, y, z, I") : f (x, y, z) * I,,g(x, Y, z)

The problem then becomes one of finding the critical points of the func-tion F of the four variables x, y , z, and )t. The method is used in the fol-lowing example.

sor.urroN: Using the variables x, A, and z and the constant V as definedin the solution of Example 2,let

S: f (r , A,z) : xy * 2xz * 2yz

and

g(x, y, z) : ryz - V

We wish to minimize the function / subject to the constraint that

g(x, y, z) : o

Let

F(x, y, z, \) : f (r, y, z) + lrg(x, y, z)

: x y I Z x z * 2 y z + L ( x y z - V )

Finding the four partial derivatives F*, Fo, F, andfunction values equal to zero, we have

Fr(x, A, z, ) , ) : y * 2z * \yz- 0

Fo(x, y , 2 , t r ) : x + 2z * \xz- 0

Fr(x, U, 2,, ̂,) :2x * 2y * \xy - 0

F ^ ( x , y , z , t r ) : x y z - V - 0

Subtracting corresponding members of Eq. (L2)(11) we obtain

y - x + ) \ z ( y - ) c ) - 0

from which we get

Q - t r ) ( L + | , 2 ) - 0

giving us the two equations

y : ) c

and(15)

(15)

givingSubsti-

Substitutingz : 0, which

from Eq. (15) into Eq. (L2) we get x + 2z - x : 0,is impossible because z is in the interval (0, +m).

Fr and setting the

( 1 1 )

(r2)

( 13)

(r4)

from those of Eq.

r__1z

203 EXTREMA OF FUNCTIONS OF TWO VARIABLES

tuting from Eq. (15) into Eq. (13) gives

2 x * 2 x * \ x 2 - 0

x(4 * Ir) - 0

and because x * 0, we get

96s

(17)

If in Eq. (L2) we take I - -41x, w€ have

x*z r - I .Qz) -o

x i 2 z - 4 2 - 0

)cz r _ F l

z-

Substituting from Eqs. (15) and (17) into Eq. (14), we get L.x3-V:0,from which it follows that r:{2V. From Eqs. (15) and. (17) It follows

that y:lfl27 and z: ^W. These results agree with those found in the

ExAMPLr' 4: Find the relative

extrerna of the function / if

f (*,A, z) - )cz + Yz and the Point(x, y , z) lies on the intersection

of the surfac es )c2 * z2 - 2 and

y z : 2 -

solution of Example 2.Note in the solution that the equation F1(r,

as the constraint given by the equationV: xyz.A, z, L) - 0 is the same

If several constraints are imposed, the method used in Example 3

and p. for which

F(x, y, z, \ , p) : f (* ,y, z) + t g(x, V, z) + p'h(x, y, z)

The following example illustrates the method.

soLUrIoN: We form the function F for which

F(x, y, z, I', p) : x,z * yz + \(* * z2 - 2) + t'c(yz - 2)

Finding the five partial derivatives and setting them equal to zero, we

have

Fr(x,U, z , t r , p) : z + 2\x - 0 (18 )

(1e)

(20 )

(2t)

Fo(x, y, z, tr, p) : z +

F"(x, ! , z ,

F^( r , ! , z ,

t r , p ) - ) c * y * 2 ) r z * p A - 0

t r , F r ) - x 2 * 2 2 - 2 - 0

966 DIRECTIONAL DERIVATIVES, GRADIENTS, APPLICATIONS OF PARTIAL DERIVATIVES, AND LINE INTEGRALS

F*(x , A , z , I r , p ) : yz - 2 :0

From Eq. (19) we obtain p - -l and z: 0. We rejectthis contradicts Eq. (22). From Eq. (18) we obtain

Substituting from (23) and p: -t into Eq. (20),we get

x *

and so

x 2 : 2 2

(22)

0 because

(23)

(24)

Substituting from Eq. (24) into (21.), we have 2* - I- 0, or *: L. Thisgives two values for r, namely 1 and -1; and for each of these values ofr we get, from Eq. (24), the two values 1 and -1 for z. Obtaining the cor-responding values for y fuomEq. Q2), we have four sets of solutions forthe five Eqs. (18) through (22). These solutions are

Y : 2 z - 1

Y : -2 z : - t

x , : - t y : 2 z - |

x : - t y : - Z z : - l

Exercises 20.3

In Exercises L through6, determine the relative extrema of f ,if there are any.

x - 1

x - |

r - -+ p-- rr - + p-- rr - + p--Lr- -+ p--L

The first and fourth sets'of solutions give f(x, y, z):3, and the secondand third sets of solutions give f (x, y , z) : 1. Hence, / has a relative maxi-mum function value of 3 and a relative minimum function value of 1.

1. f (x, A) : l8x2 - 32y' - 35x - l28y - 110

3. f(x, A) : sin(r * y) * sin x + sin y

5. f (x, A) : 4xY' - 2x'y - x

2 . f ( x ' y )

4 . f ( x , y )

6 . f ( x , y )

1 6 4 ,- - T X yx y

: x 3 * y ' - I 8 x y

_ 2 x * 2 y t Lx 2 * y ' + t

In Exercises 7 through 12, use the method of Lagrange multipliers to find the critical points of the given function subjectto the indicated constraint.

7. f(r, y) : f * 2ry * y' with constraint x - y : 3

8. f (t, V) : vz -t xy * 2y2 - 2x with constraint r - 2y -t 1:0

9' f(r, y) : Zs - f - y2 withconstraint * -f y' - 4y : o

10.

LL .

L2.

L3.

14.

15.

17.

L8.

19.

20.

21.

24.

25.

27.

28.

22.

23.

20.4 SOME APPLICATIONS OF PARTIAL DERIVATIVES TO ECONOMICS

f (x, y) :4* * 2y2 1- 5 with constraint * + y' - 2y :0

f (x, y, z) : * * y' I z2 with constraint 3x - 2y * z - 4 : 0

f(x, y, z) : f * y' * z2 with consilaint a2 - * : I

Prove that f u(n, y) : 0 if. f o(xn, y6) exists and / has a relative maximum value at (xo, y).

Prove Theorem20.3.6 when f(rs, yo) is a relative minimum value.

Proye Theorem 20.3.8(ii). 16. Prove Theorem 20.3.8(iii).

Find three numbers whose sum is N(N > 0) such that their product is as great as possible.

Prove that the box having the largest volume that can be placed inside a sphere is in the shape of a cube.

Determine the relative dimensions of a rectangular box, without a top, to be made from a given amount of material

in order for the box to have the Sreatest possible volume.

A manufacturing plant has two classifications for its workers, A and B. Class A workers earn $14 Per run, and class B

workers eam $15 p", *tt. For a certain production run, it is determined that in addition to the salaries of the workers,

i f rc lassAworkersandyclassBworkersareused, thenumberofdol lars inthecostof therunisys*f -Bry+600'How many workers of each class should be used so that the cost of the run is a minimum if at least three workers of

each class are required for a run?

A circular disk is in the shape of the region bounded by the circle I I y' :1. If T degrees- is the temperature at any

point 1r, y) of the disk and i:2xz + yi - !, tind the hottest and coldest points on the disk'

Find the points on the surface yu - xz:4 that are closest to the origin and find the minimum distance.

Find the points on the curve of intersection of the ellipsoid I + 4y2 * 422:4 and the plane x- 4y - z:0 that are

closest to the origin and find the minimum distance'

A rectangular box without a top is to be mad,e at a cost of $10 for the material. If the material for the bottom of the box

costs l5c per square foot and the material for the sides costs 30g per square foot, find the dimensions of the box of

greatest volume that can be made.

A closed rectangular box to contain 16 ff is to be made of three kinds of material. The cost of the material for the top

and the bottom is 99 per ff, the cost of the material for the front and the back is 8c per !f , and the cost of the material

for the other two sides is 6c per f8. Find the dimensions of the box so that the cost of the materials is a minimum'

Suppose that T degrees is the temPeratufe at any point (x, y, z) o1 the sphere f * y" I zz : 4' andT:700xy22'Find

the points on the sphere where the temperature is ttre greaiest and also the points where the temperature is the least'

Also find the temperature at these points.

Find the absolute maximum function value of f it f(x,y,z\: * * y'* z2 wilh the two constraints x-y + z:0 and

25f + 4y2 + 20z,2 = 100' Use Lagrange multipliers'

F indtheabsoluteminimumfunct ionvalueof f i t f (x ,a,z) :* -1 3y2*2z2wi th thetwoconstra ints x-2y-z:6

and x - 3y * 2z: 4. Use Lagrange multipliers.

20.4 SOME APPLICATIONS In Sec. 5.6 we discussed a demand equation giving the relationship

OF BARTIAL DERMTMS between x andp, where p dollars is the price of one unit of a commodity

TO ECONOMICS when r units are demanded. In addition to the price of the given com-

modity, the demand often will depend on th9 prices of other commodi-

ties 1."1ut"d to the given one. In particular, let us consider two related

commodities for wiich p dollars is the price per unit of r units of the

25.


first commodity and 4 dollars is the price per unit of y units of the sec-ond commodity. Then the demand equations for these two commoditiescan be written, respectively, as

o (x ,p ,q ) -0 and FU ,

ot, solving the first equation

x - f (p , q)and

y - s(p, q)

p ,q ) : 0

for x and the second equation for !, as

( 1 )

(2)

The functions / and g in Eqs. (1) and (2) are the demand functionsand the graphs of these functions are surfaces. Under normal circum-stances x, A, p, and q are nonnegative, and so the surfaces are restricted tothe first octant. These surfaces are called demand surfaces. Recalling that pdollars is the price of one unit of r units of the first commodity, we notethat if the variable 4 is held constant, then r decreases as p increases and rincreases as p decreases. This is illustrated in Fig. 20.4.1,, which is a sketchof the demand surface for an equation of type (1) under norrnal circum-stances. The plane 4 : b ,intersects the surface in section RST. For anypoint on the curve RT, q is the constant b. Referring to the pointsM(pr, b, rt) andN(pr,b, xr),we seethatr, ) xrif. andonly if p, < pr;thatis, r decreases as p increases and r increases as p decreases.

When 4 is constant, therefore, as p increases, I decreases; but y mayeither increase or decrease. If y increases, then a decrease in the demandfor one commodity corresponds to an increase in the demand for the other,and the two commodities are said to be substitutes (for example, butter andmargarine). Now if, when 4 is constant, y decreases as p increases, then adecrease in the demand for one commodity corresponds to a decrease inthe demand for the other, and the two commodities are said to be comole-mentary (for example, tires and gasoline).

o rLLUsrRArror L: Figures 20.4.2 and 20.4.3 each show a sketch of a de-mand surface for an equation of type (2). In Fig.20.4.2, we see that when qis constant, y increases as p increases, and so the two commodities aresubstitutes. In Fig. 20.4.3 the two commodities are complementary be-cause when 4 is constant, y decreases as p increases. .

Observe that in Figs.20.4.1 and20.4.2, which show the demand sur-faces for equations of types (1) and (2), respectively, the p and q ixes areinterchanged and that the vertical axis in Fig. 20.4.1is labeled x and inFi9.20.4.2 it is labeled y.

It is possible that for a fixed value of q, y may increase for some valuesof. p artd decrease for other values of p. Fot example, if the demand surfaceof Eq. (2) is that shown in Fig. 20.4.4, then for q:b,whenp:auy isincreasing, and when p : az, y is decreasing. This of course means that

Figu re 20.4.1

Figu re 20.4.2

Figu re 20.4.3

20,4 SOME APPLICATIONS OF PARTIAL DERIVATIVES TO ECONOMICS

if the price of the second commodity is held constant, then for someprices of the first commodity the two commodities will be substitutesand for other prices of the first commodity the two commodities willbe complementary. These relationships between the two commoditiesthat are determined by the demand surface having equation y : 8(p, q)will correspond to similar relationships determined by the demand sur.face having equation x: f(p, q) for the same fixed values of p and q'An economic example might be one in which investors allocate funds

- 4 between the stock market and real estate. As stock prices climb, theyinvest in real estate. Yet once stock prices seem to reach a crash level,investors begin to decrease the amount of real estate purchases with anyincrease in stock prices in anticipation of a collapse that will affect thereal estate market and its values.

The demand functions are now used to define the (partial) "marginaldemand."

Figure 20.4.4

20.4.1 Definition Let p dollars be the price of one unit of x units of a first commodity and 4dollars be the price of one unit of y units of a second commodity. suppose

that / and g are the respective demand functions for these two commodi-

ties so that

x : f (p , q ) and y : g (p , q )

Then

(i) *gives the (partial) marginal demand of x with respect to p;d p "

(ii) + gives the (partial) marginal demand of x with respect to q;d q -Au

(iii) = gives the (partial) marginal demand of y with respect to p;dp

, . , 0 a(i") i

gives the (partial) marginal demand of y with respect to q'

Because in normal circumstances if the variable 4 is held constant, I

decreases as p increases and r increases as p decreases, we conclude that

AxlAp is negative. Similarly, AylAq is negative in normal circumstances.

Two commodities are said to be complementary when a decrease in the

demand for one commodity as a result of an increase in its price leads

to a decrease in the demand for the other. So when the goods are com-

plementary and. q is held constant, both dr/Ep < 0 and AylAp < 0; and

when p is held constant, then Oxl0q < 0 as well as Ay I Aq < 0. Therefore, we

can conclude that the two commodities are complementary if and only if

both 6xl0q and aylaP are negative'


When a decrease in the demand for one commodity as a result of anincrease in its price leads to an increase in the demand for the other com-modity, the goods- are substitutes. Hence, when the goods are substitutes,because \xldp is always negative, we conclude that Ayl\p is positive; andbecause AylAq is always negative, it follows that 0xl0q is positive. Conse-quently, the two commodities are substitutes if and only If. 0xl6q anddyl6p arc both positive.

If. Axl\q and AylAp have opposite signs, the commodities are neithercomplementary nor substitutes. For example,lf. dxl\q < 0 and 0yl0p > 0,and because 0xl6p and AylAq are always negative (in normal circum-stances), we have both 0xl0q ( 0 and AylAq < 0. Thus, a decrease in theprice of the second commodity causes an increase in the demands of bothcommodities. Because 0xl0p < 0 and AylAp ) 0, a decrease in the price ofthe first commodity causes an increase in the demand of the first commod-ity and a decrease in the demand of the second commodity.

solurroN: If we use Definition 20.4.1,. the four marginal demands aregiven by

EXAMPLE 1: Suppose that pdollars is the price per unit of xunits of one commodity and qdollars is the price per unit of yunits of a second commodity. Thedemand equations are

) c - - 2 p * 3 q + L 2 ( 3 )

and

y : -aq *p+8 (4 )

Find the four marginal de-mands and determine if the com-modities are substitutes or com-plementary. Also draw sketches ofthe two demand surfaces.

Y -3 aY -1aq op

Because 0xl0q > 0 and AylAp > 0, the two cot'nmodities are substitutes.A sketch of the demand surface of Eq. (3) is shown in Fig. 20.4.5. To

draw this sketch we first determine from both equations the permissiblevalues of. p and 4. Because x and y must be positive or zero, p and 4 mustsatisfy the inequalities -2p * 3q * 12 > 0 and-Aq * p * I > 0. Also, p andq are each nonnegative. Hence, the values of p and q are restricted to thequadrilateral AOBC. The required demand surface then is the portion inthe first octant of the plane defined by Eq. (3) which is.above AOBC. Thisis the shaded quadrilateral ADEC in the figure. In Fig. 20.4.6 we have a

_ _ _/)^ adp

Figu re 20.4.5 Figu re 20.4.6

ExAMPLE 2: If p dollars is the

price of one unit of x units of a firstcommodity and q dollars is theprice of one unit of y units of a sec-ond commodity, and the demand

equations are given by


sketch of the demand surface defined by Eq. (4). This demand surface isthe shaded quadrilateral BFGC, that is, the portion in the first octant of theplane defined by Eq. (4) which is above the quadrilatenl AOBC.

For nonlinear demand functions of two variables, it is often more con-venient to represent them geometrically by means of contour maps than bysurfaces. The following example is such a case.

solurroN: Let the two demand fudctions be / and g so that x: f (p , q) :

Slpq and y:g(p, q):l2lpq.sketches of the contour maps of f and g

showing the level curves of these functions at the required numbers are

shown in Figs. 20.4.7 and20.4.8, respectively.

L2pq2 p'q

ayapand

8pq Because \xl 0q

q

< 0 and aylAp ( 0, the commodities are comPlementary.q

draw sketches of the contour maPs

of the two demand functions

showing the level cun/es of each

function at 6,4,2,1, and*. Are the

commodities substitutes or com-

plementary?

EXAMPLE 3: The demand equa-

tions for two related commodities

are

*c-Ae-pq and y :8eo-q

Determine if the commodities are

complement dr4, substitutes, or

neither.

SOLUTION:

6x - -4Pe-P(taq

and #-

8e--q

Because 0xl0q < 0 and aY IAP ) 0, thementary nor substitutes.

commodities are neither comPle-

If the cost of producing r units of one commodity alnd y units of

another commodity is given by c(x, y), then c is called a ioint'cost func-tion.Thepartial d"rin"ti*'"t of c are called marginal cost functions.

Figu re 20.4.7

v:+

y - - 6

Figure 20.4.8


EXAMPLE 4: A monopolist Pro-duces two commodities which aresubstitutes having demand equa-tions

x - 8 - p + qand

y - - 9 + p - S q

where 1,000r units of the first com-modity are demanded if the priceis p dollars per unit and 1000yunits of the second commodity aredemanded if the price Ls q dollarsper unit. It costs $4 to produceeach unit of the first commodityand $2 to produce each unit of thesecond commodity. Find the de-mands and prices of the two com-modities in order to have thegreatest profit.

suppose that a monopolist produces two related commodities whose

demand equations are )c: f (p, q) and y : g(p, q) and the joint-cost func-tion is C. Because the revenue from the two commodities is given bypx * qy, then if S is the number of dollars in the profit, we have

S: px * qy - C(x , y )

To determine the greatest profit that can be earned, we use thedemand equations to express S in terms of either p and q or r and y aloneand then apply the methods of the preceding section. The following ex-ample illustrates the procedure.

solurroN: When 1000r units of the first commodity and 1000y units ofthe second commodity are produced and sold, the number of dollars in thetotal revenue is 1000pr * L000qy, and the number of dollars in the total costof production is 4000r *2000y. Hence, if S dollars is the profit, we have

S:L000px+1000qy - (4000r +2000y)

S : 1000p(8 - p -r 4) + 10004(9 + p - 5q)- 4000(8 - p * q) - 2000(e + p - 5q)

S : 1000(-p' * Zpq - 5q' * l0p + LSq - 50)

$ : roool -2p *2q +t0) 9 :1000(2p-70q +75)dp dq

Setting 0Slap :0 and 0Sl6q: 0, we have

- 2 p * 2 q * 7 0 : 0 a n d 2 p - 1 0 q * 1 5 : 0

from which it follows that

p : g a a n d q : T

Hence, (3F, +) is a critical point. Because

+?: -2ooo ?'l : -1o.ooo S-: 2ooodp- dq' dq dp

we have

azs azs _ ( a'S ,'0p, 0q, \aq ap)

: (-2000) (-10'000) - (2000)2 > 0

Also, 02Sl0p' 10, and so from Theorem 20.3.8(ii) we conclude that S hasa relative maximum value at (S, +5).

Because x, U, p, and q must be nonnegative, we know that

8 - p * q > 0 9 * p - 5 q - 0 p > 0 q > 0

From these inequalities we determine that the region of permissible valuesis that shaded in Fig. 20.4.9.It follows that S has an absolute maximum

/ \

Figu re 20.4.9

Figu re 20.4.10

ExAMPLE 5: Suppose that theproduction of a certain commod-ity depends on two inPuts. The


value at the point (6F, ?), and the absolute maximum value of S is'1,4,062.5. From the demand equations we find that when p : $a and q: T,x : 3 a n d y : t .

We therefore conclude that the greatest profit of.$14,052.50 is attainedwhen'3000 units of the first commodity are produced and sold at $8'12* perunit and 1500 units of the second commodity are produced and sold at

$3.12+ per unit.

o rl-I.usrnArrox 2: In the preceding example, if the demand equations aresolved for p and 4 in terms of r and y, we obtain

p - + (49 -u r - y ) and q :+(L7 - x - Y)Because q and p must be nonnegative as well as x and y, we know that17 - x- y > O, 49 - 5x- y > 0, x = 0, andy > 0. From these four inequa-lities we determine that the region of permissible values is'that shaded inFig.20.4.10. The problem can be solved by considering r and y as the in-dependent variables. You are asked to do this in Exercise 12. .

Suppose that in the production of a certain commodity, r is thenumber of machines, y is the number of man-houts, and z is the number ofunits of the commodity produced, where z depends on r and y, andz : f (x, y) . Such a function / is called a production function. Other factors ofproduction may be working capital, materials, and land.

Let us now consider, in general, a commodity involving two factors ofproduction; that is, we have a production function f of two variables. If theamounts of the inputs are given by r and y, and, z gives the amount of theoutlrut, then z : f(x, y). Suppose that the prices of the two inputs are adollars and b dollars per unit, respectively, and that the price of the outputis c dollars (a, b, and c are constants). This situation could occur if therewere so many producers in the market that a change in the outPut of anyparticular producer would not affect the price of the commodity. Such amarket is called a perfectly competitiae market. Now if P dollars is the totalprofit, and because the total profit is obtained by subtracting the total costfrom the total revenue, we have

P : c z - ( a x + b y )

and because z: f(x, y),

P: cf(x, y) - ax- bY

It is, of course, desired to maximize P. This is illustrated by an exam-ple.

soLUrIoN: If P dollars is the proftt,

P - e(1002) - 4(100r) - L00y

974 DIRECTIONAI- DERIVATIVES, GRADIENTS, APPLICATIONS OF PARTIAL DERIVATIVES, AND LINE INTEGRALS

amounts of these are given by 100rand L0Ay,whose prices per unitare, respectively, $4 and $L. Theamount of the output is given by1002, the price per unit of which is$9. Furthermore, the productionfuncti on f has the function values

f (x, A) :5 - Llx- UY. Determinethe greatest profit.

Letting z - f(x, y), we get

P -eoo (s - + -+) - oox - Looy\ x y /

x and y arc both in the interval (0, +o). Hence,

and g: 9oro _ 1oody y'

azP 1800 a2P nV T ayax:uAP I 0y - 0, we have

AP0x

Also,

:ry-4oo

a2P0xz

1800x3

EXAMPLE 6: Solve Example 5by the method of Lagrangemultipliers.

Setting AP lAx - 0 and

ry-4oo:o and ry-loo:ofrom which we obtain 5 : $ and y : 3 (we reject the negative resultbecause x and y must be positive). At (t,3)

azP azP / a2P \2 /1800\ / -^^^,

;a' ar- \i*) :- (F/ \-T)- (o)' > oFrom the above and the fact that at (*, 3) azPlaf < 0, it follows fromTheorem 20.3.8(ii) that P has a relative maximum value'at (8, 3). Becausex and y are both in the interval (0, +-1, and P is a negative number whenx and y are either close to zero or very large, we conclude that the relativemaximum value of P is an absolute maximum value. Because z: f (x, V),the value of. z at (*,9) is f(2,3)

-5 - B - *:4. Hence, from Eq. (5)

P,no : 900 ' 4 - 400 ' 8 - fOO' 3 : 2700

The greatest profit then is 92700.

solurroN: We wish to maximize the function P defined by Eq. (5) sub-ject to the constraint given by the equationz :5- Ux- 1/y,which we canwrite as

g(x , y , z )

F(x, A, z , ) , ) - P(x, y , z) * Ig(x , y , z)

- 9002 - 400x - I00y + I

We find the four partial derivatives F, Fs, F, and F1 and set themequal to zero.

- 1+1*z -5 -oxy

20.4 SOME APPLICATIONS OF PARTIAL DERIVATIVES TO ECONOMICS 975

F *(x, A , z, Ir) - -4oo- ) :o Fo(x, A,z, ) , ) : -100 - h-

0

F,(x, r) F ^ ( x , A , z , ) ' )

2 . x - 5 - 2 p * q , y - 6 * p - q

4 . x : 9 - 3 p * q , y : L 0 - 2 p - 5 q

6 . x - - p - 3 q * 6 , y - - 2 q - p + 8

8. pqx - 4, p'qy : L5

-L +1 *z -5 -ol c y

Solving these equations simultaneously, we obtain

) r : - 9 0 0 x : 8 , A : 3 z : 4

The values of x, y,and z agree with those found previously, and P is shownto have an absolute'maximum value in the same way as before.

In some applications of functions of several variables to problems ineconomics, the Lagrange multiplier )t, is related to marginal concepts, inparticular marginal cost and marginal utility of money (for details of thisyou should consult references in mathematical economics).

Exercises 20.4

In Exercises 1 through 9, demand equations for two related commodities are given. In each exercise, determine the four

partial marginal demands. Determine if the commodities are complementary, substitutes, or neither. In Exercises 1 through

6, draw sketches of the two demand surfaces.

1 . x - 1 4 - P - 2 q , Y : L 7 - 2 P - q

3 . x - - 3 p + 5 q + 1 5 , A : 2 P - 4 q * I 0

5 . x - 6 - 3 p - 2 q , y - 2 * P - 2 q

7. x - P-o'4qos, A -

Po'4q-rs

9 . px : q , qy : p '

10. From the demand equations of Exercise 8, find the two demand functions and draw sketches of the contour maps of

these functions showing the level curves of each function at5,4,3,2,1,+,+.

11. Follow the instructions of Exercise 10 for the demand equations of Exercise 9.

12. Solve Example 4 of this section by considering x and y as the independent variables. (Refer to Illustration 2.)

13. The demand equations for two commodities that are produced by a monopolist are

x : 6 - 2 p + q a n d y : 7 + p - q

where 1.00r is the quantity of the first commodity demanded iI the price is

of the second commodity demanded if the price is q dollars Per unit. Show

If it costs $2 to produce each unit of the first commodity and $3 to Producethe quantities demanded and the prices of the two commodities in order tothe independent variables.

14. Solve Exercise 13 by considering r and y as the independent variables.

15. The production function f for a certain commodity has function values

Rf (x ,v ) :4 - ;

p dollars per unit and l00y is the quantitythat the two commodities are substitutes.each unit of the second commo dity, findhave the greatest profit. Take p ar.d q as


The amounts of the two inputs are given by 100r and 100y, whose prices per unit are, respectively, gl0 and $5, andthe amount of the output is given by 7002, whose price per unit is $20. Determine the greatest profit by two methods:(a) without using Lagrange multipliers; (b) using Lagrange multipliers.

16. Solve Exercise 15 if

f ( x ' v ) : x * * y - * f - * y ' - *the prices per unit of the two inputs are $4 and $8 (instead of $10 and $5, respectively), and the price per unit of theoutput is $16.

20.5 OBTAINING AFUNCTION FROM

ITS GRADIENT

We consider the problem of how to obtain a function if we are given itsgradient; that is, we are given

Yf (*, y) : f ,(x, y) i + f u(*, y) i

and we wish to find f (x,y).

o TLLUSTRATTON 1.: Suppose

Yf (*, A) : (y, * 2x * 4)i + (Zxy * 4y - 5)i

Then because Eq. (1) must be satisfi€d, it follows that

f " ( x , y ) : y2 *2x+4and

f ok ,y ) : zxy * 4y - 5

By integrating both members of Eq.

f (x,y) : yzx * x2 * 4x * g(V)

Observe that the "constant" of integration is a function of y and inde-pendent of r because vve are integrating with respect to r. If we nowdifferentiate both members of Eq. (5) partially with respectto y,we obtain

f oG, y) :2xy + g'(y) (5)

Equations (4) and (6) give two expressions for fu@,y).Hence,

2 x y * 4 y - 5 : 2 x y + g ' ( y )

Therefore,

8 ' ( Y ) : 4 Y - 5

g(Y) :2Y2 - 5Y * c

Substituting this value of g(y) into Eq. (5), we have

f (x , y ) - y ' x * * - l4x*2y ' -5y + C o

(1)

(2)

(3)

(4)

(3) with respect to x, we have

(s)

EXAMPLE

vf(x, V)

'1,: Find f (x, y) it

: gu2 cos xi + Zyeu' sin

Because Eq. (1) must hold, w€ have

: guz cos rxl

SOLUTION:

f " (x, y)

20.5 OBTAINING A FUNCTION FROM ITS GRADIENT

and

f o@,y ) - Zye " s i n r

Integrating both members of Eq. (8) with respect to A, we obtain

f (x, y) -- su2 sin x * S(x)

where S@) is independent of y. We now partially differentiate both

bers of Eq. (9) with respect to r and get

f * ( x ,Y ) : su2 cos x * 8 'Q )

By equating the right members of Eqs. (7) and (10), w€ have

eu' cos x: ea' cos x * 8' @)

g ' ( r ) - o

s(r) - c

We substitute this value of g(r) into Eq' (9) and obtain

f ( x , y ) : sa2 s in x * L

All vectors of the fotm M(x,y)i * N(r' y)i ate not necessarily gradients'

as shown in the next illustration'

o rLLUsrRArroN 2: We show that there is no function f such that

977

(10)

Vf (x, U) - 3Yi- 2xi

Assume that there is such a function. Then it follows that

f *(x, Y) - 3Y

and

f o@,Y) : -2x

we integrate both members of Eq. (12) with respect to r and obtain

f ( x , y ) - 3 x A * g Q )

Wepartiallydifferentiatebothmembersofthisequationwithrespecttoyand we have

fo@,Y) :3x+g ' (Y )

Equating the right members of Eqs. (13) and (14), w€ obtain

-2x :3x * g ' Q )_Sx: g , (a )

If both members of this equation

must follow that

- 5 : 0

(11)

(12)

(13)

(14)

are differentiated with resPect to )c, it


which, of course, is not true. Thus, our assumption that 3yi - 2xi is agradient leads to a contradiction. .

we now investigate a condition that must be satisfied in order fora vector to be a gradient.

Suppose that Mu and N' are continuous on an open disk B in

M(x, y) i + N(r , y) i

is a gradient on B, then there is a function / such that

f " (x, y) - M(x, y)

and

fu?,y) : N(x, y) G7)for all (x, y) in B. Because Mu@, y) exists on B, then from Eq. (15) it followsthat

R2. If

(15)

(16)

(18)

(17) that

(1e)

MaQ, y) : f ,uQ, y)

Furthermore, because Nr(x, y) exists on B, it follows from Eq.

N"( x, y) : f *ok, y)

20.5.L Theorem

Because Mn and N, are continuous on B, their equivalents f cu and, f o, arealso continuous on B. Thus, from Theorem 1.9.7.1 it follows

'that f ,u(i, y1:

fo'@' y) at all points in B. Therefore, the left members of Eqs. 1rs) and 119)are equal at all points in B. we have proved that if M, and N, are contin-uous on an open disk B in R2, a necessary condition for vector (15) to be agradient on B is that

Mo(x, y) : N,(x, y)

This equation is also a sufficient condition for vector (15) to be a gradienton B. However, the proof of the sufficiency of the condition involves con-cepts that are beyond the scope of this book, and so it is omitted. we havethen the following theorem.

Suppose that M and N are functions of two variables x and.u defined on anopen disk B((xo, yr); r) in R2, and Mu and N, are continuorrr o'B. Then thevector

M(x, y) i + N(x, y) i

is a gradient on B if and only if

Mo(x, y) : N,(x, y)

at all points in B.

. rl-I.usrRArrorrr 3: (a) we apply Theorem 20.5.1 to the vector in the rightmember of Eq. (2) in Illustration 1. Let M(x, y): y, * 2x * 4 and N(r, yj:

20.5 OBTAINING A FUNCTION FROM ITS GRADIENT 979

2xy * 4y - 5. Then

Mak, y) - 2Y and N,(x, Y) - 2Y

Thus, Mo(x,a): Nr(r, a), and therefore the vector is

(b) If we apply Theorem 20.5.1 to the vector in

Eq. (11) in Illustration 2, with M(x,a):3y and N(r,

Ma(x, y) - 3 and N"( x, Y) - -2

Thus, Mo(x,y) + N*(x,a), and so the vector is not a

a gradient.the right member of

A) : -2x, we obtain

gradient. o

EXAMP rn 2: Determine if the

following vector is a gradient

Yf (*, A), and if it is, then tind

f @,v ) t(e-o - 2x)i - (xe-o * sin V)i

MaQ,y ) : - e -u and N" ( x , Y )

Therefore, Mo(x, A): N'(r, Y), and

Yf (x, y). Furthermore,

f " (x , Y) : s-u - 2x

and

f o@,y ) xe -a - s i n Y

: -e-U

so the ,given vector is a gradient

solurroN: We apply Theorem 20.5.1. Let M(x, U): e-u -2x and N(r' y) :

-xe-a - sin y. Then

(20)

(21)

Integrating both members of Eq'

f (x, y) : xe-a - )cz * g (V)

(20) with resPect to x, we obtain

where S(fi is indePendent of x'

bers of Eq. (22) with resPect toWe now partially differentiate both

A, and we have

f , (x , y) : -xe-a + g ' (Y)

we equate the right members of Eqs. (23) and (21) and get

-xe-a * 8' Q) : -xe-a - sin Y

g' (Y) : -sLn Y

S Q ) : c o s Y + C

We substitute this exPression for g(y) into Eq. (22) and we have

+Cf (x,Y) : xe-a - xz * cos Y

we can extend Theorem 20.5.1 to functions of three variables'

Let M,N, and R be functions of three variables x' y' and z defjned on an

open ball B((xo, yo, zo); r) inf,-11d Ma' M"' N'' N" R'' and R' are con-

tinuous on B' Then iii,e l,"cto' M(x, y' z)i I ru(r' y' z)i + \!*'v' z)k is

a e rad ien t on B i f u r ra o" iy ' t iu7- , y ; z ) :N ' (x 'y ' z ) ' M" (x 'y 'z ) :

I.l!, y, z) , and N"(x, Y , z) : Ru(x, Y ' z) '

20.5.2 Theorem

980

ExAMPlr 3: Determine if the fol-lowing vector is a gradientYf (x, A, z), and if it is thenfind f (x, U, z):

(e' sin z * Zyz)i + (2xz + 2V)i

+ (e* cos z * Zxy * 322)k

Mu(x , y , z ) : 22

N"( r , y , z ) :22

Therefore, Mo(x, y, z) - At(r , y, z),Ro(x, y, z). Thus, the given vector

f *(x, y, z) : sr sin z * 2yz

f u(x, y , z) :2xz * 2y

R*(x, y, z) : e, cos z * 2y Ro(x, U, z) : 2x

M r ( x , y , z ) - s r c o s z t 2 y

N r ( x , y , z ) : 2 x

Mr(x, y, z)- Rr(r, A, z), andNr(r, A, z) :is a gradient Yf (x, A, Z).Furthennore,

(24)

(25)

(2e)

Eq. (27),

(30)

Eq. (30).

(31)

(32)

DIRECTIONAL DERIVATIVES, GRADIENTS, APPLICATIONS OF PABTIAL DERIVATIVES, AND LINE INTEGRALS

The proof of the "only if" part of rheorem 20.5.2 is similar to the proofof the "only if" part of rheorem 20.5.1 and is left as an.exercise (see Exer-cise 23).The proof of the 'tif" part is beyond the scope of this book.

solurroN: We apply Theorem 20.5.2. Let M(x, y, z): e, sin z*2y2,N(x, y, z) : 2xz * 2y, and R(x, y, z) : e, cos z * 2iy * 322. Then

f"(x, y, Z) : e'cos z * 2xy * 3zz e6)Integrating both members of Eq. e4) with respect to x, we have

f(x, y, Z) : e, sin z * 2xyz* Sg, z) ez)*-Ett g(y, z) is independent of r. we partial differentiate both membersof Eq. (27) with respect to y and, obtain

fu(x, y, z) : 2xz * gu(y, z) (28)Equating the right members of Eqs. (2g) and (25), we have

2xz * gn(y, z) :2xz * 2y

S o ( Y ' z ) : 2 Y

we now integrate both members of this equation with respect to y, and getgQ, z) : yz + h(z)

where h is independent of tr and y. Substituting from Eq. (zg)intowe obtain

f(x, y, z) : sr sin z + 2xyz * y, + h(z)We now Paftial differentiate with respect to z bothmembers ofWe get

fr(x, U, Z) - er cos z + 2xy + h,(z)Equating the right members of Eqr. (31) and (26),we have

et cos z + 2xy + h, (z) - sr cos z + 2xa * jzz

h' (z) : 322

h ( z ) : 2 3 + C

20.6 LINE INTEGRALS 981

(30), and we obtain

* 2 3 + C

We substitute from Eq.

f (x, y , z) -- sr sin z

Exercises 20.5

In the following exercises, determine

1. axi - 3yi

3. (6x - sy)i - (5r - 6V')i

if the vector is a gradient.

5. (6x'y' - l4xy + 3)i + (4xty '7xz - 8)i

7. (Zxy * y '+ L) i + (x ' * Zxy + r) i

s (L*1)r +(ry\ .Y ' \7 - , ' , \ yo

- ) l

11. (2x cos y - l)i - x2 sin Yi

13. (eu - Z*y)i * (xeo - x')i

15. (2V - Sz)i + (2x * 8z)i - $* - 9Y)U

17 . (4xy * 3yz - 2)i + (2x' * 3xz - Szz)i + Gxy - l}yz *

L8. (2y + z) i * (2x - 3z * 4yz)i + @ - 3y * 2y')U

19. z tan yi + xz secz yi + x tan Yk

20. e' cos ri + z sin yi + (e" sin x- cos y)k

ZL. e,(e, - ln i l i+ (eo ln z- e'A-t) i + (s '+" * tu7-r)k

2 2 . t

i - x - z ' x + a

y+z -

( y+4 'z t - f f iK

23. Prove the "only tf" part of Theorem 20.5.2.

If it is, find a function having the given gradient

y'i + lx'i

(4y' * 6xy - 2)i + (3x' * 8xY + l)i

3(2x2 * 6xy) i + 3 (3x2 + 8V)i

2 x - L . , x - x 2 .y

' - y ' I

(2x * rn y)i + (y' +i) i\- yl

(sin 2x - tan y)i - x secz yi

4x3i1-9y ' i -Zzk

(Zxy * 7zs)i + ( xz + 2y' - 3z)i + (21'xz2 - Ay)U

(32) into

* 2xyz *

Eq.

v'

2.

4.

6.

8.

10.

L2.

1,4.

1,6.

1 ) k

20.d LINE INTEGRALS In Chapter 7 we used the concept of area to rnotivate the definition of thedefinite integral. To motivate the definition of an integral of a vector-valued function, we use the physical concept of work.

We have seen that if a constant force F moves an obiect along a straightline, then the work done is the product of the component of F along theline of motion times the displacement. We showed in Section 17.3 that ifthe constant force F moves an object from a point A to a point B, then ifW is the measure of the work done,

W:F 'V @3) (1)

982

Figure 20.6.1


suppose now that the force vector is not constant and instead of the mo-tion being along a straight line, it is along a curye. Let the force that isexerted on the object at the point (x, y) in some open disk B in R2 be givenby the force vector

F(x, y) : M(x, y) i + N(x, y) i

where M and N are continuous on B. The vector-valued function F iscalled a force field on B. Let C be a curve lying in B and having the vectorequation

R ( f ) : f ( t ) i + S @ i a < t < b

We require that C be smooth, that is, that f, and g, be continuous on[a,bf . we wish to define the work done by the varia6re force F in movingthe object along C from the point (f(a), g(a)) to (f(b), S@)). At a poinl(f(t), S(t)) on C, the force vector is

F( / ( t ) , s ( f ) ) - M( f ( t ) , s ( f ) ) i + N( / ( t ) ,8 ( f ) ) i

Let A be a partition of the interval fa, bf :

c t : fo ( t r

On C let [$: th" point (ri, At): (f(t), g(f)). Refer to Figure 20.6.1. Thevector V(Pi-lPi) : R(tr) - R(4-r); therefore,

v (P;7i) : f (f,)i + s(tt)t - lf (t,-,) i + s ( f,-, )i lor, equivalently,

v (+7,) : lf(t,) - f(t*r) li + ls(t,) - s(ro-,) liBecause f' and g' are continuous on la, bJ, it follows from the mean-vhluetheorem (4.7.2) that there are numbers ci and rdi in the open intewal(t*t t) such that

f (t,) - f (t,-r) : f ' (c) (t, - tr-r)and

g(t) - g(f,-r) : g' @,) (t,- tr_r)

Letting \t : t1- t;1 and substituting from Eqs. (4) and (5) into Eq. (3),we obtain

v(PFi) : l f ' (c)i * s,(d)i l aot

For each i, consider the vector

Fr : M(f (ct) , Bk) )i + N(/( d,) , S@iliEach of the vectors Fr (i: 1,2, . , n) is an approximation to the forcevector F(f(t), g(f)), given by Eq. (2), along the arc of C from p;1to p1.Observe that even though c1 and di are in general different numbers inthe open interval (trr, tr), the values of the vectors F(f (t),g(f) ) are close to

(2)

(3)

(4)

(s)

(5)

(7)

R ( f ; - t )

20.6 LINE INTEGRALS

the vector F1. Furtherlnore, we approximate the arc of C from Pi-l to Pi bythe line segment Pr-rP,. Thus, we apply formula (1) and obtain an approxi-mation for the work done by the vector F(f(t), g(f)) in moving an objectalong the arc of C from Po-, to Pi. Denoting this approximation by A1W,we have from formula (1), and Eqs. (7) and (6),

AiW : lM(f (c), 8(c,) ) i + N(/(d,) , S(do) ) j l ' l f ' (c) i * I ' G)j l Al t

or, equivalently,

AlW: lM(f (c,) , g(c)) f ' (c)7 Att + lN(f (d) , s@))S'(4) l A,t

An approximation of the work done by F(f (t),g(f)) along C is i A1W or,equivalently, o:t

n n

2 tu(f (c,), ski))f'(c)l air + > tN(/(4) , B(d))s'@)l \t

,""ri l, these sums is a Rieman" J;. The first summation is a Riemannsum for the function having function values M(f(t), g(t))f'(t) and thesecond summation is a Riemann sum for the function having functionvalues N(/(0, S(0)S'(0. If we let n-->*a, these two sums approach thedefinite integral:

lM( f ( t ) , s ( t ) ) f ' ( t ) + N ( f ( t ) , s ( f ) )s ' ( f ) I d t

We therefore have the following definition.

20.5.L Definition Let C be a curve lying in an open disk B in R2 for which a vector equationof C is R(0:/(t)i+g(f)i, where f' andg'are continuous onfa, b]. Fur-thermore, let a force field on B be defined byF(x, y) : M(x, y)i +N(x, y)i,where M and N are continuous on B. Then if W is the measure ofthe work done by F in moving an object along C from (/(a), g(n)) to( f (b) ' s (b) ) '

or, equivalently, by using vector notation,

( M ( f ( t ) , g ( t ) ) , N ( / ( t ) , g ( f ) ) ) . ( f ' ( t ) , 9 ' ( f ) ) d t

I"

(8)

(e)

. R ' ( f ) d t

or, equivalently,

(10)


sot,urroN: Parametric equations of the parabola are

x : t a n d U : t z - 1 < t < z

Thus, a vector equation of the parabola is

R ( f ) : t i + t ' i

Because /( t) - t, gU) : t2, and F(x, U) : Q' t A, , Sxry> ,then

F ( / ( t ) , 9 ( f ) ) : F ( t , t ) - ( t , + t 4 , g t 4 l l

If W in.-lb is the work done, then from formula (10) we have

f2

w - | nG, P) . R ' ( f ) d tJ-r

f2: | ( t 2 + t 4 , g t 4 > . ( 1 , 2 t > d t

J - t

f2: I G r+ t4+6 t ) d t

J-,

ts 15 12:?+;* ru j_,

:8+s#+64-( -+-++1)_ 3 6 3

1.

ExAMPLE L: Suppose an objectmoves along the parabola y : x2from the point (-1 , 1) to thepoint (2, 4).Find the total workdone if the motion is caused bythe force field F(x,A) : @' * y')i +3x'yi. Assume the arc is meas-ured in inches and the force ismeasured in pounds.

Therefore, the work done is 3€3 in.

d a "line integral." A common notationlle

heen

call, i S

dy

F t hthe

rfl.

vtl

)n

i) is. (8)

V ) l

d b :iu),ritio

1q.

x , !

ted

8(fini

iq. (8)rf Eq.

N ( x , :

5esteCy 8 (detin

'.q,

lf

N t

ef

Eo

lggd 9a l <

ral integral

) d x 1

is sui',t) an(forma

d x *

' su8andrmal

tegral: integ

: , y ) a

on is= f(t)ng for

intrn e :

t (x ,

atio

nrin

e l nlint

M( :

rtatiE X :

owi

IC I]

t At _t a

nolaretllo

Ththe

t -f a

s n (I ar,foll

1S

Ct f

for

Thio f (the

fact that becausedx : f

,(t) dt andparametric equations

dy - g'(t) dt. We have

20.5.2 Definition Let M and N be functions of two variables r and y such that they arecontinuous on an open disk B in R2. Let C be a curve lying in B and havingparametric equations

x - f ( t ) y : g G ) a < t = b

such that f ' and g' are continuous on la,

M(*, y) dx * N(x, y) dy over C is given bybl. Then the line integral of

20,6 LINE INTEGRALS 985

by using vector

. ILLUsrRArror 1: In Example 1, W is given by the line integral

f

| (r'+ y') dx + 3fy ilYJc

where C is the arc of the parabola A : xz from (-1, 7) to (2, 4)' o

I f a n e q u a t i o n o f C i s o f t h e f o r m y : F ( r ) , t h e n r m a y b e u s e d a s aparameter in place of f. similarly, if an equation of c is of the form x:G(y),

then y may be used as a Parameter in place of f.

. rLLUSrRArroN 2: In Example 1 and Illustration 1., the equation of c is

a : xr, which is of the form y : F(x). Therefore, we can use r as a Param-eter instead of f. Thus, in the integral of Illustration 1 we can replace y

by xz and dy by 2x dx, artd we have

This integral is the same as the third one appearing in the solution of

Example 1., except that the variable is r instead of f. o

If the curve c in the definition of the line integral is the closed interval

fa, bf on the x axis, then y :0 and dy :0. Thus,

f f b - .

I tv t ( r ,y) dx+ N(r , y) dy: I M(x,O) dxJ c - J s

Therefore, in such a case, the line integral reduces to a definite integral.In the definition of a line integral we required that the functions /

and g in the parametric equations defining C be such that /' and g' are

continuous. Such a curve C is said to be smooth. If the curve C consists of

a finite number of arcs of smooth curves, then C is said to be sectionally

f2* :

J_, (x2 * xa) dx * 3x2xz (2x dx)

r2: J_, @' * x4 + 6x5) dx

Figu re 20.6.2

or, eq

Figure 20.6.3


smooth. Figures 20.6.2 and 20.6.3 show two sectionally smooth curves.we can extend the concept of a line integral to include curves that aresectionally smooth.

20.6.3 Definition C consist of

il dx * N(r,

M(x, y) dx * N (x, y)

Cr, Cr, . , Cn. Then the lineis defined by

M(x, y) dx + N (x, y) dy)

Let th,integra

rI

J c

e curyeI of M(x,

ExAMPrn 2: Evaluate the lineintegral

rI 4*y dx * (2x' - 3*y) dy

J C

over the curye C consisting ofthe line segment from (-3, -2)

to (1, 0) and the first quadrantarc of the circle x2 * y' : 1, from(1, 0) to (0, 1), traversed in thecounterclockwise direction.

solurroN: Figure 20.6.4 shows the curve C composed of arcs Cr and G.The arc Ct is the line segment. An equation of the line through (-9,-2)and (1., 0) is r - 2y : 1. Therefore , C1c?rtbe represented parametricallyby

x : 7 * 2 t y : t - 2 < t < 0

The arc Cr, which is the first quadrant arc of the circle I * yr:1, can berepresented parametrically by

r : c o s t y : s i n f 0 = t - + n

Applying Definition 20.6.2 f.or each of the arcs C1 and C2 we have

I

J",n*A dx + (2f - 3xy) dy

4(1 + 2t)t(2 dt) + l2(1 + 2t)2 - 3(1 + 2t)tl dt

(8f +'1.6t2 + 2 + 8f + 8t2 - 3t - 5t2) dt

(r8t2 + 13t + 2) dt

- 6F + +t'+ 2t10-l-z

: _ ( _ 4 9 + 2 6 _ 4 )

: 2 6

dx t (2x' - 3xy) dy

4 cos f sin f (-sin t dt) + [2 cosz s- 3 cos f sin f ](cos t dt)

n12

( - 4 c o s f s i n 2 t + 2 c o s 3 t - 3 c o s 2 f s i n f ) d t

r- J-,ro: J_,ro- J-,

Figure 20.6.4

: l,--L

20.6 LINE INTEGRALS 987

f n12: | [ -4cos ts inz t *2cos t (1 - s in2 f ) -3 cos2 ts in t ] d t

Jo

f nl2: | ( 2 c o s f - 5 c o s f s i n 2 f - 3 c o s z t s i n t ) d t

JO

: 2 sin f - 2 sin3 f + cos"l'''l o

: 2 - 2 - ' L

- - 1

Therefore, from Definition 20.6.3,

f

| +xy dx+ (2*-3rY) dY:26 + ( -1)Jc

ExAMPr,n 3: An obiect traversesthe twisted cubic

The concept of a line integral can be extended to three dimensions.

20.6.4 Definition Let M, N, and R be functions of three variables, x, y, and z such that theyare continuous on an open disc B in R3. Let C be a curve, lying in B, andhaving parametric equations

x : f ( t ) y : g U ) z : h ( t ) a < t < b

such that f ',

S', and h' are continuous on la,bf . Then the line integral ofM(x, y, z) dx -t N(x, y, z) dy + R(x, !, z) dz over C is given by

or, equivalently, by using vector notation

We can define the work done by a force field in moving an objectalong a curve in R3 as a line integral of the form of that in Definition20.6.4.

solurroN: Because f (t) : t, S$) : F, h(t) : f , M(x, y, z) : e' , N(x, y ' z) :

xe",ar] ldR(x,y,z):r s in ry2,thenM(f(t) ,g(t) ,h(t)) :et ,N(f( t) ,g(t) 'h(t))--tet", and R(f (t), g(t), h(t)) : f sin rf . Thrts, if W in.-lb is the work done,we have from the line integral of Definition20.6.4,R ( f ) : t i + t ' i + P k 0 < t


Find the total work done if themotion is caused by the forcefield F(x, y, z): e'i * xe"i +r sin ny'k.Assume that the arcis measured in inches and theforce is measured in pounds.

ntn) (3t' dt)lf t

w - | t @) Ut) + (tt") (2t dt) + (t sinJo

fr- | t et * 2t2et" + gt} sin ntn) dt

Jo

: et +? eF -* .o, ntnlJ +7T lo

- e + ? e - Lnn "o, n - 1 - I +fi .o,

5 , 3 5: - O +

3 " ' 2 z r 3

Therefore, the work done is I tr - I) +

*]

(2, 0) to (2, 2) .

( 0 , - L ) t o ( 1 , - 1 ) .

Exercises 20.5

In Exercises 1 through 20, evaluate the line integral over the given curve.

,. I e + ry) dx + (y'- xy) dy;C: the line y: x fromthe origin to the point (2, 2).Jc

2. The line integral of Exercise 1; C: the parabola * :2y from the origin to the point (2, 2).

3. The line integral of Exercise 1; C: the x axis from the origin to (2, 0) and then the line r:2 from

a. I V* dx l - (x+y) dy;C: the l ine ! : -x f rom the or ig in to the point (1, - l ) .J c '

5. The line integral of Exercise 4; C: the curve y : -rs from the origin to the point (1, -1).

6. The line integral of Exercise 4; C: the y axis from the origin to (0, -1) and then the line / : -1 from

,. I V dx* x dy;C: R(t ) : t i+ f j ,O < t < ] . .J c -

f8 . | ( r - 2 y ) d x * x y d y ; C : R ( t ) : 3 c o s t i * 2 s i n 4 , 0 = t - t r ,

Jc

f9 . | ( x - y ) ax+ Q+x ) dy ;C : t he en t i r e c i r c l e l t y ' : 4 .

JC

rc. [ ,y d, - y' d.y; C: R(t) : l2i + fsi, from the point (1, L) to the point (4, -8).Jc

-

Itt. | (xV + f) dx * * dy; C: the parabola y :2rz from the origin to the point (1, 2).

J C

f12.

JcV sin x dx - cos r dy; C: lhe line segment from (Lr,0) to (2, 1).

20.7 LINE INTEGRALS INDEPENDENT OF THE PATH 989

n. I g t y) dx + (V + z) dy 1- (x + z) dz; C: theline segment from the origin to the point (1,2,4).Jc

U. I gy - z) dx * e' dy * y dz; C: the line segment from (1, 0, 0) to (g, 4, 8).J c ' "

15. The l ine in tegra lof Exerc ise 14; C: R( f ) : ( t+ l ) i+ tg j + fsk,0 < t <2.

rc. I V' i lx * z2 dy + f dz; C: R(t) : (t -1)i + (t + l)i + Fk, 0 < t < 7.J c '

V. I z dx * x dy -t y dz; C: the circular helix R(t) : a cos fi * a sin fi + tk; 0 < t < 2n.Jc

rc . I Zxy dx * (6y2 -xz ) dy * l 0zdz ;C : t he tw i s tedcub i cR( t ) : t i + t z i+ t sk ,0 < t <7 ., J c

19. The line integral of Exercise 18; C: the line segment from the origin to the point (0,0,l); then the line segment from(0,0, 1) to (0,1, 1) ; then the l ine segment f rom (0,1, 1) to (1,1, 1) .

20. The line integral of Exercise L8; C: the line segment from the origin to the point (1,l, l).

In Exercises 21 through 34, find the total work done in moving an object along the given arc C if the motion is caused by thegiven force field. Assume the arc is measured in inches and the force is measured in pounds.

27. F(x, y) :2xyi + (* + y')i; C: the line segment from the origin to the point (1, 1).

22. The force field of Exercise 2\; C: lhe arc of the parabola A' : x fuom the origin to the point ( 1, 1) .

23. F(x, V) : fu - r)i * *yj; C: the line segment from the point (1, 7) to Q, a\.

24. The force field of Exercise 23; C: the arc of the parabola ! : f from the point (l , l) to Q, $ .

25. The force field of Exercise 23; C: Ihe line segment from (1, l) to (2,2) and then the line segment from (2, 2) to Q,0.

26. F(x, V) :-*Vi * 2yi; C: the line segment lrom (a,0) to (0, a).

27. The force field of Exercise 25; C:R(t): a cos fi * a sin 4,0 =t =ir.

28. The force field of Exercise 26; C; the line segment ftom (a,0) to (a, a) and then the line segment fuom (a, a) to (0, a).

2 9 . 8 ( x , l , z ) : ( y + z ) i * ( r * z ) i + Q + y ) k ; C : t h e l i n e s e g m e n t f r o m t h e o r i g i n t o t h e p o i n t ( L , 1 , 7 ) .

30. F(r, A , z) : zzi * yzj l xzk; C: the line segment from the origin to the point (4, 0, 3) .

31. F(r, A, z) : e"i * eui I e"k C: R(t) : ti + tzj + tsk, 0 < t < 2.

32. F(x, I, z\ : (ry" * x)i I (xzz + V)i + (*y + z)k; C: the arc of Exercise 31.

33. The force field of Exercise 32; C; lhe line segment from the origin to the point (1,0,0); then the line segment from(1, 0, 0) to (1,1,0) ; then the l ine segment f rom (1, 1, 0) to (1,1,7) .

34. F(x, y, z): xi* yj + (yz-r)k; C: R(f :211 + t2j + 4tsk,0 < f < 1.

20.7 LINE INTEGRALS We learned in Sec. 20.6 that the value of a line integral is determinedINDEPENDENT OF THE PATH by the integrand and a curve C between two points P, and Pr. However,

under certain conditions the value of a line integral depends only on theintegrand and the points P1 and P, and not on the path from P, to Pr.Such a line integral is said to be independent of the path.

9OO DIBECTIONAL DERIVATIVES, GRADIENTS, APPLICATIONS OF PARTIAL DERIVATIVES, AND LINE INTEGRALS

. rLLUsrRAflow 1: Suppose a force field

F(x, y) : (y, * 2x * 4)i + (2ry + ay - 5)imoves an object from the origin to the point (1, 1). We show that the totalwork done is the same if the path is along (a) the line segment from theorigin to (1, 1); (b) the segment of the parabola a - * from the origin to(1,1); and (c) the segment of the curve x: ys from the origin to (1, 1).

If W in.-lb is the work done, then

f* :

J, (y' * 2x * 4) dx * (Zxy * 4y - 5) dy

(a) An equation of C is A : x.We use ras the parameter and let y : x arrddy : dx in Eq. (1). We have then

ftr : Jor-Jo

- "r3- r L

( x ' * 2x *4 )

(3x ' * 6x - 1 )

- 3

(b) An equation of C is A: *. Again taking x as the parameter and inEq. (1) letting U : * and dy - 2x dx, we have

lrw: | (f +2x + 4) dx + (2f + 4* - 5)2x dx

JO

fr: l ( 51+8x3 -8x+4 )dx. J o

-3

We take y as the parameter and in Eq.We have

(Zyn * 4y - s) dy

12y ' * 4y: 5)

(1)

d x * ( 2 x ' * 4 x - 5 ) d x

dx

: r5 + 2f - 4x2. *],

(c)we

Anlet

w

,on:rn(

y 2 -

6y'

(y

(6

luati- y s

I(I(

Teqx -

dy+

= y3-dy.

ly' t

y ' t

4 ) 3

l S T- 3y

+4

r+

r o f (d d x

*2y

t+5

, f c i

d x -

2yt '

- Syo

3 +

(1)

: yG * yu * 4yt * 2yr- ty],

- 3

dy


Thus, in parts (a), (b), and (c) the work done is 3 -lb.

In Illustration 1., we see that the value of the line integral is the same

over three different paths from (0, 0) to (1, 1). Actually the value of theline integral is the same over any sectionally smooth curve from the origin

to (1, 1), and so this line integral is independent of the path. (This fact isproved in Illustration 2, whibh follows Theorem 20.7.1'.)

We now state and prove Theorem 20.7.'1,, which not only gives condi-

tions for which the value of a line integral is independent of the path but

also gives a formula for finding the value of such a line integral.

20.7.1 Theorem Suppose that M and N are functions of two variables x andy defined on anopen disk B((xo, yo); r) in R2, Mu and N' are continuous on B, and

VQQ, y) : M(x, y) i + N(x, y) j

Suppose that C is any sectionally smooth curve in B from the point (xr, yr)to the point (r2, y2). Then the line integral

M(x, y) dx * N (x, y) dy

is independent of the path C and

pRooF: We give the proof if C is smooth. If C is only sectionally smooth,then we consider each piece separately and the following proof appliesto each smooth piece.

Let a vector equation of C be

R ( t ) : f ( t ) i + S @ j t 1 < t < t 2

where (h, At):(f(tt), g(tJ) and (x2, yz):(f(t"),8(fr)). Then from thedefinition of a line integral,

M(x, y) dx * N (x , y) v$ ( f@,gG) ) .R ' ( f ) d t

We proceed to evaluate the definite integral in the right member of Eq. (2).Because YQ@, y): M(x, y)i * N(r, y)i, then 6"@, y): M(x, y) and

6',(x., y'S: N(r, Y). Therefore,

d6@, y) : M(x, y) dx + N(x, y) dy

for all points (x, y) in B. Then If (f (t), g(t) ) is a point on C,

d6(f G) ,8(f) ) : tM(f (t) , s?))f '(t) + N(/(f ) , g(t))g'(t)l dt

or, equivalently,

I,

(2)I,

(3)

t for which

(4)


df(f(t), 8(f) ) : tvd (f@, 1GD . R' (f) I dt

for t1 < t 3 tz.Consider the function G of the single variable

0( f ( t ) , s ( f ) ) : G( f ) hThen

d+(f (t) , 8G)) : G'(t) dt (s)From Eqs. (3) and (5) it follows that

v6(ff),8(f)) . R' (t) dt : G' (t) dt

Hence,

ft" ftz

J, ,VOUG), S(t)) . R'(t) dt: J,, ,-

G'{t) dt

But from the fundamental theorem of the calculus (Theorem 7.6.2) the nghtmember of this equation is G(fr) - G(D. Furthermore, from Eq. (4),G(t,) - G(D : 6(f $r), S(t)) - 6(f (t,), S(r,) ). Therefore, we have

ftz

I vf( f@, s(t)) .R'(t) dt: Q(fG), sG)) - eUG), s(f ,)) (6)Jt ,

From Eqs. (2) and (5) and because (f (D, S(tr)): (xr, yr) and (f (t1),g(h)):(xr, yr), we obtain

[ ,k, f i dx * N(x, y) 4y: e@r, y,) - 6@r, y,)Jc

which is what we wished to prove. I

Because of the resemblance of Theorem 20.7.1, to the fundamentaltheorem of the calculus, it is sometimes called the fundamental theoremfor line integrals.

. rLLUsrRATrow 2: We use Theorem 20.7.1 to evaluate the line integralin Illustration 1. The line integral is

fI (V '+2x*4) dx* (2xy+ay-5) dy

J C

From Illustration 1, Sec. 20.5, it follows that if 6@, y) - yrx * x2 + 4x +2y2 - 5y, then

Ybk, y) : (y, * 2x * 4)i + (2xy + ay - 5)i

Thus, the value of the line integral is independent of the path, and C canbe any sectionally smooth curve from (0, 0) to (1, 1). From Theorem 20.7.'J..,we have, then,

I,20.7 LINE INTEGRALS INDEPENDENT OF THE PATH

(y' * 2x * $ dx * (Zxy * 4y - s) dy - 0(1, t) - 0(0, 0)

- J -0

-3

993

(7)

(8)

This result agrees with that of Illustration L.

If F is the vector-valued function defined by

F (x , Y) : M(x , Y) i + N( r , V) iand

F (x, y) : V0@, Y)

then F is called a gradient field, $ is called a potential function, and $(x, y)

is the potential of F at (r, y). Furthermore, if F is a force field satisfying

Uq. (g), then F is said to be a conseroatiae force field. Also, when a function

F,- defined by Eq. (7) has a potential function {, then the expression

M(x, y) dx * N(x, y) dy is called an exact differential; this ter'minology is

reasonable because M(x, y) dx * N(x, y) dy : d6.

. rLLUsrRArrow 3: In Illustration 1, we have the force field

F(x, y) : (y' * 2x * 4)i t (2xY + aY - 5)i

In Illustration 2, we have the function { such that

6k, y) : yLx * xz -f 4x * 2yz - 5y

Because

96G, y) -F(x, y)

it follows that F is a conservative force field, and q| is a potential function

of F. In particular, because Oe, D:11, the potential of F at (2, 1) is 11'

Furthermore,

(y' * 2x * 4) dx * (2xY * 4Y - 5) dY

is an exact differential and is equal to d0.

r E x

x2 l

apl

2x)

SOLUTION: IN

Y (xe-a - :

Therefore, we

rI t - 4

I te-a -J C

dy

Pl

ISco

lv

* c c

)Ply

) d x

2,

' ) =

.eo

(xt

PIE

sy

Th

at

in

I t h

* s

/ ) =

, ?r)

ed

- a -

,a

0,

'e-a

'(x,

0(0

.ov'.xe

Q(

=+

5, we sh,

Zx)i - (

t."1, with r

, y ) dy :

1.7

in

20

a

20.

sir

- a

2(

- S

Sec.

= (e-

'em

- a +

: l

)rel

a - U

v ) i: X e - a - X 2 + cos y.

- O(n, 0)

: c o s 7 r - ( n - n 2 * I )

: Tr2 - Tr -2

e


ExAMpw 2: Show that the valueof the followirg line integral isindependent of the path andevaluate it:

C is any sectionally smoothcurye from the point (5, - 1) tothe point (9,-3).

If the value of a line integral is independent of the path, it is notnecessary to find a potential function 0. we show the procedure in thenext example.

solurroN: Let M(x, A): Uy and N(x, -xly'. Then

Ma(x, y) - and N "(x ,

y)

Because Mo(x, y):N,(r, y), it follows from Theorem 20.5.1 that (Uy)i-(xlV')i is a gradient, and therefore the value of the integral is independentof the path. We take for the path the line segment from (5,-1) to (9,-3).An equation of the line is x * 2y: 3. Thus, parametric equations of theline are

x : 3 * 2 t y : - t l . < t < 3

We compute the value of the line integral by applying Definition 20.6.2.

(M( (3 + 2t) , N( (3 + 2t) , - f ) ) . (2, -1) dt

1,

v'

The results of this section can be extended to functions of threevariables. The statement of the following theorem and its proof areanalogous to Theorem 20.7.1,. The proof is left as an exercise (see Exer-cise 31).

20.7.2 Theorem Suppose that M, N, and R are functions of three variables x, y, and, zdefined on an open ball B((xr, !o, zo)) r) in R3; Ms, M", N' N, R' and R,are continuous on B; and

. Vd (x, y, z) : M(x, y , z) i * N(x, y, z) i + R(x, y, z)k

Suppose that C is any sectionally smooth curve in B from the point(rr, yr, z) to the point (r2, Az, z2). Then the line integral

(x, y, z) dz

.........'...11.1111..1..l.i.ill..liir.,.il.,.i.riiiii.l.liiiii itt'i',,,'.l..,.iin..tt.t'"t*l+i'i'li+Hil+i+l"ilit+ii#-iffi

i*: i:iiiiit+:+lii il#*+'ifi +#*+l:ffi ii i$*+ii il+ii:iuilil+:itt ttl'nt' ffi ffi Lir'#titilintlii-ittiii'iffi ir'tii+tnitirllli

,5, we showed that the vector

c o s z * 2 x y * 3 2 2 ) k

+C

curue from (0, 0, 0) to (1, -2, n)the line integral

ly + (e' cos z * 2xy * 322) dz

is

4 n + 4 + 7 r 3 ) - 0

fI tvt(x,y, z) dx * N(r, y, z) dy * R(r, lJc

is independent of the path C and

lifflffift i. ll.. $+ #.ru' i#J.iri#.l.il.i'*flEi#ft#.lii .*H#..'it r€.'"i++f+'t"+ .1i+.lrti!;ffir r'il rijr,i"r,,r;riiir+.itl;.;l.,r.ti.rt.=ffiii.

o TLLUSTn^lrroN 4: In Example 3, Sec. 20.5, rn

(e' sin z * 2yz)i + (Zxz * 2y)i + (e' cos

is a gradient V/(x, U, z) and

f ( x , A , z ) - e ' s i n z * 2 x y z * y ' * z B + C

Therefore, if C is any sectionally smooth curthen it follows from Theorem 20.7.2 that the

f

I t e' sin z * 2yr) dx * (2xz * 2y) dY +J c

is independent of the path and its value is

f ( t , - 2 , n ) - f ( 0 , 0 , 0 ) : ( e s i n n - 4 n

20,7 LINE INTEGRALS INDEPENDENT OF THE PATH 995

- r f - 4 r * 4 o

As with functions of two variables, If F(x, y, z) : Y6(x, y,z), then Fis a gradient field, { is a potential function, and f(r, y, z) is the potential

of F at (x, y, z).If F is a force field having a potential function, then F is aconservative force field. Also, if the function F, defined by F(x, !, z):M(x, y, z)i * N(r, y, z)i * R(x, y, z)k, has a potential function, then theexpression M(x, y, z) dx * N(x, y, z) dy + R(x, y, z) dz is an exact differ-

ential.

solurroN: To determine if F is conservative, we apply Theorcm 20.5'2

to find out if F is a gradient. Let M(x, y, z): z2 +'l ', N(x, y, z):2y2, andR(x, y, z):2xz * y2. Then

Mu(x , y , z ) :0 M, (x , ! , z ) :22 Nr (x , Y , z ) :0

N"(x, y, z) :2y R,(x, y, z) - 2z Ru(x, Y, z) :2Y

Therefore, Mu(x, y, z) : N r(x, y, z), M"(x, a, z) : Rr(x, y, z), and N r(x ; y, z) :

Ru(x, y, z). Thus, by Theorem 20'5.2, F is a gradient and hence a conserva-tive force field. We now find a potential function f such that F(r, y, z\:YQ(x, y, z); hence,

Q r ( x , y , z ) : z ' + ' ] - , 6 u ( x , A , z ) : 2 y z Q " ( x , y , z ) : 2 x z * y 2 ( 9 )

Integrating with respect to r both members of the first of Eqs. (9), we have

6(x, y, z) : xzz * x * g(Y, z) (10)

EXAMPLE 3: If F is the force fielddefined by F(x, y, z) - (z'+ 1)i +2yzi * (Zxz * y')U Prove that F isconseryative and find a Potentialfunction.

g9O ] DIRECTIONIAL DERIVAT]VES, GRADIENTS,'APPLICATIONS OF PARTIAL DERIVATIVES, AND LINE INTEGRALS

we now partial differentiate with respect to y both members of Eq. (10)and we obtain

6o(x, y, z) : go(y, z) (tt)

Equating the right members of Eq. (11) and the second of Eqs. (9), we have

go(Y, zl :zvz

We now integrate with respect to y both members of this equation, andwe get

Sg, z) : yzz + h(z)

Substituting from Eq, (I2) into Eq. (10), we obtain

Q@' Y, z) :1cz2 * x * Y'z + h(z)

We partial differentiate with respect to z both members of Eq. (13) and get

Q"(x, y, z) :2xz * yz t h'(z) (14)

We equate the right members of Eq. (1a) and the third of Eqs.(9) and wehave

2xz * y' + h' (z) :2xz * y'

h ' ( z ) : o

h (z ) : C

Substituting from Eq. (15) into Eq.function 0'such that

' O (x ,Y , z ) : Jczz + ) c+Y 'z + C

(12)

(13)

(1s)

(13), we obtain the required potential

Exercises 20.7In Exercises L through 10, prove that the given force field

1. F (x, A) : yi-+ xi

conservative and tind a potential ftrnction.

2. F(x, A) : xi + yi

t a n z - s e c z ) k

1 S

3. F(x, A) : e' sin W + e' cos yi

4. F(x, A) : (sin y sinh x * cos y cosh r)i * (cos y cosh x - sin y sinh r)i

5u F (x, !) : (2xy' - yr)i + (2xzy - 3xy, + 2)i

6 . F(x, A) : (3x '* 2y - y 'e") i * (2x- 2ye") i

7. F(x, y, z) - (x2 - V)i - (r - 3z)i + k + 3y)k

8. F(x, y,z) : (2y'- 8xi2)i * (5xy, + l)i - (8xzz * lzz)k

9. F(x, y ,z) - (2x cos y - 3) i - (x , s in y * zr ) i - QVz- 2)k

L0. F(x, y, z): (tan y + 2xy sec z)i * (r secz y * x2 sec ")i

+ sec z(x,y

13.

L4.


In Exercises 11 through 20, use the results of the indicated exercise to prove that the value of the given line integral is inde-

pendent of the path.-Then evaluate the line integral by applying either Theorem20.7.7 or Theorem 20'7.2 and using the

potential function found in the indicated exercise. In each exercise, C is any sectionally smooth curve from the point A

to the point B.

f11. I y dx * x dy ; A ts (L, 4) and B is (3, 2); Exercise l

J C

12.LI,I,I,I,I,I,L

18.

x dx * y dy; A is (-5 , 2) and B is (L, 3); Exercise 2

e" siny i lxl e" cos y dy; A is (0,0) and B is (2, iz); Exercise 3

(s inys inh r *cos ycosh x ) dx * ( cosycoshx -s inys inh x ) dy ;A i s (1 ,0 ) andB is (2 ,n \ ; Exe rc i se4

(2xy'- A\ dx + (zfv - 3xy2 * 2\ dy; A is (-3, -1) and B is (1, 2); Exercise 5

(3 f +2y -y ' e " ) dx - f (2x -2ye ' ) dy ;A i s (0 ,2 ) andB is (1 , -3 ) ;Exe rc i se6

@ - il dx - (x - 3z) dy * (z I 3y) dz; A is (-3, l, 2) and' B is (3, 0, 4); Exercise 7

(2y3 - 8xz2) dx -t (6xy2 + l) dy - (Sfz 1 322) dz; A is (2,0, 0) and B is (3, 2, 1); Exercise 8

(2xcosy-3) dx- (x2 s in y tzr ) dy- (2yz-2) dz; A is ( -1,0,3) and B is (1, zr ,0) i Exerc ise 9

L5.

1,6.

17.

1,9.

ZO. [ ( t anv - l 2xusecz )dx * ( xseczy tx2secz )dy *secz (x2y tanz -secz )dz ;A i s (2 , *n ,0 )andB is (3 ,n ' , 2 ) ;Exe rc i se l0J c '

In Exercises 21 through 30, show that the value of the line integral is independent of the path and compute the value in

any convenient manner. In each exercise, C is any sectionally smooth curve from the point A to the point B.

I21. | (2v - x) dx + (y' t 2x) dy; A is (0, -1) and B is (L' 2)

J c ' -

zz . I On x *Z t i dx * (eu +2x ) dy ;A i s (3 , 1 ) and B i s (1 ,3 )J c '

z . I an y dx ' r x se i y dy; Ais ( -2, 0) and Bis G, tn)Jc

2a . [ " s i ny

dx * ( s i ny * xcosy ) dy ; A i s ( -2 ,0 ) andB i s (2 , * r r )

l 2 a ' +,t. 1"67r*dx +

1rft g,dv; A is (0, 2) and B is (1, 0)

25. [ *J- a, + --=!- av + --J* dz; A is(1, 0, 0) and B is (1,2, g)J c r r y - t z - * - l y ' ! 2 2 * v ' f + y ' a " "

z z . I f u 1 2 ) d x * ( r * z ) d y + ( x + y ) d z ; A i s ( 0 , 0 , 0 ) a n d B i s ( l , r , r )J c . -

998 DIRECTIONAL DERIVATIVES, GRADIENTS, APPLICATIONSOF PARTIAL DERIVATIVES, AND LINE INTEGRALS

f28.

J" Qz + x) dx I (xz i y) dy + (xy + z) dz; A is (0, 0, B i s ( 1 , 1 , L )

r2 9 .

J r ( e ' s i n y + y z ) d x * ( e ' c o s y * z s i n y * x z ) d y + ( x y - c o s y ) A i s ( 2 , 0 , ' 1 . ) a n d B i s ( 0 , n , 3 )

f30. I Qx ln yz - Sye") dx - (5e" - x'y-t) dy + (x27-r * 2z

J CA i s ( 2 , 1 , 1 ) a n d B i s ( 3 , 1 , e )


Reaiew Exercises (Chapter 20)

In Exercises 'J. and,2, find the value of the directional derivative at the particular point Po for the given function in thedirection of U.

7. f(r, y, z) : xy"z - 3xyz * 2x*; U : -*i + 3i - *k; Po : (2, l, 1)

z. f(r , y) : tan-1 Y,v : ;d- fr i ,po:

(4, -4)

In Exercises 3 and 4, find the rate of change of the function value in the direction of U at P.

3. f(x, a) : f ,r" (x' * u\;u - l , * + i ; P- ( r, L)

4. f(x, y, z) : yz - y2 - xz;U - 9i + +i + ?k; P - (1, 2, g)

In Exercises 5 and 6, determine if the vector is a gradient. If it is,

5 . y ( c o s ) c - z s i n r ) i * z ( c o s ) c + s i n y ) i - ( c o s y - y c o s r ) k

6. (e" tan y - sec y)i - sec AQ tan y - er sec y)j

then find a function having the given gradient.

In Exercises 7 and8, find an equation of the tangent plane and equations of the normal line to the given sur{ace at the indi-

an

dz;

o)

x!

1 )

cated point.

7 . z 2 + 2 y + z - - 8 ; ( 2 , 1 , 2 )

In Exercises 9 and 1.0, determine the relative extrema of f ,

8 . z : x 2 + Z x y ; ( L , 3 , 7 )

if there are any.

9 . f ( r , Y ) : P - t Y 3 t 3 x Y

10. f(x, V) : Z* - 3xy -f 2y2 * l}x - lly

In Exercises 11 and 12, evaluate the line integral over the given curve.

t t . I p* *3y) dx*xy d .y ;C: R( t ) :4 s in f i -cos t i ,o < t - * rJ c .

I12 . I xea i l x - xe " dy * e " dz ;C :R( t ) : t i + P i + t 3k ,0 < f < L

J C

In Exercises 13 and 14, find the total work done in moving an object along the given arc C if the motion is caused by thegiven force field. Assume the arc is measured in inches and the force is measured in pounds.

1 3 . F ( r , y , z ) : ( x y - z ) i + y i + z l q C : t h e l i n e s e g m e n t f r o m t h e o r i g i n t o t h e p o i n t ( 4 , 7 , 2 )

REVIEW EXERCISES 999

14. F(x, y) : xyzi - x'yj; C: the arc of the circle x' + y': 4 from (2, 0) to (0, 2)

In Exercises 15 and L6, prove that the value of the given line integral is independent of the path, and compute the value inany convenient manner. In each exercise C is any sectionally smooth curve from the point A to the point B.

I

15. | 2xeu dxl feu dy; Ais (1,0) and B is (3,2)J C

rc. I z sin y dx I xz cos y dy + x sin y dz; A is(0, 0, 0) and B is (2, g, +n)Jc

17. lf f (x, !, z) : 3xy' - 5xz2 - 2xyz, find the gradient of f at (-7, 3, 2).

18 . l t f ( x , l , z ) : s i nh ( r * z ) coshA ,hnd the ra teo f changeo f f ( x , y , z )w i t h respec t tod i s tance inRsa t thepo in tP (1 ,1 ,0 )in the direction PQ if Q is the point (-L , O, 2) .

19. Find the dimensions of the rectangular parallelepiped of greatest volume that can be inscribed in the ellipsoid rf +

9y2 -f z2: 9. Assume that the edges are parallel to the coordinate axes.

20. Find equations of the tangent line to the curve of intersection of the surface z:3* f' y' * L with the plane x:2 at

the point (2,-1,74).

21. The temperature is T degrees at any point on a heated circular plate and

22.

Mt :

f + a , + 9

where distance is measured in inches from the origin at the center of the plate. (a) Find the rate of change of the tem-

perature at the point (3, 2) in the direction of the vector cos 6zi * sinfzri. @) Find the direction and magnitude of the

greatest rate of change of T at the point (3,2).

The temperature is T degrees at any point (x, y) of the curve 4x2 * L2y2: L and T : 4f * 24y2 - 2x. Find the points

on the curve where the temperature is the greatest and where it is the least. Also find the temPerature at these points.

A monopolist produces two commodities whose demand equations are

x : 1 6 - 3 p - 2 q a n d A : 1 l - 2 p - 2 q

where 100r is the quantity of the first commodity demanded if the price is p dollars per unit and 100y is the quantity of

the second commodity demanded if the price is 4 dollars per unit. Show that the two commodities are complementary.If the cost of production of each unit of the first commodity is $1 and the cost of production of each unit of the secondcommodify is $3, find the quantities demanded and the price of each commodity in order to have the greatest profit'

24. Find the value of the line integral

f -Y ) ^ - , x

J , x '+ v 'dx* f iavif C is the arc of the cir c\e xz * y' - 4 from (rt, \D to

Given the force field F such that

F (x, y ,' z) : 22 sec2 xi r 2ye3J * (3y2e3" 4 2z tan x)k

prove that F is conservative and find a potential function.

26. A piece of wire L ft long is cut into three pieces. One piece is bent into the shape of a circle, a second piece is bent into

the shape of a square, and the third piece is bent into the shape of an equilateral triangle. How should the wire be cut

so that (a) the combined area of the three figures is as small as possible and (b) the combined area of the three figures is

as large as possible?

23.

25.

o ) .

1OOO DIRECTIONAL DERIVATIVES, GRADIENTS, APPLICATIONS OF PARTIAL DERIVATIVES, AND LINE INTEGRALS

27 - Use the method of Lagrange multipliers to find the critical points of the function / for which /(r, A, z) : y * xz -2f - yz - z2 subject to the constrainl z:35 - r: y. Determine if the function has a relative maximum or a relativeminimum value at any critical point.

28. In parts (a), (b), and (c) demand equations of two related commodities are given. In each part, determine if the commod-i t ies are complementary, subst i tu tes, or nei ther : (a) r - -4p- l 2q*6, A:5p-q + 10; (b) x :6-3p-2q, A:4+2p - q; (c) r : -7q - p * 7, y : 18 - 3q - 9p.

29. Determine the relative dimensions of a rectangular box, without a top and having a specific surface area, if the volume isto be a maximum.

Multiple integration

2l.l

MULTIPLE INTEGRATION

THE DOUBLE INTEGRAL

(b t , b 2 )

( t i , y t )

L i y

The definite integral of a function of a single variable can be extendedto a function of several variables. We call an integral of a function of asingle variable a single integral to distinguish it from a multiple integral,which involves a function of several variables. The physical and geo-metric applications of multiple integrals are analogous to those givenin Chapter 8 for single integrals.

In the discussion of a single integral we required that the functionbe defined on a closed interval in Rl.}or the double integral of a func-tion of two variables, we require that the function be defined on a closedregion in R2. By a closed region we mean that the region includes itsboundary. In this chapter, when we refer to a region, it is assumed to beclosed. The simplest kind of closed region in R2 is a closed rectangle,which we now proceed to define. Consider two distinct points A(ar, ar)arrd B(bt, bz) such that a1 = b1 and az - bz.These two points determinea rectangle having sides parallel to the coordinate axes. Refer to Fig.2L.1,.1,. The two points, together with the points (br, ar) and (ar, bz), arccalled the aertices of the rectangle. The line segments joining consecutivevertices are called the edges of the rectangle. The set of all points interiorto the rectangle is called an open rectangle, and the set of all points in theopen rectangle, together with the points on the edges of the rectangle, iscalled a closed rectangle.

Let the closed rectangular region of Fig. 21,.1.1 be denoted by R andlet / be a function defined on R. The region R can be considered as aregionof integration. Our first step is to define a partition, A, of R. We draw linesparallel to the coordinate axes and obtain a network of rectangular sub-regions which cover R. The norm of this partition, denoted by llAll, is de-termined by the length of the longest diagonal of a rectangular subregionof the partition. We number the subregions in some arbitrary way and letthe total be n. We denote the width of the ith subregionby A4x units andits height by LtV units. Then it LA square units is the area of the lth rec-tangular subregion,

AA: L'P A'gLet ((1, yr) be an arbitrary point in the ith subregion and f({i,7i) be thefunction value there. Consider the product f(€u y) A/. Associated witheach of the n subregions is such a product, and their sum is

n

) f f €i , r i ) A,A

There are many sums of the form (1) because the norm of the partitioncan be any positive number and each point (f;, 7;) can be any point inthe ith subregion. If all such sums can be made arbitrarily close to onenumber L by taking partitions with sufficiently small norrns, then L isdefined to be the limit of these sums as the norm of the partition of Rapproaches zero. We have the following definition.

Let / be a function defined on a closed rectangular region R. The number

Figu re 21 .1 .1

( 1 )

(a , , a r )

21.1*1 Definition

21 .1 THE DOUBLE INTEGRAL 1OO3

f (€,, Tr) L,A if L satisfies

> 0 such that

ii : l

6

L is said to be the limit of sums of the form

the property that for any € ) 0, there exists a

l n I

l ) / ( t i , r i ) ̂ ,A- L l ' €

for every partition A for which llAll < 6 and for all possible selections ofthe point (tu y) in the ith rectangle, i:7,2, . . . , n. If such a numberL exists, we write

n

lim ). f(ti,y) L1A:Ll l A l l - o { : 1

If there is a number L satisfying Definition 21.1.'1., it can be shownthat it is unique. The proof is similar to the proof of the theorem (2.1,.2)regarding the uniqueness of the limit of a function.

21.L.2 Definition A function / of trvo variables is said tobe integrable on a rectangular regionR if / is defined on R and the number L of Definition 21.1.1 exists. Thisnumber L is called the double integral of / on R, and we write

ffiju'dffi* #ffi e,

The following theorem, which is stated without proof, gives us acondition under which a function of two variables is integrable.

21..1.3 Theorem If a function f of two variables is continuous on a closed rectangular regionR, then / is integrable on R.

The approximation of the value of a double integral by using Defi-nition 21.1.3 is shown in the following example.

. : : . . , . : : . : r . . : . : : . : . : t . , . . : . : r : :

. : : . t . . : :

Other symbols for the double integral in (2) are

r r f f| | f (x ,y) dx dy and | | f (x ,y) dy dx

JnJ JnJ

solurroN: Refer to Fig. 21..1.2, which shows the region R partitionedinto six subregions which are squares having sides one unit in length.So for each i, LoA:l. In each of the subregions the point ({i,y) is atthe center of the square. Therefore, an approximation to the given doubleintegral is given by

f f

| | tz*-3y) ae- fe+,*) .1+ f (+, t ) .1+f(9,* ) .1JnJ

+ fG, t ) .1+ f (+ , i l . t+ f ( -+ , i l .L

10n5ular regit) andn o f R- 0 t X : - '

:tan- L , '

titice s r

rect(- :

)artiiner

S

PI

tneficere a lthe

is therticrrke ay thr

E I

rlueI

(2

( i svertakbv

PLE

val

^I3 Rr g vt . T

;

(AMF

ate r

Jn

herelvinlt , 3 ) .rrne(

EXA

mat

wht

hav(2,forr:

(2x' - 3y)dA

ngL

.orx =

Find an approxi-of the double inte gral

t ,

1OO4 MULTIPLE INTEGRATION

and y - 2, and take (€u, yr)center of the ith subregion.

Figu re 21 .1 .3

3)

1 )

(2,

(2,

(-+ ,*)(+,+)G,*)

(-+,+)G,i) G,*)

o

The exact value of the double integral in Example 't is-24, as will beshown in Example '1.

of Sec. 27.2.We now consider the double integral of a function over a more gen-

eral region. In Definition 8.10.2 we defined a smooth function as one thathas a continuous derivative, and a smooth curve is the graph of a smoothfunction. Let R be a closed region whose boundary consists of a finitenumber of arcs of smooth curves that are joined together to form a closedcurve. As we did with a rectangular region, we draw lines parallel to thecoordinate axes, which gives us a rectangular partition of the region R.We discard the subregions which contain points not in R and consider

r only those which lie entirely in R (these are shaded in Fig. 21.1,.3). Lettingthe number of these shaded subregions be n, we proceed in a manneranalogous to the procedure we used for a rectangular region. Definitions21.1.1 and 21.1,.2 apply when the region R is the more general one de-scribed above. You should intuitively realize that as the norm of the par-tition approaches zero, n increases without bound, and the area of theregion omitted (i.e., the discarded rectangles) approaches zero. Actually,in advanced calculus it can be proved that if a function is integrable ona region R, the limit of the approximating sums of the form (L) is the sameno matter how we subdivide R, so long as each subregion has a shape towhich an area can be assigned.

Just as the integral of a functiorgeometrically as the measure of the a:tegral can be interpreted geometricallthree-dimensional solid. Suppose thaclosed region R in R2. Furthermore, Isume that/( x,y) is nonnegative on R.'is a surface lying above the ry plane

Figu re 21.1 .2

Figu re 21 .1 .4

21.1 THE DOUBLE INTEGRAL 1OO5

shows a particular rectangular subregion of R, having dimensions of mea-sures Aal and A1y. The figure also shows a rectangular solid having thissubregion as a base and f((p yr) as the measure of the dtitude where(€t, y) is a point in the ith subregion. The volume of the rectangularsolid is determined by

LrV:f(€o,y) L{: f ( t r ,y) L,p A,g (3)

The number given in (3) is the measure of the volume of the thin rectan-gular solid shown in Fig. 2L.1.4; thus, the sum given in (L) is the sum ofthe measures of the volumes of n such solids. This sum approximates themeasure of the volume of the three-dimensional solid shown inFi9.21..1..4which is bounded above by the graph of / and below by the region R inthe xy plane. The sum in (1) also approximates the number given by thedouble integral

v) dA

It can be proved that the volume of the three-dimensional solid of Fig.21.L.4 is the value of the double integral. We state this in the followingtheorem for which a formal proof is not given.

21.1..4 Theorem Let / be a function of two variables that is continuous on a closed regionR in the xy plane andf (x,A) > O for all (x,y) inR.If y(S) is the measureof the volume of the solid S having the region R as its base and having andtitude of measure f (x, y) at the point (x, y) in R, then

solurroN: The solid is shown in Fig. 21..1..5. The rectangular region Ris the rectangle in the xy plane bounded by the coordinate axes and thelines r : 3 and A : 2. From Theorem 2'1,.1..4, if V cubic units is the volumeof the solid,

ExAMPrn 2: Approximate thevolume of the solid bounded by

the surface

f ( x ,a ) :4 - *x ' -#y 'the planes x- 3 and y -2, and

the three coordinate Planes. To

find an approximate value of the

double integral take a Partitionof the region in the xy Plane bY

drawing the lines x - L, x : 2,

and y - L, and take (tt, yr) at the

center of the ith subregion.

(4 - tx' - #y') dA

Figure 21:1.5 shows R partitioned into six subregions which are squareshaving sides of length one unit. Therefore, for each i, LA:1.. The point(tu y) in,each subregion is at the center of the square. Then an approxi-mation of V is given by an approximation of the double integral, andwe have

t/ : f(+, +) . 1 + f(9, +) . L + f(8, il .

: (4- f f i ) + (4-#) + (4- f f i )1 + f(+, g) . L + f(9, +) 1 * f(8, "r) . L

+ (4- f f i ) + (4-H) + @- +++)

1006 MULTIPLE INTEGRATION

21.1.5 Theorem

21.1.6 Theorem

: 24* 88t: 21.59

Thus, the volume is approximately 2L59 cubic units.

The exact value of the volume in Example 2 is shown to be 21.5 cubicunits in Example 2 of Sec. 21.2.

Analogous to properties of the definite integral of a function of asingle variable are several properties of the double integral, and the mostimportant ones.ue given in the following theorems.

If c is a constant and the function / is integrable on a closed region R,then c/ is integrable on R and

l f r f

JJ 4@,v) ae: , J*J f@, y) dA

If the functions / and g are integrable on a dosed region R, then the func-tion f + I is integrable on R and

r f f f

J .J l f @,y) * s (x,y) l dA: J.J f @,y)dA+ I.l s@,y) dA

The result of Theorem 21.1.5 can be extended to any finite number offunctions which are integrable. The proofs of Theorems 21.1.5 and,2']-..1.6follow directly from the definition of a double integral. These proofs areleft as exercises (see Exercises 13 and 14).

21.1.7 Theorem If the functigls / and g are integrable on the closed region R andf (x,y) >gk, y) for all (x, y) in R, then

f I t r

J.J f@,y) dA- JJ t@,v) ae

Theorem 21.7.7 is analogous to Theorem 7.4.8 for the definite inte-

Figure 21.1 .5

21.1 THE DOUBLE INTEGRAL 1OO7

gral of a function of a single variable. The proof is similar and is left asan exercise (see Exercise 15).

21.1.8 Theorem a closed region R, and suppose thatm = f (x ,y ) < M fo r a l l (x ,y ) in R.of region R, we have

The proof of Theorem 21.1.8 is left as an exercise (see Exercise 16). Theproof is similar to that of Theorem 7.5.2 and is based on Theorem 2L.1..7.

21.1.9 Theorem

The proof of Theorem 21.1..9 is also left as an exercise and dependson the definition of a double integral and limit theorems (see Exercise 17).

Exercises 2L.1-

In Exercises 1 through 6, find an approximate value of the given double integral where R is the rectangular region havingthe vertices P and Q, A is a regular partition of R, and (€t, y) is the midpoint of each subregion.

f ll . | | ( f + y ) d A ; P ( 0 , 0 ) ; Q G , 2 ) ; A : t , : 0 , x 2 : 7 , x s : 2 , x r : 3 , A r : 0 , A " : 1 , .

JRJ

f tz . I I Q - , - y ) d A ; P ( 0 , 0 ) ; Q G , 4 ) ; A ; x , : 0 , x 2 : 2 , x t : 4 , l r : 0 , y z : 2 .

JEJ

r fl. | | (xV * 3y2) dA; P (-2, 0) ; QG, 6) ; A: x, : -2, xz: 0, x": 2, y, : 0, lz : 2, Ar: 4.

JRJ

f f+. | | (xy -f 3y2) dA; P(0, -2) ; Q(6, D ; A: r, : 0, xz: 2, x": 4, Ar:

-2, y": 0, ys: 2.JnJ

-

t fs' | | (x'za - 2xy') M; P (-3, -2) ; Q0, 6) ; A: \ : -3, xz: -1, lt : -2, !z: 0, !t: 2, y o: 4.

JRJ

t f6 . | | G V - 2 x y ' ) d A ; P ( - 3 , - 2 ) ; Q Q , 5 ) ; L : x t : - 3 , x 2 : - 2 , h : - l , x a : 0 , y ' : - 2 , A 2 : - 1 , A s : 0 , y n : 1 , A s 7 2 ,

JnJVe : 3 ' l z : 4 ' ! e : 5 -

In Exercises 7 through 10, find an approximate value of the given double integral where R is the rectangular region havingthe vertices P and Q, A is a regular partition of R, and (€u y) is an arbitrary point in each subregion.

7. The double integral, P, Q, and A are the same as in Exercise l; (€', y) : (i, i); (€2, yr) : (1,0); (fa, ya) : G, D,Gn ,y ) : G , r ) ; (& , y " ) : G ,D ; ( f o , yu ) : G , i l ; G , , r t ) : G ,z1 ; ( f a , ye ) : ( 3 ,1 ) .

Let the function / be integrable onm and M arc two numbers such thatThen if A is the measure of the area

mA = I.l f @,y) dA < MA

Suppose that the function / is continuous on the closed region R and thatregion R is composed of the two subregions R1 and R, which have nopoints in common except for points on parts of their boundaries. Then

f I f f f f| | f ( x ,v ) de :1 | f k ,y ) dA+ | | f ( x ,y ) dAJnJ

' - Jn,J Jn J

1OO8 MULTIPLE INTEGRATION

S . T h " d o u b l e i n t e g r a l , P , Q , a n d A a r e t h e s a m e a s i n E x e r c i s e 2 ; ( € r , y ) : ( t , t ) ; ( t r , l ) : ( 3 , 1 ) ; G s , y s ) : ( t , t ) ;Gr, y) : (2,2) ; ( fs , %) : (2,2) ; ( fe , ye ) - (5, 3) .

9 . T " d o u b l e i n t e g r a l , P , Q , a n d A a r e t h e s a m e a s i n E x e r c i s e 3 ; ( € r , y r ) : ( - t , t ) ; ( € r , y ) : ( l , t ) ; ( € " , f ) : ( g , 2 ) ;G+, y): (-*, t); (€0,7s) : (0, 3); (f . , ye) : (4,4); (€r, rr1 : 1-i , i l ; Gr, y") : | i , .El i tA, yrl : ts, *1. '" '

10. The double integral, P, Q, and. A are the same.as in Exercise 3; (tr,yr): (-2,0); (€r,^yr): (0,0); (6r, ys): (2,0);(tu y) : (-2,2); ({0,7s) : (0, 2); (t", y) : (2,2); (tr, rr1 : 1-2,4); (fs, ye) : 10, +li (fs, ys) : lii +1.

11. Approximate the volume of the solid in the first octant bounded by the sphere * * y" I z2:64, the planes r: 3 and! : 3, and the threecoordinate planes. To find an approximate valui of the double intbgral take a partition of the regioninthexy p lanebydrawingthel ines x- l ,x :2,y : l ,andy:2,arrd. take ( f1,n; a i t t recente iof thei thsubregion.

12. Approximatethevolumeof thesolidboundedbythesurfacelfi)z:3OO-2Sf -4y2,theplanesr: -7,x:3,y:-g,and y :5, and the ry plane. To find an approximate value of the double integral i"t"

" ia"tition of the re6oi in thexyplane bydrawingthel ines x:1,y-- l ,y :1,and A:3,and, take ( f1,y , ; u t t t r "ce-nterof thei thsub-region.

13. Prove Theorem 2L 1.5.

L6. Prove Theorem 2L 1.8.

2I.2 EVALUATION OFDOUBLE INTEGRATS AND

ITERATED INTEGRALS

L4. Prove Theorem 2L.L.6.

17. Prove Theorem 21.'1..9.

15. Prove Theorem 2L.1,.7.

For functions of a single variable, the fundamental theorem of calculus

one, and we use the geometric interpretation of the double integral as themeasure of a volume. we first develop the method for the doubie integralon a rectangular region.

is the measure of the volume of the solid between the surface and theregion R. we find this number by the method of parallel plane sections,which we discussed in Sec.8.4.

Let y be a number in la2, br]. consider the plane which is parallel tothe xz plane through the point (0, y,0). Let A(y) be the measure of thearea of the plane region of intersection of this plane with the solid. Bythe method of parallel plane sections, as discussed in sec. g.4, we expressthe measure of the volume of the solid bv

x,

F igu re 21 .2 .1

f", etvl dv

z : f(x, Y)

Because the volume of the solid also is determined by the double integral,

21.2 EVALUATION OF DOUBLE INTEGBALS AND ITERATED INTEGRALS 1OO9

we have

fbz

v) dA: I efvl dy- J a z

By using Eq. (1) we can find the value of the double integral of the func-tion f on R by evaluating a single integral of A(y). We now must find

A(y) when y is given. Because A(y) is the measure of the area of a plane

region, we can find it by integration' In Fig.21'.2.L, notice that the uPPer

bolndary of the plane region is the graph of the equation z = f (x' y) when

r is in lar, brl. Therefore, A(y) : [l: f k, !) dx. If we substitute from thisequation into Eq. (1), we obtain

( 1 )

LI

A sufficient condition for formulas (3) and (4) to be valid is that the func-

tion be continuous on the rectangular region R.

Evaluate the double soLUTIoN: ar: -l , br = 2, az: 1, andbz: 3' So we have from formula (3)

(2f - 3y) dA (2x' - 3y) dx dU(2x' - 3y) dA

The integral on the right side of Eq. (2) is called

utly the brackets are omitted when writing an

write Eq. (2) as

When evaluating the "inner integral" in Eq. (3), remember

variable of integration and y is considered a constant. This

to considering y as a constant when finding the partial

f (x, y) with respect to x.By considering plane sections

velop the following formula, which

(2x' - 3il d.f dy

(2)

an iterated integral. Usu-iterated integral. So we

(3)

that r is theis analogous

derivative of

parallel to the yz plane, we can de-interchanges the order of integration.

dy dx (4)H r@,v) dA: f:,f:,r@,v)

LIEXAMPLE

integral

if R is the region consisting ofall points (x,y) for which -L

x < 2 a n d 1 = y < 3 .

: -24

1O1O MULTIPLE INTEGRATION

EXAMPLE 2: Find the volume ofthe solid bounded by the surface

f ( x , y ) - 4 - i x ' - #y ,

the planes x:3 and y : 2, andthe three coordinate planes.

x

Figu re 21 .2.2

Y : 6 r @ )

4r@)

x i-r ti xiVL t x

Figu re 21 .2.3

In Example 1 of sec. 27.1wefound an approximate value of the doubleintegral in the above example to be -25.

soLUTroNr Figure 21.2.2 shows the graph of the equation z: f(x,y) inthe first octant and the given solid. rf y cubic unitjis the volume or tn"solid, we have from Theorem 21..1.4

u: ,,lit. i fc,, vt) aA: LI f@, v) itAf3 f2: JoJo @- +* - /rY2) dY dx

F f 1 2: J"Lor- t*v- tv" ]odxf3: Jo

(#-&*) dxl3: #x_ tx" lJo

:2L.5

The volume is therefore 2!.5 cubic

In Example 2 of. sec. 21.1 an approximate value of the volume in Ex-ample 2 was found to be 2L.59 cubic units.

suppose now that R is a region in the xy plane which is bounded bythe l ines x :aar rdx :b , where a1b, andby thecurves y :e1@) an iy : 6r(x), where 0r and $, arc two functions which are continuous onthe closed interval la,bl;furthermore,6r@) = 6r@) whenevera < x < b(see Fig. 21.2.3). Let A be a partition of the interval fa,bf defined byA:a : .xo < . r1 <divided into vertical strips with widths of meisure Alx. A particular stripis shown in the figure. The intersection of the surface ,J f(*, y) and aplane r: f1, where xr-t s €t = xr., is a curve. A segment of this curve isover the ith vertical strip. The region under this curve segment and abovethe xy plane is shown in Fig. 2't.2.4, and the measure o1 the area of thisregion is given by

f G,,a) dy

the volume of the solid bounded above by the surfacebelow by the ith vertical strip is approximately equal to

Iy) dV I Lix

J

r::,:The measure ofz - f ( x , y ) a n d

| fazgilLJo,tE,r f(€"

21.2 EVALUATION OF DOUBLE INTEGRALS AND ITERATED INTEGRALS 1011

Figu re 21 .2.4 Figu re 21 .2.5

If we take the limit, as the norm of A approaches zero, of the sum of thesemeasures of volume for n vertical strips of R from x: a to x:b, we ob-

tain the measure of the volume of the solid bounded above by the surfacez: f (x, y) and below by the region R in the xy plane. (See Fig. 21'.2.5.)

This is the double integral of / on R; that is,

(s)

Sufficient conditionson the closed region

for formula (5) toR and that rf1 and

be valid are that f b" continuous

Q, be smooth functions.

ExAMPLE 3: Express as both a

double integral and an iteratedintegral the measure of the vol-

ume of the solid above the xYplane bounded by the elliPtic

Paraboloid z - x2 + 4y', and the

cylind er xz * 4y' - 4. Evaluate

the iterated inte gral to find the

volume of the solid.

soLUrIoN: The solid is shown in Fig. 21'.2.6. Using properties of sym-metry, we find the volume of the portion of the solid in the first octantwhich is one-fourth of the required volume. Then the region R in the

ry plane is that bounded by the x andy axes and the ellipse * + 4yz : 4.This region is shown in Fig. 27.2.7, which also shows the fth subregionof a rectangular partition of R, where (tu yi) is any point in this ith sub-region. If v cubic units is the volume of the given solid, then by Theorem21.1.4 we have

t (€o '* Avo\ LA- A | |= J*J

Y : d t (x ) Y - + r (x ) v : 4 t ( x ) Y : Qr@)

liml l a l l - o

(x'+ 4y') dA

( t i , Y i )


Figure 21 .2.6

f f\rytz ILJ,

(€; * 4v') dy )

Lix

t2 @, + 4yr) dy dx

Figu re 21 .2.7 Figu re 21 .2.8

To express the measure of the volume as an iterated integral wedivide the region R into n vertical strips. Figure 21.2.g shows the re-gion R and the ith vertical strip having width of L6x units and length of+IT=E units, where x*r s €i = xt. Using formula (5) we have

V : 4

- 4

Evaluating the iterated inte gral, we have

v:4 t:l*,, + +v'])o*'' o*r2:n Jo l t * I+ -P

++(4 - x :2 )3 t21 dx

( f +z) tR dx

- -*r (4 - xz)3t2 * 2x\tr4 + 8 sin: 4 r

Therefore, the volume is 4n cubic units.

If the region R is bounded by the curves r: trr(y) and.l: ),r(y) andthe lines y: c andA: d, where c < d, and ),, and )r, are two functionswhich are continuous on the closed interval lc, d] forwhich Ir(y) < Ir(y)whenever c < y - d, then consider a partition A of the intervJ lc, dl-iadivide the region into horizontal strips, the measures of whose widthsare Ag. see Fig. 27.2.9, which shows the lth horizontal strip. The inter-section of the surface z: f (x, y) and a plane A = lt, where Ai_r = Tt = Ut,is- a curve, and a segment of this curve is over the ith horizontal strip.Then, as in the derivation of formula (5), the measure of the volume ofthe solid bounded above by the surface z- f(x, y) and, below by the ith

- ' * ' ] ,

x : | rz (U)


vertical strip is approximately equal to

Taking the limit, as llAll approaches zero, of the sum of these measuresof volume f.or n honzontal strips of R from y: c to U: d, we obtain themeasure of the volume of the solid bounded above by the surface z:

f (x, y) and below by the region R in the xy plane. This measure of volumeis the double integral of / on R. Hence, we have

Sufficient conditions for formula (5) to be valid are that trt and )t2 besmooth functions and / be continuous on R. In applying both formula(5) and formula (6) sometimes it may be necessary to subdivide a regionR into subregions on which these sufficient conditions hold.

solurroN: Again we find the volume of the solid in the first octant andmultiply the result by 4. Figure 2L.2.70 shows the region R in the xy planeand the ith horizontal strip whose width has a measure of. Ag, and whoselength has a measure of 2!1- y;2. Then by formula (5) we have

V :4 l im $l la l l -o e

- -+r y(1 - yz)s/z 1 sytffi+ 8 sin-l y]',

: 4n

Hence, the volume is 4n cubicExample 3.

t/ff1 r@,v,) d'lL,a

(6)

Figu rc 21 .2.9

ExAMPLE 4: Express the volumeof the solid of Example 3 by aniterated integral in which theorder of integration is the reverseof that of Example 3. Computethe volume.

F igu re 21 .2 .10

U:* @'*4vr) d.f Liu

(x'* 4y') dx dy

ft t- 12\n42- 4 | l * r ' * 4yzx l dyJ o L

e J o\v ?,

which agrees with the answer of


From the solutions of Examples 3 and 4 we see that the double inte-f f

gril J*J

(* + 4y2) ilA can be evaluated by either of the iterated integrals

f2 f\Etrl2 fr fzt/r-az

J,J, (f + +y2) dy itx or

J,J, -

(* + 4yz) dx ity

If in either formula (5) or (6), /(r ,A) :1 for all x andy,then we obtaina formula that expresses the measure A of the area of a-region R as a doubleintegral. We have

A- I^l itu dx: [.] dx dy (7)

ExAMPLE 5: Find by doubleintegration the area of the regionin the xy plane bounded by thecun/es y -- x2 and y : 4x - x2.

Figure 21.2.11

Exercises 21 .2

In Exercises L through 8, evaluate the given iterated integral.

solurroN: The region is shown in Fig. 21..2.1,1. Applying formula (Z),we have

_ 8- 3

Hence, the area of the region is I square units.

A- I.l dy dx: I:l;-" dy d*

: I: @x - x2 - x2) dx - 2x2 - 3*]',

4.3.

8.7.

ttta dx dyTI:'[[ l * - y l dy dx

14.

15.


In Exercises 9 through 14, find the exact value of the double integral.

9. The double integral is the same as in Exercise 1 in Exercises 2l'.1.

10. The double integral is the same as in Exercise 4 in Exercises 21.1.

I. [ [ sin x dA;R is the region bounded by the lines y : 2x, y : !x, and x :'r'JaJf l

lZ. | | cos(r * y) dA; R is the region bounded by the lines y: x and x: r, artd the x axis.JRJ

t f13. | | f \/r- d.A; R is the region bounded by the cirde I t y':9.

JnJ

t l u 2

J.J b dA; R is the region bounded by the lines y : r and A :2, and the hyperbola xy :1.

Find the volume of the solid under the plane z : 4x and above the circle 12 I y':15 in the ry plane. Draw a sketch

of the solid.

1 6 . F i n d t h e v o l u m e o f t h e s o l i d b o u n d e d b y t h e p l a n e s x : y + 2 2 * 1 , x : 0 , ! : 0 , 2 : 0 , a n d 3 y * z - 3 : 0 ' D r a w asketch of the solid.

17. Find the volume of the solid in the first octant bounded by the two cylinders * * y2: 4 and x2 + z2: 4. Draw a sketch

of the solid.

Find the volume of the solid in the first octant bounded by the paraboloid z:9 - f - 3y'' Draw a sketch of the solid.

Find the volume of the solid in the first octant bounded by the surfaces xl z2:1, x:y, and x: y2.Dtaw a sketch

of the solid.

20. Find by double integration the volume of the portion of the solid bounded by the sphere * I y' 'l z2 :16 which lies

in the first octant. Draw a sketch of the solid'

In Exercises 21 through 24,use double integrals to find the area of the region bounded by the given curves in the xy plane.

Draw a sketch of the region.

2 L . y - - r B a n d A : x 2 22. y ' :4x and x2 :4a 23 . y : x2 - 9 and y -9 - x2 2 4 . x 2 + y ' : L 5 a n d A ' : 6 x

25. Express as an iterated integral the measure of the volume of the solid bounded by the ellipsoid

* * a ' * Z ' : 1a 2 ' b 2

' c 2

25. Given the iterated integral

fq fr

l f ! a 2 - * d y d xJo Jo

(a) Draw a sketLh of the solid the measure of whose volume is represented by the given iterated integral; (b) evaluate

the iterated integral; (c) write the iterated integral which gives the measure of the volume of the same solid with the

order of integration reversed.

27. Given the iterated integral

1 1a 1t/F=it

i | | (2x-r y) dy dxJ J o J o

The instructions are the same as for Exercise 26.

18.

19.


28. Use double integration to find the area of the region in the first quadrant bounded by.the parabola lz: x,the circle* -t y': 5, and the r axis by two methods: (a) Integrate first with respect to x; (b) integrate first with iespect to y. Com-pare the two methods of solution.

29. Find" by two methods, the volume of the solid below the plane 3r -l 8y -l 5z:24 and above the region in the ry planebounded by the parabola A2 :2rc, the line 2x -l 3y: 10, and the r axis: (a) Integrate first with respe-ct to r; (b) lnt"grutefirst with respect to y. Compare the two methods of solution.

In Exercises 30 and 3L, the iterated integral cannotorder of integration. Reverse the order of integration

r l r l

30. I I e"' dx dyJ o J a

32. Use double integration to find the volume of the solid conunon to two right-circular cylinders of radius r units, whoseaxes intersect at right angles. (See Exercise 8 in Exercises g.4.)

21.3 CENTER OF MASS AND In Chapter g we used single integrals to find the center of mass of a homo_MOMENTS OF INERTIA geneous lamina. In using single integrals we can consider only laminae

of constant area density (except in special cases); however, with doubleintegrals we can find the center of mass of either a homogeneous or anonhomogeneous lamina.

Suppose that we are given a lamina which has the shape of a closedregion R in the xy plarre. Let p(x, y) be the measure of the area densityof the lamina at any point (r, y) of R where p is continuous on R. To findthe total mass of the lamina we proceed as follows. Let A be a partitionof R into r rectangles. lf (t0,1) is any point in the ith rectangle havingan area of measure Arr4, then an approximation to the measure of the massof the lth rectangle is given by p(tr, y) AtA, and the measure of the totalmass of the lamina is approximated by

PGt' Yi) LoA

Taking the limit of the above sum as the norm of A approaches zero, weexpress the measure M of the mass of the lamina by

The measure of the moment of mass of the lth rectangle with respect tothe r axis is approximated by fp(€u y) AA.The sum of the measures ofthe moments of mass of the n rectangles with respect to the r axis is thenapproximated by the sum of n such terms. The measureM" of the momentof mass with respect to the r axis of the entire lamina is given by

be evaluated exactly in terms of elementary functions by the givenand perform the computation.

n

i : l

( 1 )

(2)

21 ,3 CENTER OF MASS AND MOMENTS OF INERTIA 1017

measure Ma of its moment of mass with respect to the

The center of mass the lamina is denoted by the point (I, y) and

solurroN: Choose the coordinate axes so that the vertex of the right tri-angle is at the origin and the sides of length a ft of. the triangle are alongthe coordinate axes (see Fig. 21,.3.1).Let p(x, y) be the number of slugs/ffin the area density of the lamina at the point (r, y). Then p(x, y):k(xz * y2) where k is a constant. Thereforc, if M slugs is the mass of thelamina, we have from formula (1.)

M : lisr L o$f + yf) LtAl la l l -o f !

(x' * y') dA

* y') dy dx

(*at - azx * 2ax2 - txt) dx

: k(taa - laa I &aa - *aa): gkaa

To find the center of mass, we observe that because of symmetry itmust lie on the line y: r. Therefore, if we find x, we also have y. Usingformula (3), we have

n

M,: .!iln ). kf,(f,, + :|12) LtAl lA l l -o i= i

* '* y ' ) dA

x(x' * y') dy dx

(3)

ExAMPLE '/..:

A lamina in theshape of an isosceles right tri-angle has an area density whichvaries as the square of the dis-tance from the vertex of the rightangle. If mass is measured inslugs and distance is measured infeet, find the mass and the centerof mass of the lamina.

Figu re 21 .3.1

lr*'+ +u'] i-" o*

1 a - r

txvt l axv J o


(*atx - a2)c2 * 2ax3 - *xn) dx

- k(tat - tas * ta; -

- +ka,Because MI: My, we haveI - ?a. Therefore, the center

#a')

MI - +ka'; andof mass is at the

because M - tkan, we getpo in t (?a ,?a) .

21.3.1 Definition The moment of inertia of a particle,defined to be mrz slug-ft?, wherethe particle to the axis.

whose mass Ls m slugs, about an axis isr ft is the perpendicular distance from

If we have a system of n particles, the moment of inertia of the sys-tem is defined as the sum of the moments of inertia of all the particles.That is, if the ith particle has a mass oI m7 slugs and is at a distance ofri ft from the axis, then I slug-ft2 is the moment of inertia of the systemwhere

I _n

i :1

ttltT12 (4)

Extending this concept of moment of inertia to a continuous distributionof mass in a plane such as rods or laminae by processes similar to thosepreviously used, we have the following definition.

21..3.2 Definition Suppose that we are given a continuous distribution of mass occupyinga region R in the xy plane, and suppose that the measure of area densityof this distribution at the point (x, y) is p(x, y) slugslfP, where p is con-tinuous on R. Then the moment of inertia I" slug-ff about the r axis ofthis distribution of mass is determined by

(s)

SimilarLy, the measuregiven by

la of the moment of inertia about the y axis is

and the measure /0 of the moment of inertia about the origin,axis, is given by

(6)

or the z

The number Ie of formula (7) is the measure of what is called the polarmoment of inertia.

(7)

EXAMPLE 2: A homogeneousstraight wire has a constant lineardensity of p slugslft. Find themoment of inertia of the wireabout an axis perpendicular tothe wire and passing throughone end.

EXAMPLE 3: A homogeneousrectangular lamina has constantarea density of p slugslfP. Findthe moment of inertia of thelamina about one corner.

Figure 21 .3.2

21.3 CENTER OF MASS AND MOMENTS OF INERTIA 1019

solurroN: Let the wire be of length a ft, and suppose that it extendsalong the x axis from the origin. we find its moment of inertia about they axis. Divide the wire into n segments; the length of the ith segment isL,ix ft. The mass of the ith segment is then p Aax slugs. Assume that themass of the ith segment is concentrated at a single point fi, where xa-,r s€i = x* The moment of inertia of the ith segment about the y axis liesbetween pxi-12 Lg slug-ff and pxf Ap slug-ff and is approximated byptt" A& slug-ff, where x*r s & = xr.If the moment of inertia of the wireabout the y axis is I, slug-fP, then

lo: , l ip ) o t , '- l l a l l - o i : r

Therefore, the moment of inertia ts ! pa3 slug-ft2.

solurroN: Suppose that the lamina is bounded by the lines r : d, A : b,the r axis, and the y axis. See Fig. 21.3.2.If 16 slug-ff is the moment ofinertia about the origin, then

nIo: ,liF ) p(f,, + y?) ArA

l l A l l - o ; =

p(x' * y') dA

(x' + y') dx dy

pxz dx - *pot

- + pab (a' + br)

The moment of inertia is therefore * pab (a2 r bz) slug-ftz.

It is possible to find the distance from any axis L at which the mass of alamina can be concentrated without affecting the moment of inertia of thelamina about L. The measure of this distance, denoted by r, is called the"radius of gyration" of the lamina about L. That is, if the mass M slugs of alamina is concentrated at a point r ft from L, the moment of inertia of thelamina about L is the same as that of a particle of mass M slugs at a dis-tance of r ft from L; this moment of inertia is Mrz slug-ft2. Thus, we havethe following definition.

If 1 is the measure of the moment of inertia about an axis L of a distri-bution of mass in a plane and M is the measure of the total mass of the

( ( i , y i )

21.3.3 Definition

1O'IO MULTIPLE INTEGRATION

EXAMPLE 4: Suppose that alamina is in the shape of a semi-circle and the measure of the area

density of the lamina at anypoint is proportional to the mea-

sure of the distance of the point

from the diameter. If mass ismeasured in slugs and distanceis measured in feet, find theradius of gyration of the laminaabout the r axis.

Figure 21.3.3

distribution, then the radius of gyrationmeasute r, where

of the distribution about L has

solurroN: Choose the r and y axes so that the semicircle is the top halfof the cirde f * y': a2. See Fig. 21.3.3. The area density of the laminaat the point (r, y) is then ky slugslfP. So if M slugs is the mass of thelamina, we have

M: lim $. t", Aal la l l -o f i

r f: l l kuMJnJ

fa f{E--V: l | _kydxdy

J o J _\ta2_a2

fa I lroz=F:k I l v , l -dYJo L )-\,td-Yz

f@:zk I y !a2-yz dyJ o -

: &kas

If I, slug-fP is the moment of inertia of the lamina about the r axis, then

I": lim i.yrt(h'y,) Mllal l-o i=i

f t:J)W"ava'fd NE=F

: f | . kV "dUdxJ-aJ0

fa t 1{Fst:k J_"1+r.1, dxla

: *k | (a+-2az f + f ) dxJ-a

: *k(2ai - $aa * ?a5)

_ftkas

: -?k(a, - ,r'rt,rfi

21.3 CENTER OF MASS AND MOMENTS OF INERTIA

Therefore, tf r ft is the radius of gyration

,Lka' 2 q,12__ff i_to,

and so r - t{t}a. The radius of gyration is therefore *ffia tt.

1021

4.

5 .

6.

7.

8.

Exercises 2L.3

In Exercises 1 through 10, find the mass and center of mass of the given lamina if the area density is as indicated. Mass ismeasured in slugs and distance is measured in feet.

1. A lamina in the shape of the rectahgular region bounded by the lines r : 3 and y : 2, arrd the coordinate axes. Thearea density at any point is ry2 slugslfP.

2. A lamina in the shape of the region in the first quadrant bounded by the parabola A : * , the line y : 1. , and the y axis.The area density at any point is (x * y) slugs/ff.

3. A lamina in the shape of the region bounded by the parabola f :8y, the line y:2, and the y axis. The area densityvaries as the distance from the line y : -1.

A lamina in the shape of the region bounded by the curve y : e', the line l: L, and the coordinate axes. The areadensity varies as the distance from the r axis.

A lamina in the shape of the region in the first quadrant bounded by the circle I * y': a2 arrd, the coordinate axes.The area density varies as the sum of the distances from the two straight edges.

A lamina in the shape of the region bounded by the triangle whose sides are segments of the coordinate axes and theline 3r I 2y :18. The area density varies as the product of the distances from the coordinate axes.

A lamina in the shape of the region bounded by the curve y: sin r and the r axis from r : 0 to x: r. The area den-sity varies as the distance from the r axis.

A lamina in the shape of the region bounded by the curve y : fE and the line t : x. The area density varies as thedistance from the y axis.

A lamina in the shape of the region in the first quadrant bounded by the circle f * y' :4 and the line r * y :2.Thearea density at any point is xy slugslfP.

A lam ina in theshapeo f the reg ionboundedby thec i r c l e f l y ' : 1 and the l i nes r : l andy : l .Thea readens i t yat any point is xy slugslfP.

In Exercises 11 through 16, find the moment of inertia of the given homogeneous lamina about the indicated axis if thearea density is p slugs/ff, and distance is measured in feet.

11. A lamina in the shape of the region bounded by 4y : 3x, x: 4, and the x axis; about the r axis.

12. The lamina of Exercise 11; about the line r: 4.

L3. A lamina in the shape of the region bounded by a circle of radius a units; about its center.

14. A lamina in the shape of the region bounded by the parabola f :4- 4y and, the r axis; about the x axis.

15. The lamina of Exercise 1.4; about the origin.

16. A lamina in the shape of the region bounded by a triangle of sides of lengths a ft, b ft., and c ft; about the side of lengtha ft.

9.

10.


In Exercises 17 through 20, find for the given lamina each of the following: (a) the moment of inertia about the r axis; (b)the moment of inertia about the y axis; (c) the radius of gyration about the r axis; (d) the polar moment of inertia.

17. The lamina of Exercise 1.. 18. The lamina of Exercise 2. 19. The lamina of Exercise 7. 20. The lamina of Exercise 8.

21. A homogeneous lamina of area density p slugs/ff is in the shape of the region bounded by an isosceles trianglehaving a base of length b ft and, an altitude of length ft ft. Find the radius of gyration of the lamina about its line ofsymmetry.

22. A lamina is in the shape of the region enclosed by the parabola ! :2x - f and the r axis. Find the moment of inertiaof the lanina about the line y: 4 if the area density varies as its distance from the line y : 4. Mass is measured inslugs and distance is measured in feet.

2I.4 THE DOUBLE INTEGRALIN POLAR COORDINATES

Figu re 21 .4.1

We now show how the double integral of a function on a closed regionin the polar coordinate plane can be defined. We begin by consideringthe simplest kind of region. Let R be the region bounded by the rays0 : c and g : B and by the circles r : a and r : b. Then let A be a parti-tion of this region which is obtained'by drawing rays through the poleand circles having centers at the pole. This is shown in Fig. 2'1..4.1.. Weobtain a network of subregions which we call "crJrved" rectangles. Thenorm llAll of the partition is the length of the longest of the diagonals ofthe "curyed" rectangles. Let the number of subregions be n, and let {r4be the measure of area of the ith "curved" rectangle. Because the areaof the ith subregion is the difference of the areas of two circular sectors,we have

AA - +riz (0i - 0r-r) - trr-rz (0t - 0r-r)

: t ( r t - r i - ) ( r r* r i - ) (0 i - f l i - r )

Letting 7i: t(k l rr-) , Ltr : rt - ri-r, and A,0 - 0i - 0a-1, we have

LtA: 11 Llr L10

Take the point (71, 0i) in the ith subregion, where 0*r =di = 01, and formthe sum

|rr',,E,) LtA: ,i f G,,a,)Fi a^s L,o

It can be shown that'it 1is continuous on the region R, then thethe sum in (1), as llAll approaches zero, exists and that this limitthe double integral of f on R. We write either

(1 )

limit ofwill be

( r i , o r )

Observe that in polar coordinates, dA- r dr d0.

r : 6 z ( 0 )

r :

i , 0 i )

(o) r

( r

6 ,

21 ,4 THE DOUBLE INTEGRAL IN POLAR COORDINATES 1023

The double integral can be shown to be equal to an iterated integralhaving one of two possible forms:

f(r , 0) dA: f(r, 0)r dr d0 f ( r ,0 ) r d0 dr

We can define the double integral of a continuous function / of twovariables on closed regions of the polar coordinate plane other than theone previously considered. For example, consider the region R boundedby r: dr(O) and r:6"(0), where @r and S2 are smooth functions, andby the lines d : a and e: B. See Fig. 21..4.2. In the figure, 6t@) - 6r@)for all d in the closed interval lo, Fl. Then it can be shown that the doubleintegral of / on R exists and equals an iterated integral, and we have

f ( r , 0 ) dA : f(r , 0)r dr d0

rTLI

rr;::

LT::

(4)

r :

F igu re 21 .4.2

r i

r : r .t - L

o

Figu re 21 .4.3

o : o i - t

0 : o t

( r i , o i )

o : o i - . t0 : X { r )

0 : * o

If the region R is bounded by the cun/es

Xr and Xz are smooth functions, and byshown in Fig. 21.4.3, where Xt?) = Xr!)la , b f , then

(5)

0 - Xt?) and 0 -

Xr?), wherethe circles r: a and r- b, asfor all r rn, the closed interval

LI

LIf ( r , 0 ) dA: f(r , 0)r d0 dr

We can interpret the double integral of a function on a closed regionin the polar coordinate plane as the measure of the volume of a solid byusing cylindrical coordinates. Figure 2'1..4.4 shows a solid having as itsbase a region R in the polar coordinate plane and bounded above by thesurface z: f (r, 0) where / is continuous on R and f (x, y) > 0 on R. Takea partition of R giving a network of n "crlrued" rectangles. Construct then solids for which the ith solid has as its base the lth "cvwed" rectangleand as the measure of its altitude f (fo, 0) where (7r, 0r) is in the ith sub-region. Figure 21..4.4 shows the ith solid. The measure of the volume ofthe ith solid is

f ju,0) AoA: f( l i ,0)n Alr Lg

The sum of the measures of the volumes of the n solids is

(6)

n

sLJi : l

I f V i s

f (7r, |r)Fi Air Ai?

the measure of the volume of the given solid, then

0 - xr(r)

z : f ( r , o)

EXAMPLE L: Find the volume

the solid in the first octant

solurroN: The solid andformula (7) with f (r, 0) :

the ith element are shownr, we have, where V cubic

in Fig. 21.4.5. Usingunits is the volume


bounded by the cone z - r andthecy l i nde r r -3s in0 .

z

t :L ,

0 : O

Figure 21.4.5

EXAMPLE 2: Find the mass ofthe lamina in the shape of theregion inside the semicircle r -

a cos 0, 0 < 0 = tn, and whosemeasure of area density at anypoint is proportional to the mea-sure of its distance from the pole.The mass is measured in slugsand distance is measured in feet.

Figure 21.4.6

of the

7i L,ir Lrfl

dr d0

rz dr d0

de

sins 0 d0

--9 cos 0+ 3 coss

- 5

The volume is thereforc 6 cubic units.

solurroN: Figure 21.4.5 shows a sketch of the lamina and the I'th"crrwed" rectangle. The area density at the point (r, 0) is kr slugslff ,where k is a constant.lf M slugs is the mass of the lamina, then

M _

: +kas

: &kag

Therefore, the mass is 3 kas slugs.

en

lima l l -

-I'zrl2

0 3

'nl27r lz

0

ft

J o

ve

lil l a

Lt,t,9

gr

:

v

sln 0

d,

, f i

r,

r,

l 3t t

Jo

fi

)li,n

i :7

r3

so

\o f

n 2

- o E

r2

f,[+

7t

ImA l l -

-It

kag

1il la-

Jn

it<

t(= l

2 d r

'a co

0

n12(

iin

f,r

cos

>' i : l

'cr2

,tr['i,

o i

kr

'12 t

')Ii A,ir A,$

le

r2 dr d0

s 3 0 d 0

1nl- +s i n30 fJ o

(k7

rd '

l s d

col

e -

7f

EXAMPLE 3: FiNd thc

mass of the lamina inSOLUTIoN: Let the

lamina be i and y,axis and the y axis

coordinates of the centeris custom dr4, the r axis isLn axis. Let the cartesian

center ofExample 2.

cartesianwhere, dsalong the

of maalongcoordi

21.4 THE DOUBLE INTEGRAL IN POljR COORDINATES 1025

resentation of the point (r1, 0r) be (ir, !).Then if M" slug-ft is the mo-ment of mass of the lamina with respect to the x axis,

M":,fffilo iv,ttnln a,y a,g

Replacing |rby it sin 0r, we get

M": .ti.r.n i. *rlt sin o1 A1r A1o' Iltl l 'ofi

: [ [ k f s i n o i t r i t oJnJ

lTl2 faco60: f Jo Jo

tBs in9d r i l |

ft12- lkaa I cosa 0 sin 0 dAJ O

- -fikaa.or' 0lo"Jo

:,itka4

If M, slug-ft is the moment of mass of the lamina with respect to the yaxis, then

t,: ,,Til' 2r,<ooto a,rr L$

Replacing fuby rlcos Oil we have

Mr: tim i. tr,t cos 0r Arr Ar0" llall-o ;!!

: I lkf cos o itr doJnJ

filz lacoA0:k Jo Jo

f cos 0 ilr il|

It12:*kaa Jo

cosi 0 il9

: lkaafsin e - & sins o + + rir,t o];"

: &ka4

Therefore,

+ _Mu _ {tka4 _ 3* : fr :w:5o


Ms _#kao 9Y : M : & k a s : 4 o o

Hence, the center of mass is at the point (Ea, #a).

The area of a region in the polar plane can be found by double inte-gration. The following example illustrates the method.

the ith "cufrred" rec-then

Sometimes it is easier to evaluate a double integral by using polarcoordinates instead of cartesian coordinates, Such a situation is shownin the following example.

we have

SOLUTION: Figure 2'1,.4.7 shows the region andtangle. If A square units is the area of the region,

A- l im $ aoAlltll-o

"?:t

: lim 9. r, Air L,gl la l l - o ,? : t

r r: | | r d r d |

JnJfnlS fsin g0

: l I r d rdT, J o J o

- + f'' sin2 so do

- +o -# sin 6of"''Jo

: frni

Hence, the area is #n square units.

soLUTroN: Because Jcz * y' : y2, and dA: r dr d0,

r f f f| | s-(tz*u,) dA- | | e-r, r dr d0

JnJ JnJ

-- f'' [" ,-" r dr doJ O J O

- -+ r"'' l r-,'f" d0- J o L J o

f rrl2= - + | ( e - o ' - 1 ) d 0

J O

: +n(l - e-o')

EXAMPLE 4: Find by doubleintegration the area of the regionenclosed by one leaf of the roser - sin 30.

0 : 0 i

(,, ,0 -

o 0 : O

Fig u re 21 .4.7

Evaluate the double

0 , )

0 i-t

in the firstd by thethe co-

1S

dend

dA

r R)un2 A <

ronbor= a 2

,-1rz*U2)

e regioand bry ' : ,axes.

g-(tz+

le regt and* 7 t 2 =

PLE

ral'*l 'e t hrantx:2 -

ate

rtegr€

rJn,

rhereuadreircle rrdina

XAMP

ntegri

rJn

rhere

luadriircle :,rdina

EX

in

wqrci:OI

21.4IHE DOUBLE INTEGFAL IN POLAR COORDINATES 1027

Exercises 21.4

In Exercises 1 through 5, use double integrals to find the area of the given region.

1. The region inside the cardioid r:2(l * sin 0). 2. One leaf of the rose r: a cos 20.

3. The region inside the cardioid r: a(l * cos 0) and outside the circle r: a.

4. The region inside the circle r: 1 and outside the lemniscate 12: cos 20.

5. The region inside the large loop of the limagon r:2 - 4 sin d and outside the small loop.

6. The region inside the limagon t:3 - cos 0 and outside the circle r: 5 cos d.

In Exercises 7 through 12, find the volume of the given solid.

7. The solid bounded by the ellipsoid z2 19r2:9'

8. The solid cut out of the sphere z2 * 12: 4by the cylinder r: 1.

9. The solid cut out of the sphere z2 I t2 : 16 by the cylinder r: 4 cos 0.

10. The solid above the polar plane bounded by the cone z:2r artd. the cylinder r:1 - cos 0.

11. The solid bounded by the paraboloid z: 4 - 12, Ihie cylinder r: L, and the polar plane.

12. The solid above the paraboloid z: r2 and below the plane z:2r sin 0-

In Exercises 13 through 19, find the mass and center of mass of the given lamina if the area density is as indicated. Mass

is measured in slugs and distance is measured in feet.

13. A lamina in the shape of the region of Exercise 1. The area density varies as the distance from the pole.


15. A lamina in the shape of the region inside the limagon r:2- cos 0. The area density varies as the distance from

the pole.

15. A lamina in the shape of the region bounded by the limagon r:2* cos 0, 0 - 0 < tr, and the polar axis. The area

density at any point is k sin 0 slugs/ff.

17. The lamina of Exercise 15. The area density at any point is kr sin 0 slugsiff'



In Exercises 20 through 24, frnd. the moment of inertia of the given lamina about the indicated axis or point if the area

density is as indicated. Mass is measured in slugs and distance is measured in feet.

20. A lamina in the shape of the region enclosed by the cirde r: sin 0; about the *z axis. The area density at any point is

k slugs/ff.

21. The lamina of Exercise 20; about the polar axis. The area density at any point is k slugs/fP.

22. A lamina in the shape of the region bounded by the cardioid r: a(l - cos 0); about the pole. The area density at anypoint is k slugs/ff.

23. A lamina in the shape of the region bounded by the cardioid t : a(7 f cos 0) and the circle r: 2a cos 0; about thepole. The area density at any point is k slugs/ff.

24. A lamina in the shape of the region enclosed by the lemniscate 12: a2 cos 20; about the polar axis. The area densityvaries as the distance from the pole.

25. A homogeneous lamina is in the shape of the region enclosed by one loop of the lemniscate 12 : cos 20. Find the radius

of gyration of the lamina about an axis perpendicular to the polar plane at the pole.


25. A lamina is in the shape of the region enclosed by the circle r: 4, and the area density varies as the diptance from thepole. Find the radius of gyration of the lamina about an axis perpendicular to the polar plane at the pole.

27. Evaluate by polar coordinates the double integral

[ [ ,**",4aJnJ

where R is the region bounded by the circles I * y, :7 and f -l y\:9.

28. Evaluate by polar coordinates the double integral

[ [ -Laal*l 1wt-77-'

where R is the region in the first quadrant bounded by the circle rP * y":1 and the coordinate axes.

29. In advanced calculus, improper double integrals are discussed, ".ra [.- f*' f(x, y) dx dv isdefined to be

f h f h

- J o J o

l im | | f (x ,y )dxdyh - + @ J o J o

Use this definition to prove tnut f*- e-" ilx: +\/; by doing the following: (a) Show that the double integral in Ex-J O

ample 5 can be expressed ^l[ r-" 'a*] ';b) because |[*- ,-" 'dt] ' : 5* |[ ' .""drl ' ,rr" theresultof Example5LJo I '

LJo J o -+ -LJo Ito obtain the desired result.

2I.5 AREA OF A SURFACE

Figu re 21 .5.1

The double integral can be used to determine the area of the portion ofthe surface z: f(x, y) that lies over a closed region R in the ry plane. Toshow this we must first define what we mean by the measure of this areaand then obtain a formula for computing it. We assume that / and its firstpartial derivatives are continuous on R and suppose furtherthatf (x,y) > 0on R. Let A be a partition of R into n rectangular subregions. The ith rec-tangle has dimensions of measures Ag and {y and an area of measureAy'. Let (€i, y) be any point in the ith rectangle, and at the pointQ(ti, fi, fGo, y)) on the surface consider the tangent plane to the sur-face. Project vertically upward the ith rectangle onto the tangent planeand let {o be the measure of the area of this projection. Figure 21.5.1shows the region R, the portion of the surface above R, the lth rectangularsubregion of R, and the projection of the fth rectangle onto the tangentplane to the surface at Q. The number A;cr is an approximation to the mea-sure of the area of the piece of the surface that lies above the ith rectangle.Because we have z such pieces, the summation

n

) A,oi : l

is an approximation to the measure a of the area of the portion of thesurface that lies above R. This leads to defining o as follows:

n

lim ' A;0l l a l l - o e

Q ( f ; , r i , f ( t i , y ) )

z : f ( x , y )

( t i , Y i )

21.5 AREA OF A SURFACE 1029

We now need to obtain a formula for computing the limit in Eq. (1).To do this we find a formula for computing Aio as the measure of the areaof a parallelogram. For simplicity in computation we take the point (ily)in the lth rectangle at the comer (xi-r, yi-). Let A and B be vectors havingas representations the directed line segments having initial points at Qand forming the two adjacent sides of the parallelogram whose area hasmeasure A;c. See Fig.21.5.2. Then A;o: lA x Bl. Because

, A - A ; r i * f " ( { r , y ) A , 6 k

and

n: LAi * fu(t,, y) Ailk

it follows that

i k l0 f " ( t , , f i ) A# l

L'i l f"Gr,Tr) LralFigu re 21 .5.2

21.5.1 Theorem

EXAMPLE L: Find the area of thesurface that is cut from the cyl-inder x2 + 22 : L5 by the planesx - 0 , x - 2 , A : 0 , a n d y : 3 .

- - Aix LiUf "(tr, T)i - Lix Lilf o(€r, Di * A,p LiAk

Therefore,

L icr : lA x Bl :

Substituting from Eq. (2) into Eq. (1), we get

cr - liml la l ; - o l : r

(2 )

Lfc Lil

This limit is a double integral which exists on R because of the continuityof f , and fu on R. We have, then, the following theorem.

derivatives are continuous on the closedis the measure of the area of the sur-

solurroN: The given surface is shown in Fig. 21..5.3. The region R is therectangle in the first quadrant of the xy plane bounded by the lines r: 2and y :3. The surface has the equation * * z2: 15. Solvingfor z, we getz: {G77. Hence, f(x, y): t/(-p. So if o is the measure of thearea of the surface, we have from Theorem 21..5.L

1 d x d y

dx dy

(3)

* f o'(tr, Yt)

* fo'(x'

1 030 MULTIPLE INTEGRATION

ExAMPLE 2: Find the

paraboloid z -- x' + y'plane z -- 4.

: f f +dxdy- JoJo tfio - xz

area of thebelow the

The surface area is therefore 2n square

solurroN: Figure 2L.5.4 shows the given surface. From the equation ofthe paraboloid we see that f (r, y) : * * y2. The closed region in the xyplane bounded by the circle * * y': 4 is the region R. If o square unitsis the required surface area, we have from Theorem 21.5.L

t f

cr: | | lf ,'(x, y) * fu'@, y) + | dx dyJRJ

t !: l I V4( * *y r )+7dxdy

JNJ

Because the integrand contains the terms 4(f + y'), the evaluation ofthe double integral is simplified by using polar coordinates. Ttren I* y' : 12, and dx ily : dA: r dr d0. Furtherlnore, the limits for r are from0 to 2 and the limits for 0 are from 0 to 2rr. We have then

t fo: | | \/4r2 l'1, r dr ilo

JnJf2n f2: f | ! 4 r2 * \ r d rd0

J O J O

f2n f -12

: I l # (+ r r+ t )B tz l d0J o L l o

: t r ( tz t /a -D

Hence, the area of the paraboloid below the given plane is*zr(tZtE - t)square units.

solurroN: The hemisphere is shown in Fig. 21..5.5. Solving the equationof the sphere for z and setting this equal to f(x, y), we get

f (x, y) : \F7= y'

Because f,(x, y):-rl{7=7=7, flrd fr(x, y):-y1lFF=!,

* V ,

Figure 21.5.4

EXAMPLE 3: Find the area of the

top half of the sphere x' * y'* z2 -- g2.

Figu re 21 .5.3

wetheint

n

21.5 AREA OF A SURFACE 1031

defined on the circle x2 + y' : a2 which isin the xy plane. Furtherrnore, the double

notn R(3)

areegicEq,

d x t

;1oq.

,d fothe Ifrom

ndthr re

m l

r o f l

of t.e d f

f"y oine

a

rat Iarytair

thrdi

rb

t e lfl

o

10teI

is

w dy

which is improper because the integrand has an infinite discontinuityat each point of the boundary of R. We can take care of this situation byconsidering the region R' as that bounded by the circle rP I y' : b2, whereb 1a, and then take the limit as b---> a-. Furthermore, the computationis simplified if the double integral is evaluated by an iterated integralusing polar coordinates. Then we have

a: - r d | d rl / a 2 - 1 2 '

:Zrra !':-I

Figure 21 .5 .5

:2rnl im l -"rp) 'b - a - L l o

:2na lt_T_ F\ffi * al

:2rra2

The area of the hemisphere is therefore 2na2 square units.

Consider now the curve y: F(r) with a < r < b, F positive onla,bland F' continuous on fa, b]. If this curve is rotated about the r axis, weobtain a surface of revolution. From Sec.18.7 an equation ofthis surface is

y '+ z " : [ f (x ) ] ' z (4 )

Figure 21.5.5 shows the surface of revolution. In the figure we have takenthe xy plane in the plane of the pape\ however, we still have a right-handed system. We wish to obtain a formula for finding the measure ofthe area of this surface of revolution by using Theorem 21..5.1 From prop-

r erties of symmetry, the measure of the area of the surface above the xzplane and in front of the xy plane is one-fourth of the measure of the areaof the entire surface. Solving Eq. (4) for z (neglecting the negative squareroot because z = 0), we get f(x, y) : !E6JP=V. The region R in thexy plane is that bounded by the x axis, the curve y : F(x), and the linesx: a and r : b. Computing the partial derivatives of f , we obtain

f,(x,y):ffi ful,y):a@fuWe see that f,(x, y) and fu@, y) do not exist on part of the boundary of

Figu re 21 .5.6


ExAMPLE 4: Find the area of theparaboloid of revolution gen-erated by revolving the top halfof the parabola y' : 4px, with0 < .r < h, about the x axis.

y : -F(r) and when y : F(r) ). The double integral obtained(3) is

^! f f i+f f i+ ldYdx

dy dx

This double integral is improper because the integrand has an infinitediscontinuity at each point of the boundary of R where /: -F(r) andy: F(x). Hence, we evaluate the double integral by an iterated integralfor which the inner integral is improper.

c,-^l:[t, ',ff if" 'ff i]dxwhere

I:",ffi:1,_Tf",_,ffi: t,_T lrir,-' h]',(r)-e: t t_Tsin- , (1 -6)

(s)

:L r

Therefore, from Eq. (5) we have

q:2rr [' ,G){FTil]' * L itxJ a

We state this result as a theorem, where F is replacedby f.

21.5.2 Theorem Suppose that the function / is positive on fa, b] and /' is continuous onla, bf .Then if o is the measure of the area of the surface of revolutionobtainedbyrevolving the curvey: f(x),witha < x < b,abouttheraxis,

solurroN: The paraboloid of revolution is shown in Fig. 2l.S.T. Solvingthe equation of the parabola for y (y = 0), we obtain !-2prnyrn. So ifo square units is the area of the surface, from Theorem2'1..5.2, with f(r) :2prtzyrtz, we have

: t IF(x) \ / lF ' ( iP+rJ^JW

2pttzTtttz dx

21 .5 AREA OF A SURFACE 1033

- AnPtt' t

The area of the paraboloidsquare units.

of revolution is therefore &n(@ - p')

Figu re 21 .5.7

Exercises 2L.5

1. Find the area of the surface which is cut from the plane 2x * y * z: 4 by the planes x: 0, x: t, y :0, and y : 1..

2. Find the area of the portion of the surface of the plane 35x -l 1.6y l- 92:144 which is cut by the coordinate planes.

3. Find the area of the surface in the first octant which is cut from the cylinder f * y': g by the plane r: z.

4. Find the area of the surface inthefirst octantwhich is cutfromthe cone f *y':z2by the plane x*y:4.

5. Find the area of the portion of the surface of the sphere f * y' * zz : 4x which is cut out by one nappe of the coney 2 * * : * .

6. F indtheareaof theport ionof thesur faceof thespheref +y2*22:35 whichl ieswi th inthecyl inder f +yz:9.

7. Find the area of the portion of the surface of the sphere x' + y' * z2 : 4z which lies within the paraboloid f + !2 :32.

8. For the sphere and paraboloid of Exercis e 7, find the area of the portion of the surface of the paraboloid which lieswithin the sphere.

9. The line segment from the origin to the point (a, b\ is revolved about the x axis. Find the area of the surface of thecone generated.

10. Derive the formula for the area of the surface of a sphere by revolving a semicircle about its diameter.

11. Find the area of the surface of revolution obtained by revolving the arc of the catenary y : a cosh(r/a) from r: 0 tox: a about the y axis.

12. Find the area of the surface of revolution obtained by revolving the catenary of Exercise 11 about the x axis.

13. The loop of the curve lSyz : x(6 - x)2 is revolved about the r axis. Find the area of the surface of revolution generated.

14. Find the area of the surface of revolution generated by revolving the arc of the curve A : ln x from r :1 lo x: 2 aboutthe y axis.

15. Find the area of theportionof theplaner: zwhichliesbetweentheplanes/:0and A: Sand,withinthehyperboloid9 f -4y2*1622 :144 .

15. Find the area of the surface cut from the hlryerbolic paraboloid y' - * : 6z by the rylinder f + yz : 35.

'103'f MULTIPLE INTEGRATION

21.5 THE TRIPLE INTEGRAL The extension of the double integral to the triple integral is analogous tothe extension of the single integral to the double integral. The simplesttype of region in R3 is a rectangular parallelepiped which is bounded bysix p lanes: x : a1t x : a2t A:br , ! :bz, Z: c1t and z: cr , wl th a1 1 a2,br 1br, and c1 ( cr. Let f be a function of three variables and supposethat / is continuous on such a region S. A partition of this region is formedby dividing S into rectangular boxes by drawing planes parallel to thecoordinate planes. Denote such a partition by A and suppose that n isthe number of boxes. Let AlV be the measure of the volume of the rth box.Choose an arbitrary point (fi, yt, p) in the ith box. Form the sum

= f lz ,

Let

x

Figu re 21 .6.1

EXAMPLE

integralEvaluate the triple

sin yz dV

the norm of the partition approaches zero for any choices of the points(€r, ̂ yr, lr) if f is continuous on S. Then we call this limit the triple integralof f on R and write

limI la l I ' o

Analoggral, the trirectangularwe have

>-f (€u ru Pt) LoV

Refer to Fig. 2'1,.6.L, which showswith the ith box. The norm llAll ofdiagonal of the boxes. The sums

SOLUTION:

sl^n yz

( 1 )

the rectangular parallelepiped togetherthe partition is the length of the longestof the form (1) will upproach a limit as

yz dz dy dx

x(l - cos *nV) dy dx

reS,

rnte-; then S ,

ins l

ted jr S i ru s ous

erathenuou

-ite:Whtinu

:-)vti

,Y , z ) dv

ing equal to a twiceice-iterated integral.rbove, and f is con'

f t

l ltl

bed

f(T \X ,

beithrid a

t

scri

in

luades

L,V

rbles e qe d (

)

u

Tt, 1-ci

a d oegralelepil

rs tol intarall

€,,

; t oint

if(ousPlepa

n

i : l' ) Ai

rubli s e

ped

[-'cos v"]i' av a*

dv: rff r'' *, sin

: f ,(r-*sin t*) lno' ' o*

: f .(T-""sin T) *

if S is the rectangular parallel-epiped bounded by the planes,c -- zr ,'U : tr , z -- *o, and thecoordinate planes.

( o t , b t , c t )

21.6 THE TRIPLE INTEGRAL 1035

x2

2

7r-4

z : F z @ ' Y )

We now discuss how to define the triple integral of a continuousfunction of three variables on a region in R3 other than a rectangularparallelepiped. Let S be the closed three-dimensional region which isbounded by the planes x:a and x:b, the cylindets y: dr(r) and

A : 6r(x), and the surfaces z = F r(x, y) and z : Fr(x, /), where the func-tions @1, 6r, Ft, and F2 are smooth (i.e., they have continuous derivativesor partial derivatives). See Fig. 21,.6.2. Construct planes parallel to the co-ordinate planes, thereby forming a set of rectangular parallelepipeds thatcompletely cover S. The parallelepipeds which are entirely inside S or onthe boundary of S form a partition A of S. Choose some system of num-bering so that they are numbered from L to n. The norm llAll of this parti-tion of S is the length of the longest diagonal of any parallelepiped belong-ing to the partition. Let the measure of the volume of the ith parallelepipedbe {V. Let f be a function of three variables which is continuous on S andlet (fi, Ti, F) be an arbitrary point in the ith parallelepiped. Form the sum

2 f rci,Tr, F) A,Vi: l

If the sums of form (2) have a limit as llAll approaches zero, and if this limitis independent of the choice of the partitioning planes and the choices of-the arbihary points ((u yu pi) in each parallelepiped, then this limit iscalled the triple integral of / on S, and we write

It can be proved in advanced calculus that a sufficient condition for thelimit in (3) to exist is that / be continuous on S. Furthermore, under thecondition imposed upon the functions 6r, 6r, Fr, and F2 that they besmooth, it can also be proved that the triple integral can be evaluated bythe iterated integral

f(x, y, z) dz dy dx

just as the double integral can be interpreted as the measure of thearea of a plane region when /(r, U): L on Rr, the triple integral can be in-terpreted as the measure of the volume of a three-dimensional region. If

(2)

(3)

H,::I:::

9 l

x , 0

Figu re 21 .6.2

\Y : 6z@)

r-1036 MULTIPLE INTEGRATION

EXAMPTN 2:gration theExample 3

Find by triple inte-volume of the solid of

in Sec. 2L.2.

f (x, y , z) :, 1 on S, then Eq. (3) becomes

liml l a l l - o

and the triple integral is the measure of the volume of the region S.

solurroN, If V cubic units is the volume of the solid, then

liml l a l l - o

where S is the region bounded by the solid. The z limits are from 0 (thevalue of z on the ry plane) to f * 4yz (the value of z on the elliptic parab-oloid). Th" y limits for one-fourth of the volume are from 0 (the value ofy on the xzplane) tot:y'T=P (the value of y on the cylinder). The rlimitsfor the first octant are from 0 to 2. We evaluate the triple integral in (4)by an iterated integral and obtain

Hence,

V : 4 (x' * 4y') dy dx

The right side of Eq. (5) is the sametained in Example 3 in Sec. 2'1..2, andthe same.

(s)

twice-iterated integral that we ob-so the remainder of the solution is

E1AMpLE 3: Find the volume ofthe solid bounded by the cyl-inder x2 + y' :25, the planex + y + z - 8 and the xy plane.

solurroN: The solid is shown in Fig. 21,.6.3. The z limits for the iteratedintegral are from 0 to 8 - x- y (the value of z on the plane). The y limitsare obtained from the boundary region in the xy plane which is the circle* + y':25. Hence, the y limits are from -\/25=7 to t/x=E The rlimits are from -5 to 5. If V cubic units is the required volume, we have

v- rim f a,u: f fl *l l a l l - o f i J f

:f,ml:-"-"dzdvdxfs f\ffi: J-rJ -'*-,, (8 - x - Y) dY dx

fb f l rE"e: J_, L(r

- x)y - tr,)_*ud*

21.6 THE TRIPLE INTEGRAL 1037

(8- i t f f i dx

:16 f \E4dx* f \84(_2x) dxJ-s J -s

: L6(ix\84 + z*sin-r Ex) + |eS - xzr',r]:,

:200n

The volume is therefore 200n cubic units.

F igure 21 .6 .3

EXAMPLE 4: Find the mass of

the solid above the xY Planebounded by the cone 9xz * z2: Y'and the Plane Y : 9 if the mea-

sure of the volume densitY at any

point (x, y , z) in the solid is

proportional to the measure of

the distance of the Point from the

xy plane.

solurroN: Figure 2']-..5.4 shows the solid. Let M slugs be the mass of

the solid, and let the distance be measured in feet. Then the volume

density at any point (r, y, z) in the solid is kz slagslfts, where k is a con-

stant. Then if (tr, yt, p1) is any point in the ith rectangular parallelepiped

of the partition, we have

M- l im Zkpl la l l - o i : 1

fs fals h@ifr:ZkJ ,J , J ,

zdzdxdY

:zk [n v ' dy -Tk- e J o

The mass is therefore ryk slugs.

( ( i , y i , p i )

Figu re 21 .6.4


Exercises 21,.6

In Exercises 1 through 4,evaluate the iterated integral.

- f f F * f + u z r a 1 2 f u , f r n r1' JrJ, Jr"

x dz dY dx 2' JrJ" J, Ye' dz dx dY

In Exercises 5 through 10, evaluate the triple integral.

r r f5.

J J J y dV if S is the region bounded by the tetrahedronformed by the plane t2x + 20y * l5z: 50 and the coordinate

"or! dy dx dz 4 fff '#dxdzdy

oiur,ur.

6. I i l @z+22) dvirsis the same region as in Exercise 5.

l t l7. | | | z dV if S is the region bounded by the tetrahedron having vertices (0, 0, 0), (7,1, O), (1, 0, 0), and (1, 0, 1).

J J J

i,,8. | | | yz dV if S is the same region as in Exercise 7.

J J J -

i,,9. l l t (n+32) dV i t S is the region bounded by the cy l inder f lz2:9 and the p lanes x*y:3, z :0, and y:0,

J J Js

above the ry plane.

r f t10 . l f f xyz i [V , i f S i s thereg ionboundedbythecy l inders * *y " :4 and f I z2 :4 .

J J J

tr, Uir.ire, 11 through 21., use triple integration.

11. Find the volume of the solid in the first octant bounded below by the xy plane, above by the plane z: y, andlaterallyby the rylinder y': r and. the plane x: 1..

12. Find the volume of the solid in the first octant bounded by the rylinder f + z2: 16, the plane r * y:2, and the threecoordinate planes.

13. Find the volume of the solid in the first octant bounded by the cylinders * * yr: 4 and x2 + 2z: 4, and. the three co-ordinate planes.

14. Find the volume of the solid bounded by the elliptic cone 4* I9y" - 3622 :0 and the plane z : 1.

15. Find the volume of the solid above the elliptic paraboloid 3* * y2: z and below the cylinder f * z:4.

15. Find the volume of the solid enclosed by the sphere * * y, * 22: a2.

17. Find the volume of the solid enclosed by the ellipsoid

* u 2 2 2_ + L _ l _ _ : 1a2 b2 c2

Find the mass of the solid enclosed by the tetrahedron formed by the plane 100r + 25y * 762: 400 and the coordinateplanes if the volume density varies as the distance from the yz plane. The volume density is measured in slugs/ftr.

Find the mass of the solid bounded by the rylinders x: z2 andA: *, and the planes x:l,y:0, andz:0. The vol-ume density varies as the product of the distances from the three coordinate planes, and it is measured in slugslfd.

18.

L9,

20. Find the mass of the solid bounded by the surface z: 4 - 4xz - !2 and the xyof the solid is p slugs/ff and p:3"111'

2L. Find the mass of the solid bounded by the sutf.ace z - xy, and the planes r:

at any point of the solid is p slugs/ff and p: 3V* + y2.

21.7 IHE TRIPLE INTEGRAL IN CYLINDRICAL AND SPHERICAL COORDINATES 1039

plane. The volume density at any point

L, y - 1,, and z - 0. The volume density

If a region S in R3 has an axis of symmetry, triple integrals on S are easier

to evaluate if cylindrical coordinates are used. If there is symmetry withrespect to a point, it is often convenient to choose that point as the originand to use spherical coordinates. In this section we discuss the triple inte-gral in these coordinates and apply them to physical problems.

To define the triple integral in cylindrical coordinates we constructa partition of the region s by drawing planes through the z axis, planes

perpendicular to the z axis, and right-circular cylinders having the z axis

ls axis. A typical subregion is shown in Fig. 2L.7.1. The elements of the

constructed partition lie entirely in S. We call this partition a cylindricalpartition. The measure of the length of the longest "diagonal" of any of

the subregions is t:ne norm of the partition. Let n be the number of sub-

regions of the partition and A1v be the measure of the volume of the ithsubregion. The measure of the area of the base is li L1r A,60, whete 7i:

t?r * ri-). Hence, if. A,A is the measure of the altitude of the lth subregion,

L4r A^i0 Liz

2I.7 THE TRIPTE INTEGRALIN CYLINDRICAL AND

SPHERICAL COORDINATES

( , , , 0 , , , , )

II 1

z n0

x

F igu re 21 . 7 . 1

L1V -

Let f b" aChoose a

Note that in cylindricalthe triple integral in (2)suppose that the region

coordinates, dV - r dr d0 dz. We can evaluate

and (3) by an iterated integral. For instance,

S in R3 is bounded by the planes 0 : a and

function of r, 0, and z, and suppose that / is continuous on S.

point (Vr, 0r, Zi) in the i th subregion such that 0r-r, = 0i = 0i and

zr-t < Zi s zr Form the sum

f Gr, 0r, 2r) LrV - f (7r, 0i, 2r)7, A^1r Li? Aiz

As the norm of A approaches zeto, it can be shown, under suitable con-

ditions on S, that the limit of the sums of form (L) exist. This limit is called

the triple integral in cylindrical coordinates of the function f on s, and we

write

n

i : l

n

i : l

( 1 )

(2)

(3)

1O4O MULTIPLE INTEGRATION

ExAMPLE 1: A homogeneoussolid in the shape of a right-circular cylinder has a radius of2 ft and an altitude of 4 ft. Findthe moment of inertia of thesolid with respect to its axis.

0:B, with a <F, by the cylinders /:I1(d) and r:tz(0), where 11and )t2 are smooth on [a, B] and Ir(0) < tq(0) f.or q < 0 < B, and, bythe surfaces z:F1(r,0) and z:Fr(r, g), where F1 and F2 are functions 1of two variables that are smooth on some region R in the polar planebounded by the curves r: I,a(0), r : \2(0), 0: d, and 0 : B. Furthermore,suppose that F{r, 0) < F2(r,0) for every point (r, g) in R. Then the tripleintegral can be evaluated by an iterated integral by the formula

I il ,r,, 0, z)r dr d0 dz: ff::[::: r(r, 0, z)r itz dr it. (4)s

There are five other iterated integrals that can be used to evaluate thetriple integral in (4) because there are six possible permutations of thethree variables t, 0, and z.

- -Triple integrals and cylindrical coordinates are especially useful infinding the moment of inertia of a solid with respect to ihe z uxis becausethe distance from the z axis to a point in the solid is determined by thecoordinate r.

solurroN; choose the coordinate planes so that the xy plane is the planeof the base of the solid and the z axis is the axis of the-solid. Fig:uu;e)l.z.zshows the portion of the solid in the first octant together with the ithsubregion of a rylindrical partition. using cylindrical coordinates andtaking the point (Fo, 0r, z1) in the ith subregion with k slugs/fp as thevolume density at any point, then if I, slug-ff is the tt o*urrt of inertiaof the solid with respect to the z axis, we hlve

There are six different possible orders of integration. Figure 21.7.2 showsthe orderdz dr d0.Using this order, we have

L: f f [ or, i tzr i trdo:4k f ' ' f [ '* i tzdrdoJ J J J o J o J o,s11 the first integration, the blocks are summed from z:0toz:4;theblocks become a column. In the second integration, the columns aresummed from r: 0 to r:2; the columns become a wedge-shaped sliceof the cylinder. In the third integration, the wedge-shaped-slice is rotatedfrom d : 0 to 0: Ini this sweeps the wedge about the entire three-dimen-sional region in the first octant. we multiply by 4 to obtain the entirevolume. Performing the integration, we obtain

t": tilk ["'' [' * a, ao :54k f'' ar: 32krJ o J o J o

Hence, the moment of inertia is 32kn slug-ff.

I ':rr1t, 2rru a,v: III krz dv

Figure 21.7.2

ExAMPLE 2: Solve Example L bytaking the order of integration as(a) dr dz d0; (b) d0 dr dz.

ExAMPLE 3: Find the mass of a

solid hemisphere of radius a ft if

the volume densitY at any Pointis proportional to the distance of

the point from the axis of the

solid.

21.7 THE TRIPLE INTEGRAL IN CYLINDRICAL AND SPHEBICAL COORDINATES 1041

soLUrIoN: (a) Figure 21.7.3 reptesents the otder drdzd0' It shows theblock summed from r: 0 to r: 2 to give a wedge-shaped sector' We thensum from z : 0 to z: 4 to give a wedge-shaped slice. The slice is rotated

from 0 : 0 to 0 = *n to cover the first octant. We have, then,

(b) Figure 2']...7.4 represents the oider d|ilrilz. It shows the blocks

summed from 0 : 0 to 0: tr to give a hollow ring inside the cylinder.

These hollow rings are summed from r: 0 to r: 2 to give a horizontal

slice of the cylinder. The horizontal slices are summed trom z: 0 to z: 4.

Therefore, we have

f d0 dr dz: 32kn

Figu rc 21.7.3 Figure 21.7.4

soLUrIoN: If we choose the coordinate planes so that the origin is at thecenter of the sphere and the z axis is the axis of the solid, then an equation

of the hemispherical surface above the xy plane is z:{72777-

Figure 21.7.5 shows this surface and the solid together wit! the ith sub-t"liot of a cylindrical partition. An equation of the hemisphere in rylin-drical coordinates is z: \/F7.lf (ti, ei, zl) is a point in the ith sub-region, the volume density at this point is kr6 slugs/ft3, where k is aconstant; and if M slugs is the mass of the solid, then

M: lim $ n, a,Yl la l l -o f i

t,: nn f'' ffi: r dr dz do:32kn

: lil krdv-k

I:l:13 r2dzdrdofzn fa: o Jo J ,

,21 f f i ' d rdo

fzn I: O Jo l-Lnr(az

- rz)Btz * tazrlprz * taosin

f2n-#kaan I d0J O

- + kAan'

The mass of the solid hemisphere is therefore tkann2 slugs.

1O:42 MULTIPLE INTEGRATION

Figure 21.7.5

ExAMPLE 4: Find the center ofmass of the solid of Example 3.

solurroN: Let the cartesian-coordinate representation of the center ofmass be (i

" , , Z) . Because of symmetry , i: !: 0. We need to calculateZ.,rt M,u slug-ft is the moment of mass of the solid with respect to the ryplane, we have

M,u: lig, i zt(ktt) Atvilail_o Gi

f f r: l l l kz r i lVJ J J

s

fzT fa f\/a'=Fz: k l , I z r z i l z d r i l |J o J o J o

fzn fa: tk I | (a2 - r2)r2 ilr do

J o J o

l2n- J3ka5 I deJ o

: frkaizr

Because M2: Mrn, we get

; _ Mri _ #ka,rr L6. -T:W:Go

The center of mass is therefore on the axis of the solid.at a distance of'l,6al75zr ft from the plane of the base.

( v i , o i , o )

We now Proceed to define the triple integral in spherical coordinates.

\---\.. --..t \ \

\

21.7 THE TRIPLE INTEGRAL IN CYLINDRICAL AND SPHERICAL COORDINATES 1043

A spherical partition of the three-dimensional region S is formed byplanes containing the z axis, spheres with centers at the origin, and cir-cular cones having vertices at the origin and the z axis as the axis. A tyPi-cal subregion of the partition is shown in Fig. 27.7.6.If A1V is the measureof the volume of the l'th subregion, and (Fo, 0r, dr) is a point in it, we canget an approximation to A1V by considering the region as if it were a rec-tangular parallelepiped and taking the product of the measures of thethree dimensions. These measures are p; sin 6, Lol, Fi LrQ, and Ap. Fig-ures 2'1,.7.7 and 21..7.8 show how the first two measures are obtained, andFigure 21.7.6 shows the dimension of measure dp. Hence,

LtV: ,'z sin {; Aip A$ A1Q

The triple integral in spherical coordinates of a function / on S is given by

Observe that in spherical coordinates, N: p2 sin $ dp d0 dS. The tripleintegrals in (5) or (6) can be evaluated by an iterated integral. Sphericalcoordinates are especially useful in some problems involving spheres,as illustrated in the following example.

pi s in$i

o

Figu re 21 .7 .6

(5)

p; sin $, V i L i 6

( i , , 0 , , 6 , )

Figu re 21 .7 .8

\\


ExAMPTE 5: Find the mass ofthe solid hemisphere of Example3 if the volume density at anypoint is proportional to the dis-tance of the point from thecenter of the base.

EXAMPLE 6: A homogeneoussolid is bounded above by thesphere p: a and below by thecone Q, - a, where 0

solurroN: rf (po, at, 6) is a point in the ith *bt"gior of u sphericalpartition, the volume density at this point is kpl slugslfts, where k is aconstant. lf M slugs is the mass of the solid, then

M- l im Zkp, L,Vl la l l -o i : - r

p3 sin 6 dp dO d0

sin O d0 d0

- $aakn

Hence, the mass of the solid hemisphere is taakzr slugs.

- It- is- interesting to compare the solution of Example 5 which usesspherical coordinates with what is entailed when using^cartesian coordi-nates. By the latter method, a partition of s is formed by dividing s intorectangular boxes by drawing planes parallel to the coordinate planes. If(!u Ti, g.) is any point in the lth subregion, and because p : {frV + i,then

M: ,trs, 9rt/€7 + yFTE n,vlltll-o

"?=t

: [[ [ rtF+j + zz avJ J J

,sfa f\/F-zz fla=uz-F: no JoJ" J,

l.F+T+z dx dy dz

The computation involved in evaluating this integral is obviously muchmore complicated than that using spherical coordinates.

21.7 THE TRIPLE INTEGRAL IN CYLINDRICAL AND SPHERICAL COORDINATES 1045

if I, slug-ff is the moment of inertia of the given solid about the z axis,

then

; (pr sin 6)'k L,V

kp'sin2 + dV

Find the moment of inertia of thesolid about the z axis.

fa fzn- +ka ' l l s i n30d0d0J o J o

fa- ?ka'n Jo

sins O dO

l - . 1 "- ?ka\zr l -cos O + * coss 0 |

L l o

- + k a s n ( c o s 3 a - 3 c o s a * 2 )

moment of inertia of the solid about

#ka 'n (cos3 o -3cos a*2) s lug- f f

lit4 fa frcosl ^ J_ )- )a ltl4 fzcos9

['"'"t ,, cos 0 dz dr d0t J; JJ. t secs o dz dr d0 '' Jo J"",^,Jo

lnl4 lzacoso fzt lnt2 16 facsco

t. f:^ I:'""'["" o' "in S de dp dg n'

J,^J), ps sin2 0 sin @ dp d0 dS

5. Find the volume of the solid enclosed by the sphere x' * y' * z2 : a2 by using (a) cylindrical coordinates and (b) spheri-

cal coordinates.

- k fl: f

(p2 sin' Q)p2 sin Q dp do dO

Figu re 21 .7 .9

The the z axis is therefore

Exercises 21.,7

In Exercises L throu gh 4,evaluate the iterated integral.

6. If s is the solid in the first octant bounded by the sphere f * y" I z2 : 16 and the coordinate planes, evaluate the triple

integral III

*"dv by thtee methods: (a) using spherical coordinates; (b) using rectangular coordinates; (c) using

scylindrical coordinates.

In Exercises 7 through 10, use cylindrical coordinates'

7. Find the mass of the solid bounded by a sphere of radius fl ft if the volume density varies as the square of the distance

from the center. The volume density is measured in slugs/ff'

g. Find the mass of the solid in the first octant inside the cylinder * l- y' : 4x and un-der the sphere f t y' -f z2:76'

The volume density varies as the distance from the xy plane, and it is measured in slugs/ft3'

.bi;ii.. ;i


9. Find the moment of inertia with respect.to the z axis of the homogeneous solid bounded by the rylinder r: 5, thecone z: r, and the ry plane. The volume density at any point is k slugs/fts.

10. Find the moment of inerlia of the solid bounded by a right-circular cylinder of altitude h It and,radius a ft, with re-spect to the axis of the cylinder. The volume density varies as the distance from the axis of the rylinder, and it is mea-sured in slugs/ff.

, In Exercises 11 through L4, use spherical coordinates.

11' Find the center of mass of the solid bounded by the hemisphere of Example 5. The volume density is the same as thatin Example 5.

12. Find the moment of inertia with respect to the.z axis of the homogeneous solid bounded by the sphere f * y, + z2 : 4.The volume density at any point is k slugs/fts

13' Find the moment of inertia with respect to the z axis of the homogeneous solid inside the cylinder f I yt - 2x: 0,below the cone t2 * y":22, and above the xy plane. The volume d-nsity at any point is k slugs/ff.

14' Find the mass of a spherical solid of radius n It if the volume density at each point is proportional to the distance ofthe point from the center of the sphere. The volume density is measured in slugs/ff.

In Exercises 15 through 18, use the coordinate system that you decide is best for the problem.15. Find the center of mass of the solid inside the paraboloid f * yr: z artd. outside the cone * * y": 22. The constantvolume density is k slugs/fts.

16. Find the moment of inertia with respect to the z axis of the homogeneous solid of Exercise 15.17' Find the moment of inertia about a diameter of the solid between two concentric spheres having trdii a ft and,2a ft.The volume density varies inversely as the square of the distance from the center, and it i, *"urore-d in slugs/ft3.18' Find the mass of the solid of Exercise 17. The volume density is the same as that in Exercise 12.In Exercises 19 through 22, evaluate the iterated integral by using either cylindrical or spherical coordinates.

Reaiew Exercises (Chapter 21)In Exercises 1 through 8, evaluate the given iterated integral.

r. ['l* *a av a, , IJZ_xy itzc dy 3.

n. f I"*"^'r, sin 0 dr ttl u. I:f"f" eceyez dx ity dz 6.

, f'f,:Ip3 sin { cos e dp de d0 t

f f'["* zre-n dr d0 dzIn Exercises 9 through 12, evaluate the multiple integral.

t LI

ru dA;R is the region in the first quadrant bounded by the circle 12 * yr: L and the coordinate axes.

REVIEW EXERCISES 1UI7

t f10. | | (x + y) dA; R isthe region bounded by the curve y : cos r and the r axis from x:-in to x: in.

JRJ

t t l

11. f f f z2 dV; S is the region boundedbythe cylinders x2*z:l andy2*z:1, and the ry plane.J J J

t j f f y c o s ( x

t z ) d V ; S i s t h e r e g i o n b o u n d e d b y t h e r y l i n d e r x : A ' , a n d t h e p l a n e s x I z : i z r , V : 0 , a n d z : 0 .J J J "

rS. f:raluate by polar coordinates the double integral

14.

In

15

In

17.

19.

20.

t t 1

JrJ 7i v'dAwhere R is the region in the first quadrant bounded by the two circles f * y': 1 and x2 + yz : 4.

Evaluate by polar coordinates the iterated integral f [* h(rz * yz) dx dy.JoJ1 l } r

Exercises 15 and L6, evaluate the iterated integral by reversing the order of integration.

I:Isinyz dy dx

ft fcos tu

L 5 . J r J ,

r s i n r d x d y

tFI#dzdvdxf2 6m 1m

18' J,J. J,

z@ dzdv dx

Exercises 17 and 18, evaluate the iterated integral by changing to either cylindrical or spherical coordinates.

21,.

22.

Use double integraticin to find the area of the region in the first quadrant bounded by the parabolas *: 4y and I :

8 - 4y by two methods: (a) Integrate first with respect to x; (b) integrate first with respect to y.

Use double integration to find the volume of the solid above the xy plane bounded by the cylinder f -t y': 16 andthe plane z:2y by two methods: (a) Integrate first with respect to r; (b) integrate first with resPect to y.

Find the volume of the solid bounded by the surfaces x2:4A, !2:4x, and. *: z- !.

Find the mass of the lamina in the shape of the region bounded by the parabola A: x2 and the line x- y + 2:0 ifthe area density at any point is fy2 slugslfP.

23. Find the area of the surface of the cylinder f * yz :9lying in the first octant and between the planes x: z and3x: z.

24. Find the area of the surface of the part of the cylinder * t y' : a2 that lies inside the cylinder yz * zz : az.

25. Usedoubleintegrationtofindtheareaoftheregioninsidethecircler:landtotherightoftheparabolar(1 *cos0):1.

26. Find the mass of the lamina in the shape of the region exterior to the limagon r: 3 - cos d and interior to the circler: 5 cos 0 if the area density at any point is 2lsin 0l slugs/ff.

27. Find the center of mass of the rectangular lamina bounded by the lines r : 3 and A : 2 and the coordinate axes if thearea density at any point is xyz slugslfP.

28. Find the center of mass of thelaminaintheshape of theregionboundedbytheparabolasf :4*4A andl:4-8yif the area density at any point is kf slugslfP.

29. Find the mass of the lamina in the shape of the region bounded by the polar axis and the curve / : cos 20, where0 < 0 < *zr. The area density at any point is r0 slugs/ft2.

30. Find the moment of inertia about the r axis of the lamina in the shape of the region bounded by the circle r3 * y': a'if the area density at any point is klFl! slugs/fP.

10.18 MULTIPLE INTEGRATION

31. Find the moment of inertia about the r axis of the Lamina in the shape of the region bounded by the curve y : ea, theline r: 2, and, the coordinate axes if the area density at any point is xy slugslff .

32.

33.

Find the moment of inertia of the lamina of Exercise 31 about the y axis.

Find the moment of inertia with respect to the *z nxis of the homogeneous lamina in the shape of the region boundedby the curye 12:4 cos 2d if the area density at any point is k slugi/ff.

Find the mass of the lamina of Exercise 33.

Find the polar moment of inertia and the corresponding radius of gyration of the lamina of Exercise 33.

Find the moment of inertia about the y axis of the lamina in the shape of the region bounded by the paraboluy : , - *and the line r * I :0 rt the area density at any point is (r * y) slugs/fp.

Find the mass of the solid borrnded by the spheres f * y, * ?r:4 and. f * y, * zz:9 if. the volume density at anypoint is kll*W shgs/fF.

Find the moment of inertia about the z axis of the solid of Exercise 32.

The homogeneous solid bounded by the cone z2:4* * 4y2 between the planes z: 0 and z: 4hasa volume densityat any point of k slugs/ff. Find the moment of inertia about the z axis forthis solid.Find the center of mass of the solid bounded by th9 sphele * * y, * z2 - 6z:0 and the cone * * yr: zz, and abovethe cone, if the volume density at any point is *z slugi/ft

Appendlx

Table 7 Powers and roots

n nnB wn2 \E n3n2 \E %

L23455

, 789

10L 112L3l4L5L51718t92021222324252527282930313233343535373839404L424344454647484950

25 2.236 I2535 2.449 21649 2.546 34364 2.929 5L281 3.000 729

100 3.',1.62 L,000LzL 3.317 L,33L144 3.464 '1,,729

L69 3.606 2,L97196 3.742 2,744225 3.973 3,375255 4.000 4,095289 4.t23 4,913324 4.243 5,932351, 4.359 5,959400 4.472 9,000441 4.593 9,25L484 4.690 L0,649529 4.796 12,L57576 4.ggg 13,924525 5.000 15,625676 5.099 t7 ,576729 5.L96 Lg,6g3784 5.291 2L,952841 5.395 24,389

1.000 5L1.260 521.442 531.587 54L.7L0 551.817 561.9L3 572.000 5g2.080 592.154 602.224 61'2.289 622.35L 632.4L0 642.466 552.520 662.571 672.621 5g2.568 692.714 702.759 7r2.802 722.844 732.884 742.9242.9623.0003.0373.0723.1073.1413.1753.2083.2403.27r3,3023.3323.3623.391

3.4203.4483.4763.5033.5303.5573.5833.6093.6343.6593.584

1 L.000 14 1.414 g9' 1.732 27

L6 2.000 64

900 5.47796L 5.559'/',,024

5.6571,,089 5.745L,L55 5.931

27,00029,79132,76835,93739,304

757677787980818283848585878889

909L9293949595979899

100

"1,,225 5.9"1,6 42,9751,296 6.000 46,6551,369 5.093 50,553L,444 6.154 54,972L,52L 6.245 5g,3LgL,600 6.325 64,000l,6g'j, 6.403 6g,92lL,764 6.481 74,099L,849 5.557 79,507"1,,935 5.533 g5,lg4

2,025 6.708 91,L252,115 5.782 97,3362,209 6.855 103,8232,304 6.928 Lt},5922,401, 7.000 tL7 ,5492,50Q 7.07L L25,000

2,60L 7.14L2,704 7.21r2,809 7.2802,9'1,6 7.3483,025 7.4L63,',1,36 7.4833,249 7.5503,354 7.5L63,48L 7.68"1,3,500 7 .7463,72L 7.8103,844 7.8743,969 7 .9374,095 8.0004,225 8.0624,356 8.L244,489 8.1.854,624 8.2454,761 8.3074,900 8.3575,04/., 8.4255,',1,84 8.4855,329 8.5445,475 8.6025,525 9.5505,776 g.7Ig5,929 8.7755,094 9.9326,241 g.ggg

6,400 9.9446,55L 9.0006,724 9.0556 ,ggg g. 1L07 ,055 9.1657 ,225 9.2207 ,396 9.2747 ,569 9.3277 ,744 g.3gt7 ,921 9.434

9,100 9.497g,28L 9.5398,464 9.5928,,549 9.5438,835 9.6959 ,025 9.7479 ,2L6 9.7989 ,4W 9.8499 ,604 9.8999,801 9.950

L0,000 L0.000

L32,65'1. 3.709'1,40,509 3.732

148,977 3.755t57,464 3.790L66,375 3.903175,61,6 3.926185,L93 3.949Lgs,Ll2 3.971205,379 3.99321,6,000 3.91,5226,99'j. 3.936238,329 3,959250,047 3.979262,"1.44 4.000274,625 4.021297,496 4.041300 ,763 4.0523'1,4,432 4.092329,509 4.1.02343,000 4.12L357,9'1,"1, 4.L4L373,249 4.160389 ,0L7 4.179405 ,224 4.lgg421,875 4.217439,976 4.236456,533 4.254474,552 4.273493,039 4.29'1,5L2,000 4.30953'1,,441. 4.327551,358 4.34457L,797 4.352592,704 4.39061.4,L25 4.397636,055 4.4L4658,503 4.43I68'1,,472 4.449704,969 4.455729,000 4.491753,57'1, 4.499778,688 4.5L4904,357 4.53L830 ,584 4.547.857,375 4.563884,735 4.579912,673 4.59594L,L92 4.6L0970,299 4.626

L,000,000 4.542

r

Table 2 Natural logarithms

N I8755432I0

L . 01 . L1 . 21 . 3'1,.4

1 .5r .6t .7L .81 .9

2.02 . I2.22.32.4

2.52.62 .72.82 .9

3 .03.1.3 .23 .33 .4

3 .53 .63 .73 .83.9

4.04 .L4.24.34.4

4.54 .64 .74 .84 .9

5 .05 . L5 .25 .35 .4

0000 01000953 L044L823 19062624 2704336s 3436

4055 4L2L4700 47525306 53655878 s9336419 647L

693L 698L74L9 74677885 79308329 83728755 8796

9L63 9203955s 95949933 996e

r.0296 03320647 0682

1..0985 L019L3L4 13461632 L6631939 19692238 2257

L.2528 25562809 28373083 31103350 33763510 363s

L.3863 38884LL0 4L344351, 43754586 460948L6 4839

1.504L 50635261 52825476 54975686 57075892 59L3

L.5094 6LL46292 63L25487 55066677 66966864 5882

0198 0296 03921133 L222 13101989 2070 2I5I2776 2852 29273507 3577 3546

4t87 4253 43L84824 4886 49475423 5481 55395988 6043 60986523 6575 6627

7031, 7080 7L2975L4 756L 75087975 8020 8065841,6 8459 85028838 8879 8920

9243 9282 93229632 9570 9708

*0006 *0043 *0080

0367 0403 043807L5 0750 0784

L053 L085 LLL9L378 L4r0 L442L594 1725 17562000 2030 20602296 2326 2355

2585 2613 254L2865 2892 29203137 3164 31913403 3429 3455356L 3686 3712

39L3 3938 39624L59 4183 42074398 4422 44464633 4656 4679486L 4884 4907

5085 5L07 5L295304 5326 53475518 5539 55505728 5748 57695933 5953 5974

6134 5L54 6L746332 635L 537r5525 5544 55636715 6734 67525901 69L9 6938

0488 0583L398 L484223L 23rL3001 307537L6 3784

4383 44475008 5058s596 56536L52 62066678 5729

7L78 72277655 770L8L09 8L548544 85878961, 9002

936L 94009746 9783

*0LL6 *0152

0473 05080818 0852

1151. rL84L474 L5051787 1,8172090 2Lr92384 24L3

2569 25982947 297532L8 32443481 35073737 3762

3987 40L2423L 42554469 44934702 47254929 495L

51.51 51735369 53905581 56025790 58105994 601,4

6194 62L46390 64095582 56016771 67906956 6974

0577 0770 0862L570 L555 17402390 2469 254631,48 3221, 329338s3 3920 3988

45LL 4574 46375L28 5188 52475710 5756 58226259 6313 63666780 5831 5881

7275 7324 73727747 7793 78398198 8242 82868629 8571, 87139042 9083 9L23

9439 9478 95179821 9858 9895

*018g *0225 *0260

0543 0578 06130886 09L9 0953

L2L7 L249 L2821537 L569 15001848 1878 DAg21,49 2179 22082442 2470 2499

2726 2754 27823002 3029 30563271, 3297 33243533 3558 35843788 38L3 3838

4036 4061 40854279 4303 43274516 4540 45634748 4770 47934974 4996 5019

5L95 5217 523954L2 5433 54545623 5644 55655831 585L 58726034 6054 6074

6233 6253 62736429 6448 64675620 5539 66585808 6827 58456993 70Lr 7029

Table 2 (Continued)

use ln 10 :2.30259 to find logarithms of numbers greater than 10 or less than l. Example:ln 220 : ln 2.2 * 2 ln 10 : 0.7885 + 2(2.90259) : 5.3937.

0 1 2 3 4 5 6 7 8 9

5.55 ,65 .75 .85.9

6 .06 . 16 .26 .36.4

6 .56 .65 .75 .86.9

7.07 .L7.27.37.4

7.57.67.77.97.9

8 .08 . L9 .28.38.4

8 .58.58 .78 .98 .9

9 .09.'/',9 .29 .39 .4

9 .59 .69 .79 .9

1.7047 70567228 72457405 74227579 75967750 7766

"l..7gLg 79348083 80998245 82528405 84218563 8s79

l.g7lg 97338871 8886902L 90369t69 9L849315 9330

L.9459 94739601 96159741, 97559879 9892

2.0015 0029

2.0L49 01520281 029504t2 04250541 05540669 0681

2.0794 090709L9 09311041 10541,163 1,1751282 L294

2.140L t4l21518 1529L633 1645L748 17591861 1872

2.L972 1gg32083 20942L92 22032300 23tl2407 24t8

2.2513 252326L8 2628272L 27322824 28342925 2935

7084 7102 71207263 7281 72997440 7457 747576L3 7630 76477783 7800 7817

795L 7967 79848Lt6 8L32 81488278 8294 83108437 8453 84698594 8610 8625

8749 8764 87798901 991,5 gg31

9051 9065 90819t99 9213 92289344 9359 9373

9488 9502 95169629 9643 96579769 9782 97969905 9920 99330042 - 0055 0069

0175 0189 02020308 0321 03340438 0451 04640567 0580 05920594 0707 0719

0819 0832 08440943 0955 09581056 L078 10901,187 llgg 121,11305 1318 1330

1424 1435 L448L54L L552 1554L656 L668 1679L770 L782 17931883 L894 L905

1994 2005 20L72105 21,1,5 212722L4 2225 22352322 2332 23432428 2439 2450

2534 2544 25552638 2649 26592742 2752 27622844 2854 286s2946 2956 2955

7L38 7155 71747317 7334 73527492 7509 75277664 7591, 76997834 7851 7857

8001 80L7 80348165 8181 81978326 8342 83588485 8500 85168641, 9656 9672

8795 8810 88258946 896L 89769095 9110 91259242 9257 92729387 9402 94L5

9530 9544 95599671 9685 96999810 9824 98389947 995L 99740082 0096 0109

02L5 0229 02420347 0360 03730477 0490 05030605 0518 05300732 0744 0757

0857 0859 08820980 0992 1005L102 tLL4 L1261223 1235 1,2471342 1353 1365

L459 L47L 1483Ls76 1587 1599L69L 1702 17131804 18L5 L827t9t7 1928 1939

2028 2039 20s02138 2148 21592246 2257 22682354 2364 23752460 2471, 24gl

2565 2576 25862670 2680 26902773 2783 27932875 2885 28952976 2986 2996

7L92 72107370 73877544 756177L6 77337884 7901,

80s0 806682L3 82298374 83908532 85478587 8703

8840 88568991 90069140 91559285 93019430 9445

9573 95879713 97279851 98659988 *000L

0122 0136

0255 02680386 03990516 05280543 06550759 0782

0894 09051017 L029L138 1L501258 1,2701377 1389

1494 1505L61,0 1,6221725 17361838 1,9491950 t96L

2061 20722170 2191,2279 22892386 23952492 2502

2597 2607270L 27Ll2803 28L42905 29L53005 3015

Table 3 Expoiential functions

e t logro(e") e-"

0.00 1.00000.0L L.0L01.0.02 L.02020.03 1.030s0.04 1.04080.05 1.05130.06 1.06180.07 L.07250.08 1.08330.09 r.09420.1.0 L.10520. 11 1 .11630.L2 L.L2750.L3 1.13880.14 1.15030.15 '1,.1618

0. L5 L.L7350.L7 1.18530. L8 1.19720.1g L.20920.20 L.22r40.2r 1.23370.22 L.246L0.23 L.25860.24 r.27120.25 1.28400.25 L.29590.27 L.31000.28 7.323r0.29 1,.33640.30 1.34990.31 l.36340.32 L.377L0.33 1.39100.34 r.40490.35 '1,.4191.

0.35 1.43330.37 r.44770.38 1.46230.39 L.47700.40 L.49L80.41 L.50580.42 r.52200.43 1,.53730.44 1,.55270.45 1,.55830.45 1.58410.47 1.60000.48 L"6L6L0.49 L.63230.50 L.6487

0.00000 1.000000.00434 0.990050.00859 .gg}Lgg.01303 .970446.01737 .950789

0.0217"t 0.951229.02606 .94L765.03040 .932394.03474 .9231.1.6.03909 .gL3g3L

0.04343 0.904837.04777 .895834.05212 .886920.05646 .978095.06090 .g6935g

0.06574 0.860708.05949 .852744.07383 .843655.07817 .835270.08252 .826959

0.08685 0.818731.09120 .810584.09554 .802519.09989 .794534.1,0423 .786;,528

0.10857 0.778801,.L1292 .77L052.LL726 .763379.12160 .755784.12595 .748264

0.13029 0.74081 8.L3453 .733447.1,3897 .7261,49.L4332 .7L8924.L4766 .71,L770

0.15200 0.704688.15635 .597576.16069 .690734.15503 .683851,.L6937 .677057

0.17372 0.670320.17806 .663650.L8240 ,557047.18675 .550509.19109 .544036

0.19543 0.637628.19978 .631.294.20412 .625002.20846 .618783.21290 .6L2626

0.2L7L5 0.605531

e'

0.50 L.64870.51 1.66530.52 r.68200.53 1,.59990.54 1,.71600.55 1.73330.56 L.75070.57 L.76930.58 'j..7960

0.59 1.90400.60 L.82210.5'1. 1,.84040.62 1.85890.53 L.87760.64 1.89650.65 1 .91550.55 L.g34g0.57 L.95420.58 1.97390.59 L.99370.70 2.01380.71, 2.03400.72 2.05440.73 2.075L0.74 2.09590.75 2.11700.76 2.L3830.77 2.L5gg0.78 2.19"1,50.79 2.20340.80 2.22550.81 2.24790.82 2.27050.83 2.29330.84 2.31,540.85 2.33950.86 2.36320.87 2.38690.88 2.4'1,090.89 2.435L0.90 2.45960.91 2.48430.92 2.50930.93 2.53450.94 2.56000.95 2.59570.96 2.61,170.97 2.63790.98 2.66450.9e 2.69121.00 2.7L83

logto(e") e-"

0.2L7L5 0.506531.22149 .600496.22583 .594521,.230L8 .588505.23452 .582748

0.23886 0.576950.24320 .57L209.24755 .565525.25't89 .559899.35623 .554327

0.26058 0.548812.26492 .543351..26926 .537944.2736r .532592.27795 .527292

0.28229 0.522046.28653 .51685L.29098 .51L709.29532 .506617.29966 .50L576

0.30401 0.495585.30835 .491644.3L269 .486752.37703 .481909.321,38 .4771,1,4

0.32572 0.472367.33006 .467666.3344L .463013.33875 .458406.34309 .453845

0.34744 0.449329.35178 .444858.35612 .440432.36046 .436049.36481, .43171,1

0.359L5 0.4274L5.37349 .4231,62.37784 .418952.38218 .414783.38652 .4L0656

0.39087 0.405570.39521, .402524.39955 .39851,9.40389 .394ss4.40824 .390528

0.4L258 0.38674L.41,692 .382893.42127 .979083.4256L .3753LL.42995 .371,577

0.43429 0.367879

Table 3 (Continued)

x e' logro(e") e-" x e' logre(e") e-"

1.00 2.7t93L.0L 2.74561.02 2.77321..03 2.90LLL.04 2.92921.05 2.95771.05 2.9964L.07 2.9154L.08 2.94471.0g 2.97431.10 3.00421.1L 3.0344L.lz 3.0649L. L3 3.0957t.t4 3.L2691.15 3.1592l. l5 3. Lg99L.L7 3.2220L. L8 3.2544l. l9 3.29711.20 3.3201t.2l 3.35351.22 3.3972"1,.23 3.42121.24 3.45561.251.261.271.28'/.,.29

3.49033.52543.56093.59663.6329

0.43429.43864.44298.44732.45L57

0.4560L.46035.46470.45904.47338

0.47772.48207.4864L.49075.495L0

0.49944.50378.508L2.51247.5158t

0.52115.52550.52984.53418.53853

0.54287.54721.55L55.55590.56024

0.367879.3542L9.350595.357007.353455

0.349938.346456.343009.339595.3362L5

0.332871.329559.326290.323033.3r9819

0.3t6637.3r3486.3L0367.307279.30422L

0.30L 194.298197.295230.292293.289384

0.286505.283554.280832.278037.275271

0.58630 0.259240.59054 .25566L.59498 .254L07.59933 .251579.60367 .249075

0.60801 0.246s97.61236 .244'1,43.6L570 .2417'1,4.62L04 .239309.62539 .236929

1.50 4.48L7L.5L 4.5257'1..52

4.5722L.53 4.6192L.54 4.66461.55 4.7L15L.56 4.7588L.57 4.9066L.58 4.95501.59 4.90371.60 4.9530L.5L 5.00291,.62 5.0531L.63 s.1039L.64 5.15521.65 5.20701.56 5.25931.67 5.31221.68 5.3555L.69 5.4195

1.75'/.,.76

r .77L.78"1,.79

5.75455.81245.87095.92995.9895

1.80 6.04951.81 6.Lr04'1,.82

5.1719L.83 5.23391.84 6.29651.85 6.3599L.85 6.42371.87 6.44931.88 6.55351.8g 6.61941.90 6.6959'1,.9"1,

5.7531L.92 6.92L0L.93 6.88951.94 6.95gg1.95 7.0297L.96 7.0993'1..97 7.1707L.98 7.2427t.9g 7.31552.00 7.399'j,

0.65L44 0.2231.30.65578 .2209L0.660L3 .2lg7r2.66447 .215535.66881 .21439l

0.673L5 0.212249.67750 .210135.58194 .209045.6g6tg .205975.69053 .203926

0.59487 0.20L997.5992'1. .lggggg.70356 .197899

I

.70790 .19s930 I

.7L224 .1g3gg00.71659 0.192050.72093 .1901.39.72527 .lgg247.72961 .196374.73395 .194520

1.70 5.4739 0.73930 0.1,92594L.7l 5.5290 .74264 .1909551.72 5.5945 .74699 .L79066"1,.73 5.6407 .75L33 .1772941.74 5.5973 .75567 .175520

0.75002.76435.76870.77304.77739

0.173774.172045.L70333.1.68639.156960

1.30 3.6693 0.56459 0.272532L.3L 3.7062 .55993 .2699201..32 3.7434 .57327 .257135L.33 3.7810 .5776t .26M771.34 3.81.90 .58L95 .25L9461.35 3.9574t.35 3.99521.37 3.93541.38 3.9749L.39 4.01491.40 4.0552'/.,.4'1, 4.09601.42 4.137L"1..43 4.L7971.44 4.2207L.45L.46'/.,.47

l.4gr.491.50

4.2531 0.62973 0.2345704.3060 .63407 .2322364.3492 .6384r .2299254.3929 .64276 .2276394.437t .647t0 .2253734.4817 0.651,44 0.2231,30

0.79173 0.165299.79607 .153554.79042 .'1,62026.79476 .L604L4.7ggl0 .159917

0.80344 0.157237.90779 .L55673.81213 .154L24.91647 .152590.82082 .r5L072

0.92515 0.L49569.92950 .1490g0.93395 .L46507.83819 .L45L4g.84253 .143704

0.84697 0.142274.85t22 .14095g.85555 .139457.85990 .139069.86425 .L35595

0.86959 0.135335

(Continued)

2.00 7.3ggL2.01 7.45332.02 7.53932.03 7.614'1,2.04 7.69052.05 7.76792.05 7.84502.07 7.92492.08 9.00452.09 g.0g4g

2.lo 9.t6622.lt 9.24922.12 9.33112.13 g.4l4g2.1,4 g.4gg4

2.15 8.58492.16 8.571'1,2.L7 8.75832.18 8.84632.L9 8.93522.20 9.02502.21 9.11572.22 9.20732.23 9.29992.24 9.39332.25 9.48772.26 9.583L2,27 9.67942.28 9.77572.29 9.87492.30 9.97422.31 L0.0742.32 10.1752.33 10.2782.34 10.3812.35 "1.0.485

2.35 L0.59L2.37 L0.5972.38 L0.8052.39 L0.91.32.40 1,L.0232.4L Lt.L342.42 L1.2462.43 11.3592.44 11.4732.45 11.5882.46 11.7052.47 1L.8222.48 LL.94L2.49 L2.06L2.50 L2.r82

t2.L82 t.09574L2.305 1.0900912.429 L.0944212.554 1.0987712.580 1. L03LLt2.807 t.t074512.936 L.LLlTg13.056 1.1,161,413.197 1,.1,2049L3.330 L.12482L3.464 ',1,.12917

13.599 1,. L335LL3.736 L.13785'1,3.874 l. l421gL4.0L3 1.1,4654L4.I54 1.1508814.296 ],.15522L4.440 L.L595714.585 'l.,."1.6391

14.732 L.L682514.880 1.17260L5.029 1,.17594L5.L80 1.18128L5.333 r.L8562L5.487 L.1899715.643 1,.L943115.800 L.19865L5.959 L.20300r6.tL9 L.20734'1,6.281 '1,.2L168

16.445 1.2160216.510

'1,.22037'1,6.777 1.2247116.945 '1.22905

17.116 1.2334017.288 1.2377417.462 r.2420817.637 1.2454317.81.4 r.2507717 .993 L.255LLL8.r74 L.25945L8.357 1,.26380L8.541 1,.2681,418.728 r.272491g.9L5 1.27693L9.r06 L.28Lr719.298 1.2955L19.492 1.2898s19.688 1,.2942019.886 1.2985420.085 L.30288

g-r

0.082085.081269.080460.079659.078966

0.078082.077305.076536.075774.075020

0.074274.073535.072803.072078.071351

0.070651.069948.069252.058553.067881,

0.067206.056537.065875.0552L9.064570

0.063928.053292.052652.a62039.051,421,

0.050810.060205.059606.059013.05u26

0.057u4.057259.056699.055L35.055575

0.0s5023.054476.053934.053397.052866

0.052340.051819.05L303.050793.050287

0.049787

e ' log1s(e"') e-" x

0.96959 0.135335 2.50.97293 .1,33989 2.51.87727 .132655 2.52.88152 .131336 2.53.88595 . L30029 2.54

0.89030 0.L28735 2.55.89465 .L27454 2.56.89899 .L26L86 2.57.90333 .124930 2.58.90755 .123587 2.59

0.91202 0.122455 2.60.9L636 .t2L238 2.6L.92070 .L20032 2.62.92505 .Lt8837 2.63.92939 .1,17655 2.64

0.93373 0.1,16494 2.65.93808 .1.L5325 2.66.94242 .tL4l78 2.67.94675 .'1,13042 2.68.95LL0 .lllgl7 2.69

0.95545 0. L 10803 2.70.95979 .10970L 2.71.96413 .L08509 2.72.95848 .107528 2.73.97282 .106459 2.74

0.9771,6 0. L05399 2.75.98L5L . L04350 2.76.98585 .L033L2 2.77.990L9 .L02284 2.78.99453 ."1,01266 2.79

0.99888 0. L00259 2.801.00322 .09926L 2.8L1.00756 .098274 2.821.01191 .097296 2.831,.0L625 .096328 2.84L.02059 0.095369 2.85L.02493 .094420 2.851,.02928 .09348'1, 2.87I.03362 .09255L 2.88L.03796 .09L530 2.89L.0423L 0.090718 2.90L.04665 .089815 2.9L1.05099 .088922 2.921.05534 .088037 2.931.05968 .08716t 2.941.06402 0.086294 2.95L.06836 .085435 2.96L.07271 .084585 2.97L.07705 .083743 2.981.0g139 .0829L0 2.99L.09574 0.082085 3.00

e' logrs(e")

Table 3 (Continued)

x e' logro(e") e-" x e' logt6(e") e-"

3.00 20.0853.01. 20.2873.02 20.4913.03 20.5973.04 20.9053.05 21.'/',153.05 21.3293.07 21.5423.08 21.7593.09 21,.9773.10 22.Lgg3.L1. 22.4213.r2 22.6453.L3 22.9743.I4 23.L043.15 23.3363.15 23.5713.L7 23.9073. L8 24.0473.79 24.2993.20 24.5333.21 24.7793.22 25.0293.23 25.2903.24 25.5343.253.263.273.283.29

25,79026.05A26.3L126.57626.843

L.30288L.307231.3L r57r.3L59LL.320261.324601.328941.333281.337631,.34197r.346311.350661..35500t.35934L.36368L.368031.37237L.35571L.38L061.38540

0.042952.042426.042004.0415g5.041172

L.58517'/...58952

1.593851.59820t.60255

1.629601,.63295'/.,.63729'/.,.64L53

1.54598

0.025991.025733.025476.025223.024972

0.02351g.023294.023052.022923.422595

0.049787 3.50.049292 3.5L.04880L 3.52.048316 3.53.047835 3.54

33.1.L533.44833.78434.r2434.467

0.047359 3.55 34.8L3.046988 3.56 35.1.63.046421 3.57 35.517.045959 3.59 35.874.045502 3.59 35.234

0.045049 3.60 35.599.04460L 3.6L 36.966.044157 3.62 37.339.043719 3.63 37.713.043293 3.64 39.092

1.52003 0.030L97L.52437 .029897L.52872 .0295991.53305 .0293051.53740 .0290t31,.54175 0.0287257.54609 .028439L.55043 .028L56L.554n .0278761.559L2 .0275981,.55346 0.027324L.56780 .027052t.572t5 .026793L.57649 .0265161.58083 .026252

3.553.563.673.583.69

1.38974 0.040764 3.70l.3g40g .040357 3.71"1,.39843 .039955 3.721.40277 .039557 3.73l.407ll .039164 3.74

40.447 1,.50699 0.02472440.854 '1..6LL23

.02447941.264 1.5155g .02423441.679 t.6lgg2 .02399342.098 L.52426 .023754

38.47538.86139.25239.64640.045

l.411461..4L5801.420141.42449'/.,.42883

0.038774.038388.039005.037629.037254

3.753.753.773.783.79

42.52142.94843.38043.81,5M.256

L.44317 0.036993 3.90 44.70LL.4375L .036516 3.91 45. L50l.44lg6 .035L53 3.92 45.6041.44520 .035793 3.93 45.063L.45054 .035437 3.94 46.5251.45489 0.035094 3.95 46.9931.45923 .034735 3.96 47.4651.45357 .034390 3.97 47.942'1,.46792

.034047 3.gg 49.4241.47225 .033709 3.gg 4g.gt1,1.47650 0.033373 3.90 49.4021.48094 .033041 3.gt 4g.ggg1.48529 .0327L2 3.92 50.4001.48963 .032397 3.93 50.907L.49397 .032065 3.94 5t.4tg1,A9932 0.031746 3.95 51.9351.50256 .031430 3.96 52.457L.50700 .031. Ll7 3.97 52.9951.5L 134 .030907 3.gg 53.517l.51569 .030501 3.gg 54.0551.52003 0.030197 4.00 54.599

1.55032 0.0223711,.65456 .022L491.55900 .02lg2g'j,.56335

.0217101.65769 .021494r.57203 0.0212g0r.67639 .021069r.68072 .0208581.68505 .020651.1.68941 .0204451.69375 0.020242t.69g0g .0200411.70243 .0199401.70679 .0tg6$|.7LLLT .0lg44g1.71546 0.019255l.7lg8.l, ,019063'/.,.724L5

.01.9973'1,.72849

.0196961.73293 .01g500'/,,.73718

0.01.9316

3.30 27.1133.31 27.3g53.32 27.6603.33 27.9393.34 2g.2lg3.35 29.5033.36 2g.7gg3.37 29.0793.38 29.3713.39 29.6553.40 29.9643.41 30.2653.42 30.5593.43 30.9773.44 3l.lg73.45 31.5003.46 3'l,.gl73.47 32.1373.48 32.450

,3.49 32.7963.50 33. L 15

Table 3 (Continued)

x e, logro(e") e-" x ea logrs(e") e-"

4.00 54.598 1.737184.01 55.147 1.741524.02 55.701 L.745964.03 56.261 L.750214.04 56,825 "1..75455

4.05 57.397 t.7sggg4.05 57.974 1.763244.07 58.577 1.757584.08 59.145 l.77rg24.09 59.740 t.77625

60.340 L.7806160.947 1.784956L.559 '1,.78929

62.178 1.7935452.803 t.79798

4.104 . t14.124.134.L44.15 63.434 1.802324.16 54.072 t.806674.17 64.715 1.81.1014.18 55.366 1.8L5354.19 56.023 1.819694.20 66.686 1.824044.21 67.357 1.828384.22 58.033 1.832724.23 58.717 1.837074.24 69.408 '1,.84141

4,25 70.1.05 1.845754.26 70.810 1.850094.27 7t.522 1.954444.28 72.240 1.g5g7g4.29 72.956 1,.8531.24.30 73.700 t.857474.3'1, 74.440 l.87l8l4.32 75.tgg l.g76L54.33 75.944 1.9g0504.U 76.709 1.884g44.35 77.479 L.gg91g4.36 79.257 l.gg3524.37 79.044 L.gg7g74.38 7g.g3g l.g022l4.39 90.540 1.905554.40 91.451 l.glOga4.41 92.269 l.gl5244.42 93.095 1.g1g5g4.43 93.931 L.g23g24.44 94.775 l.g2g274.45 95.627 l.g326L4.46 g6.4gg l.936954.47 97.357 l.g4l304.48 99.235 L.945644.49 99.12'j, l.949gg4.50 90.017 L.95433

4.50 90.0174.51 90.9224.52 gL.g364.53 92.7594.54 93.691,4.55 94.6324.56 95.5934.57 96.5444.58 97.5'/.,44.59 98.494

"1,.95433 0.01.110gL.95867 .0L099g1 .95301 .010999L.96735 .0107811.97L70 .0L06731.97604 0.0L05671.98038 .010462L.98473 .0L03581.98907 .0L0255L.99341 .010L53

4.60 99.484 L.99775 0.0L00524.61 L00.48 2.00270 .0099524.52 10L.49 2.00544 .0098534.63 102.51 2.01078 .0097554.64 103.54 2.01513 .0096584.55 104.58 2.01947 0.0095624.66 '1,05.64 2.0239L .0094664.67 L06.70 2.02916 .0093724.68 L07.77 2.03250 .0092794.69 L08.85 2.03684 .009L874.70 109.95 2.04'1,18 0.0090954.7L L11.05 2.04553 .0090054.72 '1,'1,2.17 2.04987 .0089154.73 LL3.30 2.05421 .0088264.74 1"1,4.43 2.05956 .0087394.75 115.58 2.05290 0.0086524.76 LL6.75 2.06724 .0085564.77 tl7.g2 2.07L58 .0084804.78 119.10 2.07593 .0083954.79 1.20.30 2.08027 .0083L24.804.814.824.834.844.954.864.874.884.894.904.9L4.924.934.94

!21.5L 2.08461 0.008230122.73 2.08896 .008L48123.97 2.09330 .008057125.21 2.09764 .007987L26.47 2.t0199 .007907127.74 2.10633 0.007928L29.02 2.LL067 .007750130.32 2.71,501, .0076731,31,.63 2.71936 .007597t32.95 2.L2370 .007521134.29 2.12804 0.007477L35.64 2.13239 .007372"t37.00 2.13673 .007299L38.38 2.L4L07 .007227L39.77 2.L4541 .007L55

0.018315.0L8133.0L7953.0t7774.017597

0.017422.017249.0L7077.016907.0L6739

0.0L6573.01,5408.016245.0L5083.015923

0.0L5764.01.5608.015452.0t5299.0L5L45

0.014995.014846.0"1,4699.01.4552.014408

0.014264.014122.01.3982.013843.013705

0.0L3569.013434.013300.013158.01.3037

0.012907.0L2778.0L265L.012525.0L240L

0.412277.0L2L55.012034.0119'/.,4.0LI795

0.01L679.011552.011447.01 L333.0"1,"1,221

0.01.1109

4.95 L4l.L74.96 1,42.594.97 144.034.98 L45.474.99 146.945.00 148.41

2.1,4976 0.0070932.t54L0 .0070132.'1,5844 .0069432.16279 .0069742.16713 .0069052.L7147 0.006739

Table 3 (Continued)

e * logls(e')

5.00 '/.,49.41

5.01 L49.905.02 L5t.4L5.03 L52.935.04 154.475.05 L55.025.05 157.595.07 159.175.08 L60.775.09 '1,52.39

5.10 '1,64.02

5.11 L65.675.L2 L67.345.13 159.025.L4 L70.725.L5 172.435.L5 174.1,65.L7 L75.915.1.8 L77.685.L9 t79.475.20 L81.275.21 193.095.22 Lg4.g35.23 Lg6.7g5.24 L88.575.25 190.575.26 Lg2.4g5.27 194.425.28 196.375.29 rgg.345.30 200.345.3L 202.355.32 204.395.33 205.445.34 209.5L5.35 2L0.615.35 212.725.37 2L4.865.38 217.025.39 2L9.205.40 22L.4L5.41 223.635.42 225.995.43 229.155.44 230.445.45 232.755.46 235. L05.47 237.455.48 239.955.49 242.265.50 2M.59

e-r

2.17147 0.0067382.L7582 .0066712.18015 .0056052.18450 .0055392.18884 .0054742.L9319 0.0054092.19753 .0053462.20L87 .0062822.20522 .0052202.2L056 .0061592.2L490 0.0060972.2L924 .0050352.22359 .0059762.22793 .0059L72.23227 .0059592.23562 0.0057992.24095 .0057422.24530 .0056952.24955 .0055292.25399 .0055722.25833 0.005s172.26267 .0054622.26702 .0054072.27L36 .0053542.27570 .0053002.28005 0.0052492.28439 .0051952.28873 .005L442.29307 .0050922.29742 .0050422.30L76 0.0049922.30610 .0049422.31045 .0049932.31479 .0049442.31913 .0047962.32348 0.0047492.32782 .00470r2.33215 .0045542.33550 .0045092.34085 .0045622.345L9 0.0045172.34953 .0044722.35388 .0044272.35822 .0043932.35256 .0043392.36690 0.0042952.37L25 .0042542.37559 .0042L12.37993 .004L692.38428 .0041,292.38862 0.004087

e' log16(e")

5.50 244.69 2.399625.55 257.24 2.410335.50 270.43 2.432055.55 284.29 2.453755.70 298.87 2.475485.75 3L4.Ig 2.4g7Lg5.80 330.30 2.51,99L5.85 347.23 2.540525.90 365.04 2.562345.95 393.75 2.594056.00 403.43 2.605776.05 424.L1 2.627495.10 445.96 2.649205.15 469.72 2.670916.20 492.75 2.592635.25 519.016.30 544.576.35 572.496.40 601.855.45 532.70

2.714342.735062.757772.779482.80120

6.50 665.14 2.92291,6.55 699.24 2.944636.60 735.10 2.966346.55 772.79 2.ggg066.70 8L2.4L 2.909775.75 954.05 2.g3l4g6.80 gg7.g5 2.953205.85 g43.gg 2.974926.90 992.27 2.996536.95 1043.1 3.018357.00 L096.67.05 1,L52.97.10 12L2.07.L5 L274.L7.20 1339.47.25 1409.17.30 14g0.37.35 L556.27.40 L636.07 .45 L719.9

e- t

0.0040868.0038875.0035979.0035175.0033460

0.0031828.0030276.0028799.0027394.0026058

0.0024788.0023579.0022429.0021335.0020294

0.0019305.0018353.0017467.00L56L5.001.5805

0.0015034.0014301.0013604.0012940.0012309

0.001L709.0011138.0010595.0010078.0009s85

0.0009119.0008674.000825L.0007849.0007465

0.0007L02.00057s5.0006426.00061 L3.0005814

0.0005531.0005261.000s005.0004760.0004528

0.0004307.0004097.0003898.0003707.0003527

0.0003355

3.040063.06L783.083493.',1,05213.126923.1.49633.170353.L92063.2L3783.23549

7.50 1909.0 3.2572L7.55 1900.7 3.279927.60 lggg.2 3.300647.55 2L00.6 3.322357.70 2209.3 3.344077.75 232L.5 3.365797.80 2440.6 3.387507.85 2565.7 3.409217.90 2597.3 3.430937.95 2935.5 3.452648.00 2ggL.0 3.47436

Table 3 (Continued)

e ' logrc(e") e-" x e* logro(e") e-c

8.00 298L.09.05 3L33.88. L0 3294.59.1.5 3453.48.20 354L.09.25 3827.59.30 4023.98.35 4230.29.40 4447.18.45 4675.18.50 49L4.89.55 5L56.88.50 543L.78.55 57L0.09.70 5002,99.75 5310.78.80 6634.28.85 6974.49.90 7332.08.95 7707.99.00 8L03.1

3.47436 0.0003355 9.003.49607 .0003L9L 9.053.51779 .0003035 9.103.53950 .0002887 9.153.55121 .0002747 9.203.58293 0.0002513 9.253.60464 .0002495 9.303.62636 .0002364 9.353.64807 .0002249 9.403.56979 .0002L39 9.453.69L50 0.0002036 9.503.7L322 .0001935 9.553.73493 .0001.841. 9.603.75565 .000L751. 9.553.77836 .0001.656 9.703.80008 0.00015g5 9.753.82179 .000L507 9.803.8435L .0001434 9.853.86522 .000L364 9.903.88694 .0001297 9.953.90865 0.000L234 10.00

8L03. L 3.9085585L8.5 3.930378955.3 3.9520894L4.4 3.973799897.1 3.9955L

10405 4.0172210938 4.03894L1499 4.0606512088 4.08237L2708 4.L040813360 4.1258014045 4.t475It4765 4.t692315522 4.L9094L6318 4.2L255L7154 4.2343718034 4.2560918958 4.2778019930 4.2995220952 4.32L2322026 4.34294

0.0001234.0001174.0001r17.0001062.000L010

0.000095L.0000914.0000870.0000827.0000787

0.0000749.00007t2.0000577.0000544.00005L3

0.0000583.0000555.0000527.0000502

0.00004770,0000454

Table 4 Hyperbolic functions

)c sinh r cosh r tanh r

00.1.0 .20.30.40.50 .60 .70 .80.91 .01 . 11,.2L .3L .41..5'/-,.6

L . 7L .8t . 92.02 . 12.22 .32.42.52.52 .72 .82.93 .03 . 13 .23.33.43 .53.63 .73 .83 .94.04 .L4.24,34.44.54.54 .74.84.95.0

.00000

.100L7

.20t34

.30452

.4L075

.52Lr0

.63565

.75858

.888L L'/.,.0265'j.,,1752

L.33561.5095L.69847.90432.12932.37552.64562.94223.26823.62694.02L94.457r4.93705.46525.05026.69477.40538.19L99.0596

10.018rL.076L2.24613.s38t4.96516.543L8.28520.zLL22.33924.69127.29030.L6233.33536.84340.71945.00349.73754.95960.75t67.1,4174.203

1.00001.00501.020L1.04531 .081 11.12761.18551,.25521.3374"1,.433"1,

1.54311.6685'l.,.8107

1,.97092.L5092.35242.57752.82833.10753.4L773.76224.1,4434.55795.03725.55596.13236.75907.47358.25279.L146

10.058Ll.l2212.2871,3.575t4.99916.573L8.3L320.23622.36224.7LL27.30830.17833.35136.85740.73245.0L449.74754.97860.75957.14974.210

.00000

.09967

.r9738

.29L31

.37995

.462L2

.5370s

.50437

.66404

.7L630

.76159

.80050

.83365

.86172

.88535

.905L5

.92L67

.93541

.94681,

.95624

.96403

.97045

.97574

.980L0

.98367

.9865'1.

.98903

.99"1,0'/.,

.99253

.99396

.99505

.99595

.99568

.99728

.99777

.99818

.9985L

.99878

.99900

.999L8

.99933

.99945

.99955

.99963

.99970

.99975

.99980

.99983

.99986

.99989

.99991

Table 5 Trigonometric functions

Degrees Radians Sin Cos Tan Cot

0L234

56789

101 11213L4

1516t718L9

2021,222324

2526272829

303L323334

3536373839

4041424344

45

0.0000 0.0000 1.0000 0.00000.0L75 0.0L75 0.9998 0.01754.0349 0.0349 0.9994 0.03490.0524 0.0523 0.9985 0.05240.0598 0.0598 0.9975 0.0699

0.0873 0.0872 0.9962 0.08754.t047 0. L045 0.9945 0.10510.L222 0.t219 0.9925 0.L2280.1395 0.1392 0.9903 0.14050.L57'/', 0.1554 0.9877 0. 1584

0.t745 0.1735 0.9848 0.t7530 .1920 0. 1908 0.981.6 0.19440.2094 0.2079 0.978L 0.21250.2269 0.2250 0.9744 0.23090.2443 0.24L9 0.9703 0.2493

1.570857.290 L.553328.535 1.53591.9.081 1.5L8414.30L 1.501.0

11.430 L.48359.5'1,44 L.465'j,8.L443 L.44857.1154 1.431,25.3138 '1,.4137

5.6713 r.39635.1446 1.37884.7045 L.36r44.33L5 L.34394.01.08 1.3255

3.7321 L.30903.4874 L.29L53.2709 '1..274'1,

3.0777 1,.25662.9042 L.2392

2.7475 1.22172.505L L.20432.4751 1. L8682.3559 L.L5942.2460 '/.,.1519

2.L445 1,.13452.0503 t. t t701.9626 "1,.0996

L.8807 1..08211.8040 r.0647

L.732L 1.04721.6543 L.0297L.5003 L.0123r.5399 0.9948L.4825 0.9774

I.4281 0.9599t.3764 0.9425L.3270 0.9250L.2799 0.90761.2349 0.8901.

L.1918 0.87271.. L504 0.85521..71.05 0.9378L.0724 0.9203L.0355 0.8029

1..0000 0.7854

0.26L8 0.25880.2793 0.27560.2967 0.29240.31.42 0.30900.3315 0.3255

0.349L 0.34200.3565 0.35840.3840 0.37460.40L4 0.39070.4lgg 0.4067

0.9659 0.26790.9613 0.28670.9563 0.30570.9511 0.32490.9455 0.3443

0.9397 0.36400.9336 0.38390.9272 0.40400.9205 0.42450.9L35 0.4452

0.4353 0.4226 0.9053 0.45630.4538 0.4384 0.8988 0.48770.47'1.2 0.4540 0.8910 0.50950.4887 0.4695 0.8829 0.531.70.5061 0.4848 0.8746 0.s543

0.5235 0.5000 0.8660 0.57740.5411 0.5150 0.8572 0.50090.5585 0.5299 0.8480 0.52490.5750 0.5446 0.8387 0.64940.5934 0.5592 0.8290 0.6745

0.6109 0.5736 0.8192 0.70020.6283 0.5878 0.8090 0.72650.5458 0.5019 0.7986 0.75360.6632 0.6L57 0.7880 0.78130.5807 0.5293 0.777L 0.8098

0.5981. 0.6429 0.7650 0.83910.7L56 0.6561. 0.7547 0.86930.7330 0.6691, 0.743"1. 0.90040.7505 0.6820 0.73L4 0.93250.7679 0.5947 0.7L93 0.9657

0.7854 0.7071, 0.7071 1.0000

9089888786

8584838281

8079787776

757473727L

7059685766

656463626L

605958)- l,

c/

56

5554535251

5049484746

45

Cos Sin Cot Tan Radia Degrees

A-13

Table 6 Common logarithms

N 0 1 2 3 4 5 6 7 8 9

t011L213t4

L5L5t71,8L9

2021,222324

2526272829

3031323334

3536373839

404L424344

4546474849

5051525354

0000 004304t4 04s30792 08281139 IL731461 1492

L76L 17902041, 20682304 23302553 25772788 28L0

3010 30323222 32433424 34443617 35363802 3820

3979 39974150 4L56431,4 43304472 44874624 4639

477L 478649L4 49285051 505ss185 51985315 5328

5441 5453s553 55755582 55945798 58095911 5922

5021 50316L28 51385232 5243533s 53456435 64/,4

6532 55426528 5637672L 67306812 682r5902 59rr

6990 59987076 70847150 71587243 725r7324 7332

40L4 403L 40484183 4200 42L64346 4362 43784502 4518 45334654 4669 4583

4800 48L4 48294942 4955 49595079 s092 5105521,1 5224 52375340 5353 s356

0253 02940645 06821004 10381335 L367L544 1,673

t93r 19592201 22272455 24802595 27182923 2945

3L39 31503345 3355354L 35503729 37473909 3927

4082 40994249 42554409 44254564 45794713 4728

4857 48714997 50115L32 51455263 52765391. 5403

55L4 55275635 56475752 57535866 58775977 5988

5085 50956L9r 620L6294 63046395 54055493 6503

5590 65996684 66935775 67856866 68756955 6964

7042 70507L26 71357210 721,87292 73007372 7380

0334 03740719 07551072 1106L399 1430L703 1732

1987 201,42253 22792504 25292742 27652967 2989

3181 3201,3385 34043579 35983755 37843945 3952

4L1,5 4133428L 42984440 44554594 46094742 4757

4885 49005024 50385159 5L725289 53025415 5428

5539 555L5558 55705775 57865888 s8995999 5010

6107 5lL76212 62226314 63256415 54255513 6522

6609 66185702 67126794 58036884 68936972 6981,

7059 70677L43 7L527226 72357308 73167388 7395

0085 0128 0170 02120492 0531 0569 06070864 0899 0934 09591206 L239 I27L 13031523 1553 1584 1,514

1818 L847 1875 19032095 2t22 2L48 2L752355 2380 2405 2430260L 2625 2648 26722833 2856 2878 2900

3054 30753263 32843464 34833555 36743838 3856

3096 31183304 33243502 35223692 37LL3874 3892

40554232439345484698

48434983511952505378

546555875705582L5933

5478 5490 55025599 561,1. 552357L7 5729 57405832 5843 58555944 5955 5966

6042 60s3 6064 607s6L49 6L60 6L70 61805253 6253 6274 62846355 6365 5375 53856454 5454 5474 6484

5551 656L 657L 55805545 6656 5565 66755739 5749 5758 67675830 5839 6848 6857692A 6928 6937 6946

7007 70L5 7024 70337093 7I0L 7L10 7LL87177 7185 7193 72027259 7257 7275 72847340 7348 7356 7364

Table 6 (Continued)

N 0 r 2 3 4 s 6 7 8 9

5555575859

506r625364

6556676859

707T727374

7576777879

8081828384

8586878889

9596979899

909L929394

7404 74127482 74907559 75657634 76427709 7715

7782 77897853 78607924 793L7993 80008062 8069

8129 91,368195 8202826L 82578325 83318388 8395

845L 84578513 85198573 85798633 86398592 8598

875L 87568808 88148865 8871892L 89278976 8982

9031 90369085 90909138 91,439I9I 91969243 9248

9294 92999345 93509395 94009445 94509494 9499

9542 95479590 95959638 96439685 96899731 9736

9777 97829823 98279868 987299L2 99L79956 995L

74L9 74277497 75057574 75827649 76577723 7731

7796 78037868 78757938 79458007 80148075 8082

81.42 81.498209 82158274 82808338 83448401 8407

8463 84708525 85318585 8591864s 86518704 .8710

8762 87588820 88258876 88828932 8e388987 8993

9042 90479096 9101,9L49 9154920L 92069253 9258

9304 93099355 93609405 94L09455 94609504 9509

9552 95579600 96059647 96529694 9699974t 9745

9786 9791,9832 98369877 9881992L 99269955 9969

7435 7443 745L751,3 7520 75287589 7597 75047664 7672 76797738 7745 7752

7810 78t8 78257882 7889 78967952 7959 7966802L 8028 803s8089 8096 8102

9155 81,62 81,598222 8228 82358287 8293 82998351 8357 836384L4 8420 8426

8476 8482 84888537 8543 85498597 8503 85098657 8663 8669871,5 8722 8727

8774 8779 87858831 8837 88428887 8893 88998943 8949 89548998 9004 9009

9053 9058 905391,06 91,1,2 91,179L59 9L55 91709212 9217 92229263 9269 9274

93L5 9320 93259365 9370 93759415 9420 94259465 9469 947495L3 9518 9523

9562 9566 957L9609 96L4 95L99657 9661 95559703 9708 971,39750 9754 9759

9795 9800 98059841 9845 98509886 9890 98949930 9934 99399974 9978 9983

7459 7466 74747536 7543 7551,7512 76L9 76277686 7694 7701,7760 7767 7774

7832 7839 78457903 7910 79177973 7980 79878041 8048 805581.09 8115 8L22

8176 81,82 81898241, 8248 82548305 83L2 83198370 8376 83828432 8439 8445

8494 8500 850685s5 8561 8567851.5 862L 86278675 868L 86868733 8739 8745

879L 8797 88028848 8854 88598904 8910 89158960 8965 89719015 9020 9025

9059 9074 90799122 91,28 9L339175 9L80 91869227 9232 92389279 9284 9289

9330 9335 93409380 9385 93909430 9435 94409479 9484 94899528 9533 9538

9575 958L 95859624 9628 95339671 9575 95809717 9722 97279753 9758 9773

9809 991,4 gglg

9854 9859 98639899 9903 99089943 9948 99529987 999L 9996

Table 7 The Greek alphabet

a alpha

B betay gamma6 delta€ epsilon( zeta'q eta0 thetac iotaK kappaI lambdap m u

v n u

€ x io omicrontr pip rho(, sigmar tauu upsilon

0 PhiX chi,lt psi@ omega

Answers toodd-humbered exercises

Exercises L.1 (Page 14)

1 . {0 . , 1 , , 2 ,4 ,6 ,8 } 3 . { 2 ,4 ,8 } 5 . { 0 ,1 ,2 ,3 ,4 ,6 ,8 ,e } z . A e . { 0 ,1 . , 6 } 7 r . ( - 2 ,+@) 13 . ( -o , { ; t s . [ 4 ,8 ]17. (-*, Sl 1e. (-e,-t) u (0,+-) 2r. (-a,-2) U (2,+q) 23. (-o,-s) U (3,+o) zs. t-t,Ll zz. (-s,i)2e. (-a, -1) u (+,3) 31. (t +il u (3, +-)


1. 1, -E 3. -3,8 5. -* ,4 7. -&,+ 9. +,3 11. [9 ,+o) 13. ( -a,1] u [a,+o) ts . ( -11,3) 17. Ig ,2]19. ( -a,1) u (4,+@) 2r . ( -o,31 u [ ro,+-) 23. [ -g,g] 25. ( -a, * ) u (2,+*) zs. lx l> la l sr . lx-2 l>235. lcl < a- b

Exercises'1..3 (Page 27)

1. (a) (1, 2); (b) (-1, -2); (c) (-7, 2); (d) (-2, 1) 3. (a) (2, -2); (b) (-2, 2); (c) (-2, -z); (d) does not appty 5. (a) (-1, 3);(b) (t) -3); (c) (1,3); (d) (-3, -l)

Exercises 1.4 (Page 32)

s. *163;L l -$et \ /B .z .X t t . -2,8 13. ( -2+t \ /s , t+2\ /1) and(-2- t \ /5 , t -2\ f r ) ls .77; l *e1xy*2ryz-r22x-66y-55:0 17. ( -2,* ) ; $ ,* ) ; ( t * ,T) le . (1,+) 2r . ( - t ,D 2s. (s , -7) 2s. (a)2x+y:s


1 . - 1 3 . - + 5 . 4 x - y - t 7 : 0 7 . x : - 3 9 . 4 x - g y + 1 2 : o t t . t 6 l - y + ( 2 \ 6 - s ) : o 1 3 . x r y : g

rs . f +$ :1 l e . ( a ) yes ; (b )yes ; ( c )no ; (d )no 2 t . 2x *3y+z :0 ; lE zz . @-$ ;1u ) . t g - ^g ; ( d )Bx -Ay :s

2 s . y : ' t ; 9 x - 4 y - 1 1 : 0 ; 9 x * 4 y - 1 9 : 0 2 7 . 2 x * 3 y - t 2 : 0 ; ( 2 + z t / Z ) x + ( S - t t 4 1 y - 1 2 : 0 ; ( Z - z t / Z ) x + ( g + S { | ) V- 72 : 0 29. (a) x : t; (b) t : l; (c) x : l; (dl 2x + y - 3 : 0; (e) r - 2y + 1 : 0; (I) x + y - 2 : O


L . ( N - 4 ) ' + ( y + . 3 ) ' : 2 f ; * + y , - 8 x + 6 y : g 3 . ( r + 5 ) 2 + ( y + t 2 7 2 : g ; * + f * 1 0 r * 2 a y + 1 , 1 ; 0 : O S . * * y z - 2 j - 4 y-8 :0 7 . * * y2+6x* lOy +9 :0 9 . f * y2 -U-4y -2 :0 t L . @,D ;4 13 . (0 , - i ) ; * 15 . c i r c l e LZ . rheemptyset 19. c i rc le 21. x*y+5:0 23. 3x1-4y-19:0

A-17

A.18 ANSWERS TO ODD-NUMBERED EXERCISES


1. domain: (--, *-); range: (-o,4o; 3. domain: (-., *-); range: [-5, *o) 5. domain: [*, +*); range: [0, +o)

7. domain: all r not in (-2,2); range: [0, +-; 9. domain: (--, **); range: [0, +o) 11. domain: (-*, *-); range: -2 and 2

13. domain: (--,4-); range: all real numbers except 3 15. domain: (--, *-); range: [-4, +-) 17. domain: all real numbers

except -5and-1 ; range: a l l rea lnumbersexcept -7and-3 19 . domain :a l l rno t in ( -1 ,4 ) ; range: [0 ,+- ; 21 . domain :a l l rea l

numbersexcept2;iange: [0, +*; 23. domain:allrealnumbersexcept-5;range: [-5, +o) 25. domain: (--, *.);range: [1, +*)

(see Fig. EXl.7-251 27. domain: (-*, +-); [0, 1) (see Fig. EXl.7-27) 29. domain: (--, **); range: the nonnegative

integers (see Fig. EX1..7-29) 31. domain: all real numbers except 0; range: (--, -11 U {0} U (}, 1l (see Fig. EX1.7-31)

33. domain: (--, +*), range: {1, 3} (see Fig. EX1.7-33)

Figure EXl .7-27

Figure EXl .7-25Figure EX1 .7-29

Exercises 1.,8 (Page 61)

1. (a) -5; (b) -6; (c) -3; (d)

+ 5 h - 6 ; $ ) a x + 2 h + s

(e) 2h2 + th (f) 8f * 10x2 - 3; (d zt' - 7x2; (h)

(c) 2; (d) 3 rt; @) (f)

(e) - l i f x > 0, t i f x = 0; (f) 1 i f x > -"1., -1, i f x <-1; (g) 1; (h) -1

( -m, +@); (c ) f - 5x2 - x * 5 ,domain : ( - - , + - ) ; (d ) ( r - 5 ) l (x , - L ) ,

domain: all real numbers except ( f ) domain: ( -* , *o) ;

real numbers except 0 and- 2 L 1

(b) ffi,

domain: all real numbers except 0 arid

-2 J-. -(d)

}f, domain: all real numbers except 0 and

( f f i+ t f f i )7. (a) x '* x- 6, domain: ( -* , +co); (b) -xz * x- 4, domain:

domain: al l real numbers except -1 and 1; Id (x'- 1)/( x - 5),

( g ) x , - 1 . 0 r + 2 4 , d o m a i n : ( - * , + . o ) 9 . @ i w , d o m a i n : a l l

x * ' / ' ,(r)

fi, domain: all real numbers except 0 and

all real numbers except -L, 0, and l; (f)#,domain: al l real

2 l,E

Ilt

i l'E

(b)(a)

2 x 2 * 4 h x * 5 x * 2 h 2 + 5 h

2

Figure EXl .7-33

(i) Zxz * 5r + 2h2

(b) -1; (c) (d) -1; (s. (a)

numbers except 0 and

13. (a) even; (b)

(b) t-11 + [r'];

<e\ft, domain: all real numbers except -l and 1 11. 1a) lF - | + t/x- t, domain: [t, +-;,

D vETT, domain: [t, +-;, Gl tlx + t, domain: (1, +-;, C) t#T, domain: (1, +*;,

(--, \tr) and l\8, +@)

(c) neither; (d) even; (e) (f) even; (g) odd; (h) neither 17. (a) [r' + 2] + [0];

(c) [ l + 3] * [x ' - xl ; (d) t0] (") [fil].1#l; tr ) tttzxt+ lx - 1l + lr + + + i l x - 1 1 - l r + 1 l l

L9. (a) even; (b) odd; (c) even; (d) even 21,. domain: (-@, +oo); range {0, 1}

appear in Figs. EXt .8-23 through EX1.8-33.)

ANSWERS TO ODD-NUMBERED EXERCISES A-19

(Sketches of the graphs for Exercises 23-33

o

Figure EXl .8-23

3s. (f " S) (x) :

37. (f " S) (r) -

Figure EXl .8-29

< 0- x = i( - x < t

> L

[lll

i f r < 0

i f 0 < x < 7

i f x > ' / . ,

rf. x

i f 0if i -

i f x

Figure EXl .8-25 Figure EX1 .8-27

v

Figure EX1.8-31

Figure EX1.8-33

39. S( r ) - x - 3 ; S( r ) : ' / . . - x

g . a : i 1 1 . D : e 1 3 . E : m i n ( l , * e )5

23 . D: min (1 , (2 tE + 3)e)

- #,

domain: all real numbers excep ,-Xand-1; (g) #,

domain: all real

Reoieat Exercises for Chapter 1. (Page 63)

1 . ( * , g l 3 . ( - + , 1 ) s . t - # , * l 7 . ( a ) ( x - 3 ) " + ( y + s ) ' z : 0 ; ( b ) ( r - 3 ) ' + ( y + s ) ' < 7 6 ; ( c ) ( x - 3 ) " + ( y + 5 ) " - 2 59. x :0 ,3x*4A:0 11 . k : -+ rh :+ 73 . (3 , -6 ) ; ( -2 , -5 ) 75 . 0 <k <2 . 17 .7x2*7yr* l l x -1 ,9y-6 :0 19 . l2x*3y-2 :021. domain: [-8, +*); range: [0, 4o) 23. domain: all real numbers except -B and 2; range: all real numbers except ] and 5

25.(a) f * 4x-7,domain: (--,**); (b) l-4r- 1., domain: (-o,4o;. (c) (rP- a)Gx-3), domain: (--,**); @)#,domain:

2 , A r - aall real numbers except i; @ ffi, domain: all real numbers except -2 and 2; (f ) L6xz - 24x * 5, domain: (--, *-); G) a* - 19,

domain: (--, *-) 27. (a) #h,

domain: all real numbers except -1 and 3; (b) G4+#$,

domain: all real numbers

except -1 and 3; (c) G=;T;TT;

domain: all real numbers except -1 and Z; @) #*,domain: all real numbers except 0, -1, and 3;

x 2 - 3 x(e)

,, + I , domain: all real numbers except -1 and 3; (f)

numbers except 2 and 3 35. (+s, +a); (9,7) ; (2,3)

Exercise 2.1. (Page 73)

1. 0.005 3. 0.005 5. rtr 7. 0.01

79 . E : m in ( l , * e ) ' 2L E - m in ( l , * e )

15. 6 - min (L, te) 1 7 . 0 - m i n ( * , * e )

A-20 ANSWERS TO ODD.NUMBERED EXERCISES

23. (a) 0

a) 0; (b) 0; (c) 0 e. (a) 1; (b) -1; (c) does not

-3) 7. -3,2; h(-3) and h(2) do not ex ist .

15. 2; lim Sk) does not exist. 17. all integers;r - z

O, jli f Q) does not exist, lim f (x) does not

31.. / is continuous everywhere.

2 15. all real numbers except "1., -3, and -4

i n ( 0 , +@) 25 . a l l x i n ( - 1 ,0 ) and (0 , 1 )

7. + e. 1, ir. +\r0 13. *{z 1s. + 17. ++

exist 3. (a) 7; (b) 7; (c) 7 5. (a) 5; (b) s; (c) s 7. (a)

does not exist L3. (a) 2; (b) 1; (c) does not exist

7 . - @ 9 . * m 1 1 . - @ 1 3 . - @ 1 5 . * r c

3. -3; F(-3) does not exist. 5. -r, jt11; g(r) + Sei

11. continuous everywhere 1,3. 3; f (3) does not exist.

any integer. t9. O, t"tT f G) does not exist. 2L -1, 0

25. removable; 0 27. removable; * 29. essential

. all real numbers 13. all real numbers except 2 and -2

. all x not rn l-4,4] 21,. all real numbers 23. all x ir


1 . 7 3 . - # s . 1 2 '


1. (a) -3; (b) 2; (c) does not eexist 11. (a) -2; (b) 2; (c)


1 . f r c 3 . * o 5 . - @


l. 4; l im f (x) does not exist.t - 4

9. 0; lim f (x) does not exist.r - 0

-9] and [8, * *) 7. continuous on 13, 4] 9 ' 6 - *e 11. E - min (1, e)

Reaipw Exercises for Chapter 2 (Page 1-08)

1. 9 3. -+ 5. tontinuous on (-o

13. 6 - min (1, 5e) 17. + 19. -@ 29. (b) all real numbers;

(c) any number that is not an integer

Exercises 3.1 (P age 114)

1. -2xt 3. -6 - 2x, 5. 3xr2 - 3 7. 12xr2 - 26x, I 4

x * 6 y - 5 2 : 0 1 3 . 2 x * 5 y - 1 7 : 0 ; 5 r - 2 y * 3 0 : 0

l Z x * y - 9 8 - 0 1 9 . 8 x - y - 5 - 0 2 L . 2 x - y - 2 - 0- 4\F0: 0

Exercises 3.2 (P age 1.20)

1. 6tr' - gtr + 2 3. 6fi L8 5.

moving to right; changes direction

right; t > -1 + \6, moving to lefq

15. (a) (20h + 24) ftlsec; (b) $ sec


23. -+ 25. (a) yes; (b) no 27. f (x)

9 . 8 r * y + 9 - 0 ; x - 8 y * 5 8 : 0 1 1 .

15 . 2x * 3y - L2 : 0 ; 3x- 2y - 5 : 0 t7 .

2 3 . ( 1 2 - 2 \ m ) x - y - 3 0 + 4 { 3 0 : 0 ;

6 x - y - 1 6 - 0 ;

x - 1 2 y + 1 6 - 0 ;

( 1 2 + z @ l x - y - 3 0

1 . L . , - 5

. _ 5 q t < .2 t f f i 1 ' n

t ' ( 5 f 1 + 5 ) t t z ' 6 4

" 9 \

w h e n t : - 3 a n d t : l 1 , 1 . t < - L - \ 6 ,

changes direction when | - -l + \R 13.

-3, moving to right;-3 < t

moving to left; -1. - tE < t

(a)-32 ftlsec; (b) -64 ftlsec;

1 l, moving to left; t > l,

(c) 4 sec; (d) -128 ftlsec

11 . - 6 13. -#-6 15. z 17. 3 19. -127 -4 - 'f,

1 . 8 r * 5 3 .

2r. -+ 23. s@)


1. (b) yes; (c) +1,-1; (d) no 5. (b) yes; (c) 0, 1; (d) no 7. (b) (d) yes e. (b)3. (b) yes; (c) -1, +1; (d) no

11. (b) yes; (c) neither exisU (d) no

does not exist


13. (b) yes; (c) -6, (d) yes 17 . /i (0) '1.9. a:2 and(c) does not exist, 0; (d) no

b : -t 21. (a) 0; (b) 1; (c)

Exercises 3,5 (P age 137)

1 . 3 x 2 - 6 x * 5 3 . x 7 - 4 f

- tsxz - 6 L7. - t@ + t )L "

( x - l ) t

3L . 3 (2x2 * r * L )2 (4x + 1 )

x * 8 Y - 2 : 0


5 . S z - t 7 . 4 r r r z g . 2 x * 3 - ? ,

ts . - , t= ,= 21 , .1 !1 , -=2 f i ] z i .(x - 1 ) ,

3L ' (1 + 2x ' )2

-

33. x * (4\re - I})y - 8\re + 2L : 0; x -

13. 3 6sz - 2 6s 15. 70x6 + 60/.

6 (r2 .+ 10r-f L ) 29. 2(ix + 2) (6x, * 2x- 3)( r * 5 )2

8 \ r e * 2 L : 0 3 5 . x * 8 y * 2 - 0 ;

1 . 6 ( r + z ) ( x z + 4 x - 5 y z 3 . 2 ( 8 1 3 - 2 1 , t 2 + 2 ) ( 2 t 4 - 7 t s + 2 t - L ) 5 . - 2 ( x l - 4 ) - 3 7 . 6 ( 3 u - l ) ( 3 u 2 + 5 ) 2 ( L 2 u 2 - 3 u + 5 )

s .2 ( - t2x+17) (2x-5) - ' z (4 r+ l ) -a 11 . 18(y -7 ) (y+2) -3 13 . -2 ( l4x+3) (7 f +3x-L) -2

15. i ( r , + 1) ' (2r + 5) (8r ' * 15r + 2) 17. 2z(zz - 5) ' (z ' 122)(22 * 4; -e !

19.4(4x-1.)2(*+2)3(2txa-3xf + 49x2-4x+30)(31 +5)-3 ". f f i f t lsec;0

f t lsec;f f i f t lse. x * l 6 y - 3 5 : 0

25. (a) 3xa; (b) 615

Exercises 3.7 (P age

1,. 2(3x * S;-tra 3.

145)

L7

r * 5

5. Z,-rtz - EX-t,, 7.6 x 2 - 1 0 x * 1

ztTzx-s\@Tl)Tx(5x2 - 1 )

23.

35.

- A ^ | 1 ELJ' 6\tr-1\7(fry

Lv' yffi\?reWt 7 . L g . 4 x - 5 Y + 9 : 0

+t/g + \ffitFxr v \ i I r /

/ v 2 - 4 \(a) 0; (b) 4 (c) no value of t 2s. z,lffi) 27. sxlxl sg. *(3r + 2)3(f - 1)-trs(L2f + 2x - e)

+(t3 - 2t + 1)tt2 (t2 + f + 5) -r3(31 t4 + 29t3 + ]i09P - 18t - 88)

Exercises 3.8 (P age 1.50)

x ^ gy_ l x z t r _y , 7 _1 r t z . t _ r z x - x t t z 3x2 -4u1. -; 3.

3yz_8x i. -b 7. -yttzx-ttz 9

_,1 u Lr

7;-_r,xs + 8tr3

1 g . _ # 2 1 ' . 2 x i a : 4 2 5 . ( a ) f , ( x ) _ 2 \ f f i . , d o m a i n : . t r > _ 2 ; f , ( x )!x'y - 4x"

d o m a i n : x ) 2 ; f L ( x ) - - ( x - 2 ) - t t 2 , d o m a i n : x ) 2 ; ( e ) l ; 0 x - y - ! - 0 ; x + y -v

L : 0 27. (a) f ,(r) - \E= 9, domain:

fr(x):-{V4, domain: lxl > S, (d) /i(r) : x(f - 9)-1/2, domain: lrl > 3; fLG):-x(x" -9)-trz, domain: lxl > 3; (e)

i;

v * 2 x ( x - i l ' y + 4 @ 3 x 2 - v 3i 3 . i 5 . f f i ; 7 .&

- -z; domain i )c > 2; (d) f 'r(x) - (x - z)-ttz,

x = 4; fr(x)_ - \FPT| * 2, domain:

doma in : - 2 I x 1a ; ( e ) = ;

( f ) y - 5 : 0 ,v

( f ) 5 x * 4 y * 9 : 0 ; 5 x _ a y * 9 : 0 2 g . ( u ) f , ( x ) : \ m i + 2 , d o m a i n : - 2 <

-2 < x < 4 ; ( d ) ( l - x ) (8 - x2 +Zx ) - t n , doma in ; -2 1x < -4 ; ( x - 1 ) (8 - x2 * ' 2x ) - t r z ,

y + 7 : 0


1. (a) 8.6; 1b) 8.3; (c) 8.1; (d) 8 3. (a) -'t, -*; (b) -rlr, -* 5. (a) 18,750 gallmin; (b) 17,s00 gallmin

(b) not profitable s. oa:-E 11. 2.7 mi/min

Exercises 3.10 (Page 1,56)

L 3 ftlsec a. *ftlmin 5. * ftlmin T. f r,lr". s. #ftlmin

1.5. rta(3 \E + 97) ftlsec - 0.65 ftlsec L7. 22 ff lmtn

7. (a) profitable;

11. 1800 lblft2 per min L3. '1.4

ftlsec

A.22 ANSWERS TO ODD.NUMBERED EXERCISES

Exercises 3.L7 (Page 160)

t . f ' ( t c ) : 5 f - 6 * * 7 ; f " ( r ) : 2 0 f - 1 2 x 3 . 9 ' ( s ) : 8 s 3 - 1 2 s 2 + 7 ; S " ( s ) : 2 4 s 2 - 2 4 s s . f ' ( x ) : x ( i ? + 1 ) - t t 2 '

f , , ( x ) : ( f +11 - t r , 7 . F ' ( x ) :W2-5 .F " ( x ) : t l a r r z 9 . G ' , ( x ) : - 2x (3+2x2 ) -3 t2 .G" (x ) : ( 8 r r -5 ) (3+2x2 ) -5 t211. D,1y: ll,y L3. D,8f (x): 6(3 + r) (t - a;-s 15. D,ay : *(+trt2 | lsstz - tx-ttz1 19. D,2y: -lstxly-z 21,. -t;5

23. v : t3-6P+8t ;a:3f -Lzt+8; a:0 when f :0 and t :4 i toward the or ig in:2< t < 4; away f rom the or ig in:0 <t <2

and t > 4 25. t; #; -+ 27. sr; tfi; ?\re 29. f ' (x)

31. f ' (r) - 4l. f l , domain: (-*, +*) ) f " (x) -

+ (f ' " g) (x)g" (x)domain: (--, +o) 33. f " ' (x) :24lx l 35. h" (x) -

Review Exercises for Chapter 3 (Page 161)

domain: (-*, +*) ; f " (x) : r4rdomain:x '

(-*, o) u (0, +*;

(f " " s) (r) (g' Q))'

str(**+)'(#)(xn * x)'l ' l3x' * 2(xa * r) (4xt + 1) I 15. (a)

n "1 , -2x3

o 1 ," gxzts(xt *

' ! .)+rt ' ' \F@- \F>

t : 3 and f : 8 ; ( b ) when t : 3 , 7 ) - - 15 , and

11 x (4x2 - 13) \Fl l

\E__4particle is moving to the

2lefq when t: 8, o: 40, and particle is moving to the right 17. 5x - 4y - 6: 0; 4x + 5y - L3 : 0 79. \Fg

27. 29. (a) 32 ftlsec; (b) 255 (c) 7 sec; (d) -128 ftlsec

7 . x = 3 , x : - 3 ; y - 4 9 . x : - 2 , x : 2 ; y : 0

1 7 . y - - 1 , y - 1 t 9 . x - 3 ; y - - L , y : 1 2 7 .

and k : 4

21,. z|x+ 1l- l'l , (ffi-fr) zs. (b) 0; (c) 0 2s. -1, 0)

3 1 . a : - L ; b : t

#t lblft' per min

445-1,80\/2 , E1tlZ. jffiknots - 10.6 knots 35.

frU.in./sec 37. -?units/sec 39. decreasing at a rate of

a1. ( f . s) ' (0) :2 43. f (x) : lx l , sk) :x , M. p ' ( r )


1 . ? 3 . 0 s . 1 7 . + 9. 0 11. 0 13. f o 23. f is discontinuous at 0


1 . x : 5 ; y - 0 3 . x - - 2 ; y - 0 5 . x : - 6 , x : 1 ; y : 01 1 . y - - 3 , A : 3 1 3 . x - - \ n , x - f i . 1 5 . x - t i y - &


1. continuous; discontinuous; discontinuous; continuous; discontinuous; continuous 3. continuousi continuous; continuous;

continuous; discontinuous 5. continuous; discontinuous; discontinuous; discontinuous; continuous 7. continuous; continuous;

continuous; continuous; discontinuous; discontinuous 9. continuous; discontinuous; discontinuous; discontinuous

11. discontinuous; continuous; continuous; continuous; discontinuous; discontinuous; discontinuous; continuous 73. (-a,-21'

(-2,2); (2,+a1 15. (-o,-31; [4,+oo) L7. (-a,- l) ; (-1,1); (1,+o) 19. continuous on (0, 1) and al l intervals fn,n+'] . .)

wheren is anyintegerexceptzero Zt. t /F;continuousatal l t in (0,+-; ZZ. -! .continuousatalxin (2,+o) 25. : :- ,v x - z ! x - 2 '

continuous at all r in (0,4) and (4, +o1 ZZ. !!],continuous at all r in (0, L) and (1, +o1 29. 5 31.lx - l '

33. no; / will be continuous on la, cf iflim g(r) exists and is equal to h(b).


1. -5, * 3. -3, -L,1 5. 0, 2 7. -2,0,2 9. no critical numbers 1.1. abs min: l(2) : -2

15. abs min : / ( -3 ) :0 17 . abs min : f (S) : L 19. abs min: f (+) :1 2L. abs max: f (5) :213. no absolute extrema

abs mint f (2) - 0

abs min: f (21'.:0; abs max:

35. abs min: /(-3| -- -13;

25. abs min: l(-3) : -6; abs max: /(-1) : -1p 27. hbs min: /(-2) : 0; abs maxt f (-4) : LM

f ( 3 ) : 2 5 3 1 . a b s m i n : / ( - 1 ) : - 1 ; a b s m a x : f ( 2 ) : l 3 3 . a b s m i n : / ( - l ) : 0 ; a b s m a x : / ( 1 ) -abs max: f (3) :7

23.29.

VE

ANSWERS TO ODD-NUMBERED EXERCISES A.23


1. 2500 ftz 3. * in. 5. From Ato P to C, where P is 4 miles down the river from

height - GtQ.in. 11. (a) 3.a units; (b) 9.4 units 13. 225 15. 400 17. (a)

1 0 . t 5 . . r . rs q u a r e : # f U ( b ) r a d i u s o f c i r c l e - 5 f t a n d t h e r e i s n o S q u a r e .

7T

Exercises 4.7 (Page 200) i

1. 2 3:".. (a,t \fr) or *(2 - rt) 5.7. + s. # 11. +(3 + 4\E) B.satisfied 21. (b) (ii) not satisfied

satisfied 33. (ii) not satisfied

Reaiew Exercises f or Chapter 4 (P age 202)

L. 4 Z. -i 5. y:5; x:2, x:-2 7. (f . ilQ): ,/x=1' continuous at all r in (1, +o;; continuous on the closed interval

[1 ,+-1 9 . ( f "S) ( r ) :sgn( f -1 ) ;con t inuousata l l rea lnumbersexcept -1 and l 11 . absmin : l ( -5 ) :6 13 . absmax:

f ( O ) : 9 ; a b s m i n : / ( 3 ) : o 1 5 . a b s m a x : / ( 5 ) : 3 5 1 ; a b s m i n : f ( V 6 ) : 0 L 7 . 6 a n d 6 1 9 . l e n g t h : 1 8 o i n . ; w i d t h : d e p t h :

* in . 25 . y :2 ,y : -z ;x :7 , x : -L 27 . 12 lo r r i f rom po in ton bank neares tA 29 ' (a ) f (x ) :sgn r ;

B. T. g mi g. radius _ 3tD.in.,

radius of circle - 5

ft and length of side ofr s G + a )

1-t,-*), c:-t-t .VD;.t8,+f I :-r1+t/6 or c:-1 -trt /w; (-+,+), ' :-r+*t/er. 1s. (0, (ii), (iii) satisiied; (c) (*, -*v6) 17. (b) (ii) not satisfied 19. (b) (i) not

23. (b) (i), (ii), (iii) satisfied; (c) (0, 9) 25. 4 'Zz.

s + '\\/, zs. *+\/39 31. (i) not

[ x i f r < 0 ( x r f x(b ) / (x ) : l+ i f 0 < x <1 ; (c ) f (x ) : J r+ r i f 1

L r r f ! < x t r

Exercises 5.L (P age 21'0)

< L

< x

1. (a) and (b) /(2): -5, rel min; (c) [2, +o;; (d) (-*, Z] 3. (a) and (b) no relative extrema; (c) (-*, +*); (d) nowhere 5. (a) and

(b) l(2) : -50, rel min; f(-2) : 46, rcl max; (c) (-*, -21, lz, +*'1, @) l-2,21 7. (a) and (b) no relative extrema; (c) (0' +o);

(rd) nowhere 9. (a) and (b) /(2) : 4, rcl max; (c) (--,2f; (d) [2, 3] 11. (a) and (b) /(+) : 3'59, rel max; f (1) : 0, rel min;

(c) (--, *1, [r, +-;, (d) t*, 1] 13. (a) and (b) l(4) :t, rer max; (c) (--, +); (d) [4, +o) 15' (a) and (b) l(4) :9, rel max;

(c) (-*,41; (d) [4, +-) 17. (a) and (b) f(-2):5, rel max; /(0) : 1, rel min; (c) (-*,-z], [0, +-;, (d) [-2, 0] 19' (a) and (b)

f ( 2 ) : - g , r e l m i n ; ( c ) 1 2 , + @ ) ; ( d ) ( - - , 2 1 2 1 . ( a ) a n d ( b ) / ( - 1 ) : 2 , r c l m a x ; l ( 0 ) : l , r e l m i n ; f ( 2 ) : 5 , r e l m a x ; ( c ) ( - * ' - t l ,

t 0 , 2 1 ; ( d ) t - 1 , 0 1 , f 2 , + * 1 2 3 . ( a ) a n d ( b ) f ( - g ) : - 8 , r e l m i n ; / ( - 4 ) : - 5 , r e l m i n ; f ( 2 ) : - 7 , r e l m i n f ( - 7 ) : - 4 , r e l m a x ;

f (O) : -g , re lmax; (c ) t -9 , -71 ,1-4 ,01 ,12 ,+*1 ; (d ) ( - - , -s l , l -7 , -4 l , lO ,21 25 . (a )and(b)nore la t i veex t rema; (c ) [0 ,+- ; ;

(d) nowhere 27. (a) and, (b) /(-1) : Q rel max; /(1) : -94, rel min; (c) (--, -11, 11, +*); (d) I-1, 1l 29. (a) and (b) f (4) : +f/4'

r e l m a x ; ( c ) ( - 4 , 4 1 ; ( d ) ( - - , - 4 , l a , + * ) 3 r . a : - 3 , b : 7 3 3 . a : - 2 , b : 9 , c : - 1 2 ' d : 7


1 . / (+ ) :3 , re l m in 3 . f (g ) :T , re l max; f ( -1 ) : -11 , re l m in 5 . f (4 )

rel min IL. f ( l ) :8, rel min 13' F(27):9, rel max


rel min 7. G(3) :0, re l min 9. h(-2)

1. /(0) :0, abs min 3. no absolute extrema 5. 8(+) :*, abs min

by 60 yd. 1,1. Depth is one-half the length of the base. 13. \n

*(4 + n) . L7. If R is the radius of the cylinder, breadth : R; depth :

7 . / (0) :0, abs min; f ( f r ) : +\ /5, abs max 9.

15. Ratio of height of rectangle to radius of semicircle

rfgR. rg. 30( \m + 1) ft by 44({m + 1) ft 21'.

a5 yd

rsR : f

23. 2\/2


1. concave downward ftor x 10; concave upward for x ) 0; (0, 0) pt. of infl. 3. concave upward everywhere 5. concave

downwardforr <-1 and0 ( r ( 1; concaveupwardfor-1 < r ( 0 and x>'!.; (0,0) pt. of infl. 7. concaveupward for x<2;

concave downward for x ) 2; (2, O), pt. of infl. 9. concave upward for x < 0; concave downward fot x ) 0; (0' 0) pt. of infl.

l l . a : - l , b : 3 13 . c : 2 ,b : - 6 , c :O , i l : 3 (Ske t cheso f t heg raphs fo rExe rc i ses15 -25appea r i nF igs .EX5 .4 - l 5 t h rough

EXs.4-2s.)

A-24 ANSWERS TO ODD-NUMBERED EXERCISES

c

Figure EX5.4-21

c

Figure EX5.4-15


Figure 8X5.4-17 Figure EX5.4-19 Figure EX5.4-25

(Sketches of the graphs appear in Figs. 8X5.5-L through EX5.5-33.)

1. f(-1):5, rel mn<;f( l) :-3, rel min; (0, 1), pt. of inf l ; / increasing on (-o,-11 and I l ,+*1;/decreasing on [-1, 1]; graph

concave downward for x ( 0; graph concave upward for r > 0. 3. f(D:-1+, rel min; (0, 0), (1, -1), pts. of infl.; f increasing on

B,+-);/decreasingon (-m,*J; graph concaveupwardforr < 0andx > 1; graphconcavedownwardfor0 ( r ( 1 5. /(-3) -t

rel max; f (-+) : -9, rel min; (-*,,H, pt. of infl.; f increasing on (-o, -31 and [-*, +-); / decreasing on [-3, -*J; graph concavedownward for r < -*; graph concave upward for r > -* 7. f (0) :1, rel min; (t,lil , O,2) , pts. of infl.; / decreasing on (-@, 01.

/ increasing on [0, +o); graph concave upward for r < * and r > 1; graph concave downward for t < x < 7 9. f (-l): r1, rel min;

f (O) : t , re lmax; f (2 ) : - t , re lmin ;p ts . o f in f l . a t r : i ( l = lV) ; /decreas ingon ( - rc , -11 and [0 ,2 ] ; l inc reas ingon [ -1 ,0 ] and

12,+*1t graphconcaveupwardforr < +(l- fr) andr > *(1 + .y'7); graphconcave downwardfor*(1 - 'y 'V) <r<t([+ \ /7)

11. l(0) : 0, rel min; no pts. of infl.; / decreasing on (-o, 0]; / increasing on [0, ao); graph concave upward everywhere 13. no

relative extrema; (0, 0) pt. of infl. \Mith horizontal tangent; / increasing on (--, *-); graph concave downward on (-oo, 0); graph

concaveupwardon (0 ,+o ; tu . /G4) : '+# , re lmax; f (2 ) :O, re lmin ;p ts .o f in f l .a t ( -1 ,0 )andr : r t (s ' r3 {6) ; f inc reas ing

on( -o , f land [2 ,+@); /decreas ingon[$ ,2 f ;gnphconcavedownward for r ( -1 and+(8-3V6)<r< :b(8+3rG) ;g raph

concave upward for - l < r < rb(8 -s lq andr > lb (8+3\6) . 17 . f ( -+ ) : { f , re lmax; / (0 ) :0 , re l . m in ; ( -1 ,2 ) ,p t .o f in f l . ;

I increasing on (-o, -*] and [0, +-;; f decreasing on [-*, 0]; graph concave downward for r < -1; graph concave upward for r > -1

19. f (0): 0, rel min; l(1) : 1, rel max; / decreasing on (-o, 0) and [1, 4a); /increasing on (0, 1]; graph concave downward forr < 0

and r ) 0 21. /(-*) : -8, rel min; (0, 0), (+, *.92), pts. of infl.; f decreasing on (-o, -*l; I increasing on [-L ao); graph concave

upward for r ( 0 and r > *; graph concave downward for 0 < r < I 23. no relative extrema; (3, 2), pt. of infl.; f increasing on

(-* , **); graph concave downward fior x > 3; graph concave upward for r < 3 25. no relative extrema; (3, 2) , pt. of infl. with

horizontal tangent; /increasing on (-o, 1o); graph concave downward for r( 3; graph concave upward forr> 3 27. f(-l):3,rel min; / decreasing on (-m, -11;

/ increasing on [-L, 1o); graph concave upward for all r 29. f (O): 0, rel min; l(f) :1X\/5,

rel max; pt. of infl. at r: +(48 - 8{6); / decreasing on (-co, 0l and [1F, 4]; / increasing on [0, r*J; graptr concave upward for

r < 'i(4U - 816); graph concave downward for *(rtt| - 8\6) < x < 4 31. f(-1) : Q rel min; /(l) :2, rcl max; Pts. of infl. at r: 0

and x: d5; / decreasing on (-o, -1] and [1, +-;, / increasing on [-1, 1]; graph concave upward for -16 < r < 0 and r > 16;graph concave downwardforr ( -rEand0 < r < tE n.fei l :{ /6, relmax;/ increasingon (-o,-31; /decreasingon [-3,0];graph concave downward for r (

(c, f(cl)

(c, f(c)l

Figure EX5.5-1

Figure EX5.5-3

Figure EX5.5-5

Figure 8X5.5-7


Figure EX5.5-9

Figure EX5.5-L3 Figure EX5.5-17

Figure EX5.5-15

Figure EX5.5-19 Figure EX5.5-21 Figure EX5.5-23

Figure EX5.5-31

Figure EX5-5-27

Figure EX5.5-25Figure EX5.5-29 Figure EX5.5-33


r . ia leo-4; @)$29.99;(c)g2e.9s78 3. (a)50centspergal lon;(b)25 5. (a)Q(r) :x- I -4+i ; (b)c ' (x) :2a14;(c)Q'( r ) :L-7, tr r \ s , w

x 2 , \ - , - - - - - ' ' \ .

(d) $9.55 7' (a) f2,+a1;(b) C,(r) : f _ 4x* 5; (c) decreasing on [0,2] and increasing on [2,.to); (d) graph concave downward for, . 1 - \ : 2 - - L 4 n n '

;-:;:r,r;";;;:"""upwardforr > 2,pt.of.infl. at (2,i*), infl. tangent:3x-3v +14:0 e. (a)c(r):3r*rt00;

(b )Q( r ) -a *$ , ( c )C ' ( r ) : 3 ; (e )e53 11 . (a )3 - * x ; (b )3x - * * ; ( c )3 - * r 13 ' ( a )P ( r ) : t /Fa ;& )R( r ) : r 00x { f i =7 ;


(c )R ' ( r ) -3500 -200 f , r . 1 )$1800 15 . (a )S ( r ) : - ? f + t6x -2 ; ( c )$3000 ; (d )R , ( r ) : 14 -2x ,C , ( x ) : 2a -2 12 . (a ) [0 ,8 ] ;y36- * ' ' -

(b) R'(r) : 3x2 -32x * 64; C' (r) - 18 - 2x; (c) +(15 - tffi l - t.Bg

Reaiew Exercises for Chapter S (Page 240)

. t E 7 F1. ; 5. B sec; velocity of horizontal particle -'!. ftlsec; velocity of

19. (a) 1625; (b) 67.5 cents; (c) $3Zg.tZ 21,. g7.S cents

3. 1a) ffi - fr; b) #; (c) - (ztfil-,( \E _ tmi,

vertical particle - 3 ftlsec

f (4 ) :0 , re lmin ;p ts .o f in f l . a tx : -2andr : * (g -+S\ /6 ) ; f inc reas ingon (_oo,T l and [a , * * ) ; /decreas ingon [g ,4 ] ;g raphconcave upward lor-2 < r < *(8 -316) and g(8 +lt/6,1 < x < +o; graph concave downward for_o ( x 1_2and,*(8 - 3V6) < r < *(8 + g\/6) 9. no relative extrema; pt. of infl. at x:3;/ increasing on (-o, 4o); graph concave upwardfor3 ( I < +€; graphconcave downward for-e < r < 3 15. (a) increasing ata rate of$10100 peryear; (b) increasingat a rateof $800peryear 77. radius:*r in.; alr i tude:3ft in. 19. (a) p(r) :TY;15) R(r) -3600t:400x2.

(c) R,(r) : - aoo(r + 9) (x - 3). (d) 12oo( x * 3 ) 2 21,. 1000, $11 23. (hut t7 ,zrz)stz f1


1. (a) 3x2 Ax * 3x(Ar)2 + (Lx)t; (b) 3xz Ax; (c) 3x(Ar), + (Ar)t

5. (a) (5r f *6x) Ax+(6x+3)(Ar) ,+z(Lx)B;(b)(Gf *6x) L,x ; (c) (5r+3)(Ax) , +2(Lx)s z. @)0.030e;(b)0.03;(c)0.00099. (a)-0.875; (b)-1.5; (c) 0.625 11. (a)-0.00248; (b)_0.00250; (c) 0.00002 rg. O.tzS 15. 3.009 77. O.Oggg lg. .!,h21. (a) 6.75 in.3; (b) 0.3 in.2 23. Io ff 25. 10 fts 27. 4Vo


1. 6(3x - 1) (3x'- 2x + 1)2 dx 3. ""(7x, * 9x)(2x * 3)-zrs n* 5 . (x - 1 . ) - r rz (x * l ) -sn O, 7. 2(x' + 1) -st2 dx

11. -yttzx-rtzt e . ( 2 t - 1 ) [ 1 - 9 ( y z - t + A ) - u z f

21, . 6ts - 4tB - 2t - 9t( t2 + 1) r tz + 6t( tz * l ) -uz


7 . E r t + C 3 . * t t - t 2 + g t + C 5 . F L 35 x - p - ; + c 7. e\E - {zx+ C s. B(x ' - 1)Btz + C 11. +\AF+ 1 + C

1 / 1 \ 3 / 213 . - i ( l+ i r l +c 15 . - t r rs l - f x . -gas*#r ,+ c t7 .2 [+( r+3)3 / , -3 ( r+3)u , ] +c+#(3- x )s rz -ur t - x )s rz*c 21 . . &( f +3 f * l y rz*c 23 . * (3_y1nr "_18(3_y ;v :1627. (a)2f +4f +3**x ' lC;(b) iQx+1)4+ C zg. g isnotd i f ferenr iableon(-1, 1)


i t r 9 x 2 - Z x y * 2 y 2 - 6 xr rr 17. Lgf5 _ 55t3 + 3g,^ v '

x 2 - 4 x y * g y r - i

19. -+(3 - x)zrz

25. t(r ' , t + 2)5 + C

1 . ! : x B * x 2 - 7 x + C * 3 . i * r y + C y t 2 : 0'1.1.

,x2:4y' 13. ! :3xa * 4x3 * 2x2 * 2x 15.2 1 . r c z + 2 y , - C ,


5. z\fTtr : ixz + C. 7. I2y - Sxa * 6x2 * Crx * C2a : x2 - 3x * 2 L7. 3y :-2x3 * 3x2 * 2x * 6 19.

9 . 3 Y : 1 3 - 3 x 2 - 1 2 x + 1 812y - -€c4 *.5x2 - 20x + 2Z

l . o :2 I5 t - t2 ; s :2 t * *P-+ t t 3 . 1600s :o2 *1200 5 . *sec ; 20 f t l sec ; *sec ; ? f t9. a2:-64s+iooo,24ft lsec 11. i&ft lseC 13. r*sec;r#! i t rs.

$f2minrExercises 5.6 (Page 272)

1. R (r) -'1.2x - *x, 3. px(x + 5) _ - L8x - 4xz

7 . 3.4 sec; 99 ftlsec

5 . x i n [ 0 , 4 ] ; R ( r ) : ' ! . 6 x - x3 ; p -16 - xz 7 . C ( r ) : 1 3 * 4 x 2 * 4 x * 6 9. $32s


ANSWERS

Reoiew Exercises for Chaptet 6 (Page 272)

t . -gx-uat( g. * (2+3f1srz*c 5. a#/Gx+3)sD(30rP-78x1-79)+c z ' tQ' -

11. (a) -0.15; (b) -0.64 13' y:rux-2* -e ls' 0 t'' #:W=t " )(W)

" Olf;+{+f * c 21.. .y:#(x* 4)it2+*(6- s2\/r)x--#(75-na\/r) 23.'t sec;80 ftlsec

27. anenorof fain.allowed 29' anerrorof approximatelyn2l'l'6L0tsec 33' R(x):-h-


1.51 S. i+ 5. + l . & 77 ' 10,400 19 ' 10(10"-1) 21 ' +8+ 23 ' na- In3-3n' - tn

TO ODD-NUMBERED EXERCISES A.27

1 ) - t : x - r + C 9 . Y * Y - t - - x + C

( f + 1 ) 3 , - .19 . (a ) g *L ;

25. 25\tr sec : 43.3 sec

ac x i a ; p :

x + b

1,1,. * tq units

ZS. gzn _ +(3-rn) - #n - I

13. tm(b' - a') sq units

19. 38" sq units

s. +\730 11. -2 + \8. 13. o

15. + 17. "# 19. \m 21'.

+(3 - \n) 31. ry 35. $4,s3s 3s

+ +) 13. +t ls. s+s

25. 22= 9-4 zs. $tre6.6zbv'

175

StfZ sq units 11. # sq units t3. +8 sq units

23. t?e sq units 25. 64 sq units 27. (a) ttp' sq units;

9. trr' cu units 1,1. L+s?r cu units

').2

1. I sq units 3. 15 sq units 5. fl*a units 7' 9 sq units 9' B tq units

L5. +h(bt + b) sq units


t ' W 3 . + # 9 8 5 , 0 ' 8 3 5 2 . & s . a F 1 1 ' . 6 6 1 r 3 . 4 1 5 . 2 0 s q u n i t s 1 7 . * s q u n i t s

21. f' *' d* 23. f' * o*J o J o x '


g. 0; 64 11. 0; 27 13. 0; 576 L5' -3; * L7 ' 0; 6

Exercises 7 .5 (P age 310)

l. c : tr(1+ \G) 3. c : -4 5. / discontinuous at -2 7. f drscontinuous at L

15. 32t; 32 17. rr


1. zy 3. t+s 5. -8 7. 2- ffi. g. t€ 11. + 13' +\n

2g. ? sq,units 25. # sq units 27. +(40 \tr - 20) sq units

Reuiew Exercises for Chapter 7 (Page 320)

t. -+ 3. 4,100,656,560 7. g. ?$m- \D 11' h21\m

19. 4\fr = I:

\ffi dx = e zL. L8 sq units 23. %" sq units


L. # rq units 3. %z sq units 5. * tq units 7.9 tq units 9'

15. # sq units L7. ++ sq units 19.3 sq units 21. 12squnits

(b) 32 2e. - -q-m4

Exercises 8) (Page 335)

1. 64n cu units 3. zfn cu units 5. atran cu units 7.

13. tffen cu units 15. #n cu units 17. 1804r cu units


9. tn cu units 11. #n cu units

23 * x z

2e. *;

a$€o zr cu units

19. #n cu units

13. 8zr cu units 15. #rr cu units L7. #nAB crt units 19. tn cu units 21,. #n cu units


Exercises 8,4 (Page 344)

1. *tfgrg cu units 3. 8r3 cu units 5. tabc cu units


1. 180 in.-lb 3. 12,000 in.-lb 5. 6562.5w ft-lb 7.

Exercises 8,6 (Page 350)

1. 320w lb 3. 64ut lb 5. 2.2\at lb T. r+r! w lb


1. 4 3. 6 5. 171 slugs; 5.92 in. from one end 7.4 4 2 0 o r11. f i * 'slugs/ft


7. l9M in.8 9. trt in.t

256nw ft-lb 9. 100,000 ft-lb 11. 5500 ft-lb 13. Ztfr tt 15. #84,l sec

9. 96w lb L1,. 1,L,250fiw lb lg. 2SO\@w lb

t g . ( 2 , 0 )

2s. tr"(n + *)

25 slugs; 5.33 ft from left end 9. 16 slugs; #ft from one end

1. (2, +) s. (0, g) 7. (0, +) e. (+9, t+) 11. (+, -*) 1s. *p rS. 2.25w rb 17. 100,000p ft_lb23' The point on the bisecting radial line whose distance from the center of the circle is 4/3zr times the radius.Exercises 8.9 (Page 371)

1,. (0, g, o) 3. (+, o, o) 5.17. on axis , th units from vertex


(+, o, 0) 7. (2, +, O)19. ({q, o, o;

11. (0, +, 0) 13. (#, +, o1 15. (2, Ep , 0)9. (0, t , 0)

1. +A 3. +8 5. #(gT',, - tzl) 7. tz g.8a' - (q'*-??') 't ' 11. 4{g _ *8(a' - bz)

Reaiew Exercises for C.hapter I (Page 379)

l. i# ,q units 3. '1,024 in.s 5.

13'900 wn ft-lb

17.28 s lugi l I : f 19. Tn cu uni ts 21.90 f tg29. 756 lb 31. *nrzh cu units 33. (0, +)

Exercises 9.L (P age 390)

7 . 8 tq units 9. 3n cu units

23. Wwn ft-lb 25. 400 ft-lb

11. +q\6 13. (+9, i+) 1s. (9, +)

27. B#ot Q251,)3t2 - +(1,0999)3/21

, 8x , - x q 1+ ln l , j r ,t.771#; 3. 4=e xlnx .. f +t 9. +(1 + laly'r tt. t(x1 + t)-6/5(8xg -4f + 151 + t0)

13. *(3r+a)[(x+1,)(x*2)]-srz tu. ff i f .# v.-## re. lnl3 -2xl-uz*, 21. ln2 3 . * l n 8 3 r * C 2 5 . * + 4 l r . l f - 4 1 + C 2 7 . t n l 2 9 . t t n *


11 . 2 -2 |n2 -x -2A 13 . l n4squn i t s 15 . 2000 l n2 lb l f t z

Exercises I .3 (P age 404)

l. f-t(r) -ffx; domain: (--, +*) 3. no inverse1

5. f-t(r) :r=; domain: (-rc, 2; U (2, +ay T. no inverse

9. (a) f'(r) : t/9=7; f"(x) : -{T-x'; (b) neither has an inverse; (c) D,y : -i,o"r: -X ,r. <^l f k) :11t) /-'(r) : }; do^ain,

(-*,0) u (0,+-;; (")D,v:-X,o"r:- i 13. (a)/(r): 'z!#,(b)noinverse;(c)D,t:+,o,r:f ,

f C

17. ln x - -+ 19. ln 1,6 slugs; ;: ,'fr - f


(0, +-; 19. domain: (-@, +*); range: (-o, +-)

121,.f(-rt ' (r): f f i 23.

15. domain: [0, +m); range: [4, +-;

Exercises 9.4

5. -6xe-3*'

+m); range:

1

r < 8 1

81

4s - 11 . 2x 13 . r ' ( ln r * 1 )/ " (e " * e - " )z

" '

Z t . e r - e - , * C 2 3 . # + C

2 5 . e t - 3 l n @ + 3 ) + C

35. ik, - e-z - 4) cu units : 1.627 cu units 37. -9.17 lblftz Per sec

27. ez

39.

27. "t--' +' " L * l n a

day; (b\ 2.26 sales

19. -tsQ-sx) + C

2

domain:

43. Lnw(e-'- e-t) ft- lb

Exercises 9.5 (P age 4L9) logro 1g. (5 ln 3)3'" 11. 25'.34' '2(5 ln 2 * 8x ln 3) 13'

; 15'

a2r/

' l \ *

21,. x"e'(t" trl + :) 23. 4sxe3',r2(5x * 5ln2) 25. ffi+ c

33. (4 ln 2+ 1,)x- (8 tn 2+ 4)y * 4:0 35. (a) 61 sales per

t7.( logro e)2

(r * 1) log's(r + 1)1 ort

c 2e .5 i " i 6+c

per day 37. 2.999

lg. x\E-tn) (1 + i ln lrl)

t t #+c


1. 8000 rt : 11,300 3. 43'9 g

15. 15.9 years 17 . 15,000 Years

5. 68.4 years 7 . 69 .9 9 ' 1'02 (b\ 42.1" 11. (a) (b) 66T0 13. L1,8.7

Reoiew Exercises f or Chapter 9 (P age 427)

, . n \ * z . s 5 . 0 7 . ( x ) , n , - " " | 1 * e " l n 2 | r | + 1 " ' r ' . | ' | ] 9 . * l n ( 1 * e 2 ' ) * C;

r ' f - - -2 \ Lx x

l logro H0- x ' )

,r. I (e**6-)*c rr. #(3' 2'+ 4)ttz *, rs. *(es-1) tz' *rn 4 Ls' -f+## 21' rn5 srugs;

5# ft from one end 21. o: et - e-t + t; s: st * e-, * t 25. tn(7 - e-'o); tr 27. 82L2 yeats 29' g(x)

(-*, *-) 31. /-r(r): !+:,domain: (--, 0) u (0, +-) 35' 3000 ln * in'lb 37' 8'66 yeats 39' sgn t (1- e

Exercises L0.1- (P age 437)

7 . s in 3 f :3 s in t -4 s ins t 9 ' (a )

(c) -cos f; (d) cos (Lrrr - t) 15' (a)

integer; (b) t : 2nzr, where n rs any

any integer

Exercises 1.0.2 (P age 445) ,-

1 . 4 3 . + s . + 7 . L 2 g . 6 c o s 2 x 1 1 . 3 c o s 5 r t 3 . - #

zL. (sin x2)4r (+n[sin x ' l +8t '#) 23' (cos tr) ' in" (- * * cos x ln

\

27. . sr-n(x --y) - 35. decreasing at the rate of t rad/sec 37. ll\EW

s r n ( I - A ) - t

g{i _&; b) ,\/2 +4 tt. (a) t{-z + :V-s; <r,t +^/z -ffi 13. (a) -cos t; (b) -cos(}z -

c o s ( t z r - t ) ; ( b ) - c o s r ; ( c ) - c o s ( * z r - f ) ; ( d ) - c o s t L 7 ' ( a ) t : ( 2 n * i ) a ; w h e r e z i s a n y

integer 19. (a) f : (2n+*)rr or (2n*$)n, where r is any integer; (b\ t: (2n t*)2, where n r s

1 n 1 * 2 c o s r ' l g . 3 c o s x c o s 2 xt / . - ; L 7 .

(2 + cos x ) "

c o s a x * 2 s i n z x c o s z x * 2 s i n 2 x * 1

cosa x\M

A.3O ANSWEBS TO ODD-NUMBERED EXERCISES

Exercises 10.3 (Page 45L)

1 . 2 s i n r - 3 c o s r * C 3 . - e c 6 c + C 5 . c o s ( c o s r ) * C Z . * c o s c r - c o s r * C g . 8 r - * s i n b * # s i n 4 x * C7 7 . t x + t s i n r * C 1 3 . - $ c o s 3 r * f c o s 5 r - { c o s ? r * C 1 5 . * f - E t s i n L 2 f * C l z . } s i n 2 / r 3 r - } s i n 8 i s 3 x * C

1 9 . r L s i n T x * t s i n x l C 2 ' l ' . x - s i n r - * s i n 4 r * * s i n 5 r - g s i n 6 r * C 2 g . 1 . 2 s . 3 z z . * 3 5 . 4 v o l t sn

37. 2 sq units 39. *r2 cu units

Exercises L0.4 (Page 460)

3 . 2 x s e c f t a n f 5 . 2 t a n 2 x ? . - W 9 . 2 s e C x - s e c x l ' 1 . . - 3 t z c s c ( t s + 1 ) c o t ( , s + 1 ) l . 3 . 4 c o t f c s c z f2Ycot3r

1 5 . 5 s e c 5 x 1 7 . 3 x s e c r ( l n 3 * t a n r ) 1 9 . ( s i n r ) t " n " ( s e C r l n l s i n r l + r ) 2 1 . - + ( s e C r * s i n , r - 3 c o s 2 r )Vl - cotz t

23 . -cs f (x* .y ) 25 . sec2r tanrs in2y tany 31 . . y : {2 (x+1- In ) 33 . 5VEf tExercises 1.0.5 (Page 465)l . ( a ) * ; ( b ) e , ; ( c ) - i ; ( d ) & t ( e ) t g . 2 T , 4 s " , t o g " 5 . 3 x - y + z : 0 ; x * 3 y - 1 1 : 0 9 . _ + _ + \ A m13. (+ (1 *4n)n , *V2) ,wherez isanyeven in teger ; ( i0 ,+4n) r , -+ {2) ,wherez isanyodd in teger ;109.30 ,


1 . * l n f s e c 2 r l + C 3 . r b l n l c s c 5 r 2 - c o t 5 l l + c 5 . - * c s c r - l n l s i n r l * c 7 . - t c o t 6 x * c 9 . i t a n z r * * t a n e r * c11. - | . co tS3r - r tco t53r*C 13. ( tan3r -co t3r ) +C 15. r - 2 tan lx*&tans* r f C 17 ._#cof 2 f * * co l2 t_ lco t2 t- t+C 19. u -2 tan tu lC 21 . - *csCr*C ' 23 . * ln 2 25 . +* 22 . '+ 31 . ln (2+ \6 ) 33 . gzrcuun i ts


7' (a) -to; (b) q (c) -*n; (d) tr; (e) n (r) -in 3. (") &17; @) i{1; () 2t/2; (d) *V4 (e) 3 5. (a) -*V+ .o) +\6; (c) -}; (d) rfs;(e) -*V5 7. @) 13; b) +\/n e. @) !r; (b) -*a.; (c) z; (d) *a 11. +++ 13. +(1 + V10) 1s. Bb(48 _ 2s\/j)17. #(4\/fr+ \/5)


?. # s. sin-r2y*& 11. csc-,L+ !+F 13. 2\R,

cot- l t 23. -er -# 25. 27. #nmi/min

Exercises L0.9 (Page 486)1 v t/i t/lnz. i tan-' !+c e. f sin-r # r*c 11. *sin-1(|72;-.,u6 rt. +tan-' f f i +c 15. 2 tan-l t / i+ c

t -+t " " - ' (?d)+c 1e. cos- , (?) . . 21. cos- , (+) - \ /s-z i - '+c 23. s in- ' ( \ f r ) - \ /4-2 i - '+c

ZS. t ( r+2)2++ln(2f -4x+3) -* {2tan- ' t /1(x- t ) +c 27. * t * i ln2 29. tn 3. t . tan- t e-or 33.*n

*. ,:n*' 37. 2tr 39. s: 5 cos 4f; amplitude:5, period: +,r,

Reoiew Exercises for Chapter 10 (Page 488)

t. * 3- 4 t df,!fu e. co, 1+1 ,ir, L n. f# ,r. (tan x)li"'(r sec'z--r-12-tan r lri ltan rl)

s . f f i 77 . * x_ f r s i n3 r -$s i n2 r * c |g . f t t ^n - , # *c21 ' . t _ * t an * t+c

ANSWERS TO ODD.NUMBERED EXERCISES A-31

23. * cos 2x - t cos 8x * C 25. +89 27'

all the way 37 . tfr - tn 39. Ln - 1'

4 5 . T : 2 n , P : 2 0 0 4 7 . \ m

Exercises 1L.2 (P age 497)

1 . r ( l n x - 1 ) * C 3 . r t a n r * l n l c o s r l * C 5 . r s i n - l x + t / t = T + C 7 . i t a n - t x ( x 2 + 1 ) - i x + C

9 . *e " ( cos r *s in r ) *C t t . - * ^ / t=F -3 (1 - l z1 t t ' *C 13 . (2 -x ' ) cos I+2 rs in l *C 15 ' * secs r tan r

+ g [ s e c r t a n r + l n l s e c r + t a n x l ] + C t Z . & 1 9 . i ( 3 e a + 1 ) 2 1 ' $ ( g 3 t r a * l ) Z S ' ' , ' - 1 5 2 5 ' ( e ' z + 1 )

ZT. tzr(Je4*1)cuuni ts 29.2( l -e-6)s lugs; f f in f romoneend 31' (4- I r )wlb ag '* (1 -gs-ahoi f t - lb

Exercises 1'L.3 (Page 503)

sq units

3s. fi6320

?n * t [g- z 29. ln (2 + v3) 31. L06",90", ' ! . ' ! .2",52" 33. : | .rr hr; he walks

41. 7r2:4(5 - s) (s - 3) ; a: aG - s) 43. (a) 120 tadlhr ; (b) 60 rad/hr

7 . i ( s i n - r u+u \ f f i ) +C

9.32 s in- l (*x1 - +(16 - xz) 1/2(8r - f ) + C 11. -bx (4x'-s)- t tz + C 13' ln l r +2+ \ trRl + C 1s' # + C

17.

29.

&+c Ls . + -24 \n4\/4 - tantz x

tn *(2rE + \m - \n- 1) + \m - \n

Exercises LL.4 (Page 511)

t . 1"1# l * . 3 rn [C (x-2) ' (x+ 2) ' l

11 . # r * h l ' + 1 l - 1 t " l zx +31 + c 13 .

27. 2n(2 * 6 ln 3 - 2ln 2) cu units

Exercises 1,1,.5 (P age 515)

25. (t tn 2 + +fi"t) sq units

Exercises 1"1-.6 (Page 518)

(ln 3 - +) sq units

(&rr + 3{3)w lb

L5. - tn(3r + Z)zrs(x - L) '

t - t 25. ln 4.5 sq units

25. Y*n

t : f,sin kf

s }*lff i l , i '"1#l -+.+c s -lu+2)-2+c

17.4 ln * -g 19 . -2+ ln ' * 21 ' -4 ln 5 + 13 ln 2 23 ' ln

t , * zx - h- h l ( r - 1) (x + 3) l + c

) q / 8 l n 3 - 8 - , ' , 2 . = . \ 3 1 ' + + 3 3 ' 7 ' 4 l bL 7 ' \ 2 - h 3

' 9 ( 2 - l n 3 ) /

I '

L . + ' " ( # )3 . : h | # l . i t an_12x*C5 . l n | x -1 l * t an_1x*C

s. t " ( ; f f i ) - ; tan- , . - rx(x ' * r ; - r 11. f r r " lex '+3x*11 -#t " ls r - t l

* x(4x2+ e)-1 + C ls. ln l1 * tan rl + Lnru* (z tary- t) * a L7. 6ln 2

71( '"#) -+tan-'1(#)

.# tan - l ( # ) * . 13 * tan - ' 3x

1 9 . l n r * + g l n i * 2 1 , . * n 2 3 . * l n 2 * E n

1. t tan- , (* tan lx) -Fc 3. - ln11- tan l r l *c s . * ln l tan Lx l - r tan2lxr_C z.z t / l l " l f f i l .c

e. !,u.,-,(.""14-f r(Sffi:lJ.., '. 17.2-..,(2+tan+x) +c t3. trn3 ls. zVSh(r + Vs)

17. +\R h(l + +\r3)

Exercises 1.1. .7 (P age 52L)

j.. *f,, - 3,x+ tg \fr - 54 ln(3 + \n) + C

27.*k1n2s1ugs2g.* , . f f i_ ! , , ,+2)_ | - f t ,u^_, , , *h

'*{

3 '"1ffi| *.i - - l

5 . -2\F*+\6nlf f i l * .' ' - - - ' l V t * x - ! 2 1


_ 3 ( x - 2 ) r l o * , l n | 1 + ( I _ 2 ) 1 / s | + c g . 2 \ / 2 x + 2 \ t r + 4 + 4 ! 2 | n | W | - ,

11,.3 tan-r ff i , -L 3{x L C 13 ) tan-r(r * \f-' x+ f f i+C 13. 2 tan- l (x* \m1) +C

Le. ln # 21,. +(54 - 2{3)

1 , 7 . 4 - 2 l n 3Cx

x * 2 + z \ G i + I tCx

x * 2 - 2 V 1 , * x * x 2Exercises 11.8 (Page 525)

l. approx: 0.695; exact: ln 2 : 0.693exact:2 9. 0.248 11. 1.481, 13.21.. 0.882 23. 3.090


1. by Simpson's rule: !r; exact: g 3. by Simpson,s rule: 0.gg1; exact: ln(l + fD) - O.SS1 5.

*t{5 - 0.6044, 7. e,:Q t. - #= €, = 0 11. 0.248 13. 0.883 15. 1,.402

21,. (a) -0.0952; (b) -0.0950; (c) -0.0991

Reaiew Exercises for Chapter 11. (page 532)1 . * r - r z l ss inL6 r *c g . -2 \ /T=7+c 5 . ( r * r ) t an - r . y ' i - .Vv+c z . i r * * s i n3 r *C- ( r - t ; - , - t 6 11 . * s i n 2x - i s i n4x *C 13

| - r l 3 Is t "

l 1 f ;m l +c 15 . * t an3 r - * co t3x *3 ln

2 t+h#-#+c Ls . r - tan - , x * * rn l - l +cl r * 1 l

t r 2s in x2+ {cos x2+c 27 . he t r z (4s in 2 t+cos2 f ) +c 29 .

3. approx: 4.250; exact:4 5. approx:0.880; exact: ln(1 *3.694 15. -0.007 s €r < -0.001 1T. -0.5 = €r < 0

!2) : 0.881 T. approx: 1.954;L9. -#e s €r

by Simpson's rule: 0.G045; exact:

17. nrzh 19. inhr(rr, * rzz * r{z)

e . ln l r - 1 l - 2 (x - 1 ) - '

ltan lxl + C

17.

25.

21,. #r - r# sin 12x - r*a sins 6x * C

* tutt-t (* sin2 x) + C

23 . s i n - rG)+c

\ E - t l n ( 2 + \ 6 )

(# ,0 ,0 ) 1o t . su) tb

t r l . 2 - 1 _ \s r . . f ; \ - " o " n r * j c o s r z r - t c o s 5 n x ) + C i f n # 0

$ . - * c s c r c o t x - * c s c r c o t x * g l n l c s c x - c o t r ] + c

[ c i r n : o

35. ZLn lv - 2 l - 8(V - 2)- , - te - z) -2 + C' l - | t l4 t . i tn

l t -A +C 43. *s in - , ( *e"1 + c

37. - tan-r (cos x) + C ss.2 s in- , (+)* lU-z)(4t - t2)u2 + c

45. -+ cots 3x- $ cots 3x * C 47. *r3 sin-t x * te, + 2)\R + C

49. tan-r(cos r) + C 51. B r".- ' lz sin 3f l + C 53. -O'e,tr iT+ C

sr. #\ /z+f f iOl ;_t-4) + c , , +, .n fx "+r , l y - , * i . * '=

.=+ c i f n * -6 1 ' . I , * l

- @ r c t t n * - r 6 g .

[ * t n ' x * C i f n - - ' ! ,75. a'(tn - 8\6) 77. * ln B - tn 7s. s

91.. * ln I e3. +f es. (a) 1 .624; (b) 1 .s63

ss. \Ei_ \Fn sin-r {n+ C

, \n . / - - \D+ ; t a n - 1 ( l M x - 1 ) + t t a n - , ( t f f i + 1 ) + c

8 1 . t + t n g 8 3 . + 8 s . l , - + l n 3 8 7 . h r

e7. e\n- 3\6 * g h Q\m+ 3\6 - 4\n - 6)

65 . t+z lng 67 . +o- E \ / ' 6s . # \ t r - tn 71 , . + 73 . +_+tn2

89.

99.

103 .7 : t (# ) ;a :+ (# )105 . (a )x :300(m) ;@)35 .g4 |b

Exercises L2.1 (Page S4l,)

i ? ' 4 , r 4 x * l,r. S

sec.rt' 5 19. e"(cosh r * sinh r) 21.. Z csch Zt

31 . i l n ' cosh x *C 33 . #zs inh Z8x*#s inh I 4x * * r *C '

3 9 . * c o s h 3 2 - c o s h z + 3 4 g . a 2 s i n h T " r u n i t s 4 s . 7 r : e

23. 2x sech x2 2s. rsinh r-r(x cosh r ln lrl + sinh x)

35. t tanh 3x - $ tanh' 3x * c 37. 2 cosh tfx + c-cttzl(B - tcA) sinh t + (A - tcB) cosh fl;


a: Kf * Kra, where Kr: 1-+c2 and Kz:-cn : s-ct,r l(A _ cB * IczA) sinh


11. tn(* + +\n) 13. * ln 3

25. -csc x cot x/lcot rl


Trzr : f f i

(a) 4r sin 0 - 12 cosz

t+ (B - cA + Ic'B) cosh

4 ' ^ |1 F lL ' '

1 , - 1 6 x 2 L / '

2 x * 3 x 221.. Z* (rorh-r x2 +4\

\ \/xa - I/23. lsec r l

27. (x' * y ') ' - 4(x' - y ') 2 9 . 4 x 2 - 5 y ' - 3 6 y - 3 6 : 0

1 . s i n h - l L x + C : t n * ( r + { F + 1 ) + C 3 . ! c o s h - r x 2 + C : } l n 1 r ' ? * t / F t ) + C

' l_t::ll-l,gli: I |;| :;]:*!" lffil *. 7 sinh-' T++ c:,n (sin x+ r'---coi' x) + C

, [+tanh- ' f f i +c ntx+2t ' *J:-1=m

1{++?+r, l+c, , . r (*@uu) 13. rn3 rs. } r . , '#' l+cotn- ' f f i +c i r lx+21' *J '&^ l {o-z- ' t ' -

Reoiew Exercises for Chapter 12 (Page 552\

5. 5 cosh, 2x sinh2x z. ffiffi

9. 2w sinh-L2w t #

11. ln cosh ix'+ c

13 . * cosh rs in r - * s i nh rcos r *C 15 . t anh l ' 19 ' + t


1. (a) (-4,*a); (b) (4,-t,n);(c) (-a,-*z') 3. (a) (-2,*a'); (b) (2,-8tr);(c) (-2,-it) s. (a) (-rD,*z); (b) $/1,-in);

G) ?11,-Zd z. (3, t r ) ; ( -3, *zr) s . ( -4, -&d; @,- t r ) 11. ( -2, i f l ; Q, Izr ) 13. (a) ( -3,0) ; (b) ( -1, -1) ;

( c ) (2 , -2 {5 ) ; (d ) (0 ,2 ) ; ( d?12 , {1 ) ;$ ) (+ \ / 3 , -+ ) t 5 ' r - - l a l " ' ' : r } *

Le ' r2 :4cos20

3q stn 2021'. r:

fu 2g' (x' * y')'-- 4xA 25' (x' * y')t -- xz

Exercises L3.2 (P age 566)

4 1 . 0 : 3 n , 0 - t n 4 3 . 0 : i r , 0 - * n 4 5 . - L


i,. (t,*t); (t,Et) 3. pole; (t/7,*d 5. (fu,tzt); (-*n,-ir) 7. (1',&n); (1,it); ( i,tn); (i,3,,);

(0.22,g.82) 9. pole; Glr , tQn+\)zr) ,wheren:0,1, . . - ,7 11. pole; (VT5,cos- ' i l ; ( !15,r -

t z . 1d , i d ;$ ,Ez r ) ; ( 2 ,8 ' . ) ;Q ,J * r ) 15 . po le ; ( i l 2 ' tQn+1)z r ) , whe rez :0 ,1 ' , 2 ,3 ,4 ,5


7. trr 3. '!,53" 26' 5. 38" g' 7- En f . irr 11' *n

G\8, tn); 7go 6' at (-L\E , En)

Exercises 13 .5 (P age 57 6)

13. 0 a t po le ; tn a t (1 , n ) ; in a t (1 , 0 )

t . * r 3 .4n 5 .4 7 . &n g . * t -1 ls in - ' + -g \ / -2 t t . *n - t *15 t3 .9zr -9 15 . a2(2- i ' , )

19. 4 21.. '1.6a2n3

Reaiew Exercises for Chapter 73 (Page 576)

t . r : 9 c o s d - 8 s i n 0 3 . 4 x a * 8 x ' y ' * A y n * 3 6 x 3 * 3 6 x y 2 - 8 ' l ' y 2 - g 9 . 0 - i r r 11 . a ' (+n * +* )

pole; (0.22, 2.47);

cos-r +)

15. 0o at (0, tn); 79" 6' at

13.

L7. i@ + 1,)

g4kr - 1

4k

15.

25.

17. no points of intersection

0 : 4 ; ( b ) x ' : 4 y - 4

19. tn, tn, Ln ' 27. 16n - 24{5; 32n + 24{3


Exercises 1.4.1 (Page 582)

1 . ( 0 , 1 ) ; Y : - 1 ; 4 3 . ( - 2 , 0 ) ; x : 2 ; 8

15. y' : -6x 19. x2 : -A 21. 1,56 ft


t. e : *16; cen ter (2, 3); foci: (2 t \8,

y - + ! +\f170 5. point-circle (-8, +1


3. 16y' - 972 :36 5. 7z - A2 : 16

Reaiew Exercises for Chapter 14

1,. 9(x - 1)' + 5(y - 2)' : 405 3.

s . ( 0 , -+ ) ; y : * ; 1 7 . ( 8 , O ; ; x : - * ;

23. *? in. 25. yz - 10x - 1,0y * 20:0

g 9. Az :20x 11. x2 : -8y

27. xz + y' + '1.0y - g13.

l . x ' 2 l y ' 2 : 1 3 3 . y ' ' : 6 x ' 5 . x ' 2 * 4 y ' 2 : 4 7 . y ' : 2 7 ' t 9 . x ' 2 * 4 y ' , : 1 6 l ' ! . . 3 x ' 2 - 2 y ' " : 6 7 3 . ( - 3 , * ) ;( - 3 , - * ) ; x : -3 ; y : l 15 . ( 1 , -5 ) ; ( - i , - s1 ; y : -5 ; x : t 17 . ( t , 1 ) ; ( i a , t 7 ; y :1 ; x :3 t I e . y2+20x -By -24 :02 1 . . * + 2 x - B y + 4 1 : 0 2 3 . * - 6 x - 6 y - 3 : 0 ; * - 6 x - t 6 y + 2 1 , : 0 2 5 . y " - 4 x - 4 y - 7 2 : 0 2 7 . y " - 2 x - B y + 2 8 : 0 ;

y 2 - t l x - B y + 1 , 6 e : 0 , t ( - * , 9 # ) z r . * - 2 x * 4 y * t : 0 ; * - 2 x - r 6 y - z e : 0 ; * - 2 x - 4 y - 3 1 : 0 ; f - 2 x + 1 . 6 y

* 4 9 : 0 3 3 . x ' 3 - A ' ' : 0

Exercises 1,4.3 (P age 591)

1 . 3xz -24x * U ' : 0 3 . 8x2 -24x - y '+ ay - 4 :0* 1tz x2 ttz

a r + W : 1 ; e > t ,

o r - W : t

5. '/'.6x2 * 4xy * lgyz - l52x + t1,6y * 496: 0


5. (a) 1; (b) parabola; (c) r cos g - -2 7. (a) i; (b) ellipse; (c) r sin 0 : S

(b) hyperbola; (c) 2r sin d - -3 13. (a) 4; (b) ellipse; (c) r sin 0: -S

9. (a) ?; (b) ellipse; (c) r cos 0: -j

1 g . # 6 , s q u n i t s 2 L , ( a ) r : f f i , q @ ) 2 0 , 0 0 0 , 0 0 0 m i l e s


1. vert ices: (-f3,0);foci: (-r1,f ,0);directr ices: x:=E{-5;t: tr t /-S;endsof minoraxis: (0,-+2) 3. vert ices: (-F3,0); toci:(-r1/ j ,directrices: x:-+313; e: *{3; ends of minor axis: 10, -rV6y 5. vertices: (+2,0); foci: (+y'i}, 0); directrices: x:-rt*.V-l};t : * \ / tZ ;2a :4 ;2b :6 7 . ver t i ces : ( l * ,0 ) ; foc i : ( - r1 t r ,0 ) ;d i rec t r i ces : r : :L r t s : f l ;2a : | ;2b : * g .g f -4y2 :96l l . 16 f *25y2:1gg 13 .32 f -33y2- 380:0 15 .2x ] -3y-12 :0 tZ . Zxz-4yz :29 19 .76* f t 2 . t . 98r in . r23. the right branch of the hyperbola l6f -9y2: 111400 25. *rabz 27. 2400w ft-lb


9.

11. (a)

) ? r z - a 2 ( 1 - e z )-v"

l - e2 cosz 0

directr ices: x:2'+313 3. e:*t/-10; cenrer: (0,*); foci: (0,+*lnt/ tm); direcrr ices:

7 . s r+4y2:soo , . " *n ' ' *QtzY -1 1 r . #*

( t i r t ) ' : ,

,u. k - ,r) ' * (y1-s) '_

| 17. . t lnwtb64 39


1 . . e :13 ; cen ter : ( -3 , -1 ) ; foc i : ( -3 , -1 *3{6) ;d i rec t r i ces : y : - t *& t / -e ;asympto tes : -+x+{2y+12*S:O g . s : f f i ; cen ter :(- l ,a); foci: (-L,-3), (-1, 11); directr iceszy:0, y:8; asymptotes2-+2x+ {-Zy*Z-+.y' l :0 5. , :+t/ tg;center: (1, -2);toci:( t ,-z '+ Vi3); directr ices2 y:-2t+t{B; asymptotes: 3x*2y + 1:0, 3x-2y -7:0 7. 25x2 - t44y2:14amn Q + 1 ) ' ( x * 2 ) z 4_ _ :

1,44 g1 r 17. (3, 4)

7. 972 + 4A' :36 9. 3i'2 1,1,. 7'2 + 4y'' : 16 13. 7'z - 4y' ' : 1518

(P age 625)

3 x 2 - y 2 - 3 : 0 7. 8l - -'

| - 3 cos 05. 49(x + 9a)2 + 76(V - 3)2 :576


1 . + 3 . t n 5 . 7 7 . 2 9 . f o 1 1 . - t 1 3 .


1 . 0 3 . 0 s . t 7 . + 9 . 1 1 1 . + 1 3 . 1


1 . 1 a . - l n 1 5 5 . 1 T . d i v e r g e n t g . 2 1 1 . 1 1 3 . d i v e r g e n t 1 5 . ( a ) d i v e r g e n t ; ( b ) 0 1 7 . ( a ) 0 1 9 ' t r

21,. Ln 23. n:g; ? ln #


1. 2 3. divergent 5. divergent 7. divergent

2t. (a) divergent; (b) 0 23. n > _t, #

25' n

L5. d ivergent 17.0 L9. *r r

11. (a) e - +'(b) el l ipse; (c) r sin 0: -2 13. (a) e: g'

( -3, 8 + 2\6) ; d i rectr ices: A:818\6 17. e - tB;

5 r * y + 2 : 0 , 5 x - y * 8 _ - 0 7 9 . 2 1 ' x ' 2 - 4 9 y ' z : 7 2

29. 600 miles 31. (3 I *\ft) million miles 33. the

f . increasing 3. not monotonic 5. decreasing


ANSWERS TO ODD.NUMBERED EXERCISES A.35

(b) hyperbola; (c) 3r cos g - 4 15. e - +\re; center: (-3 , 8); foci:

center: (*1,3); foci: (- t + \m, 3); directrices: r 1 I zb ffi; asymptotes:1

21,. r : ft-e

23. 4tf3na2b 25. (x - 5)' : 8(V - 1) 27. 9n in.

left branch of the hyperbola 16xz - 9y' : 'l',440,000

l n 3 l s . + 1 7 . 1 l s . g 2 1 . 2 2 3 . - 1 2 5 - YL V I . L

15. ez 17 . e,, lg. ez 21,. e-rt\ 23. + 25. 0 27 . 1 31. 2

9. divergent 11. divergent 1'3. 0

)\ \ - t .

-' ' ( , + 1 ) t

7. not monotonic f. increasing 11. decreasing


1 . p3 ( r ) : t + i t / j ( x - * z ) - i ( r - * z ) ' - # l i ( r - * z ) ' ;& ( r ) : #s in E@- t r )a , { be tween}zand r 3 . Pn ( r ) : x * t x3 ;

Ro ( r ) : r , r ocosh f r ; , f be tween0andr 5 . Ps ( r ) : x -1 . - * ( x - 1 ) ' ++ ( t - l ) 3 ;R r ( r ) : -+ t -4 (x - l ) n , t be tween 1 'andx

7. ps(x) : - ln2- t / i (x-1" i l -2(x- in)" -+Vg(r-*z)3;Rs(r) : -areseC{tan2f *seC ( \ (x-* t )a ' f between*zandr_

9 .ps (x ) : t+8x+*x , - * r i ;& ( r ) : a *E (1 * { ) - s r zy , (be tween0andr 1 ' 1 .2 .77828 13 .0 .515 15 .0 '1823 t t ' f f i

lerrorf < a+"6\/2 2L. x: #6

Reoiew Exercises for Chapter 75 (Page 657)

1. +o 3. + 5. +- Z. | 9. 0 11. 1 13. e 15. divergent 17. + 19. divergent 27. tr 23. divergent

25 .PuQ) :1 - i xz - r t axn - t t r l ;& ( r ) : o -#os in f rT 27 . f ' ( 0 ) :+ 29 ' (a )no ; (b )0 31 ' lR3 (+ ) l < r l z -0 ' 005 33 ' *o

35. # sq units 37. $152,500 39. (b) 0


5. + 7. divergent 9. 0 11. 1


13. divergent 15. divergent 17. erts 'l '9-

1

N L1. sn-- f f i ; ,

3. sn: ln

13. divergent 15. 2 17.

s . sn: f f i , t

21,. divergent 23. g

( 3 n - 2 ) ( 3 n + 1 ) ' 3

divergent 27. +++

1 + @ 1

e . + - \ i : 0' ' ) Z J ) n '

n:2 -

29. jt&

13. divergent

+ 6

7.n : l

25.

2


1. convergent 3. convergent 5. divergent

15. convergent 17. convergent

7. convergent 9. divergent 11. convergent

I



1. divergent 3. convergent 5. convergent 7. convergent


9. convergent 11. divergent

1. convergent 3. convergent 5. convergent 7.

17 . absolutely convergent 19. absolutely convergentlRo l <+ 11 . lRn l <# 13 .

23. absolutely convergent

convergent 9.

21,. divergent

0 .113 15 .0 .406

25. absolutely convergent27. divergent


1. (--, +€) 3. l-+, +l 5. 0 7.

19. [ -1 , 1 ]


1. (a) r - t; l-r, t l; (b) y, +; r:1; (c)= t

+oo ^.2n-2(b) \ (-1) "-' lrf=\ilo ,--, +@)n : l

+ @

1 1 .n:o


r. 0.4854 3. 0.7468 5. 0.2483 7.


( - s , g ) g . ( 0 , 2 1 1 1 . ( -

[ -1, 1) 3. (a) r - t ; l -1, t ) ; (b) )n- l

( - 1 , 1 ) 5 . ( a ) r : * o i ( - - , *m) ;

13. [ -1, 1] 15. e) 17. 14, 6)

t/nx"-r; r -'1,; (c)

r - 3; (c) (-2, 4)

+ @

1,9. (a)n:0

+00

e . i ) n ( nn:2

un12

( - 1 ) n *

in !

(b) i# v 2

- l )Yn-z

+@ yn

21, . t .L t n ln : O

nTn

1.3179 g. 0.2450 11,. $ 4,-2" 2n * |

+co '-]):f-.2 5. 1 +1 3 ' 2 t 2 , n * , ' b

n:o

9 . 4 . 8 9 8 9 1 1 . 5 . 0 1 0 1 3 .

t t r . z + * ( x - 4 ) + 2 t ( - 1 ) n - r ' t " 3 ' 5 ' . ' ( 2 n - 3 ) ( x - 4 ) " .n : z 2 . 4 . 6 . ( Z n ) . 4 n '

+# (x - *n1+- ' ;R=*m 1s . ;2W

19. 0.5299 21. 1,.97435 23. -0.223L 25. 0.0415 27. 0.0048

+ @

ls. s.Ltn : O

+co

n : 1

R - 4 13. +- +\B(x - *n) - *(x - *n), + +fi(*- t"r) '

t7. (a) r * *rt + i"x'; (b) 1 * x2 * &xn; (c) tx, * #xn* *ru29 . 0 .2398 31 . a4 :3 ; as - - 5 ; az :2 ; a , l ; as : 6

Exercises 1.6.11. (Page 742)

1 . 1+ 51(2n 5(-1)n "1.

' 3 (2n - 1,)Xn+z; 1

7. r+ 5 )czn+r

nrrir* Exercises for Chapter L6 (Page 742)

n = l

0.3351

2"n!

15. 0.s082

1' 1' t' t' z; s 3. 0' +, t, i+ 1 5. 1.,3, 1,,3; no limit 7. convergent; g 9, divergent 11. convergent;4 -l 2{!13. convergent; { 15' convergent 17. divergent 19. convergent 27. divergent 23. convergent 25. divergent27. absolutely convergent 29. conditionally convergent 31. divergent 33. absolutely convergent 35. [-1, 1)

97. l-3,g1 39. x:3 4r. (-7,5) 43.0.1973 45. 5.0658 47. O.ss86 49.0.0124 Sr. T (1"1)"

rr; (-o.*o)

s3 . i ) ( -1 ) " - ' f f i v2n , ' ( -oo + o ( " v I n ) z n - r

+@) 55.'7

\ -' ' (2n - 1')l


Exercises 1,7 ,1 (P age 7 50)

1. 5 g. 2 5. \n 7. (2, -3> 9.

21. (-2, 2> 23. (7 , 3> 25. (-9, -4>(-2, -7> 1,1,. (5, 6)

27. ({2, V-g) 2e. (a)13. ( -4, 3) 15. (12, -5 ) 17 .

(1 , -z ) ; (b ) (1 , -2 ) 31 . ( -2 , 7>

( - 1 , g )

33. \mLg . (1 , - 5 )

35. \M


1. 6 i+2i 3. -1,4 i+21. i s . z{ f r 7 .5\6 s. # i+Ltmi

(b) -+i + L{zi 15. (a) 2[(cos Ln)i * (sin Ln)i]; (b) i

Exercises 17 .3 (P age 7 62)

1,. 10 3. -1, 11. #rt 13. -+"s 15. 10, -? 17. (a) 0; (b) no k

- +t/vi 23. -#i + ++i 25. 3S\60 27. ({ 2a ft-tb; (b) 24tE tt-tu


1. ( - * ,0 ) and (0 ,41 3 . [ -1 , 1 ] 5 . a l l rea l numbers no t in ( -4 ,3 )

11 . h :2 , k :3 13 . (a ) 8 [ (cos ?n) i * (s in ?n) i ] ;

21,. -+tEii + +tfrli; #ttrii

Q + t ) - ( 1 + l n f ) ( 2+ 4 t + t z )t lezt(Z + t )g

du20 2t. fr: o,

7 . t t ; t 9 .

79. 2y + 5t/3x -1 1 . - - c o t t ; - =a a ' 1 .r. ' t 1,3. (y - 1) 2 : x 15. 25xz - 9y' :225 17 ' A : 1 ; x : - 1

a : - 1 .4 "udx2 4r, #:0

25. x2t3 + yzts - a2t3 29. 3ta2


1. 4i 3. 2i 5. R',( t ; -

R" (f) : -zt(L + t2)-2i * (ln

2L . ( f l n t - t ) i ++ f3 i +C

Exercises 1,7 .6 (P age 7 85)

1. 1 ++{1tn(1 + f i l 3 .

15. ia(0 ' - I s in &0' )


1. (a)Zti+ i; (b) 2i; (c) tE; (d) 2 3. (a) i * tan ti; @) secz

(c) * {%; (d) + {W 7. @) 2i + 6i; @) 2i; G) 2t/t0; (d) 2

13. ,#r*#i 1s. (a) 3ea,625 ru (b) Yu,1,9. 40o 8' 21. 283 ftlsec


+ z - 7 . , 2 t ) f , , 1 - f '1 . r ( r ) : f f i i + f t i ; N\ t ) : f t i + f r5. T( f ) : cs in k f i * cos k f i ; N( t ) - -cos k f i - s in

N(r ) :# t -# is(Zsp'

Y - ' vo )

} t i +2 i ;R " ( t ) - z i 7 - R ' ( f ) - 2e2 t i + t r t i ; t r ( ' ( t ) : 4ez t i - t - ' i 9 '

2)r2, i 11. (2t - g) (2t2 - 6t * 5) -trz 13. 20t 15. Seat - 8e-at

2 3 . i ? n + t - * s i n z t ) i * ( - n + t + t s i n 2 t ) i 2 5 ' x 2 + ! ' : 1 ; 0

R'( f ) : (1 + tz)- ' i + ( lnZ)2t i ;

19 . ln l sec f l i - h l r l i + C

filef|a, - (13)',,1 5. 6a 7. afln cosh 2 *ln cosh 1] 9. 2tra 1'l'. 2ra 13. 12

fj; (c) t/2; G) 2 5. (a) cos fi * secz ti; (b) -sin fi + 2 sec2 t tan ti;

s. (a) 3i; @)-ai; (c) 3; (d) 4 1'1.. (a) i + {3i; (b) -fsi+i; (c) 2; (d)2

(c) 1250 {2i- 12s0\nj 1,7. (zs + {ffi) sec; (20,000 t/5 + 800\m93) ft

i 3 r(r) : #{ - #i ; N(r) - #, i * f t ,^u, i

kt i 7. T( f ) (1 + cota f ) - t tz i + (1 t tana S)- t rz i ;

# \E 13 . x : 2 * cos s ; A :3 * s i n s 15 . x = / 3a - z i t ' '= o \ g o )


Exercises 17 .9 (P age 803)

1.+rt 3.+\n s. f f i ; f i .

21. (3, s) 23. (-3 , -!) 2s. ei8\fr 27.

Exercises 1-7 .10 (P age 807)

7. + s. +rt 11. #\f7

1.

W 3r. (0, - i l ;2 33.

1. v(r) :2i+2ti; A(t):2i;rtrt:ofr= t+ffii; nto:oft r+ffii; lv(r)l :zt/TTT; e,<,1:ffi,? . . 1 1 2 - 7 1 4A ' ( r ) : f f i ;K( t ) : ro ; t " ) * ;Y(2) :z i+ a i ; r (2) : t6 t +f t i ; A(2) :2 i ; N(2) : t6 t + : . rF i , lv (2) l : z t /5 ; A, (2) :G;

, 1A* (2 ) :+ ; K (2 ) : t o i 5 3 . V ( f ) : - 15s in3 t i+15cos3 t i ;A ( f ) : - 45cos3 f r -45 s in3 f i ;T ( t ) : - s i n3 f i * cos3 f i ;

N ( f ) : - cos3 l i - s i n3 t i ; l v ( t ) l : 15 ;A r ( t ) : 0 ;A ru ( t ) : 45 ;K ( t ) : * ; v ( * r r ) : - 15 i ;A ( *z ) :a5 i ;T ( *z r ) : - i ;N ( * r ) : i ; l v ( * z ) l : 15

s .v( r ) :s t i -s - t i ;A( t ) : r t i+e- t i ; r ( t ) :#r - f f i i ;N(4:#t* f f i i ; lv ( r ) l :T ,Ar( t ) :# ,2 e t 2 d t 1 t 1 1ANG) : f f i , x ( t ) :d tw ;v (0 ) : i - i ;A (0 ) : i+ i ; r to ) : * i - f t i , N (0 ) : ib i+ f t I l v (0 ) l : t / - z ;e ,@'1 :s ,

An(o) : A ;xOl : f t 7 . R :2 i+4 i ; i t k :cons tan tspeed,v :# r *$ i , o :# i -E t , r :# r *$ i ,N :+ , -$ i ,l,2tE

A r :0 ; e r : i

Reaiew Exercises for Chapter L7 (Page 807)

r. -zsi+ 6si s. -22i+ loi s. sz ?. ikr -

ufu i s. h:-L; k:tr tt. # 13. (a) all real numbers in [0, +-)

excep t t -1 ; (b ) * i+ t i ; ( c )G# t * f f i i l s . x : r2 ; y : t 6 ; y : -16 t z . * \ / 9+*h (6+ tE ) n . l a t l

*. fl:-r*,,,?o:fi sec r cs* 27. A7-;Alie;*i m ; e*76=S7,: zs. r(t):#r* 717i,N(f) :A# r*ff i i 31. x:2(3s* r7tff i lzrs -yr; y :*{(3s* 17 : 'rTT'lzrt -75ysrz 33. h:al(oo sin c),

97. V(t):2 sinh 2fi * 2 cosh 2ti; A(t) = 4 cosh 2li * 4 sinh 2fi; lV(t)l:zt/&-et;arfrl :#;ar(r): U.#7


1 . (b ) (7 ,2 ,0 ) , ( 0 ,0 ,3 ) , ( 0 ,2 ,0 ) , ( 0 ,2 ,3 ) , ( 7 ,0 ,3 ' ) , ( 7 ,0 ,0 \ ;G ) @. 3 . ( b ) (2 ,1 ,2 ) , ( - L ,3 ,2 ) , ( - r , r , 5 ) , ( 2 , s ,2 ) , ( - 1 , , s ,5 ) ,( 2 , r , 5 ) ; G ) { E 5 . b \ 3 f r ; ( c ) ( 0 , 0 , 0 ) ; ( 1 s , t 8 , 1 2 ) , ( 1 s , 0 , 0 ) , ( 1 5 , 1 8 , 0 ) , ( 0 , 1 8 , 0 ) , ( 0 , 1 8 , 1 2 ) , ( o , o , 1 2 ) , ( t 5 , 0 , L 2 )7.(a) t ; (b) (* , -1,2) 9. (a)E; (b) (* ,+,+) l r . f ta t f t ,a , r ' , 17. spherewi thcenter at (4, -2, - l ) andr :5 19. thepoint (0,0, 3) 21. the empty set 23. f + (y - l)'t (z * 4')2: r2,lrl > 0


z. <21,-rs,-21 e. (-2s,-26,s, tt. zt/Ts-s{n B. <-re,-16,-r') rs. (-6\/-%,-s\/n,-2\/q,t t7. a:b:0

t t .#,#,# n.-#rt-#*-k B. (+.o,-*) 2s.7(-+i++i++k) 22. \ /E(hr*hi-#*)

2e. cos-r 1! o, "or-' (- +r) 3r. (+, +, +) 33. (c) r : 8, s: -7 , t: S

ANSWERS TO ODD.NUMBERED EXERCISES 4.39


5. -44 7. -468 9. <-84, t98, L24>

25. g

Exercises 1-8.4 (Page 835)

n. -++fi 13. (a) -3; (b) (2 , L, -2> ts. +\M. 2t. ?r\n zs. e\re ft-lb

' t . x - t 2 y - 3 2 * 1 : 0 3 . x - 3 y - 4 2 - 3 - 0 5 . 3 r * 2 y + 6 2 : 2 3 7 . ( l , - i , ' ? ) ; \ - & ' * ' - ? ) 9 . ( + i , a \ , - * 3 ) ; ( - r t , - f r , + 3 )1 7 . t \ I ? t \

t t . ( t r , 0 , ? r ) , f tC ,o , - f r ) 13 .5 r -3y *72+14 :0 15 .2x -y - z+L :0 17 .4v -32 - r : 0andz :1

rg. -lt 2r. +g\/6 x. tExe rc i ses lS 'S (Page840 )

, . x - 4 v *5 z -20 - x y zt . x : t + 4 t , y : 2 - 3 t , 2 : t , ' 1 o 1 : ' = a ' , r : , 3 . x : 4 + t , y : - 5 + g t , z : 2 0 - O U T : + : ; 5 ' 1 3 : = f r : 4

9 . 8 x - y - 6 6 : 0 ; l 1 x - 5 2 - 7 0 2 : 0 ; l 3 y - 4 0 2 * 4 2 : 0 l l . 4 x * y + 3 : 0 ; 3 x - z * 4 : 0 ; 3 y * 4 z - 7 : 0 1 3 . t \ 6

15. 4r* 7y-32+7:0 77. 4x*2y- 32+5:0 7g. ' l t3 : ry :+ 2 l ' * \m

Exercises 1-8.6 (P age 851)

1. (7 , 13, - \1) 3. -490 11. (9, -1, -23>

31. 5x - 2y + 7z:0 33. #


ts. 3\n 2r. \re n. e\89

g. x2 + z2:4a 11. x2 + 4y '* 422: " l '6

Exercises 1,8 .8 (P age 864)

1. ellipsoid 3. elliptic hyperboloid of one sheet 5. elliptic cone 7. elliptic paraboloid

11. elliptic hyperboloid of two sheets 13. (a) 1 < lkl < rD; (b) lkl < 1. 15. *$z 17. *tabc


9. hyperbolic paraboloid

1 . T ( r ) : ( 5 + 9 t r 1 - r z 1 i - 3 t 2 i - 2 k ) 3 . T ( t ) : * 1 6 [ ( c o s t - s i n t ) i * ( c o s f + s i n l ) j + k ] S ' t / n + * + i l n ( 4 + t / n )

z. t/3ka-D i1. R(r) :t i*eti*tetk 13. T(1) : iut/4i++\hhi++VIat;N(1):-*V255i -rir. ' /-zoeia4'../256k;

B(1) : # (3i - 3i + k);K(l) : iE \ff i 15. r(-1) : +(i + 2i - 2k) ;N(-1) : +vsi - + v5i - *V5'k;

B ( - 1 ) : - & : V - s i - + \ 6 k ; K ( - 1 ) : t \ / 5 1 7 . r ( o ) : * V s ( i + i + k ) ; N ( 0 ) : - + V 2 ( i - j ) ; B ( 0 ) : - * v 6 ( i + i - 2 k ) ; K ( 0 ) : + \ D

19 . V ( *z ) : -a i4k ;A ( tn ' ) : - a j ; l v ( t n ) l : \ / 7+7 21 . v ( l ) : 2e2 i -2e -2 i *3e2k ; A (1 ) :4e2 i+4e -2 i+8e2k ;

lv(l)l: {trVT@ 27. 2

Exercises 18.L0 (Page 876)

1. (a) (0,3, s); (b) (-1r,1r\/s,-a); (c) (cos 1, sin 1,1) 3' (a) (V6, t/2,2t/7); F) (0, zV5, 2); (4 Gl',t,-\/3)

s. (a)(2, fu, -215) ;g)(o, tn, -11) ;G)( /6, fu, {s) 7. e l l ipsoid; 12+422: ' ! .6 9. e l l ip t ic conel12cos20:322 11. sphere;

pr-gpcosd:0 13. rightcircularcylinder; p2sin2 6:9 15. (a) rightcircularcylinder; x"*A"-- 16; (b) planethroughzaxis;

!:x 17. f -y2:7t 19. (a) sphere; f lyztz2:81; (b) plane through zaxis;x: y; (c) conewith vertex atorigin, z: IFT!

zr. x!€TfTE :zy 25. 2rlF17

Reaiew Exercises for Chapter 1.8 (Pnge 878)

3. the r axis 5. the circle in the rz plane with center at the origin and radius 2 7. the plane perpendicular to the ry plane and

intersecting the ty plane in the line y : a 9. the solid of revolution generated by revolving !2 :92 about the z axis 11. the solid

of revolution generated by revolving /: r about the r axis 13' i + 26i - 15k 15' -3 17' 7\/wd 19' 16

12 7 . + ( i + i - k ) 2 e . 2 0

V 3 \ '

13. y'+ z2: sinz x 15. xz - z2:4; z axrs 17. z: f f i ; y axis

A.4O ANSWERS TO ODD.NUMBERED EXERCISES

2L . <60 , -40 ,80 ) 23 .29s 25 . ( r+2 ) r+ (y +1 ) r+ ( z -3 )2 :17 27 .3 2s .+ t \ / , 31 . r - 6v - toz+23 :0 33 . 3

3s. *\fr sr. :: L:14 _3 I ; , : 4t, y : -3t, z: t 39. 24 4 1 . v ( t r ) - _ l 2 n i + i + k ; A ( t n ) : - 2 i - t n i ; | v ( t n ) | : * \ f f i

43. (a) z : r2(7 * sin 20) * l; (b) rr(25 cos2 0 * 4 sine 0) = 1gg

Exercises 19.1 (Page 887)' 1 . . . . f

* u z . f + 2 w + u 2r ' \a ) -V ; fo lV ; : f7 ; (c )7 -AT; ; (d )0 ; (e ) theseto f a l lpo in ts (x ,y ) inR2except thoseonthe l ine x :y ; ( f ) ( - . ,+ - )

3. domain: set of all points (r, y) in R'interior to and on the circumference of the circle I + yz:21except those on the line x:0;range: (-rc, 1o) 5. domain: set of all points (r, y) in .* interior to the circle f + yz:25 and all points on the y axis except (0, 5)and (0, -5); range: (--, *-) 7' domain: set of all points (r, y) in R'z except those on the line r : y; rantei (-€, +a)9. domain: set of all points (r, y) in R'z except those on the r axis; range: (--, +.) 11. domain: set of all points (r, y) in Rz forwhich ry ) 1.; range: (--, *-) 13. domain: set of all points (r, y, z) in R3 for which lrl - I and lyl = u nnge: (-r,2r)15. domain: set of all points (x, y) in R ; range: [0, +o) X7. domain: set of all points (x, y) in Rr; range: (-@, ld] 19. domain:set of all points (/, y) in R'z for which x + y, = 10; range: [0, +-; 27. h(x, /) - sin-r \/f=F=T; domain: set of all points (r, y)in R2 interior to and on the circle f + yz:1 29, (a) 2; (b) 6; G) {V - yr; (dl lx - yl; (e\ lx - yl

Exercises 19.2 Page Bgg)

1. 6: *e 3. 6: min( l , *e) 5. 6 - min( l , *e) 17. 0 lg . Z Zl . In27 . limit does not exis t 29. limit exists and equals 0

23. * 25. limit exists and equals 0


1' continuous at every point in Rl 3. continuous at every point (r, il * Q,0) in R' 5. continuous at every point in R29. all points (x, y) in R2 which are interior to the circle * + yz:15 11. all points (x, y) inR, which are exterior to the ellipse4*+9yz :36 L3 ' a l lpo in ts (x ,y ) inR2whichare ine i ther the f i rs to r th i rdquadrant 15 . a l lpo in ts (x ,y ) inRzforwh ichlxyl > 7 17. all points in P 19. removable; /(0,0) : O 21. essential 23. continuous at every point (r, y, z) in Rs forwhich rz * y2 * zz ) 1. 25. continuous at all points in Rl


1 .6 3 .3x -2y 5 . f f i 7 . xz -6xy *22 s . xy+

t .#(r t " i - . ) Ls. f f i zr .+xy+| zs.(b) 1 31, (a) -2; (b) 0 33. 4 35. -4 d,eglft; -g deglft

,-,, 5000 ln L.06. ,r, 5000 ln 1.06 ..(c, -(1J6X ; (d) - -57


1. (a) 0.5411; (b) 0.54 3. (a)0.21,47; (b)0.214 5. (a)2(nyo-vo) Lz* (h2-2xo) Ly+.(wLx+ Lx6y) ax*2(xsa,x_ax) ay;(b) e, : ao Lx + Lx Ly; e2:2(ro Ar - Ar) 7. 1^72nyo

M + yo(Lt)-.',- n'. Lt--i7 + vo ti-;@) e' : ##q, e :Af;ffi!1 5 . 6 r - * v 1 5 6 1 7 . E z : \ / s - r 2 r . ( a ) ( y s - a ) u * 4 L y * ( 2 z o - r o ) & - L z L x * A r A y * L z L z ; ( b ) . r - - M , e z : L r ,es: Az 25. 0.L4 in.; l.4%o 27. tt#;0.325Vo 29. $1200

( ^ - - , 1 x l37. D,f(x, i l :1" sm

@t- G,l: lcosvFfi i f (x'v) # (0'0)[O if ry: (0, 0)

y t * z t 1 1 . 4 t5. -2 sin 30 sin 2f

xzetuz . &

25. -ln sin x; ln sin y zg. (a) 1;

sr (") # l##' -'], tu) 0,#36 t#e -'] - -24 4;


( L Y 1D,f(x,y) -lzv sin6,-Gifr*'tm if

[o irExercises 19 .6 (P age 932)

r. ff: 6'1 6x- 2y; (b) 16r - ros;ff: @) -2x - 4y; (b)'r0r - 6s a. ff: 1a) '$ <r, sin t - y cos t); (b) 0;

6 u . . 2 r e u t r . . - A u 6 r d * s c o s r s A u 3 r 2 e " I r c o s r s- a , . @ ) - ? ( y s i n f + 2 I c o s f ) ; ( b ) 2 d B n l s e c ' u . - u , : @ , a s : @ t

0 u 5 s a7. 7: - s inh t (xe ' - U) ;d r x ' x

( x , y ) * ( 0 , 0 )

( x , y ) : ( 0 , o )

#:*s i* f ;Qxe,-yr , ) :0 9. #=rrs inOcos 0*2y s in{s in d *2zcosf ,#:2: r -cos$cos d* 2yrcosf s in d- 2zrs in6;

A1t

# - - r r r s i n { s i n 0 * 2 y r s i n S c o s e n . ( a ) e " ( c o s t - y s i n t ) * e u ( x c o s t - s i n f ) ; ( b ) e c o 8 t ( c o s t - s i n 2 f ) + d t n ' ( c o s z f - s i n l )

r s e c 2 t - y s i n f + r y @ ) t a n f s e c t 1 , :

t x e t - y13. (a) ; (b) tan f sec t 15

@

0 z 3 v - 6 x - 4 2 E z 3 x - 2 y19. - :L ' '

d x 2 z * 4 x ' a y

2 z * 4 x

x - ' / . , n - e5 . e x - A : 0 ; _ e

: T , z : 0

n. fr:-i,fr:W##W n. fr:cos a, sinh o ff * ,,., a, cosh o fi,ffi:-"t"a, cosh u ff * "o, w sinh ofi29. decreasing at a rate of t rad/sec 3f increasing at a rate of 16 in'3/sec


1 o?_y,@r#3. (a) 4e2" sin y; (b) -e2' sin y 5. (a) 2 ""-'f;- ffi;

(b) 2 tan-r i. ffi 7. @) 3v cosh x;

(b) ,tr sinh y s. (a) 0; (b) es 17. (a) r2r +4s; (b) 4r - 30s re. ("1 (",;#}p, Ol {f${1ff 21. neither exist

23 . f p (0 ,0 ) : -1 ;6 ' (O ,O) : t 27 .6se ' -8 (2 * r ) -gs -x 29 . l 0cos20+ '8 3 ' l ' . - l 0 t s i n20

Reuiew Exercises for Chapter 19 (Page 941)

, . ? t " tU -1 * r - ! t , 3 . f 2cos s t z * td ;2s t coss t ' *d ;2 f ( cos s t2 -s t2s ins t2 ) *e " , . # r -h rh'' 3Y'' 3Y" ' 3Y"

- . . a u 6 y ( x + y ) , ̂ , , . . , , - . " , . d i l - a y G l ) 3 u l R tz. (^) i:-i#+ 3ln(f + y\;i: 2rn(f + v'l; (b) i: G, - zs) 7Vle1,* 3 ln(8s2 + rBP);

Art Rs

* : t s , - z r ) ng i sp -2 ln (8s r+18P)

9 . ( a )3 rcos r -4 (y *2 r ) s i n t ; ( b \12cos2 f -12s in2 t -32s in t "o " t 4 f : - 164t -lt-1;4,,

11. al l (r , y) such that lr l > *{-y;[0,+-) 13. al l (r , y) suchthat lr l > lVl; [O,tr) 15. al l (x, y, z) such thatz > 0; (-e, 1o;

77. alI(x,y,z) excepty:!z; (--, **) 19. 6:min(l ,ele) 21. l imitexists andequals 0 23. continuous atevery

point in R2 25. continuous at all points (x, y) in R2 not on the lines x:!2y 27. continuous at all points (r, y)

in P except (x, y) : (2n + L,2m I 7),where n and z are any integers 31' 73 cents


1. 2{Lx + stfzy 3. 3x * fry * 4z 5. -42 7. -2

13. (a) (-12 ,2,'!.4>; (b) + 17. #n * ?; *ff i 19. -?t;

25. (a) directior,of nfur-fri;{b) climbing at50 ft per ft; (c) descending at20l2ftPerft; (d) direction.f

fti* nfui o,

1 . 3 .- \m'- tmfExercises 20.2 (Page 956)

1. 2x-2Y+32:17;+:Y#:+

9. -3enla cos #n 11. (a) (-4, -4h (b) -2 - 2\f3

\n 21,. o : tan-r 3 29. (a) -1; (b) -i; 2J T I i L

x + 2 a + 4 z - 63 . 4 x * 8 y + 3 2 * 2 2 : 0 ' 4

: = g : - - 5 -


x + 8 _ y - 2 7 : r - !- 3 2 6

7 . x -y -3 :0 ; + : f f , " :z e . x !2y r2z-8 :0 , ' : r4 :+ :+ 11 . 3x-2y-62*84:0 ;. , ^ x - 2 y + 2 z: -

4 -1 20

Exercises 20.3 (Page 955'S

1. no relative extrema; (1,-2) a saddle point 3. t\/5, rel max at (*r,*n);-t1:/?., rel min at (*zr, *zr) 5. no relative extrema;(0, }) and (0, -|) saddle points 7. (t, - i l 9. (0, 0) and (0, 4) 11. (+, -+, +) 17. +N, +N, *N L9. l:w:h: r:7:*

21. hot test at ( '++aE.-*) ; co ldestat (0,1) zz. (0, f r , -#)*a (0, -#,h) , , zs.2r tbysf tby2i f t 27. s*


- A x - A x - A V _ A V _ A t c _ 0 x A V A A O x 6 x' ' * : - t , aq:-2, ur:- , , uu:-r , complementary l .

* : - l ; i :u, w:z;f i : -+; subst i tutes s.

* :- l ; f i : -z;Av Au

#: t,

#: -r, neither ,

#: -0.4p-t.tqo.s.

fi: o.rr".,u-"", #o:0.4p-o.sq-'r.s. fl:

-r.rr,.,u-2.5; substitute s s. -qp-2i p-t;

2pq- t ; -p "q- " ; subs t i tu tes '1 ,1 . x :qp- ' , y :p 'q - t 13 .250un i tso f f i r s tso lda t$7 .50perun i tand300un i tso f secondso lda t$11..50 per unit, 15. $2000


1 . f ( x , y ) : 2 f - t y z + C 3 . f ( x , V ) : 3 * - S x y - t 2 y 3 * c 5 . f ( x , y ) : 2 x " y ' - 7 * y + 3 x - 8 y + C 7 . n o t a g r a d i e n t

g . f ( x , y ) : ' f = ' V ] " - ' * C 1 1 . f ( x , y ) : x 2 c o s y - x + C 1 3 . f ( x , i l : x e u - x 2 y l C 1 , 5 . f ( x , y , z ) : 2 x y - S x z 1 8 y z * C.2nt2

7 7 . f ( x , y , z ) - - 2 f y * 3 x y z - S y z z - 2 x * z * C 1 9 . f ( x , ! , 2 ) : x z t a n y * C 2 1 . f ( x , y , z ) : e z + z + e v l n z - e " l n y * C

Exercises 20 .6 (P age I88)1 . + o 3 . s s . - # 7 . 1

23. 20* in.-lb 25. 27i_ in.-lb

1 5 . x : 4 , A : 1 6 17. surfaces are tangent

9. 8n 11. + 13. + 15. #+ 2e(e2 + 1) 17. 3n 1.9.8 21, . * in ._tU

27. #naa * a2 in.-lb 29. 3 in.-lb 31. (e, + e4 + e8 - 3) in.-lb 33. 2* in.-tb


1 . p ( x , y ) : r y + C 3 . p ( x , y ) : e , s i n y * C 5 . p ( x , y ) : f y , x y s + 2 y + C 7 .9 . q ( x , y , z ) : * c o s y - y z z - 3 x i 2 z * C L t . 2 B . e 7 5 . - 4 l Z . l S l g .

g(x , y , z ) : * rs * Lz ' - xy * 3yz + C-1,4 21,. + 23. 4 25. 0 27. 3

n . 4

Reoiew Exercises for Chapter 20 (Page 998'S

t . - € 3 . + (1+ V3 ) 5 . no tag rad ien t T .2y+Sz -L2 :0 ; x :2 , | : +

9. rel max at (-1, -1) 11. !*- 3, 13. sF in.-lb 15. 9e2 - I 17. -5i - 14i + 26k- t t 6 t r / i , + t 2 3 2Ie. 2\/3 x'-f x zt/5 21. (a) - -i- degrees per in.; (b)f vI3 aegrees per in. in the direction -

ifut -

ifrt23. 350 units of the first commodity sold at $3 and 150 units of the second commodity sold at g1.75 25. 9@, y, z): yzel" + zz tan x +27. (9,11,15), rel max 29. square base and a depth which is one-half that of the length of a side of the base


1. 50 3. 1358 s.704 7. 'F 9. 1376 71. 68.6

Exercises 2L.2 (Page 1014)

1. 42 S. * 5. + 7. + 9. 1ta 17. tr 13. sts 15. 5l2cu units 17. !$cu units 19. (ttt- lt) cu units.

21. fi sq units 23. 72 sq units * k I:J:'::ffivm=wz=W oa a, 27. (b\ &at; nr,i ffif ex + y) dx dyI


29. # cu units 31. I - cos L


1. 12 sluss; (2, B) 3. 1#tk sluss; (88, 1*) 5. 3ka3 sluss; (&a(2 + n) , &a(2 + zr)) 7. f,tcn sruss; e'#) e. * sluss; (f, 9)

il. 9p slug-ft, Lg. trpa4 slug-fP 15. **p slug-fP 1.7. (a) {a slug-ff; (b) 54 slug-fP; (c) l ViS fU rla slug-ff

19. (a) #nk slug-ft'z; (b) $r(2zr' - 3)k slug-fP; (4 *lo ft; (d) (]zd - #r)k slug-fP 21. lzb\/6 ft

Exercises 2L.4 (Page 1027)

l .6zsquni ts 3. *a2(8*zr) squni ts 5. (4r+ 12\6) squni ts 7.42 'cuuni ts S. r8s(3r-4) cuuni ts 11 ' Izcuuni ts

13. *ztkslugs; (0, ?6) 15. Ykr slugs; (-*1,0) 17. +k slugs; (B*, rifetzr) 19. tj3tk slugs; (0, -#) 21'. &tur slug-ft'z

23. I*knaa slug-ft2 25. +\/fr ft 27. zre(eg-L)

Exercises 21.5 (Page L033)

l. V6 sq units 3. 9 sq units 5. 8z sq units 7. 122 sq units 9. fiIFTT sq units LL. 2na2(L - e-r) sq units

73. 12tr sq units 15. +12 + \A h(l + {2)l sq units


l. + 3. fu- 1 5. tr ' 7. lz 9.6*! 1.L. *cuunits 13. *zrcuunits 15. 4acuunits 17. ttabc cuunits

19. 2!k slugs 21. tQlr- 1) slugs

Exercises 21-.7 (Page 7045)

f . ia" 3. nA\ 5. tnat 7 . t abnk slugs 9. l250nk slug-ftz

17. ffna}k slug-ft2 79. "1.8n 21. #nQfi- 1)

Reaiew Exercises for Chapter 21, (P age 1046)

1 . t 3 . * z 5 . E e a - I e z + e - * 7 . * r 9 . + 1 1 . t 1 3 . L n l n z 1 5 . ! ( 1 - c o s 1 ) 1 7 . 3 2 r 1 9 . * s q u n i t s

21. a+P cu units 23. 18 sq units 25. (*zr - $) sq units 27. (2, il 29. rtr?r - 7) 31. *(7ea + 1) slug-ff

33. k(zr 4€) slug-ft'z 35. 2ktr slug,ff; +\/G ft 37. 65kn slugs 99. #Tk slug-fP

1,1 , . (0 ,0 ,?a) 13 . t * t< s lug- f tz 15 . (0 ,0 ,+)

Absolute extrema of functions, 185,956-966

on a closed interval, applications in- .volving, 189-795

problems involving, 213-219Absolute maximum value, 185

of functions of two variables,956Absolute minimum value, L85

of functions of two variables,957Absolute value, t4-21Absolutely convergent infinite series, 70LAcceleration, instantaneous, 158, 786

normal component of, 805tangential component of, 805

Acceleration vector, 786, 870Accumulation point, 893Addi t ion,4

of vectors, 748, 820Additive identity, existence of for vectors,

75'1. ,822Additive inverse, 5Algebraic function, 60

differentiation of, 130-137Altemating series, 697Alternating-series test, 698Analytic geometry, 21,-55Angle, 431

between two curves, 465between two lines, 462-463between two planes, 832between two vectors, 757, 826direction of a vector, 818of inclinatron, 46'1.

Antiderivative, 253Antidifferentiation, 253 -260

application of to economics, 269-272chain rule for, 257-258and rectilinear motion, 265-268

Apollonius, 588Arc, of a curye in R3, 868

length of ,779-784as a parameter,794-795of a plane curye, 372-378rectifiable, 374

Archimedes, 31Lspiral of , 565

Area, 281-287of'a region in a plane, 324-329

of a region in polar coordinates,573-576

of a surface, 1028-1033Associative laws, for real numbers, 4

for vectors, 751,-752, 822Asymptotes, of a hyperb oLa, 597 , 614-615

horizontal, 171-173vertical , 172-173

Auxiliary rectangle of a hyperb ola, 615Average concept of variation, 230Average cost cuwe,232Average cost function , 230Average value of a function, 309-310Axes, coordinate, 22

rotation of, 620-625translation of, 583-587

Axiom, 4of completeness , 9, 669-670of order, 6

Axis, conjugate, of a hyperbola, 604major, of an ellipse, 503minor, of an ellipse, 603polar, 555principal, of a conic, 590

of a parabola,579of revolution, 330of surface of revolution, 855of symmetry,363transverse, of a hyperbola, 604

Bases for vector space,754, 824Bernoulli, Johann, 389Binomial series, 738-742Binormal vector, 869Boundary conditions, 262Bounded sequences/ 667-673

Calculus, fundamental theorem of,31,1,-319

of vector-valued functions, 772-778Cardioid,563Cartesian coordinates, rectangulat, 22

three- dimensio nal, 812Cartesian equations, 558, 764

of the conics, 599-605of a plane, 830in three-dimensional space, 855

Catenary, 540Cauchy, Augustin L., 631

Cauchy-Riemann equations, 941Cauchy's mean-value theorem, 631,-632Center, of a circle, 43

of a conic, 591of curvature, 804of mass , 352, 1016-1018of a plane region, 356-364of a rod, 351-355of a solid of revolution, 366-371,of a sphere, 815

Center-radius form, of an equation of actrcle, 44

of an equation of a sphere, 816Central conics, 59LCentral quadrics, 859Centroi d, 360Chain rule, 138-1,41

for antidifferentiation, 257 -258

general, 926-932Circle, 43-48, 588

center of, 43of curvature, 800degenerate, 607 , 609equation of, center-radius form, 44

general f.orm, 44osculating, 800point-, 607radius of., 43

Circular-disk method for volulr€, 330-332Circular helix, 866Circular-ring method for volume, 332-335Cissoid, 567Closed ball, 889Closed disk, 890Closed interval, 10

absolute extremum on, 1'89-t95Closed rectangle, "1,002

Closure Law, 4Commodities, complementary, 968

substitute, 968Common logarithms, table of , A-14-A-15Commutative laws, for real numbers, 4

for vectors, 751-752,757, 822, 825Comparison test, 685-687Complementary commodities, 968Completeness, axiom of , 669-670Components of a vector, 746,76'1., 8'l'8, 826Composite function, 58, 884

derivative of, 1.38-742

Index

Absciss a, 22

A-45

II

A-46 INDEX

Compound interest, 424-426Concavity, 220-225

applications of to drawing a sketch of

the graph of a function,227-229downward,22'l'upward,221.

Conchoid of Nicomedes, 567

Conditionally convergent infinite series,

70LCone, elliptic, 852

generator of, 588nappe of, 588right-circular, 857vertex of, 588

Conic sections, 579-627Conics, cartesian equations of ,599-505

center of., 591central, 591degenerate cases of, 588-589directrix of.,579eccentricity of , 589focus of, 579polar equations of, 592-598principal axis of, 590properties of, 588-591.vertices of, 590

Conjugate axis of a hYPerbola, 504Conjugate hyperbolas, 615Conservative force field, 993Constant, derivative of, 13LConstant function, 59Constant of integration , 3'l'6Constant terms, infinite series of , 673-683

Constant times a function, derivative of,L32

Constrained extrema, 963Constraint,963Continuity, 97-107

and differentiability, 126-130, 9'/..6of a function , 97 -107

on an interval, 177-180left-han d, 178at a number, 97-l0lon an open ball, 903right-hand, 178theorems on, L0l-107

of a function of more than one variable,900-904

of a vector-valued function , 772Continuous differentiability, 920Continuous function, 97, 88L

' Contour curve of a function, 886Contour map, 886Convergence, interval of , 7ll

radius of ,71'1.Convergent improper integtal, 643, 648

Convergent infinite series, 575Convergent sequence, 654, 670-672Coordinate axes, 22Coordinates, cylindrical, 872-87 6

and triple integrals, 1039-1042left-handed, 811polar, 555-577

and double integrals, 1022-1'026rectangular cartesian, 22, 812spheric al, 872-876

and triple integrals, L043-1045Cosecant functi on, 452-450

derivative of, 456integrals involving powers of, 458-470inverse, 475

derivative of, 481Cosine function, 431,-437

derivative of, 438-445integrals involving powers of, 447-451'integration of rational function of,

516-518inverse, 472

derivative of , 478Cosines, direction, 819Cost function, average, 230

joint,97"l,margin al, 23!, 971marginal avera ge, 23'1.total, 230

Cotangent function, 452-460derivative of, 455integrals involving powers of, 468-470inverse , 474-475

derivative of, 479-480Critical number, 184Critical point, 958Cross product of vectors, 842-851'Cross section of a surface in a plane, 854Cubic function, 60Curyaturc,796-803, 858

center of., 804circle of, 800radius of, 800vector, 799, 868

Curve(s), 23angle between two, 455equipotential, 886

' generating, 855plane, length of arc of.,372-378in R3, 864-87Lsmooth, sectionally, 985

Cycloid,769Cylinder, 341., 852-854

directrix of , 852elliptic, 853generator of, 852height of ,34lhyperbolic, 853parabolic, 853of revolution, 855right,34'1.right-circular, 341ruling of ,852as a solid, 341

as a surface, 852Cylindrical coordinate s, 872-87 5

and triple integrals, 1039-1042Cylindrical partition, 1035Cylindrical-shell method for volumes,

336-339

Decay, and growth, laws of,420-426natural, law of, 42'1.

Decimals, nonrepeatin9, 6nonterminating, 5repeatin9, 6terminatrng, 5

Decreasing function, 205Decreasing sequence, 667Definite integr al, 27 6-322

applications of, 324-380definition of, 291,294properties of, 296-305

Degenerate cases of the conic sections,588-589, 620, 624

Degenerate circle, 607, 509Degenerate ellipse, 588-589, 620, 624Degenerate hyperbola, 589, 620, 624Degenerate parabola, 620, 624Del (v) ,947Delta (A) notation, 29Delta (6), 67Demand, marginal, 969-971,Demand curve,235Demand equation,234Demand surface,968Density, linearr 353Dependent variable(s) , 49, 883Derivative(s), L2L-L63

applications of, 204-242to economics, 230-238

of a composite function, 138-142of a constant, 131of a constant times a function, 132of cos u,438-445of cos-r u, 478of cosh u, 539of cosh-t u,546of cot u, 455of cot-l u, 479-480of coth u, 539of coth-r u, 546o f csc u ,456of csc-r u, 48'j.of csch u,539of csch-r u, 547definition of, l2'/-.directional, 945-951of exponential functions, 408-4@, 4L5

first through nth, 157ofa func t ion ,12 l ' -126of higher order, 157-1'62-introduction of concept of,25lof inverse functions, 401-403

of inverse hyperbolic functions,546-547

of inverse trigonometric functions,477-482

from the left, 128of logarithmic function , 417of natural logarithmic function , 384notation f.or, 250-251,one-sid ed, "1,27

ordinary, 908partial, 905-91,2

applications of to economrcs, 957 -975

definition of, 905higher order, 934-939

of power function for rational expo-nents, 1,42-1'46

of the product of two functions, L35of the quotient of two functions,'1.36as a rate of change, 150-L53from the right, 127of sec u ,455of sec-r u, 480-48Lof sech u,539of sech-t u, 547of sin u, 438-445of sin-l u, 478of sinh u, 539of sinh-r z 546of the sum of two functions, 133of tan u, 454-455of tan-l u, 479of tanh u,539of tanh-t u, 546total, 93Lof trigonometric functions, 438-445,

454-460of vector-valued functions, 772, 866See also Differentiation

Descartes, Ren6,, 2LDifference, 5

of functions, 57of vectors,749, 82'1.

Differentiability, 125, 9L5and continu ity , "1,26-130, 9'1.6continuous,920of vector-valued functions, 77 4

Differential(s), 244-249exact, 993formula s, 249-252total, 921.-924

Differential equations, first-ord er, 26'1.partral,94l.second- order, 262with variables separable, 261'-264

Differential geometry, 870Differentiation, of algebraic functions,

130-137implicit, 146-tSAinverse of ,253-260logarithmic, 389, 419

partial, 905of power series, 7'1,3-721of trigonometric functions, 438-445,

454-460See also Derivative(s)

Dimension of vector space,754Directed distance, 28, 81,3Directed line segment, 746Direction of a vector, 747, 8'1"8Direction angles, 81.8Direction cosines, 8L9Direction numb ers, 837Directional derivative, 945-951,Directrix, of a conic, 579

of a cylinder, 852of a parabol a, 579

Discontinuity, 99essential, 99, 90'!.removable, 99, 90'1,

Discontinuous function , 97, 900Discrimtnant, 623Displacement vector, 7 6LDistance, between two points, 28,874

directed , 28, 8j-,3from a point to a line, 220, 827from a point to a plane, 834-835undirected, 29

Distributive law, for real numbers, 4for vectors, 752, 757 , 822, 825, 844

Divergent improper integral , 643, 648Divergent infinite series, 675Divergent sequence, 664Divisiofl, 5Domain of a function , 49, 58, 881Dot product of vectors, 756-762, 825-828Double integral , 1002-1007

definition of, 1003evaluation of, 1008-1014in polar coordinates, 1022-1026

Drawing the sketch of a graph of a func-tion, 227-229

e (the base of natural logarithms), 407,4L6-417

Eccentricity of a conic, 589Economics, applications of antidifferen-

tiation to, 269-272applications of derivatives to, 230-238applications of partial derivatives to,

967 -975

Edges of a rectangle, 1002Elernents of a sequence, 660Ellipse, 588, 602-603, 606-672

degenerate,5SS-589 , 620, 624major axis of, 603minor axis of, 603point-, 609

Ellipsoid, 858of revolution, 856

Elliptic cone, 862

INDEX A-47

Elliptic cylinder, 853Elliptic hyperboloid, of one sheet, 858

of two sheets, 859Elliptic integral, 808Elliptic paraboloid, 861Empty set, 3Endpoints of an interval, 10Epicycloid, 809Epsilon (e), 67Equation(s), cartesian, 558, 764, 865

of a circle, 44of the conics, cartesian, 599-605

polar, 592-598differential, 261,-264

of a graph,25graph of , 23-27 , 8'1,5linear, 38, 835of motion, L15parametric, 764, 837, 865of a plane, 830of rotating the axes/ 627-623of a sphere, 816of a straight line in a plane, 33-40

intercept form, 37point-slope form, 36slope-intercept form, 36two-point form, 35

of a straight line in R3, 836-840parametric, 837symmetric, 837

of translating the axes, 584vector, 764, 865

Equilateral hyperb ola, 614Equipotential curves, 886Equipotential surfaces , 95IEssential discontinuity , 99, 90'1.Euler, Leonhard, 407Euler's number (e), 407Even function, 59Exact differen tral, 993Existence theorem, 67'1.Explicit function, 146Exponential function (s), 405- 41'2

to base a, 4'1.4-419definition of, 4'1,4

definition of, 405derivative of, 408-409 , 415table of, A-5-A-11

Exponents, rational, derivative of powerfunction for, 742-1'46

Extrema, absolute, 1.85, 956-966applications involvin g,'1.89 -195,

21,3-219constrained, 963free, 963of functions of two variables,956-966relative , 182,957

first-derivative test for, 206-209

second-derivative test fot, 2'l '1'-213Extreme value theorem , 1,87 -L88

A-48 INDEX

Extreme value theorem (Continued)for functions of two variables,957

Field(s), force,982gradient,993

Field axioms, 5Finite sequence, 660First derivative, 157First-derivative test for relative extrema,

206-209First derived function, 157First-order differential equations, 267Fixed cost, 238Focus, of a conic, 579Force field, 982

conseryative, 993Fractions, 5

partial, use of in integration, 504-515Free extrema, 953Function(s), 48-60

absolute extrema of ,185, 956-966applications involvin g, 189 -195,

213-219algebrarc, 60

differentiation of ,'1,30-137antiderivative of , 253average values of, 309-310composite, 58, 884

derivative of, 1,38-142constant, 59continuity of ,97-107

on an interval, 177-180left-han d, 178with more than one variable, 900-904at a number,97-101,on an open ball, 903right-hand, 178theorems on, l0L-107

continuously differentiable, 920contour curves of, 885cosecan t, 452-450, 457 -470

cosine, 431-437, 447 -451

cotange nt, 452- 450, 466-470cubic, 60decreasing,20Sdefinition of, 48, 881derivative of, 121-126

total,931difference of , 57differentiability of , 913-920

and continuity,'1.26-130differentiable, L25discontinuous, 97, 900domain of' 48,58, 881,even, 59explicit, 146exponential, 405-412

to base a, 4'1.4-419extrema of, absolute, 185

relative , 182

first derived, 157gradient of , 947 -951, 976-981graph of,48-55, 885

sketching of,227-229greatest integer, 54hlryerbolic, 535-553

derivatives of, 539integrals of, 539-540

identity, 60increasing, 205integrable, 290-291, 1003inverse,395-404inverse cosecant, 475inverse cosine, 472inverse cotangent, 47 4-475inverse hlryerbolic, 535-553

integrals yielding, 548-551inverse secant, 475inverse sine, 472inverse tangent, 474inverse trigonometric, 471,-488

integrals yieldi ng, 484-488level curves of , 886level surface of., 887limit of , 66-97

involving infinity, 88-97, 165-17'1.left-frand, 85with more than one variable, 889-898one-sided, 85-88right-hand, 85theorems on, 74-83,91-96, 174-177two-sided, 85undirected, 85

linear, 60logarithmic, to base a, 41,4-419maximum value of, 181-189,956-966

absolute, L85, 955- relativ e, L82, 957

mean value of, 309minimum value of., l8L-189, 956-966

absolute, 185,957relative, 182,957

monotonic,206of more than one variable, 8&1-887

continuity of,900-904limits of, 889-898

of n variables, 881, 883, 953-966natural logarithmic, 382-394

graph of.,39t-394notation f.or, 56-57obtaining of., from its gradient, 976-981'odd, 59one-parameter f.amily, 252operations on, 57-59periodic, 434polynomial, 59, 884potenti al, 993power,'1,42product of'57

derivative of, L35

quadratrc, 60quotient of , 57

derivative of, 136range of., 48, 881rational, 60, 885

integration of, 504-515 , 516-518relative extrema of , L82, 957secant, 452-460, 467 -470

of several variables, differential calculusof, 88L-943

sign,72signum,72sine, 431-437, 447-451smooth, 374sum of,57

derivative of, 133tangent, 452-460, 468-470

application of to slope of a line,451,-465

total differential of, 921-924transcendental, 60trigonometric, 431-488

inverse, 471-488two-parameter family, 262of two variables, absolute maximum

value of.,956absolute miminum value of ,957extrema of ,956-966extreme-value theorem for, 957relative maximum value of ,957relative minimum value of,957

types of, 58-60unit step, 72vector-valued, 763, 778, 8&

calculus of , 772-778Fundamental theorem of the calculus,

311-319

Gas, ideal, law of, 910Generating curve of a surface of revolu-

tion, 855Generator, of a cone, 588

of a cylinder, 852Geometric series, 678-679Geometry, analytic, 2l-55

differential, 870Gradient field, 993Gradient of a function, 947-951

obtaining a function from, 976-987Gradient vector, 947Graph(s), of equations, 23-27, 815

equations of, 25in polar coordinates, 560-566

of a function, 48-56, 885sketching of, 227-229

intersection of, in polar coordinates,567-570

of natural logarithmic function, 391-394reflection of, 40Lsymmetry of,25

Gravity, acceleration of., 265Greatest integer function, 54Greatest lower bound of a sequence, 669Growth and decay, laws of,420-426Gyration, radius of, 10L9

Half life, 42"1.Harmonic motion, simple, M7Harmonic series, 676-677Helix, 866

circular, 866Hermite, Charles,407Higher-order derivatives' 157 *152

Higher-order partial derivatives, 934-939Homogeneous lamina, 358Homogeneous mass, 353Homogeneous rod, 353Homogeneous solid of revolution,

366-37rHooke's law, 345Horizontal asYmPtotes, l7l-173Hyperbola, 588, 604, 513-620

asymptotes of , 597 , 61'4-515auxiliary rectangle of, 515coniugate, 615coniugate axis of., 604degenerate, 589, 620, 624equilateral, 614transverse axis of', 504unit,768

Hyperbolic cosecant function ' 536derivative of, 539inverse, SM

derivatle of.,547Hlperbolic cosine function, 536

derivative of, 539inverse, 5M

derivative of, 545Hlperbolic cotangent function, 536

derivative of, 539inverse, SM

derivative of , 546Hlryerbolic rylinder, 853Hlperbolic functions, 535-553

derivatives of, 539integration of, 539-540inverse, 543-553

derivatives of., 546-547yielded bY integrals, 548-551

table of , A-12Hlryerbolic Paraboloid, 861'Hlryerbolic radian,768Hyperbolic secant function, 536

derivative of, 539inverse, sM-545

derivative of, 547Hlryerbolic sine function, 536

derivative of, 539inverse, 543

derivative of ,545

Hyperbolic tangent function, 536derivative of, 539inverse, 544

derivative of ,546Hlperboloid, of revolution, 855Hyperharmonic series, 692Hypocycloid, 77'1.

Ideal gas law,9t0Identity, 433Identity function, 50Implicit differentiation, 146-150Improper integrals, 54L-650

convergent, 543, 548divergent, 643, 648with infinite limits of integration,

64L-650Inclination, angle of , 461Increasing functions, 205Increasing sequence, 667Increment of function of two variables,

914Indefinite inte gral, 29'/.,, 3'l'6, 492Indefinite integration, 492Independent variables, 49, 883Independent vectors, 756, 824Indeterminate forms, 629 -641

Index of summation,276Inequalities, 7

nonstrict, Tstrict, 5

Inertia, moment of, 1018-1021polar, L018

Infinite sequence, 660Infinite series, 550-744

absolutely convergent, 70'l'alternating, 697alternating-series test, 698binomi al, 738-742comparison test, 685-687conditionally converg ent, 701of constant terms, 673-683converg ent, 675definition of , 673divergent, 675geometric, 678-679harmonic, 576-577hyperharmonic, 692integral test, 694-696limit comParison test, 687-690Maclaurtn, TSlp, 692partial sum of., 574bf positive and negative terms , 697 -706

of positive terms, 684-693power series, 707 -712-

differentiation of, 713-727integration of., 722-728interval of convergence of.,7llradius of convergence of',711

INDEX A-49

ratio test, 703-7A6remainder of, 699sum of , 575Taylor, 729-737terms of., 673

Infinity, limits involvin g, 88-97 , 165-L7'l'negative,

'1,0, 90

positive, 10, 90Inflection, point of , 220-225

applications to drawing a sketch of thegraph of a function, 227-229

Inflectional tang ent, 224Initial conditions, 262

Initial point,746Initial side of an angle, 43'l'

Inner product, 756-762Instantaneous acceleration, L58, 786

Instantaneous rate of change, LS"l'

Instantaneous velocity , L"l'6-117, 785

in rectilinear motion, 115-L2A

Integers, 6Integrable function, 290-291, 1003

Integral(s), definite, 276-322

applications of, 324-380definition of, 291, 294properties of, 296-305

double, 1'002-L007definition of, 1003evaluation of, 1008-1014in polar coordinates, 1022-1026

elliptic, 808of exponential functions, 409' 4'l'5

formulas, standard indefinite integra-

tion, 492of hyperbolic functions, 539-540

improp er, 641-650indefinite, 29'l', 316, 492

involving Powers of sine and cosine,

447-451involving powers of tangent, cotangent,

secant, and cosecant, 466-470

iterated, 1009line, 981-988

indePendent of the Path, 989-996

mean-value theorem fot, 306-310

multiple, 1002s ign ( [ ) , 291 'single, 1002table of , 492test, 694-696of trigonometric functions, 447 -45'l',

466-470triple, 1'034-1037

in cytindrical coordinates, 7039-1042

definition of, 1.034in spherical coordinates, 1043-1-045

yielding inverse hyperbolic functions,

548-551yielding inverse trigonometric func-

tions, 484-488

A-50 INDEX

Integral(s) (Corz tinu e d)

See also IntegrationIntegrand,29l

lower limit of , 291,upper l imit of ,291

Integration, 316constant of, 316indefinite, 492limits of,29l

infinite, 64L-646miscellaneous substitutions, 5t9-521multiple, 1002-1045by parts, 493-497

formula f.or, 493of power series, 722-728of rational functions by partial frac-

tions, 504-515denominator contains quadratic

factors, 512-515denominator has only linear factors,

504-s1.0of rational functions of sine and cosine,

516-518region of,

'J,002

techniques of, 492-534by trigonometric substitution, 498-503See also Integral(s)

Intercepts, of a line, 35of a plane, 831

Interest, compound, 424-426Intermediate-value theorem, 306-907Intersection, of graphs in polar coordi-

nates, 567-570'of sets, 4

Interval, 9closed, 1.0continuous on, L78of convergence of power series, T'!.Lendpoints of, 10half-open on the left, 10half-open on the right, 10open, 9partition of , 288

Invarian t, 624Inverse cosecant function, 475

derivative of, 48LInverse cosine function, 472

derivative of, 478Inverse cotangent function, 474-4Ts

derivative of, 479-480Inverse of a function , 395-404

definition of , 397derivative of, 401,-403

Inverse hyperbolic cosecant function , 544derivative of, 547

Inverse hyperbolic cosine fun ctioni 544derivative of, 546

Inverse hyperbolic cotangent function,544

derivative of., 546

Inverse hyperbolic functions, 543-553derivatives of, 539yielded by integrals, 548-551

Inverse hyperbolic secant function, 544derivative of ,547

Inverse hyperbolic sine function, 543derivative of , 546

Inverse hyperbolic tangent function , 544derivative of., 546

Inverse of the logarithmic function , 4'l.LInverse operations, 253Inverse secant function , 475

derivative of, 48LInverse sine function, 472

derivative of, 478Inverse tangent function , 474

derivative of, 479Inverse trigonometric function s, 47'1.-47 6

derivatives of, 477-482integrals yieldi ng, 484-488'

Irrational numbers, 6Isothermal surface, 95LIsothermals, 886Iterated integrals, L009

Joint-cost functron, 971.

Lagrange, Joseph, 250, 653, 963Lagrange form of the remainder, 553Lagrange multi plierc, 97 5

' Lamina, 358Laplace's equa tion, 940Latus rectum of a parabola, 58LLaw of mass action, 509Law of natural decay, 42'1.Law of natural growth, 421Laws of growth and decay, 420-4ZGLeast upper bound of a sequence, 669Left*hand continuity, 178Left-hand limit, 85Left-handed system, 8LLLeibniz, Gottfried Wilhelm, 250, gl2Lemniscate, 567Length of arc, of curves,372-378,

779-794, g6g

as a parameter,794-795Level curye of a function, 885Level surface of a function, 887L'H6pital, Guillaume FranEois de, 630L'H6pital's rule , 630-63"1,, 534-695,

637-639LimaEon,562,567Limit(s), of a functi on, 66-97

theorems on, 74-83, 1V4-177of functions of more than one variable,

889-898involving infinity, 88-97, 165-171

theorems on, 9l-96left-hand, 85lower, of integrction, 291

of a sum,276onersided, 85-88right-hand, 85of a sequence, 662of sums, 29"1,trigonometric, 438- 44Itwo-sided, 85undirected, 85upper/ of integratton, 29L

of a sum,275of a vector-valued function ,772, 866

Limit comparison test, 687-688Line(s), equations of in R2, 33-40

intercept form, 37point-slope form, 36slope intercept form, 36two-point form, 35

equations of in R3, 836-840normal , 1!3,954parallel , 38-39perpendicular, 39-40slope of , 33

application of tangent function to,461-465

symmetry with respect to,25tangent, 1,'!."1.-'1,15, 955

Line integrals, 98L-988independent of the path, 989-996

Line segment, directed, 746midpoint of, 30-31, 815

Linear equation, 38, 831.Linear function, 60Liquid pressure, 348-350Logarithmic differentiation, 389, 4I9Logarithmic function, to bas e a, 4'!,4-419

definiti on of , 4'1,6derivative of., 417

natural, 382-394definition of, 383derivative of, 384, 41.8graph of , 39'1,-394inverse of., 4'j-,"1,

Logarithmic spfual,566Lower bound of a sequence, 668

greatest,669Lower limit, of integration, 29"1,

of a sum,276

Maclaurin, Colin, 655Maclaurin seri es, 73'1,Maclaurin's formula, 655Magnitude of a vector, 747, 818Major axis of an ellipse, 503Map, contour, 886Marginal average cost function , 231Marginal concept of variation,230Marginal cost curve, 231Marginal cost function, 231, 97'1,Marginal demand, 969-971Marginal revenue curye, 235

Marginal revenue function, 235Mass, center of, definition of ,352

of a lamina, 1016-1018of a plane region, 356-364of a rod, 351-355of a solid of revolution, 366-371'

definition of, 351, 1017homogeneous, 353moment of., 352,366

Mass action, law of, 504Maximum value of a function, L81-1'89,

956-966absolute, 185

of functions of two variables,956relative, 182

of functions of two variables,957Mean value of a function, 309Mean value theorem, 197-201

Cauchy's, 53L-632for integrals, 306-310and Rolle's theorem, 195-20'l'

Measure,ZSLMidpoint of a line segment, 30-31, 815

Minimum value of a function , 18'l'-189,956-966

absolute, L85of functions of two variables,957

relative, 182of functions of two variables,957

Minor axis of an elliPse, 503Moment, of inertia, 1018-1021

polar, 1018of mass,352, 366

Monopoly,236Monotonic function, 206Monotonic sequence, 563-673Motion, equation of, of a particle, Lls

plane, 785-791iectilinear, and antidifferentiation,

265-268instantaneous velocity in, L 15-120

simple harmonic, 447Moving trihedral,869Multiple integral, 1002Multiple integration, 1002-1045Multiplication, 4Multiplication of vectots, 842

cross-product, 842-85Ldot product, 756-752, 825-828scalar, 7 49 -750, 75'l'-755, 821

Multiplicative inverse, 5Muzzle speed, 788

n-dimensional number sPace, 881'

Nappe of a cone, 588Natural decay, law of'42lNatural growth, law of., 421

Natural logarithmic function s, 382-394

graph of.,39'l'-394Nitural logarithms, table of, A-3-A-4

Newton, Sir Isaac, 250,312Nicomedes, conchoi d of., 567Noncentral quadrics, 861Norm of partition, 288, 1002, 1022' L035

Normal component of acceleration, 805

Normal line, L13to a surface, 954

Normal vector, to a plane, 829to a surface, 953unit, 792,869

Notation, for functions, 56-57

sigma, 276-280rth derivative, 157Number line, 9Number plane, 21'-22Number space, n-dimensional, 881

three- dimensional, 8ll -8t7

Numbers, critical, L84irration al, 6rational, 6real, 4-14transcendental, 407

Numerical tables, A-2-A-15

Oblate spheroid, 858

Octants, Sl2Odd function, 59One-parameter family of functrons, 262

One-sided derivatle, 127

One-sided limit, 85-88One-to-one corresPondenc e, 7 46, 812

Open ball, 889continuity of a function on, 903

Open disk, 890Open interval, 9Open rectangle, 1002

Operations on functions, 57-59

Order, axiom of., 6Ordered fields, 7Ordere d n-tuPle, 881

Ordered pair, 21,746, 882Ordered triple, 81LOrdinary derivative, 908

Ordinate,22Orig in, 9, 2 '1,555, 8LLOrthogonal vectors, 759, 828

Osculating circle, 800

Overhead cost, 230

p series, 692Pair, ordered , 21,746, 882Pappus, theorem of., 366Parabol a, 23, 579-581', 588

axes of., 579degenerate, 620, 624directrix of , 579focus of.,579latus rectum of, 581vertex of, 580

Parabolic cvlinder, 853

INDEX

Parabolic rule, 526-53LParabolic spiral, 567Paraboloid, elliptic, 861

hyperbolic, 86Lof revolution, 855

Parallel lines, 38-39Parallel plane sections, of a solid, volume

of, 34'1.-343Parallel planes, 832Parallel vectors , 758-759 , 827 -828

Parallelepiped, rectangular, 34'l'

Parallelogram law, 749Parameter,764Parametric equations, 7 63-770

of a line, 837in R3, 865

Partial derivatives, 905-912applications of to economics, 967-975

higher-order, 934-939

Partial differential equ atrons, 94'l'

Partial differentiation, 905

Partial fractions, integration by, 504-515

Partial sum of an infinite senes, 674

Partial sums, sequence of., 674

Partition, cylindrical, 1035

of an interval, 288norrn of , 288, t002, 1022, 1035

regular, 293Percent rate of change, 153

Perfectly comPetitive matket, 973

Penod,445Periodic function, 434Perpendicular lines, 39-40Perpendicular planes, 833

Perpendicular vectors, 759

Plane curve, length of arc of , 372-378

Plane motion, 785-79'l'Plane region, center of mass of ,356-364Plane section, 341Plane(s), 829-835

angle between,832cartesian equations of, 830

definition of, 829equation of., 829parallel ,832perpendicular, 833tangent,953-956traces of, 831

Point, 2"1,,22accumulation, 893critical, 958in n-dimensional number sPace, 881

reflection of, 401saddle, 958symmetry with resPect to, 25

in three-dimensional number space,

811Point-circle, 45, 607Point-ellipse, 609Point of inflection, 220-225

4.52 INDEX

Point of inflection (Continued)applications to drawing a sketch of the

graph of a function, 227-ZZ9Point-slope equation of a line, 35Point-sphere, 816Polar axis, 555Polar coordinates, 555-577

arc length in,784area of a region in, 573-576double integrals in, 1022-1026graphs of equations in, 560-566intersection of graphs in, 56T -570

Polar curves, tangent lines of, S7'1-57jPolar equations, 558

,and the conics, 592-598Polar line, 555Polar moment of inertia, 1018Pole, 555Polynomial function, 59, 884Position representation , of a vector, 746,

818Position vector, 764, 870Potential function, 993Power function, 142

derivative of, for rational exponents,L42-1,46

for real exponents, 41.8Power series, 707-712

differentiation of , 713-722integration of, 7ZZ-728interval of convergence of , Tl"!,radius of convergence of.,7l'!,

Powers and roots, table of, A-2Pressure, liquid, 348-350Price function, 294Principal axis of a conic, 590Principal part of a function , g2j,Principal square root, 19Prismoidal formula, 530-531Product, 4

of functions, 57of two functions, derivative of, 135of vector and a scalar , 750, 920

cross/ 842-8Sldot, 756-762,825-B2Btriple scalar, 847

Production function, 9ZgProfit function, 296Projectile, motion of, T8T-7gl

muzzle speed of ,788Projection, scalar, of. vectors, T59-760, g26Prolate spheroid, 858Pythagorean theorem , 29, 40

Quadrants,22Quadratic factors in rational functions,

512-515Quadratic function, 60Quadric surfaces, 858-864

central, 859

noncentral, 86'1.Quotient, 5

of functions, 57of two functions, derivative of, LJ6

R2, equation in, 22graphs of equations in, 23-25

R3 (three-dimensional number space),gtt-8L7

curves in, 864-87'1,graph of an equation in, 815lines in, 836-840

Radian, hyperb olic, 768Radian measure, 43'!,Radius, of a circle, 43

of convergence of a power series, 211of curvature, 800of gyration, '1,9

of a sphere, 815Radius vector,764Range of a function , 49, BB'J-,Rate of change, derivative as, 150-L53

instantaneous, L5Lpercent, L53relative , 152-L5g

Rates, related, L54-lS7Ratio test, 703-706Rational exponents, derivative of power

function for, 142-146Rational function, 60, 885Rational numbers, 6Real number line, 9Real numbers, 4-"1,4Real vector space, TSg-7Ss

basis of ,754dimension ot,7S4

Reciprocal spiral, S7TReciprocals, 5Rectangle(s), auxiliary , of a hyperb ola, 6'l.5

closed, 1.002edges of,'1,002open, 1002vertices of, 1002

Rectangular cartesian coordinates, 22three- dimensio nal, B!2

Rectangular parallelepiped, 341Rectifiable arc, i74Rectilinear motion, and antidifferentia-

tion, 265-268instantaneous velocity in, 115-120

Reflectiofl, of a graph , 401of a point, 401.

Region of integration, 1002Regular partition, 293Related rates, 1,54-j,57Relative extrema, lB2, 957

first-derivative test for , 206-209second-derivative test for, 2ll-219

Relative maximum value, of a function,182

of functions of two variable s, 957Relative minimum value , of a function,

182of functions of two variables, gST

Relative rate of change, 1,52-"1.53Remainder in infinite series, 699

a f t e rn te rms ,6sgin Taylor's formula, integral form of,

655Lagrange form of, 653

Removable discontinui ty, 99, 901Representation of a vector,746Revolution, axis of, 330

cylinders of, 855ell ipsoid of,856hyperboloid of, 856paraboloid of, 856solid of, 330

center of mass of , 366-3TIvolume of , 320-Zi9

surfaces of , 854-857Riemann, Georg Friedrich Bernhar d, ZggRiemann sum, 289Right-circular cone, 857Right-circular cylinder, 34LRight cylinder,34IRight-hand continu tty, l7gRight-hand limit, g5Right-handed system, 81LRod, center of mass of, 351-355

homogeneous, 353moment of mass of, 355

Rolle, Michel, 195Rolle's theorem, L}S-Z}I

and the mean-value theorem , tg?_Z}tRose, eight-le afed, 566

five-lea fed, 566four-lea fed, 566three-le af.ed, 566

Rotation of axes, 620-625Ruling of a cylinder, BSz

Saddle point, 958Scalar, 7 49Scalar multiplication of vecto rs, 749-T50,

75r-755, 920Scalar (dot) product, T56-762, g2}-g}gScalar projection of vectors, TG\, g26Secant function, 452-460

derivative of, 455integrals involving powers of, 467-470inverse , 475

derivative of, 480-481Secant l ine, 11LSecond derivative,'J,STSecond derivative test for relative ex-

trema, 21.1,-21,9applications to drawing a sketch of the

graph of a function , 227-Z2gfor functions of two variables, g5g-963

Second order differential equations' 262

Sectionally smooth cunre, 985

Sequence(s), 660-673bounded,667-673convergent, 664, 670-672

decreastng, 667definition of, 660

divergent,664elements of , 660

finite, 650greatest lower bound of', 669

increasing, 667infinite, 660least upPer bound of-, 659

limit of, 662lower bound of , 668

monotonic, 667 -673

of partial sums, 674

strictly decreas ing, 667

strictly increasi ng, 667

upper bound of , 668

Series. See Inftnite serles

Set(s), 2intersection of., 4union of, 3

Set-builder notation, 2

Side condit ion,963Sides of an angle, 43L

Sigma notation, 276-280Sign function, 52Signum function, 52

Simple harmonic motion, 447

Simpson's rule, 526-53t

Sine function, 431-437derivative of, 438-445

integrals involving powers of, 447 -451

integration of rational functions of,

516-s18inverse , 472

derivative of., 478

Single integral , L002

Slope, of. a line, 33application of tangent function to,

46t-465of a tangent line to a graPh, 112-113

of a tangent line to a Polar curve/

560-551Slope-intercept form of equation of a line,

36Slug, definition of, 352

Smooth curve, 985Smooth function, 374Solid, with known parallel plane sections,

volume of ,341-343of revolution, center of mass of,

366-37Lvolume of ,330-339

Solution set, 11Special functions, 58-60

Speed, 117,786, 870

muzzle,788Sphere, 8L5, 858

center of, 815equation of, center-radius form of ' 8l'6

general form of , 8-16

point-, 8L6radius of, 815

Spherical coordinates, 872-87 6

and triple integrals, L039-1042

Spheroid, 858oblate, 858prolate, 858

Spiral, of Archimedes, 566, 577

logarithmic, 566paraboltc, 567reciproc al, 577

Square root, PrinciP al, 19

Squeeze theorem, 175

Stindard position of an angle, 43'l'

Stirling, James , 655

Strictly decreasing sequence, 667

Strictly increasing sequence, 667

Subsets, 2Substitute commodities, 968

Subtraction, 5of vecto rs, 7 49 , 82'l'

Sum, 4of functions, 57of infinite series, 675limits of , 29Lpartial, sequence of., 674

Riemann,289of two functions, derivative of , 133

of vecto rs, 7 48-7 49 , 820

Summation, index of ,276Surface(s), 8L5, 852

area of, 1.028-1033demand,968equipotential, 95Lisothermal, 951level, 887quadric, 858-864of revolution, 854-857

Symmetric equations of a line, 837

Symmetry, axis of.,363of a graph,25tests f.or,26

Tables, numerical, A-2-A-15Tangent functi on, 452-460

application of to slope of a line, 461'-465

derivative of, 454-455inflectional,224integrals involving powers of, 468-470

inverse , 474derivative of , 479

Tangent line, 11'1'-1'15, 955

to a graph, sloPe of ,'l'1'2-1'1'3of a polar curve, 57L-573

slope, of, 560-561

INDEX A-53

Tangent Plane, 953-956

Tan[ent vector, unit, 792, 869

i"t ["tttial component of acceleration' 805

Taylor, Brook, 651

Taylor polYnomial, 653

Taylor series, 729-737

Taylor's formula, 651-657

with integral form of the remainder'

555with Lagrange form of the remainder'

653Terminal point,746Terminal side of an angle, 431

Terms of infinite series, 673

Third derivatle, L57

Three-dimensional rectangular cartesian

coordinates, 812

Three-dimensional number space (R3 ),

8"1.1-817cartesian equations in, 865

vectors rn, 8L8-824Total cost cuwe,232Total cost function, 230

Total differential of a function,92L-924

Total derivative, 931Total revenue curve, 235Total revenue function ' 235

Traces of a Plane, 83LTractrix , 77'1.Transcendental functions, 50

Transcendental number, 407

Translation of axes, 583-587

equations of, 584

Ttansverse axis of a hYPerb ola, 604

Trcp ezoidal rule, 521'-525

Triangle inequ alttY, L9-20

Trigonometric function s, 43'l'-488

derivatives of, 438-445, 454-460

integration of, 447-45L, 466-470

inverse , 471-488derivatives of, 477 -482

integrals Yieldi ng, 484-488

limits of, 438-441table of, A-13

Trigonometric substitution, integration

by, 498-503Trihedral, moving,869Triple integral , 1034-1037

in cylindrical coordinates, 1'039 -1042

in spherical coordinates, L043-1045

Triple scalar product of vectorc, 847

Triple, ordered, 8I'l'Twisted cubic, 866Two-parameter family of functrons, 262

Two-point form of equation of a line, 35

Two-sided limit, 85

Undirected distanc e, 29Undirected limit, 85Union of sets, 3

Uniqueness theorcm,72Unit binormal vector, 869Unit hyperbola,768Unit normal vector, 792,869Unit step function,72Unit tangent vector, 792,869Unit vector, 754, 820Upper bound of a sequence, 668

Ieast, 669Upper limit, of integration,29'l'

of a sum,276

Variables,2,49dependent, 49,883independent, 49, 883

Vector(s),746acceleration, 786,870

normal component of, 805tangential comPonent of, 805

addition of, 7 48-7 49, 75'l'-755, 820analysis of , 746angle between,757 , 826associative laws for, 751'-752, 822, 844commutative laws for, 751-752, 757 ,

822,825components of ,746,76'l ', 826cross product of, 842-851'cunraturc,799, 868difference of, 749,821direction of, 747 , 818direction angles of, 818direction cosines of, 81.9displacement, 761dislributive laws fot,752,757, 822, 825,

844dot product of , 756-762, 825-828equation,764, 865existence of additive identity for,75t'

822

existence of negative for, 75L, 822existence of scalar multiplicative iden-

tity for, 751,, 822gradient,947independent, 756,824magnitude of,747, 818multiplication of , 7 49 -750, 75'1.-755,

756-762,82-1,, 842negative of, 749, 820normal, 829, 953orthogonal, 759, 828parallel, 758-759, 827 -828in the plane, 746-750position,764, 870position representation of., 746, 818product,750, 842

dot, 756-762projection of onto another vector,

759-760,826quantitres, T46radtus,764scalar multiplication of , 7 49 -750,

751-755, 820scalar (dot) product of ,756-762,

825-828scalar proiection of, 760,826space, 753-755

basis for,754, 824dimensron of ,754

subtraction of , 749, 821sum of ,748-749,820in three-dimensional sPace/ 818-824triple scalar product of ,847unit, 754,820unit binormal,869unit normal, 792,869unit tangent ,792, 869velocity ,785, 870zero,747, 9tB

Vector-valued function (s), 763-778, 864calculus of, 772-778continutty of.,772derivatives of, 765-766, 773-778,

866-867differentiability of , 77 4graph af , 764limits of ,772, 866

Velocity, instantaneous , 1,,L6-117, 785in rectilinear motion, 115-120

Velocity vector, 785, 870Vertex, of an angle, 431

of a cone, 588of a conic, 590of a parabola, 580of a rectangle, 1002

Vertical asymptotes, 172-L73Volume, of a solid having known parallel

plane sections , 341-343of a solid of revolution, 330-339

circular-disk method, 330-332circular-ring method, 332-335cylindrical- shell metho d, 336-339

Work, 344-347,76L,983

x axts,2Lr coordinate, 22,812x interc ept, 36

of a plane, 831

y axis,2ly coordinate, 22,8L2y intercept, 36

of a plane, 831

z coordinate, 8L2z intercePt, of a Plane, 83L

Zero vector, 747 , 818

7 8 9 8 7 5 5 4

vectors in three-dimensional space and solid analytic geometr

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