studyguides for exams 1 2 3 2

8/9/2019 StudyGuides for Exams 1 2 3 2

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Syllabus for Math 233 Section 8 Fall 2014

See Web-site: www.math.umass.edu/ bill/m233/233-F14.html for further infor-

mation on the course, on reviews and on the exams. It will updated throughout thecourse.

Instructor: William Meeks, 1536 LGRTEmail: [email protected], office phone 545-4239Office Hours: By appointment and on the days and hours I am in the calculus helpcenter, which are Tuesdays from 11:00 to 12:00. Normally I will have all day officehours in the Blue Wall on the days of the exams and many students find these officehours are extremely useful to them. Also there will probably be an undergraduateTA associated to this course and he/she might also be available to help you some toin his/her office hours.

Reviews for exams: I will be giving 1 review for each exam just for our section onthe night before each midterm; these reviews for exams also seem to be very helpfulto the students in this class.

Textbook: Calculus: Early Transcendentals 7e (7-th Edition customized hybridversion for Umass at Amherst) by James Stewart

Grading: Exam 1, Exam 2 and Final Exam each count 25% of your grade and

25% of your grade comes from your section grade. All scores will be scaled to a0-100 scale before averaging. Roughly, the section grade will be 15% WebAssignhomework and 10% in class exercises to be turned in to me in class; these exercisesappear throughout the lecture note handout that is given to you on the first dayof class. See the web pages and course syllabus on the course web page for furtherinformation on grading and exams.


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1 BASICS. 2

Study guide for Exams 1 and 2 and 3.by William H. Meeks III September 1, 2014

1 Basics.

First we cover the basic definitions and then we go over related problems. Note thatthe material for the actual midterm may include material from the review guidefor midterm 2. Before the exam, view the updated course web page for the exactmaterial covered on midterm 1.

Definition 1 Let n be a positive integer. Then the cartesian product ofn copiesof the real number line Ris:

Rn = R

R

. . .

R =

{(a1, a2, . . . , an

|aj

R)}

,

which is the set of all ordered n-tuples of real numbers.

Example 2 (a) R2 = R R = {(a1, a2) | ai R} is the Euclidean plane.(b) R3 = RRR = {(a1, a2, a3) | ai R3} is Euclidian three-space.

There are two standard notations for points in R3, or more generally Rn. IfP R3, thenP = (a1, a2, a3) for some scalars a1, a2, a3. The book also denotes thispoint by writing P(a1, a2, a3). The scalar a1 is called the x-coordinate ofP, a2 iscalled the y-coordinate ofP and a3 is called the z -coordinate ofP.

Example 3 The point P = (1, 0, 7) in R3 can also be written as P(1, 0, 7). Itsz-coordinate is 7.

Definition 4 (a) Given points P = (x1, y1, z1) and Q = (x2, y2, z2) in R3, then

P Q= x2 x1, y2 y1, z2 z1 denotes the arrowor vectorbased at P withterminal pointQ.

(b) If R is a scalar and v =a,b,c is a vector, then consider the new vectorv =a,b,c; if > 0, then v is the vector pointed in the direction vand has length |v|; if < 0, then v is the vector pointed in the oppositedirection ofv with length

|||

v|.

(c) Ifu = x1, y1, z1 and v= x2, y2, z2, then u + v= x1+x2, y1+y2, z1+z2.In other words, vectors add by adding their coordinates.

Definition 5 If a =x1, y1, z1 and b =x2, y2, z2, then the dot product of aand b is:

a b= x1x2+y1y2+z1z2.

Example 6 The dot product of1, 2, 3 and1, 0, 7 is

1, 2, 3 1, 0, 7 = 1 + 0 + 21 = 22.

It turns out that the length of a vector can be found by using the dot productand it satisfies some nice algebraic properties listed in the next two theorems.


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1 BASICS. 3

Theorem 7 Leta= x1, y1, z1 be vector and let P = (x2, y2, z2), Q= (x3, y3, z3)be points. Then:

1. The length ofa is|a| = a a= x21+y21+z21 .2. The distance d(P, Q) from the pointP to the point Q is:

d(P, Q) = |P Q| =

(x3 x2)2 + (y3 y2)2 + (z3 z2)2

Theorem 8 (Basic algebraic properties of dot product) Leta = x1, y1, z1,b= x2, y2, z2, c = x3, y3, z3 be vectors and let be a scalar.

1. a b= b a.2. (a + b)

c= a

c + b

c.

3. (a) b= (a b).

Definition 9 The sphere in R3 with center C= (x0, y0, z0) and radius r is the setwhere (x x0)2 + (y y0)2 + (z z0)2 =r2. Note that this sphere is geometricallythe set of points (x,y ,z) of distance r from the point (x0, y0, z0).

Example 10 Consider the subset ofR3 defined byx2 + y2 + 6y + z2 + 2z= 26. Bycompleting the square, we have

x2 + (y2 + 6y+ 9) + (z2 + 2z+ 1) = 26 + 9 + 1 = 36,

which simplifies to bex2 + (y+ 3)2 + (z+ 1)2 = 62.

So this set is the sphere centered at (0,3,1) of radius 6.

For convenience, it is useful to pick out the special unit vectors pointed respectivelyalong the positive x, y and z-axes, as given in the next definition.

Definition 11 We define the standard basis vectors for R3 as follows: i= 1, 0, 0,j= 0, 1, 0,k = 0, 0, 1. Note that the vector a,b,c can be expressed by a,b,c =ai +bj +ck.

For nonzero vectors a, b

a b= |a||b| cos(),

where [0, ] is the angle between the vectors. It follows that:1. a and b are orthogonal or perpendicular if and only ifa b= 0.2. The angle between a and b is an acute angle if and only ifa b> 03. The angle between a and b is an obtuse angle if and only ifa b< 0.4. cos() = ab

|a

||b

|.

5. = arccos( ab|a||b|) = cos1( ab|a||b|) In particular, ifa and b are unit vectors, then

= arccos(a b).


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1 BASICS. 4

Example 12 The vectors1, 2,1 and3,1, 1 are orthogonal, since1, 2,1 3,1, 1 = 3 2 1 = 0.

Definition 13 1. The scalar projection (component) ofb onto a is compab =ab|a| . In particular, ifa is a unit vector, then compab= a b.

2. The vector projection ofb onto (in the direction of) a is a projab = (abaa )a.

In particular, ifa is a unit vector, then projab= (a b)a.3. The direction cosines of the vectorb are:

(a) cos() = b|b| i,(b) cos() = b|b|j,(c) cos() = b

|b

|k,

and so,,,are the respective angles thatb makes with thex, yand z-axes.

Example 14 Consider the vectors a =1, 2, 2 and b =1, 1, 1. Since projab =abaaa, then

v= b projab= 1, 1, 1 591, 2, 2 = 4

9,1

9,1

9

must be perpendicular to a and must lie in the plane containing a and b.

Definition 15 1. The determinant of the matrix M with rows vectors v =

a, b

and w =

c, d

can be calculated by:

|M

|= a b

c d =ad

bc.

The absolute value|ad bc| of this determinant equals the area of the paral-lelogram with sides v and w.

2. The determinant of the matrix M with rows vectors a =a1, a2, a3, b =b1, b2, b3 and c = c1, c2, c3can be calculated by:

|M| =

a1 a2 a3b1 b2 b3c1 c2 c3

=a1 b2 b3c2 c3

a2 b1 b3c1 c3

+a3 b1 b2c1 c2

The absolute value of the determinant

|M| equals the volume of the paral-

lelepiped or box spanned by the vectors a, b and c.

3. The cross product a b of vectors a =a1, a2, a3 and b =b1, b2, b3 canbe calculated by:

a b=

i j ka1 a2 a3b1 b2 b3

= a2 a3b2 b3

i a1 a3b1 b3

j + a1 a2b1 b2

k.The length ofa b is given by:|a b| = |a||b| sin(), where [0, ] is theangle between a and b. Also|a b| is area of the parallelogram with sides aand b. Note that it follows that area of the triangle with vertices0, 0, 0andthe position vectors a and b is |ab|2 .


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1 BASICS. 5

Example 16 Consider the points A= (1, 0, 1), B = (0, 2, 3) and C= (1,1, 0).Then the area of the triangle with these vertices can be found by taking the areaof the parallelogram spanned by

AB and

ACand dividing by 2. Thus:

Area() =|AB

AC|

2 =

1

2

i j k

1 2 22 1 1

=

1

2|0,5, 5| =1

2

0 + 25 + 25 =

1

2

50

Example 17 Consider the vectors a =1, 0, 1, b =0, 2, 3 and c =1, 7, 0.Then the volume of the parallelepiped or box spanned by these 3 vectors is:

1 0 10 2 31 7 0

= | 21 0 + 2| = | 19| = 19

Definition 18 IfFis a force with magnitude A applied in the unit direction a|a| toan object in order to move it from the point P to the point Q, then the work W

done is: W = A|a|a P Q.

Example 19 If F is a force of 10N (10 Newtons) applied in the unit direction162, 1, 1 to an object to move it from P = (3,2, 5) to Q = (1, 2, 3), then the

work done is (length measured in meters):

W =10N

62, 1, 1 4, 4,2 =100N m

6,

where m is one meter.

Definition 20 The torqueon a rigid body with position vector a with a force ofmagnitude A in the unit direction b|b| is:

=a A b|b| .

Example 21 What is the magnitude (the length) of the torque on a rigid body withposition vectora = 1,1, 3 with a force of 10Nin the direction of b|b| = 162, 1, 1(length measured in meters m) ?

Solution:

|| = |1,1, 3 10N m62, 1, 1| = |10N m

64, 5, 3| =10N m

50

6.

Definition 22 Given a point P = (x0, y0, z0) and a vectorv = a,b,c, the vectorequation of the line L passing through P in the direction ofv is:

r(t) =P+tv= x0, y0, z0+ta,b,c = x0+at, y0+bt, z0+ct.

The resulting equations:x= x0+at,

y= y0+bt,

z= z0+ct,


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1 BASICS. 6

are called the parametric equations for L. The resulting equations (solving fort):

x x0a

= y y0

b =

z z0c

,

are called the symmetric equationsfor L.

Example 23 The vector equation for the line L passing through P = (1, 2, 3) andQ= 4, 0, 7 is given by:

r(t) =P+tP Q= 1, 2, 3+t3,2, 4 = 1 + 3t, 2 2t, 3 + 4t.

Definition 24 The plane passing through the point P = (x0, y0, z0) with normalvector n =a,b,c is given by the following equation, where (x,y ,z) denotes ageneral point on the plane:

0 =n x x0, y y0, z z0.Equivalently, we have:

a(x x0) +b(y y0) +c(z z0) = 0.Example 25 The equation of the plane passing through P = (1, 2, 3) and withnormal vector n = 3, 4, 1is:

3(x 1) + 4(y 2) + (z 3) = 0.Example 26 Find the equation of the plane passing through pointsP = (1, 0, 2), Q=(4, 2, 3), R= (2, 0, 4).

Solution: Since a plane is determined by its normal vector n and a point on it,say the point P, it suffices to find n. Note that:

n=P Q

P R=

i j k3 2 11 0 2

= 4,5,2.So the equation of the plane is:

4(x 1) 5y 2(z 2) = 0.

Given two planes with unit normal vectors n1 and n2, respectively, then the co-sine of the angle between them is the cosine of the angle between the lines determinedbyn1 and n2, which can be calculated using dot products.

Example 27 The cosine of the angle betweenx2y +2z = 1 and 2xy +2z= 10is given by

cos() = |131,2, 2 1

32,1, 2| =1

9(2 + 2 + 4) =

8

9.

Definition 28 Let r(t) be a vector valued curve in R3, where r(t) = f(t), g(t), h(t).Here t is called the parameter ofr(t). If the derivative r(t) = limh

0r(t+h)r(t)

hexists for each t, then the curve r(t) is called differentiable and r(t) is called thederivative or velocity or tangent vector field v(t) = r(t) to the curve r(t). Thelength|v(t)| is called the speed of the curve r at the parameter value t.


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2 SOME PRACTICE PROBLEMS SOLVED. 7

Theorem 29 Ifr(t) = f(t), g(t), h(t)is a differential curve in R3, then:r(t) = f(t), g(t), h(t).

Conversely, iff(t), g (t), h(t) are differentiable functions, then r(t) is differentiable.The speed function for r(t) is then:

speed(t) =

(f(t))2 + (g(t))2 + (h(t))2.

Example 30 Suppose r(t) =t, sin(2t), t2 + 1, then r(t) =1, 2 cos(2t), 2t withr(0) = 1, 2, 0. Hence, the tangent line to r(t) at t= 0 is given by:

L(t) =r(0) +tr(0) = 1, 0, 1+t1, 2, 0 = 1 +t, 2t, 1and the speed function ofr(t) is: speed(t) =

12 + 4 cos2(2t) + 4t2.

Definition 31 The lengthLof a parameterized curve r(t) in R3 on a time interval[a, b]

is the integral of the speed:

L=

ba|r(t)|dt.

Example 32 Ifr(t) = sin(t), cos(t), 2t, then r(t) = cos(t), sin(t), 2 with con-stant speed

cos2(t) + sin2(t) + 4 =

5. Hence, the length ofr(t) from time t = 1

to time t= 6 is:

L=

61

5 dt=

5t

6

1

=

5 6

5 1 = 5

5.

2 Some practice problems solved.

1. Find parametric equations for the line which containsA(2, 0, 1) andB(1, 1,1).Solution: Letv =

AB= 1, 1,12, 0, 1 = 3, 1,2.SinceA(2, 0, 1)

lies on the line, then:x= 2 3t,

y = 0 t= t,z= 1 2t.

2. Determine whether the lines l1 : x = 1 + 2t, y = 3t, z = 2

t and l2 : x =

1 +s, y= 4 +s, z = 1 + 3s are parallel, skew or intersecting.Solution: Vector part of line l1 is v1 =2, 3,1 and for line l2 is v2 =1, 1, 3. Clearly, v1 is not a scalar multiple of v2 and so these lines are notparallel. If these lines intersect, then for some values oft and s:

x= 1 + 2t= 1 +s 2t= 2 +s,y= 3t= 4 +s 3t= 4 +s.

Solving these two linear equations yields:

t= 6 and s= 14.

Plugging these values into z = 2 t = 1 + 3s yields the inequality4= 43,which means there is no solution and the lines do not intersect. Thus, the linesare skew.


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3. Find an equation of the plane which contains the pointsP(1, 2, 1),Q(1,2, 1)and R(1, 1,1).

Solution: Consider the vectors

P Q= 2,4, 0 and

P R= 2,1,2 whichlie parallel to the plane. Then consider the normal vector:

n=P Q

P R=

i j k2 4 02 1 2

= 8i + 4j + 6k.So the equation of the plane is given by:

8, 4, 6 x+ 1, y 2, z 1 = 8(x+ 1) + 4(y 2) + 6(z 1) = 0.

4. Find the distance from the point (1, 2,

1) to the plane 2x+y

2z= 1.

Solution: The normal to the plane is n =2, 1,2 and the point P =(0, 1, 0) lies on this plane. Consider the vector from P to (1, 2,1) which isv= 1, 1,1. The distance from (1, 2,1) to the plane is equal to:

|compnv| =v n|n|

= |1, 1,1 132, 1,2| = 53 .5. Let two space curves

r1(t) = cos(t 1), t2 1, t4, r2(s) = 1 + ln s, s2 2s+ 1, s2,

be given where t and s are two independent real parameters. Find the cosineof the angle between the tangent vectors of the two curves at the intersectionpoint (1, 0, 1).

Solution: After taking derivatives, we obtain:

r1(t) = sin(t 1), 2t, 4t3,

r2(s) = 1

s, 2s 2, 2s.

At the point (1, 0, 1), t = 1 and s = 1 and so, r1(1) =0, 2, 4 and r2(1) =1, 0, 2 are the related tangent vectors. Thus,

cos() = r1(1)|r1(1)|

r2(1)

|r2(1)|=

820

5=

4

5.

6. Suppose a particle moving in space has velocity

v(t) = sin(t), cos(2t), et

and initial position r(0) = 1, 2, 0. Find the position vector function r(t).Solution: Since r(t) =sin(t), cos(2t), et, then r(t) = t v(s)ds. Thus,r(t) = cos(t) + x0, 12sin (2t) + y0, et + z0 withr(0) = 1 + x0, y0, 1 + z0 =1, 2, 0.Hence,x0= 2, y0 = 2, z0= 1 and so, r(t) = cos(t)+ 2,

1

2sin (2t) +2, et 1.


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7. Find the center and radius of the spherex2 +y2 +z2 + 6z= 16.

Solution: Complete squares to obtain from x2 +y2 +z2 + 6z = 16, theequation:

x2 +y2 + (z+ 3)2 = 16 + 9 = 25.

Hence, the center is at C= (0, 0,3) and the radius is 5.

Review guide for midterm 2 in Math 233

Midterm 2 covers material that begins approximately with the definition of par-tial derivatives in Chapter 14.3 and ends approximately with methods for calculatingthe double integral of a function f(x, y) over a domainD described in thexy-plane.See the updated course web page for the exact material covered on this exam.

Definition 33 (Partial Derivatives) Iffis a function of two variables, its par-tial derivatives are the functions fx and fy defined by

fx(x, y) = limh0

f(x+h, y) f(x, y)h

fy(x, y) = limh0

f(x, y+h) f(x, y)h

.

We have the following rule for calculating partial derivatives.

1. To find fx, regard y as a constant and differentiate f(x, y) with respect to x.

2. To find fy, regard x as a constant and differentiate f(x, y) with respect to y .

Example 34 Calculate fx, fy for f(x, y) =x2exy +y2.

Solution: We apply the sum, product and chain rules for derivatives, to get:

fx(x, y) = 2xexy +x2exyy= 2xexy +x2yexy

fy(x, y) =x2exyx+ 2y= x3exy + 2y.

Definition 35 (Second Partial Derivatives) Forz = f(x, y), we use the follow-ing notation:

(fx)x= fxx= x

fx

= 2f

x2 = 2z

x2

(fx)y =fxy =

y

f

x

=

2f

yx =

2z

yx

(fy)x= fyx =

x

f

y

=

2f

xy =

2z

xy

(fy)y =fyy =

y

f

y

=

2f

y2 =

2z

y2

Example 36 Find the second partial derivatives of

f(x, y) =x3 +x2y3 2y2.


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Solution: Note:

fx(x, y) = 3x2 + 2xy3 fy(x, y) = 3x

2y2 4y.

Therefore,fxx= 6x+ 2y

3 fxy = 6xy2

fyx = 6xy2 fyy = 6x

2y 4.

Note that in the above example fxy = fyx. This is no coincidence and followsfrom the next theorem that states that under weak conditions on f(x, y), takingpartial derivatives is a commutative process.

Theorem 37 (Clairauts Theorem) Suppose f is defined on a disk D that con-

tains the point (a, b). If the functions fxy and fyx are both continuous on D , then

fxy(a, b) =fyx(a, b).

The next definition of tangent plane generalizes in a natural way the followingequation of the tangent line of a function of 1 variable:

y y0= f(x0)(x x0).

Definition 38 (Tangent Plane) Suppose f has continuous partial derivatives.An equation of the tangent plane to the surface z = f(x, y) at the point P =(x0, y0, z0) is

z z0= fx(x0, y0)(x x0) +fy(x0, y0)(y y0).Example 39 Find the tangent plane to the elliptic paraboloid z = 2x2 + y2 at thepoint (1, 1, 3).

Solution: Let f(x, y) = 2x2 +y2. Then

fx(x, y) = 4x fy(x, y) = 2y

fx(1, 1) = 4 fy(1, 1) = 2.

Then Definition 38 gives the equation of the tangent plane at (1, 1, 3) as

z 3 = 4(x 1) + 2(y 1)

orz= 4x+ 2y 3.

The next definition of linear approximation generalizes the linear approximationL(x) of a function f(x) of 1 variable at a point x0= a :

L(x) =f(a) +f(x)(x a).

Definition 40 (Linear Approximation) The linear approximation off(x, y) at(a, b) is

L(x, y) =f(a, b) +fx(a, b)(x a) +fy(a, b)(y b).


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Solution : If the dimensions of the box arex,y , andz , its volume is V =xyz andso

dV = V

x

dx+ V

y

dy+ V

z

dz= yz dx+xz dy+xy dz.

We are given that|x| 0.2,|y| 0.2, and|z| 0.2. To find the largest errorin the volume, we use dx = 0.2, dy = 0.2, and dz = 0.2 together with x = 75,y= 60, and z = 40:

V dV= (60)(40)(0.2) + (75)(40)(0.2) + (75)(60)(0.2) = 1980.

Thus, an error of only 0.2 cm in measuring each dimension could lead to an errorof as much as 1980 cm3 in the calculated volume! This may seem like a large error,but its only about 1% of the volume of the box.

Theorem 47 (Chain Rule Case 1) Suppose that z = f(x, y) is a differentiablefunction ofx and y, where x = g(t) and y = h(t) are both differentiable functionsoft. Then z is a differentiable function oft and

dz

dt =

f

x

dx

dt +

f

y

dy

dt.

Theorem 48 (Chain Rule Case 2) Suppose that z = f(x, y) is a differentiablefunction ofxandy, wherex= g(s, t) andy = h(s, t) are both differentiable functionsofs and t. Then z is a differentiable function oft and

z

s =

z

x

x

s+

z

y

y

s

z

t =

z

x

x

t +

z

y

y

t.

Theorem 49 (The Chain Rule (General Version)) Suppose thatu is a differ-entiable function ofn variablesx1, x2, . . . , xnand eachxj is a differentiable functionof the m variables t1, t2, . . . , tm. Then u is a function oft1, t2, . . . , tm and

u

ti=

u

x1

x1ti

+ u

x2

x2ti

+. . .+ u

xn

xnti

for each i = 1, 2, . . . , m.

Example 50 Ifz = x2y + 3xy4, wherex = sin 2tand y = cos t, find dzdt when t = 0.

Solution: The Chain Rule gives

dz

dt =

f

x

dx

dt +

f

y

dy

dt = (2xy+ 3y4)(2cos2t) + (x2 + 12xy3)( sin t).

Its not necessary to substitute the expressions for x and y in terms oft. We simplyobserve that when t = 0 we have x = sin 0 = 0 and y = cos 0 = 1. Therefore,

dz

dt t=0= (0 + 3)(2 cos 0) + (0 + 0)( sin0) = 6.

Example 51 The pressureP(in kilopascals), volumeV(in liters), and temperatureT (in kelvins) of a mole of an ideal gas are related by the equation P V = 8.31T.Find the rate at which the pressure is changing when the temperature is 300 K andincreasing at a rate of 0.1 K/s and the volume is 100 L and increasing at a rate of0.2 L/s.


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Solution: If t represents the time elapsed in seconds, then at the given instantwe have T = 300, dT/dt = 0.1, V = 100, dV/dt = 0.2. Since P = 8.31 TV, withPT =

8.31V and

TV =

8.31TV2

, then Case 1 of the Chain Rule gives

dP

dt =

P

T

dT

dt P

V

dV

dt =

8.31

V

dT

dt 8.31T

V2dV

dt

= 8.31

100(0.1) 8.31(300)

1002 (0.2) = 0.04155.

The pressure is decreasing at a rate of about 0.042 kPa/s.

Example 52 Ifz = ex sin y, where x = st2 and y = s2t, find zs and zt .

Solution : Applying Case 2 of the Chain Rule, we get

dzds

= zx

xs

+ zy

ys

= (ex sin y)(t2) + (ex cos y)(2st)

=t2est2

sin(s2t) + 2stest2

cos(s2t),

dz

dt =

z

x

x

t +

z

y

y

t = (ex sin y)(2st) + (ex cos y)(s2)

= 2srest2

sin(s2t) +s2est2

cos(s2t).

Example 53 Ifu = x4y + y2z3, wherex = rset,y = rs2et, andz = r2s sin yt, findthe value ofu/s when r = 2, s = 1, t = 0.

Solution: We have u

s =

u

x

x

s +

u

y

y

s+

u

z

z

s

= (4x3y)(ret) + (x4 + 2zy3)(2rset) + (3y2z2)(r2 sin t).

When r = 2, s = 1, and t = 0, we have x = 2, y = 2, and z = 0, so

u

s = (64)(2) + (16)(4) + (0)(0) = 192.

Theorem 54 (Implicit Differentiation) Suppose that z is given implicitly as afunction z = f(x, y) by an equation F(x,y ,z) = 0, i.e., F(x,y ,f (x, y)) = 0 for all

(x, y) in the domain off(x, y). Then:

z

x =

FxFz

z

y =

Fy

Fz

.

Example 55 Find zx and zy ifx

3 +y3 +z3 + 6xyz = 1.

Solution: Let F(x,y ,z) =x3 +y3 +z3 + 6xyz 1. Then, from Theorem 54, wehave

z

x= Fx

Fz= 3x

2 + 6yz

3z2 + 6xy = x

2 + 2yz

z2 + 2xy

z

y = Fy

Fz= 3y

2 + 6xz

3z2 + 6xy = y

2 + 2xz

z2 + 2xy.


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Definition 56 (Directional Derivative) Thedirectional derivativeoff(x, y)at (x0, y0) in the direction of a unit vector u = a, b is

Duf(x0, y0) = limh0

f(x0+ha, y0+hb) f(x0, y0)h

if this limit exists.

Definition 57 (Directional Derivative) The directional derivative off(x,y ,z)at (x0, y0, z0) in the direction of a unit vector u = a,b,c is

Duf(x0, y0, z0) = limh0

f(x0+ha, y0+hb, z0+hc) f(x0, y0, z0)h


Definition 58 (Gradient) If f is a function of two variables x and y, then thegradientoffis the vector functionf defined by

f(x, y) = fx(x, y), fy(x, y) = fx

i + f

yj.

Definition 59 (Gradient) For f(x,y ,z), a function of three variables,

f= fx, fy, fz = fx

i + f

yj +

f

zk.

The next two theorems give simple rules for calculating the directional derivative

of a function in 2 or 3 variables in terms of the gradient of the function.

Theorem 60 Iff is a differentiable function ofx and y, then f has a directionalderivative in the direction of any unit vector u = a, band

Duf(x, y) =fx(x, y)a+fy(x, y)b.

Theorem 61 Iffis a differentiable function ofx,y, andz, thenfhas a directionalderivative in the direction of any unit vector u = a,b,c and

Duf(x,y ,z) = f(x,y ,z) u.

By the above two theorems, we have for any unit vector u,

Duf= f u= |f||u| cos() = |f| cos().

Thus, the next theorem holds.

Theorem 62 Suppose f is a differentiable function of two or three variables. Themaximum value of the directional derivative Duf(x) is|f(x)| and it occurs whenu has the same direction as the gradient vectorf(x).

Example 63 Find the directional derivative of the function f(x, y) =x2y3

4y at

the point (2,1) in the direction of the vector v = 2i + 5j.


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Solution : We first compute the gradient vector at (2,1):f(x, y) = 2xy3i + (3x2y2 4)j

f(2,1) = 4i + 8j.Note thatv is not a unit vector, but since |v| = 29, the unit vector in the directionofv is

u= v

|v| = 2

29i +

529

j.

Therefore, by Theorem 60, we have

Duf(2,1) = f(2,1) u= (4i + 8j) ( 229

i + 5

29j)

=4 2 + 8 5

29=

32

29.

Theorem 64 SupposeSis a surface determined as F(x,y ,z) =k for k = constant.ThenFis everywhere normal or orthogonal toS. In particular, ifP = (x0, y0, z0) S, then the equation of the tangent plane to Sat p is:

Fx(x0, y0, z0)(x x0) +Fy(x0, y0, z0)(y y0) +Fz(x0, y0, z0)(z z0) = 0 (1)Example 65 Find the equations of the tangent plane and normal line at the point(2, 1,3) to the ellipsoid

x2

4 +y2 +

z2

9 = 3.

Solution : The ellipsoid is the level surface (withk = 3) of the function

F(x,y ,z) = x2

4 +y2 +

z2

9.

Therefore, we have

Fx(x,y ,z) =x

2 Fy(x,y ,z) = 2y Fz(x,y ,z) =

2z

9

Fx(2, 1,3) = 1 Fy(2, 1,3) = 2 Fz(2, 1,3) = 23

.

Then Equation 1 in Theorem 64 gives the equation of the tangent plane at(

2, 2,

3) as

1(x+ 2) + 2(y 1) 23

(z+ 3) = 0,

which simplifies to 3x 6y+ 2z= 18 = 0.SinceF(2, 1,3) = 1, 2,23, the vector equation of the normal line is:

L(t) = 2, 1,3+t1, 2,23.

Definition 66 A function of two variables has a local maximum at (a, b) iff(x, y) f(a, b) when (x, y) is near (a, b). (This means that f(x, y) f(a, b)for all points (x, y) in some disk with center (a, b).) The number f(a, b) is calleda local maximum value. Iff(x, y)

f(a, b) for all f(x, y) in the domain of f,

thenfhas an absolute maximumat (a, b). Iff(x, y) f(a, b) when (x, y) is near(a, b), then f(a, b) is a local minimum value. Iff(x, y)f(a, b) for all (x, y) inthe domain off, then f has an absolute minimumat (a, b).


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The next theorem explains how to find local maxima and local minima for afunction in two variables.

Theorem 67 If f has a local maximum of minimum at (a, b) and the first-orderpartial derivatives offexist there, then fx(a, b) = 0 and fy(a, b) = 0.

Definition 68 A point (a, b) is called a critical point of f(x, y) if fx(a, b) =fy(a, b) = 0.

The next theorem gives a method for testing critical points of a function f(x, y)to see if they represent local minima, local maxima or saddle points (a critical point(a, b) is a saddle point if the Hessian D defined in the next theorem is negative).

Theorem 69 (Second Derivative Test) Suppose the second partial derivativesoffare continuous on a disk with center (a, b), and suppose that fx(a, b) = 0 andfy(a, b) = 0 (that is, (a, b) is a critical point off). Let

D= D(a, b) =fxx(a, b)fyy(a, b) [fxy(a,b)]2.(a) IfD >0 and fxx(a, b)> 0, then f(a, b) is a local minimum.

(b) IfD >0 and fxx(a, b)< 0, then f(a, b) is a local maximum.

(c) IfD 0 and fxx(1, 1) = 12> 0, we see from case (a) of thetest thatf(1, 1) = 1 is a local minimum. Similarly, we have D(1,1) = 128> 0and fxx(1,1) = 12> 0, so f(1,1) = 1 is also a local minimum value.


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Definition 71 A subsetD R2 isclosed if it contains all of its boundary points.

Definition 72 A subset D R2 is bounded if it is contained within some diskin the plane.

Theorem 73 (Extreme Value Theorem for Functions of Two Variables) Iffis continuous on a closed, bounded set D in R2, then fattains an absolute maxi-mum valuef(x1, y1) and an absolute minimum valuef(x2, y2) at some points (x1, y1)and (x2, y2) in D .

To find the absolute maximum and minimum values of a continuous function fon a closed, bounded set D :

1. Find the values offat the critical points off in D.

2. Find the extreme values offon the boundary ofD .

3. The largest of the values from steps 1 and 2 is the absolute maximum value;the smallest of these values is the absolute minimum value.

The next theorem is stated for a function f of three variables but there is asimilar theorem for a function of two variables (see Example 75 below).

Theorem 74 (Method of Lagrange Multipliers) To find the maximum and min-imum values off(x,y ,z) subject to the constraintg(x,y ,z) =k (assuming that theseextreme values exist andg=0 on the surface g(x,y ,z) =k):

1. Find all values ofx, y , z , and such that

f(x,y ,z) =g(x,y ,z)and

g(x,y ,z) =k.

2. Evaluate f at all the points (x,y ,z) that result from step 1. The largest ofthese values is the maximum value off; the smallest is the minimum value off.

Example 75 Find the extreme values of the function f(x, y) = x2 + 2y2 on the

circle x2

+y2

= 1.

Solution: We are asked for the extreme values of f subject to the constraintg(x, y) =x2+y2 = 1. Using Lagrange multipliers, we solve the equations f=g,g(x, y) = 1, which can be written as

fx= gx fy =gy g(x, y) = 1

or as2x= 2x (2)

4y= 2y (3)

x2

+y2

= 1. (4)From (2) we have x = 0 or = 1. Ifx = 0, then (4) gives y =1. If = 1, theny = 0 from (3), so then (4) gives x = 1. Therefore,fhas possible extreme values


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at the points (0, 1), (0,1) (1, 0), and (1, 0). Evaluating f at these four points,we find that

f(0, 1) = 2 f(0,1) = 2 f(1, 0) = 1 f(1, 0) = 1.Therefore, the maximum value offon the circle x2 + y2 = 1 is f(0,1) = 2 and

the minimum value is f(1, 0) = 1.Example 76 Find the extreme values off(x, y) =x2 + 2y2 on the diskx2 + y2 1.Solution : We will compare the values offat the critical points with values at thepoints on the boundary. Since fx= 2xand fy = 4y, the only critical point is (0, 0).We compare the value offat that point with the extreme values on the boundaryfrom Example 75:

f(0, 0) = 0 f(

1, 0) = 1 f(0,

1) = 2.

Therefore, the maximum value offon the disk x2 + y2 1 is f(0,1) = 2 and theminimum value is f(0, 0) = 0.

We now start the second material for midterm 2 which concerns double integrals.For a positive, continuous function f(x, y) defined on a closed and bounded domainD R2, we denote by

Df(x, y)dA,

the volume under the graph of f(x, y) over D. This volume for a rectangle R ={(x, y)| a x b, c y d} = [a, b] [b, c] R2 can be estimated by thefollowing Midpoint Rule for Double Integrals described in the next theorem. We

also use this rule for defining the double integral when f(x, y) is not necessarilypositive.

Theorem 77 (Midpoint Rule for Double Integrals) Letm, n be positive in-tegers. Letx0 =a < x1 < x2 < . . . < xm = b be a division of [a, b] into n intervals[xi, xi+ 1] of equal width x=

bam . Similarly, let y0 = c < y1< y2< . . . < yn= d

be a division of [c, d] into m intervals [yj, yj+1] of equal widths y= dcn . Then:

Rf(x, y)dA

mi=1

nj=1

f(xi, yj) A,

where xi is the midpoint of [xi1, xi] and yj is the midpoint of [yj1, yj]. Further-more, the right-hand side above converges to the left-hand side as m, n Definition 78 Iffis a continuous function of two variables, then its average valueon a domain D R2 is:

Df(x, y)dA

Area(D) =

DdA.

Definition 79 The iterated integral off(x, y) on a rectangle R = [a, b] [c, d] is ba

dc

f(x, y)dy dx or

dc

ba

f(x, y)dx dy.

One calculates the integralbadc f(x, y) dy dxby first calculatingA(x) =

dc f(x, y) dy,

holdingxconstant, and then calculatingbaA(x) dxand similarly, for calculating the

other integral.


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Example 80 Evaluate the iterated integral.

3

0 2

1

x2y dy dx

Solution: Regardingx as a constant, we obtain

21

x2y dy=

x2

y2

2

y=2y=1

=x2

22

2

x2

12

2

=

3

2x2.

Thus, the function A in the preceding discussion is given by A(x) = 32x2 in this

example. We now integrate this function ofx from 0 to 3:

3

0 2

1x2y dy dx=

3

0 2

1x2y dy dx=

3

0

3

2x2 dx=

x3

2 3

0

=27

2 .

Example 81 Evaluate the iterated integral. 21

30

x2y dx dy.

Solution : Here we first integrate with respect tox:

21

30

x2y dx dy =

21

30

x2y dx

dy=

21

x3

3y

x=3x=0

dy=

21

9y dy= 9y2

2

21

=27

2 .

Theorem 82 (Fubinis Theorem) Iffis continuous on the rectangleR= {(x, y) |a x b, c y d}, then

Rf(x, y)dA =

ba

dc

f(x, y)dy dx=

dc

ba

f(x, y)dx dy.

More generally, this is true if we assume that f is bounded on R, fis discontinuousonly on a finite number of smooth curves, and the iterated integrals exist.

Example 83 Evaluate the double integral

R(x 3y2) dA, where R ={(x, y)|0 x 2, 1 y 2}.

Solution: Fubinis Theorem gives R

(x 3y2)dA = 2

0

2

1(x 3y2)dy dx=

2

0

xy y3y=2

y=1 dx

=

20

(x 7)dx = x2

2 7x

20

= 12.

Example 84 Find the volume of the solid Sthat is bounded by the elliptic paraboloidx2 + 2y2 +z = 16, the planes x = 2 and y = 2, and the three coordinate planes.

Solution: We first observe that S is the solid that lies under the surface z =16 x2 y2 and above the square R = [0, 2] [0, 2]. We are now in a position toevaluate the double integral using Fubinis Theorem. Therefore,

V =

R

(16 x2 2y2)dA = 20

20

(16 x2 2y2)dx dy


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=

20

[16x 13

x3 2y2x]x=2x=0dy

= 20

(88

3 y 4y2)dy =88

3 4

3 y320

= 48.

In general, for any continuous functionf(x, y) on a closed and bounded domainD R2, the integral D f(x, y) dA is defined and it is equal to the area underthe graph off(x, y) on D when the function is positive. There are two cases for D,called type I and type II, where the integral

Df(x, y)dA

can be calculated in a straightforward manner.

Definition 85 A plane region D is said to be oftype I, if it can be expressed as

D= {(x, y) | a x b, g1(x) y g2(x)},where g1(x) and g2(x) are continuous.

Definition 86 A plane region D is said to be oftype II, if it can be expressed as

D= {(x, y) | c y d, h1(y) x h2(y)},where h1 and h2 are continuous.

Theorem 87 Iff is continuous on a type I region D such that

D= {(x, y) | a x b, g1(x) y g2(x)},then

Df(x, y)dA =

ba

g2(x)g1(x)

f(x, y)dy dx.

Theorem 88 D

f(x, y)dA =

dc

h2(y)h1(y)

f(x, y)dx dy.

where D is a type II region given by Definition 86.

Example 89 Evaluate

D(x+ 2y) dA, where D is the region bounded by theparabolasy = 2x2 and y= 1 +x2.

Solution: The parabolas intersect when 2x2 = 1 +x2, that is x2 = 1, sox = 1.We note that the region D, is a type I region but not a type II region and we canwrite

D= {(x, y) 1 x 1, 2x2 y 1 +x2}.Since the lower boundary isy = 2x2 and the upper boundary isy = 1+ x2, Definition85 gives

D(x+ 2y)dA =

11

1+x22x2

(x+ 2y)dy dx=

11

xy+y2

y=1+x2y=2x2

dx

= 1

1(3x4 x3 + 2x2 +x+ 1) dx

= 3 x5

5x

4

4 + 2

x3

3 +

x2

2 +x

1

1=

32

15.


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Example 90 Find the volume of the solid that lies under the paraboloid z = x2+y2

and above the regionD in thexy-plane bounded by the liney = 2xand the parabolay= x2.

Solution 1: We see that D is a type I region and

D= {(x, y) | 0 x 2, x2 y 2x}.Therefore, the volume underz = x2 +y2 and above D is

V =

D

(x2 +y2)dA =

20

2xx2

(x2 +y2)dy dx

=

20

x2y+

y3

3

y=2xy=x2

dx=

20

x2(2x) +

(2x)3

3 x2x2 (x

2)3

3

dx

= 20x

6

3 x4

+

14x3

3

dx=x7

21x5

5 +

7x4

620 =

216

35 .

Solution 2: We see that D can also be written as a type II region:

D= {(x, y) | 0 y 4, 12

y x y}.Therefore, another expression forV is

V =

D

(x2 +y2)dA =

40

y1

2y

(x2 +y2)dx dy

= 4

0 x3

3 +y2x

x=y

x= 12

y

dy= 4

0 y

3

2

3 +y

5

2 y3

24y

3

2 dy=

2

15y

5

2 +2

7y

7

2 1396

y440

=216

35 .

Example 91 Evaluate the iterated integral10

1x sin(y

2)dy dx.

Solution : If we try to evaluate the integral as it stands, we are faced with thetask of first evaluating

sin(y2) dy. But its impossible to do so in finite terms

since

sin(y2) dy is not an elementary function. So we must change the order ofintegration. This is accomplished by first expressing the given iterated integral as adouble integral. We have

10

1x

sin(y2)dy dx=

Dsin(y2)dA,

whereD= {(x, 0 | 0 x 1, x y 1)}.

We see that an alternative description ofD is

D= {(x, y) | 0 y 1, 0 x y}.This enables us to express the double integral as an iterated integral in the reverseorder:

1

0 1

x

sin(y2)dy dx= D

sin(y2)dA = 1

0 y

0

sin(y2)dx dy= 1

0x sin(y2)

x=y

x=0 dy

=

10

y sin(y2)dy =12

cos(y2)

10

=1

2(1 cos 1).


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3 BASIC MATERIAL. 22

Review guide for the final exam in Math 233

3 Basic material.This review includes the remainder of the material for math 233. The final examwill be a cumulative exam with many of the problems coming from the materialcovered beginning approximately with chapter 15.4 of the book.

We first recall polar coordinates formulas given in chapter 10.3. The coordinatesof a point (x, y) R3 can be described by the equations:

x= r cos() y = r sin(), (5)

where r=

x2 +y2 is the distance from the origin and (xr ,yr ) is (cos(), sin()) on

the unit circle. Note that r

0 and can be taken to lie in the interval [0, 2).

To find r and when x and y are known, we use the equations:

r2 =x2 +y2 tan() = y

x. (6)

Example 92 Convert the point (2, 3 ) from polar to Cartesian coordinates.

Solution: Since r = 2 and = 3 , Equations 5 give

x= r cos() = 2 cos

3 = 2 1

2 = 1

y= r sin() = 2 sin

3 = 2

3

2 =

3.

Therefore, the point is (1,

3) in Cartesian coordinates.

Example 93 Represent the point with Cartesian coordinates (1,1) in terms ofpolar coordinates.

Solution: If we choose r to be positive, then Equations 6 give

r=

x2 +y2 =

12 + (1)2 =

2

tan() = y

x= 1.

Since the point (1,1) lies in the fourth quadrant, we can choose = 4 or = 74 .Thus, one possible answer is (

2,4 ); another is (r, ) = (

2, 74 ).

The next theorem describes how to calculate the integral of a function f(x, y)over a polar rectangle. Note that dA = r dr d.

Theorem 94 (Change to Polar Coordinates in a Double Integral) Iffis con-tinuous on a polar rectangle R given by 0 a r b, , where0 2, then

R f(x, y)dA=

b

a

f(r cos(), r sin())r dr d.

Example 95 Evaluate

R(3x+ 4y2)dA, where R is the region in the upper half-

plane bounded by the circles x2 =y2 = 1 and x2 +y2 = 4.


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Solution: The region R can be described as

R= {(x, y) | y 0, 1 x2 +y2 4}.

It is a half-ring and in polar coordinates it is given by 1 r 2, 0 .Therefore, by Theorem 94,

R

(3x+ 4y2)dA=

0

21

(3r cos() + 4r2 sin2())r dr d

=

0

21

(3r2 cos() + 4r3 sin2())dr d

=

0

[r3 cos() +r4sin2()]r=2r=1d =

0

(7 cos() + 15 sin2())d

=

0[7 cos() +15

2(1 cos(2))]d

= 7sin() +15

2 15

4 sin(2)

0

= 15

2 .

Example 96 Find the volume of the solid bounded by the plane z = 0 and theparaboloid z = 1 x2 y2.

Solution: If we putz = 0 in the equation of the paraboloid, we get x2 +y2 = 1.This means that the plane intersects the paraboloid in the circle x2 + y2 = 1, so thesolid lies under the paraboloid and above the circular disk D given by x2 +y2

1.

In polar coordinatesD is given by 0 r 1, 0 2. Since 1x2y2 = 1r2,the volume is

V =

D

(1 x2 y2)dA=

20

10

(1 r2)r dr d

=

20

10

(r r3) drd = 2

r2

2 r

4

4

10

=

2.

The next theorem extends our previous application of Fubinis theorem for type

II regions.

Theorem 97 Iffcontinuous on a polar region of the form

D= {(r, ) | , h1() r h2()}

then D

f(x, y)dA=

h2()h1()

f(r cos(), r sin())r dr d

The next definition describes the notion of a vector field. We have already seenan example of a vector field associated to a function f(x, y) defined on a domain

D R2

, namely the gradient vector fieldf(x, y) =fx(x, y), fy(x, y). In natureand in physics, we have the familiar examples of the velocity vector field in weatherand force vector fields that arise in gravitational fields, electric and magnetic fields.


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Definition 98 Let D be a set in R2 (a plane region). A vector field on R2 is afunction F that assigns to each point (x, y) in D a two-dimensional vector F(x, y).

Definition 99 LetEbe a subset ofR3

. Avector field on R3

is a function F thatassigns to each point (x,y ,z) in E a three-dimensional vector F(x,y ,z).

Note that a vector field F on R3 can be expressed by its component functions.So ifF = (P,Q ,R), then:

F(x,y ,z) =P(x,y ,z)i +Q(x,y ,z)j +R(x,y ,z)k.

We now describe our first kind of line integral. These type of integrals ariseform integrating a function along a curve Cin the plane or in R3. The type of lineintegral described in the next definition is called a line integral with respect toarc length.

Definition 100 LetCbe a smooth curve in R2. Given n, consider n equal subdi-visions of lengths si; let (x

i , y

i ) denote the midpoints of thei-th subdivision. Iff

is a real valued function defined on C, then the line integral off along C is

C

f(x, y)ds = limn

nj=1

f(xi , yi )si,


The following formula can be used to evaluate this type of line integral.

Theorem 101 Suppose f(x, y) is a continuous function on a differentiable curveC(t), C: [a, b] R2. Then

C

f(x, y) ds=

ba

f(x(t), y(t))

dx

dt

2+

dy

dt

2dt

In the above formula, dx

dt

2+

dy

dt

2,

is the speed ofC(t) at time t.

Example 102 EvaluateC2x ds, where Cconsists of the arc C1 of the parabola

y= x2 from (0, 0) to (1, 1).

Solution: We can choosex as the time parameter and the equations for Cbecome

x= x y= x2 0 x 1

Therefore, C1

2x ds=

10

2x

dx

dx

2+

dy

dx

2dx

=

1

02x

1 + 4x2 dx= 1

4 2

3(1 + 4x2)

3

2

1

0

=55 1

6 .


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Actually for what we will studying next, another type of line integral will beimportant. These line integrals are called line integrals of f along C withrespect to x and y. They are defined respectively for x and y by the following

limits: C

f(x, y)dx = limn

ni=1

f(xi , yi )xi

C

f(x, y)dy = limn

ni=1

f(xi , yi )yi.

The following formulas show how to calculate these new type line integrals. Notethat these integrals depend on the orientation of the curve C, i.e., the initial andterminal points.

Theorem 103 C

f(x, y) dx=

ba

f(x(t), y(t))x(t) dt

C

f(x, y) dy=

ba

f(x(t), y(t))y(t) dt.

Example 104 EvaluateCy

2 dx+ x dy, where C = C1 is the line segment from(5,3) to (0, 2)

Solution: A parametric representation for the line segment is

x= 5t 5, y= 5t 3, 0 t 1Then dx = 5 dt, dy= 5 dt, and Theorem 103 gives

C1

y2 dx+x dy =

10

(5t 3)2(5dt) + (5t 5)(5 dt)

= 5

10

(25t2 25t+ 4) dt

= 5

25t3

3 25t

2

2 + 4t

10

= 56

.

Example 105 EvaluateCy

2 dx+x dy, where C = C2 is the arc of the parabolax= 4 y2 from (5,3) to (0, 2).

Solution : Since the parabola is given as a function of y, lets take y as the pa-rameter and write C2 as

x= 4 y2 y= y, 3 y 2.Then dx = 2y dy and by Theorem 103 we have

C2

y2dx+x dy= 2

3y2(

2y) dy+ (4

y2) dy

=

23

(2y3 y2 + 4) dy


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=

y

4

2 y

3

3 + 4y

23

= 405

6.

One can also define in a similar manner the line integral with respect to arclength of a function falong a curve C in R3.

Theorem 106C

f(x,y ,z) ds=

ba

f(x(t), y(t), z(t))

dx

dt

2+

dy

dt

2+

dz

dt

2dt

=

C

P(x,y ,z) dx+Q(x,y ,z) dy+R(x,y ,z) dz,

where f(x,y ,z) = P(x,y ,z), Q(x,y ,z), R(x,y ,z).The next example demonstrates how to calculate a line integral of a

function with respect to x, y and z.

Example 107 Evaluatecy dx + z dy + xdz,, whereCconsists of the line segment

C1 from (2, 0, 0) to (3, 4, 5) followed by the vertical line segment C2 from (3, 4, 5) to(3, 4, 0).

Solution: We writeC1 as

r(t) = (1 t)2, 0, 0+t3, 4, 5 = 2 +t, 4t, 5tor, in parametric form, as

x= 2 +t y = 4t z = 5t 0 t 1.Thus

C1

y dx+z dy+x dz=

10

(4t) dt+ (5t)4 dt+ (2 +t)5 dt

=

10

(10 + 29t) dt= 10t+ 29t2

2

10

= 24.5.

Likewise, C2 can be written in the form

r(t) = (1 t)3, 4, 5+t3, 4, 0 = 3, 4, 5 5t

orx= 3 y= 4 z= 5 5t dz = 5 dt, 0 t 1.

Then dx = 0 =dy, soC2

y dx+z dy+x dz=

10

2(5) dt= 15.

Adding the values of these integrals, we obtainC=C1C2

y dx+z dy+x dz= 24.5 15 = 9.5.

We now get to our final type of line integral which can be considered to be aline integral of a vector field. This type of integral is used to calculate the workWdone by a force field F in moving a particle along a smooth curve C.


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Theorem 108 IfCis given by the vector equation r(t) =x(t)i +y(t)j +z(t)k onthe interval [a, b], then the work Wcan be calculated by

W = ba

F(r(t)) r(t) dt,

where is the dot product.

In general, we make the following definition which is related to the formula inthe above theorem.

Definition 109 Let F be a continuous vector field defined on a smooth curve Cgiven by a vector function r(t), a t b. Then the line integral of F along Cis

CF dr=

b

a F(r(t)) r(t) dt= CF Tds;here, Tis the unit tangent vector field to the parameterized curve C.

Example 110 Find the work done by the force field F(x, y) =x2i xyj in movinga particle along the quarter-circle r(t) = cos t i + sin tj, 0 t 2 .

Solution: Since x = cos t and y = sin t, we have

F(r(t)) = cos2 ti cos t sin tj

and

r(t) = sin ti + cos tj.Therefore, the work done is

C

F dr=

2

0F(r(t)) r(t)dt=

2

0(2cos2 t sin t)dt

= 2cos3 t

3

2

0

= 23

.

Example 111 Evaluate

CF dr, where F(x,y ,z) =xyi +yzj +zxkand C is the

twisted cubic given by

x= t y= t2 z= t3 0 t 1.

Solution: We haver(t) =ti +t2j +t3k

r(t) =i + 2tj + 3t2k

F(r(t)) =t3i +t5j +t4k.

Thus,

C

F dr= 1

0

F(r(t)) r(t)dt

=

10

(t3 + 5t6)dt= t4

4 +

5t7

7

10

= 27

28.


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Theorem 112 IfC in R3 is parameterized by r(t) and F =Pi +Qj +Rk, then

C

F

dr=

C

P dx+Qdy+Rdz.

We now apply the material covered so far on line integrals to obtain severalversions of the fundamental theorem of calculus in the multivariable setting. Recallthat the fundamental theorem calculus can be written as b

aF(x)dx= F(b) F(a),

when F (x) is continuous on [a, b].

Theorem 113 LetCbe a smooth curve given by the vector function r(t), a t b. Let fbe a differentiable function of two or three variables whose gradient vector

f is continuous on C. ThenCf dr= f(r(b)) f(r(a)).

For further discussion, we make the following definitions.

Definition 114 A curve r : [a, b] R3 (or R3) closed ifr(a) =r(b).

Definition 115 A domain D R3 (or R2) isopenif for any point p in D, a smallball (or disk) centered at p in R3 (in R2) is contained in D .

Definition 116 A domain D R3 (or R2) is connected if any two points in Dcan be joined by a path contained inside D .

Definition 117 A curve r : [a, b] R3 (or R2) is a simple curve if it doesntintersect itself anywhere between its end points (r(t1) =r(t2) whena < t1 < t2< b).

Definition 118 An open, connected regionD R2 is a simply-connectedregionif any simple closed curve in D encloses only points that are in D .

Definition 119 A vector field F is called a conservative vector fieldif it is thegradient of some scalar function f(x, y); the function f(x, y) is called a potential

function for F. For example, for f(x, y) = xy +y2,f =y, x+ 2y and so,F(x, y) =yi + (x+ 2y)j is a conservative vector field.

Definition 120 IfF is a continuous vector field with domain D, we say that theline integral

CF dris independent of pathif

C1

F dr= C2 F drfor any twopaths C1 and C2 in D with the same initial and the same terminal points.

We now states several theorems that you should know for the final exam.

Theorem 121CF dris independent of path inD if and only if

CF dr= 0 for

every closed path C in D .

Theorem 122 SupposeF is a vector field that is continuous on an open connectedregion D. If

CF dr is independent of path in D, then F is a conservative vector

field on D ; that is, there exists a function f such thatf=F.


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Theorem 123 IfF(x, y) =P(x, y)i+Q(x, y)j is a conservative vector field, wherePandQhave continuous first-order partial derivatives on a domainD, then throughoutD we have

Py

= Qx

.

Theorem 124 Let F = Pi+ Qj be a vector field on an open simply-connectedregion D . Suppose that P and Q have continuous first-order derivatives and

P

y =

Q

x throughoutD.

Then F is conservative.

Example 125 Determine whether or not the vector fieldF(x, y) = (xy)i+(x2)jis conservative.

Solution: Let P(x, y) =x y and Q(x, y) =x 2. ThenP

y = 1 Q

x = 1.

Since Py = Qx , F is not conservative by Theorem 123.

Example 126 Determine whether or not the vector field F(x, y) = (3 + 2xy)i+(x2 3y2)j is conservative.

Solution: Let P(x, y) = 3 + 2xy and Q(x, y) =x

2

3y2

. ThenP

y = 2x=

Q

x.

Also, the domain of F is the entire plane (D = R2), which is open and simply-connected. Therefore, we can apply Theorem 124 and conclude that F is conserva-tive.

Attention! You will likely have a problem on the final exam which is similar tothe one described in the next example.

Example 127 (a) IfF(x, y) = (3+ 2xy)i + (x2

3y2)j, find a functionfsuch that

F= f.(b) Evaluate the line integral

CF dr, where Cis the curve given by

r(t) =et sin t i +et cos tj, 0 t .

Solution:

(a) From Example 126 we know thatFis conservative and so there exists a functionf withf=F, that is,

fx(x, y) = 3 + 2xy (7)

fy(x, y) =x2

3y2 (8)

Integrating (7) with respect to x, we obtain

f(x, y) = 3x+x2y+g(y). (9)


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Notice that the constant of integration is a constant with respect tox, that is,it is a function ofy , which we have called g (y).

Next we differentiate both sides of (9) with respect to y :

fy(x, y) =x2 +g(y). (10)

Comparing (8) and (10), we see that

g(y) = 3y2.

Integrating with respect toy, we have

g(y) = y3 +K

where K is a constant. Putting this in (9), we have

f(x, y) = 3x+x2y y3 +K

as the desired potential function.

(b) To apply Theorem 113 all we have to know are the initial and terminal pointsofC, namely, r(0) = (0, 1) and r() = (0,e). In the expression for f(x, y)in part (a), any value of the constant Kwill do, so lets choose K= 0. Thenwe have

CF dr=

Cf dr= f(0,e) f(0, 1)

=e3

(1) =e3

+ 1.This method is much shorter than the straightforward method for evaluatingline integrals described in Theorem 103.

Definition 128 A simple closed parameterized curve C in R2 always bounds abounded simply-connected domain D. We say that C is positively orientedif forthe parametrizationr(t) ofC, the region D is always on the left as r(t) traversesC.Note that this parametrization is the counterclockwise one on the boundary of unitdisk D = {(x, y) | x2 +y2 1}.

The next theorem is a version of the fundamental theorem of calculus, since it

allows one to carry out a two-dimensional integral on a domain D by calculating arelated integral one-dimensional on the boundary ofD. There will be at least onefinal exam problem related to the following theorem.

Theorem 129 (Greens Theorem) Let C be a positively oriented, piecewise-smooth, simple closed curve in the plane and let D be the region bounded by C.IfP and Q have continuous partial derivatives on an open region that contains D,then

CP dx+Qdy=

D

Q

x P

y

dA.

An immediate consequence of Greens Theorem are the area formulas describedin the next theorem.


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Theorem 130 LetD be a simply-connected domain in the plane with simple closedoriented boundary curve C. Let A be the area ofD . Then:

A=C

x dy= C

y dx= 12C

x dy y dx. (11)

Example 131 Find the area enclosed by the ellipse x2

a2 + y

2

b2 = 1.

Solution : The ellipse has parametric equations x = a cos t and y = b sin t, where0 t 2. Using the third formula in Equation 11, we have

A = 1

2

C

x dy y dx

= 1

2 2

0

(a cos t)(b cos t) dt

(b sin t)(

a sin t) dt

= ab

2

20

dt= ab.

Example 132 Use Greens Theorem to evaluateC(3yesinx) dx+(7x+

y4 + 1) dy,

where C is the circlex2 +y2 = 9.

Solution: The region D bounded by C is the disk x2 +y2 9, so lets change topolar coordinates after applying Greens Theorem:

C(3y e

sinx

) dx+ (7x+

y4 + 1) dy

=

D

x(7x+

y4 + 1)

y(3y esinx)

dA

=

20

30

(7 3) rdrd

= 4

20

d

30

r dr= 36.


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4 PRACTICE PROBLEMS FOR EXAM 1. 32

Practice problems from old exams for math233

William H. Meeks III

September 1, 2014

Disclaimer: Your instructor covers far more materials that we can possibly fitinto a four/five questions exams. These practice tests are meant to give you an ideaof the kind and varieties of questions that were asked within the time limit of thatparticular tests. In addition, the scope, length and format of these old exams mightchange from year to year. Users beware! These are NOT templates upon whichfuture exams are based, so dont expect your exam to contain problems exactly likethe ones presented here. Check the course web page for an update on the materialto be covered on each exam or ask your instructor.

4 Practice problems for Exam 1.

Fall 2008 Exam

1. (a) Find parametric equations for the line L which contains A(1, 2, 3) andB(4, 6, 5).

(b) Find parametric equations for the line L of intersection of the planesx 2y+z = 10 and 2x+y z = 0.

2. (a) Find an equation of the plane which contains the points P(1, 0, 1),Q(1,

2, 1) and R(2, 0,

1).

(b) Find the distance D from the point (1, 6,1) to the plane 2x+y2z = 19.(c) Find the pointQ in the plane 2x+y 2z = 19 which is closest to the

point (1, 6,1). (Hint: You can use part b) of this problem to help findQor first find the equation of the line L passing through Q and the point(1, 6,1) and then solve for Q.)

3. (a) Find the volume V of the parallelepiped such that the following fourpoints A = (3, 4, 0), B = (3, 1,2), C = (4, 5,3), D = (1, 0,1) arevertices and the vertices B,C, D are all adjacent to the vertex A.

(b) Find the center and radius of the spherex2 4x+y2 + 4y+z2 = 8.4. (a) The position vector of a particle moving in space equalsr(t) = t2it2j +

12t

2kat any time t 0.Find an equation of the tangent line to the curve at the point (4,4, 2).

(b) Find the lengthL of the arc traveled from time t = 1 to time t = 4.

(c) Suppose a particle moving in space has velocity

v(t) = sin t, 2cos2t, 3et

and initial positionr(0) = 1, 2, 0. Find the position vector functionr(t).5. (a) Consider the points A(2, 1, 0), B(3, 0, 2) and C(0, 2, 1). Find the area

of the triangle ABC. (Hint: If you know how to find the area of aparallelogram spanned by 2 vectors, then you should be able to solve thisproblem.


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(b) Three of the four vertices of a parallelogram are P(0,1, 1), Q(0, 1, 0)and R(2, 1, 1). Two of the sides are P Qand P R. Find the coordinates ofthe fourth vertex.

Spring 2008 Exam

6. (a) Find an equation of the plane through the pointsA = (1, 2, 3),B = (0, 1, 3), and C= (2, 1, 4).

(b) Find the area of the triangle with vertices at points A, B, and C givenabove. Hint: the area of this triangle is related to the area of a certainparallelogram

7. (a) Find the parametric equations of the line passing through the point(2, 4, 1) that is perpendicular to the plane 3x

y+ 5z = 77.

(b) Find the intersection point of the line in part (a) and the plane 3x y+5z = 77.

8. (a) Aplanecurve is given by the graph of the vector function

u(t) = 1 + cos t, sin t, 0 t 2.

Find a single equation for the curve in terms ofx and y, by eliminatingt.

(b) Consider thespacecurve given by the graph of the vector function

r(t) = 1 + cos t, sin t, t, 0 t 2.Sketch the curve and indicate the direction of increasingt in your graph.

(c) Determine parametric equations for the line T tangent to the graph ofthe spacecurve for r(t) at t = /3, and sketch T in the graph obtainedin part(b).

9. Suppose that r(t) has derivative r(t) = sin2t, cos2t, 0 on the interval0 t 1. Suppose we know that r(0) = 12 , 0, 1.

(a) Determiner(t) for all t.

(b) Show thatr(t) is orthogonal to r (t) for all t.(c) Find the arclength of the graph of the vector functionr(t) on the interval

0 t 1.

10. Ifr(t) = (2t)i + (t2 6)j (13 t3)k represents the position vector of a movingobject (where t 0 is measured in seconds and distance is measured in feet),

(a) Find the speed and the velocity of the object at timet.

(b) If a second object travels along a path given defined by the graph of thevector functionw(s) = 2, 5, 1+ s2,1,5, show that the paths of the

two objects intersect at a common point P.(c) Ifs= t in part (b), (i.e. the position of the second object is w(t) when

the first object is at position r(t)), do the two objects ever collide?


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Spring 2007 Exam

11. (a) Find parametric equations for the line which containsA(7, 6, 4) andB(4, 6, 5).

(b) Find the parametric equations for the line of intersection of the planesx 2y+z = 5 and 2x+y z= 0.

12. (a) Find an equation of the plane which contains the points P(1, 0, 2),Q(1,2, 1) and R(2, 0,1).

(b) Find the distance from the point (1, 0,1) to the plane 2x+y 2z = 1.(c) Find the point P in the plane 2x+ y 2z = 1 which is closest to the

point (1, 0,1). (Hint: You can use part b) of this problem to help findP or first find the equation of the line passing through P and the point(1, 0,1) and then solve for P.)

13. (a) Consider the two space curves

r1(t) = cos(t 1), t2 1, 2t4, r2(s) = 1 + ln s, s2 2s+ 1, 2s2,

wheretand sare two independent real parameters. Find the cosine of theangle between the tangent vectors of the two curves at the intersectionpoint (1, 0, 2).

(b) Find the center and radius of the spherex2 +y2 + 2y+z2 + 4z = 20.

14. The velocity vector of a particle moving in space equalsv(t) = 2ti 2tj +tkat any time t

0.

(a) At the timet = 4, this particle is at the point (0, 5, 4). Find an equationof the tangent line to the curve at the time t = 4.

(b) Find the length of the arc traveled from timet = 2 to time t = 4.

(c) Find a vector function which represents the curve of intersection of thecylinderx2 +y2 = 1 and the plane x + 2y+z = 4.

15. (a) Consider the points A(2, 1, 0), B(1, 0, 2) and C(0, 2, 1). Find the areaof the triangle ABC. (Hint: If you know how to find the area of aparallelogram spanned by 2 vectors, then you should be able to solve thisproblem.)

(b) Suppose a particle moving in space has velocity

v(t) = sin t, cos2t, et

and initial positionr(0) = 1, 2, 0. Find the position vector functionr(t).

Fall 2007 Exam

16. Find the equation of the plane containing the lines

x= 4 4t, y= 3 t, z= 1 + 5t andx= 4 t, y= 3 + 2t, z= 1


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31. Find parametric equations for the line of intersection of the planesx2y+z = 1and 2x+y+z = 1.

32. Let L1 denote the line through the points (1, 0, 1) and (1, 4, 1) and let L2denote the line through the points (2, 3,1) and (4, 4,3). Do the lines L1and L2 intersect? If not, are they skew or parallel?

33. (a) Find the volume of the parallelepiped such that the following four pointsA = (1, 4, 2), B = (3, 1,2), C = (4, 3,3), D = (1, 0,1) are vertices andthe vertices B,C, D are all adjacent to the vertex A.

(b) Find an equation of the plane through A, B, D.

(c) Find the angle between the plane through A, B, Cand the xy plane.

34. The velocity vector of a particle moving in space equalsv(t) = 2ti + 2t1/2j + k

at any time t 0.(a) At the time t = 0 this particle is at the point (1, 5, 4). Find the positionvector r(t) of the particle at the time t = 4.

(b) Find an equation of the tangent line to the curve at the time t = 4.

(c) Does the particle ever pass through the point P= (80, 41, 13) ?

(d) Find the length of the arc traveled from time t = 1 to time t = 2.

35. Consider the surface x2 + 3y2 2z2 = 1.(a) What are the traces in x = k, y= k, z= k? Sketch a few.

(b) Sketch the surface in the space.

36. Find an equation for the tangent plane to the graph of f(x, y) = y ln x at(1, 4, 0).

37. Find the distance between the given parallel planes

z= 2x+y 1, 4x 2y+ 2z = 3.

38. Identify the surface given by the equation 4x2 + 4y2 8y z2 = 0. Draw thetraces and sketch the curve.

39. A projectile is fired from a point 5 m above the ground at an angle of 30degrees and an initial speed of 100 m/s.

a) Write an equation for the acceleration vector.b) Write a vector for initial velocity.c) Write a vector for initial position.d) At what time does the projectile hit the ground?e) How far did it travel, horizontally, before it hit the ground?

40. Explain why the limit off(x, y) = (3x2y2)/(2x4 +y4) does not exist as (x, y)approaches (0, 0).

41. Find an equation of the plane that passes through the point P(1, 1, 0) and

contains the line given by parametric equations x= 2+3t, y= 1t, z = 2+2t.42. Find all of the first order and second order partial derivatives of the function.


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(a) f(x, y) =x3 xy2 +y(b) f(x, y) = ln(x+

x2 +y2)

43. Find the linear approximation of the functionf(x, y) =xyex at (x, y) = (1, 1),and use it to estimate f(1.1, 0.9).

44. Find a vector function which represents the curve of intersection of the paraboloidz= 2x2 +y2 and the parabolic cylinder y = x2.

Spring 2009 Exam

45. Givena= 3, 6,2, b= 1, 2, 3.a) Write down the vector projection ofb to a. (Hint: Use projections.)

b) Write b as a sum of a vector parallel to a and a vector orthogonal to a.(Hint: Use projections.)

c) Let be the angle betweena and b. Find cos .

46. GivenA = (1, 7, 5), B = (3, 2, 2) and C= (1, 2, 3).a) Let L be the line which passes through the points A = (1, 7, 5) andB = (3, 2, 2). Find the parametric equations for L.

(b)A,B and Care three of the four vertices of a parallelogram, whileC AandCB are two of the four edges. Find the fourth vertex.

47. Consider the points P(1, 3, 5), Q(

2, 1, 2), R(1, 1, 1) in R3.

a) Find an equation for the plane containing P, Q and R.

b) Find the area of the triangle with vertex P, Q, R.

48. Find parametric equations for the line of intersection of the planesx+y+ 3z= 1 and x y+ 2z= 0.

49. Consider the parameterized curve

r(t) =

t, t2, t3

, t R.

a) Set up an integral for the length of the arc between t = 0 and t = 1. Do

not attempt to evaluate the integral.

b) Write down the parametric equations of tangent line to r(t) at (2, 4, 8).

50. a) Consider the sphereS in R3 given by the equation

x2 +y2 +z2 4x 6z 3 = 0.

Find its center C and its radius R. b) What does the equation x2 +z2 = 4describe in R3? Make a sketch.

51. Jane throws a basketball at an angle of 45o to the horizontal at an initialspeed of 12 m/s, where m denotes meters. It leaves her hand 2 m above theground. Assume the acceleration on the ball due to gravity is downward withmagnitude 10 m/s2 and neglect air friction.


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(a) Find the velocity functionv(t) and the position function r(t) of the ball.Use coordinates in the xy-plane to describe what is happening; assumeJane is standing with her feet at the point (0 , 0) and y represents the

height.

(b) Find the speed of the ball at its highest point.

(c) At what timeTdoes the ball reach its highest point.

Fall 2009 Exam

52. Let v = 2i j + 2kand w = 2i + 6k + 9j.(a) Find the vector representing the projection ofv onto w.

(b) Find cos , where is the angle between v and w.

53. Consider the points P = (0, 3,3), Q = (1, 3, 2), R = (1, 2,3).(a) Find an equation for the plane containing P,Q, R.

(b) Find the area of the triangle with vertices P,Q, R.

54. Let P1 be the plane x +y z = 0 and P2 be the plane x 2y+z = 1.(a) Find parametric equations for the line of intersection ofP1 and P2.

(b) Find the distance from the origin to the plane P2.

55. Let r(t) =t2i +t3j +t6k.

(a) Find an equation for the tangent line to the graph at the point given byt= 1.

(b) Find the unit tangent vector T to the graph at the point given by t = 1.

(c) Write a definite integral that computes the length of the graph ofr(t) fromt= 1 to t = 2, but do not attempt to evaluate it.

56. Consider a particle moving with accelerationa(t) = t, et, sin(t).(a) Find the velocity vector v(t) of the particle, assuming that v(0) =0.

(b) Find the position vector r(t) of the particle, assuming that r(0) =0.

Spring 2010 Exam

57. Consider the parallelogram with vertices A, B, C, D such that B and C areadjacent to A where A = (1, 2,1), B = (3, 5, 1) and D = (2,1, 2).

(a) Find the area of the parallelogram.

(b) Find the coordinates of the point C.

58. Consider the points A = (0, 3,3), B = (1, 3, 2), C= (1, 2,3).

(a) Find the orthogonal projection projAB

(AC) of the vector

AC onto the

vector AB.(b) Find the distanced from the point Cto the line L that contains points

Aand B .


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59. Let P1 be the plane x + 3y+z = 0 and P2 be the plane 2x+y z = 1.(a) Find the cosine of the angle between the planes.

(b) Find the parametric equations of the line of intersection between the 2planes P1 and P2.

(c) Find the distance from the planeP2 to the origin.

60. Let r(t) = cos(2t)i + sin(2t)j +tk.

(a) What is the length of the curve starting att = 0 and ending at t = 5.

(b) Find a vector equation for the tangent line to the graph at the point givenbyt= 0.

61. Show that the limit lim(x,y)(0,0)x2y2x2+y2

does not exist.

62. Consider the sphere S in R3 given by the equation

x2 +y2 +z2 2x 8y 2 = 0(a) Find the coordinates of its center and its radius.

(b) What does the equation x2 +y2 = 64 describe in R3. Make a sketch.

Spring 2012 Exam

63. GivenA = (1, 7, 5), B = (3, 0, 2) and C= (1, 2, 3).(a) Let L be the line which passes through the points A and B. Find the

parametric equations forL.

(b) A, B and C are three of the four vertices of a parallelogram, while ABand B Care two of the four edges. Find the fourth vertex.

64. Consider the points P(1, 1, 1), Q(2, 1, 2), R(1, 3, 5) in R3.(a) Find an equation for the plane containingP, Q and R.

(b) Find the area of the triangle with vertices P, Q, R.

65. Let P1 be the plane x 2y+ 2z = 10 and P2 be the plane 2x+y + 2z = 0.Find the cosine of the angle between the planes.

66. a) Find the distance from the pointQ= (1, 6,1) to the plane 2x+y2z = 19.(Hint: To do this you can use the vector projection of some vector P Q, whereP is some point on the plane.)

b) Write the parametric equations of the line L containing the point T(1, 2, 3)and perpendicular to the plane 2x+y 2z= 19.c) Find the point of intersection of the line L in part b) with the plane 2x+y 2z= 19.

67. Find the equation of the sphere with center at the point (1 , 2, 3) and whichcontains the point (3, 1, 5).

68. Make a sketch of the surface in R3 described by equation y2 +z2 = 36. Inyour sketch of this surface, include the labeled coordinate axes and the tracecurves on the surface for the planes x = 0 and x = 4.


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(You are given more values than you will need for this problem.) Supposethat x and y are functions of variable t: x= t3; y = t2 + 1, so that wemay regardfas a function oft. Compute the derivative offwith respect

to t when t = 1.Use the Chain Rule to find zv when u = 1 and v = 1, where

z = x3y2 +y3x; x= u2 +v2, y= u v2.

(b) Use theChain Rule to find zv when u = 1 and v = 1, where

z = x3y2 +y3x; x= u2 +v2, y= u v2.

4. Consider the surface x2 +y2 2z2 = 0 and the point P(1, 1, 1) which lies onthe surface.

(i) Find the equation of the tangent plane to the surface at P.

(ii) Find the equation of the normal line to the

surface atP.

5. Letf(x, y) = 2x3 +xy2 + 6x2 +y2.

Find and classify (as localmaxima, localminimaor saddlepoints) allcrit-ical pointsoff.

6. A flat circular plate has the shape of the region x2 +y2 4. The plate(including the boundary x

2

+y2

= 4) is heated so that the temperature atany point (x, y) on the plate is given by T(x, y) = x2 +y2 2x. Find thetemperatures at thehottestand the coldest points on the plate, including theboundary x2 +y2 = 4.

Spring 2008 Exam

7. Consider the equationx2 +y2/9 +z2/4 = 1.

(a) Identify this quadric (i.e. quadratic surface), and graph the portion of thesurface in the region x

0, y

0, and z

0. Your graph should include

tick marks along the three positive coordinate axes, and must clearly showwhere the surface intersects any of the three positive coordinate axes.

(b) Calculatezx and zy at an arbitrary point (x,y ,z) on the surface.

(c) Determine the equation of the tangent plane to the surface at the point( 1

2, 32 , 1).

8. Given the functionf(x, y) =x2y+yexy.

(a) Find the linearization offat the point (0, 5) and use it to approximatethe value offat the point (.1, 4.9). (An unsupported numerical approx-

imation to f(.1, 4.9) will not receive credit.)(b) Suppose thatx(r, ) =r cos and y (r, ) =r sin . Calculate f at r = 5

and = 2 .


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(c) Suppose a particle travels along a path (x(t), y(t)), and that F(t) =f(x(t), y(t)) wheref(x, y) is the function defined above. CalculateF(3),assuming that at timet = 3 the particles position is (x(3), y(3)) = (0, 5)

and its velocity is (x(3), y(3)) = (3,2).9. Consider the functionf(x, y) = 2

x2 + 4y.

(a) Find the directional derivative off(x, y) at P = (2, 3) in the directionstarting from P pointing towards Q = (0, 4).

(b) Find all unit vectorsu for which the directional derivativeDuf(2, 3) = 0.

(c) Is there a unit vectoru for which the directional derivativeDuf(2, 3) = 4? Either find the appropriate u or explain why not.

10. letf(x, y) = 23x

3

+ 13y

3

xy.(a) Find all critical points off(x, y).

(b) Classify each critical point as a relative maximum, relative minimum orsaddle; you do not need to calculate the function at these points, butyour answer must be justified by an appropriate calculation.

11. Use the method of Lagrange multipliers to determine all points (x, y) wherethe functionf(x, y) = 2x2 + 4y2 + 16 has an extreme value (either a maximum

or a minimum) subject to the constraint x2

4 +y2 = 4.

Fall 2007 Exam

12. Find the xand y coordinates of all critical points of the function

f(x, y) = 2x3 6x2 +xy2 +y2

and use the second derivative test to classify them as local minima, localmaxima or saddle points.

13. A hiker is walking on a mountain path. The surface of the mountain is modeledby z = 1 4x2 3y2. The positive x-axis points to East direction and thepositive y-axis points North. Justify your answers.

(a) Suppose the hiker is now at the point P(14 ,12 , 0) heading North, is sheascending or descending?

(b) When the hiker is at the pointQ(14 , 0,34), in which direction should she

initially head to ascend most rapidly?

14. Find the volume of the solid bounded by the surfacez = 6xy and the planesx= 2, x= 2, y= 0, y= 3, and z = 0.

15. Letz (x, y) =x2 + y2 xy wherex = s r is a known function ofr and s andy= y(r, s) is an unknown function ofr and s. (Note thatz can be considereda function ofr and s.) Suppose we know that

y(2, 3) = 3, yr

(2, 3) = 7, and ys

(2, 3) = 5.

Calculate zr when r = 2 and s = 3.


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16. Let F(x,y ,z) = x2 2xy y2 + 8x+ 4y z. This problem continues on thenext page.

(a) Write the equation of the tangent plane to the surface given byF(x,y ,z) =0 at the point (2, 1,5).(b) Find the point (a,b,c) on the surface at which the tangent plane is hori-

zontal, that is, parallel to the z = 0 plane.

17. Find the points on the ellipsex2 + 4y2 = 4 that are closest to the point (1, 0).

Fall 2006 Exam

18. (a) Let f(x, y) be a differentiable function with the following values of thepartial derivatives fx(x, y) and fy(x, y) at certain points (x, y):

x y fx(x, y) fy(x, y)

1 1 2 41 2 3 11 2 1 1

(You are given more values than you will need for this problems.) Supposethat x and y are functions of variable t:

x= t3; y= t2 + 1,

so that we may regard fas a function oft. Compute the derivative offwith respect to t when t = 1.

(b) Use the Chain Rule to find zv when u = 1 and v = 1, where

z= x3y2 +y3x; x= u2 +v, y= 2u v.

19. (a) Letf(x, y) =x2y3 + y4. Find the directional derivative offat the point(1, 1) in the direction which forms an angle (counterclockwise) of /6with the positive x-axis.

(b) Find an equation of the tangent line to the curvex2y+ y3 5 = 0 at thepoint (x, y) = (2, 1).

20. Let

f(x, y) = 2x3 +xy2 + 5x2 +y2.

Find and classify (as local maxima, local minima or saddle points) all criticalpoints off.

21. Find the maximum value off(x, y) = 2x2 +y2 on the circle x2 +y2 = 1.

22. Find the volume above the rectangle1x1, 2y5 and below thesurface z = 5 +x2 +y.

23. Evaluate the integral

10

1y

x3 + 1dxdy

by reversing the order of integration.


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These problems from older exams

24. Use Chain Rule to find dz/dt orz/u, z/v.

(1) z = x2y+ 2y3, x= 1 +t2, y= (1 t)2.

(2) z = x3 +xy2 +y3, x= uv, y= u+v.

25. Ifz = f(x, y), wheref is differentiable, and x = 1 + t2, y= 3t, computedz/dtat t = 2 provided that fx(5, 6) =fy(5, 6) = 1.

26. For the following functions

(1). f(x, y) =x2

y+y3

y2

, (2)g(x, y) =x/y+xy, (3)h(x, y) = sin(x2

y)+xy2

.

(a) Find the gradient.(b) Find the directional derivative at the point (0, 1) in the direction ofv =

3, 4.(c) Find the maximum rate of change at the point (0, 1).

27. Find an equation of the tangent plane to the surfacex2 + 2y2 z2 = 5 at thepoint (2, 1, 1).

28. Find parametric equations for the tangent line to the curve of intersection ofthe surfacesz2 =x2 +y2 and x2 + 2y2 +z2 = 66 at the point (3, 4, 5).

29. Find and classify all critical points (as local maxima, local minima, or saddlepoints) of the following functions.

(1)f(x, y) =x2y2 + x2 2y3 + 3y2, (2)g(x, y) =x3 + y2 + 2xy 4x 3y + 5.30. Find the minimum value off(x, y) = 3 + xy x 2y on the closed triangular

region with vertices (0, 0), (2, 0) and (0, 3).

31. Use Lagrange multipliers to find the extreme values of the following functionswith the given constraint.

(1) f(x, y) =xy with constraint x2 + 2y2 = 3;

(2) g(x,y ,z) =x+ 3y 2z with constraint x2 + 2y2 +z2 = 5.32. Find the following iterated integrals.

(1)41

20(x+

y) dx dy

(2)21

10(2x+ 3y)

2 dy dx

(3)10

2xx (x

2 y) dy dx

(4) 101x2x3 sin(y3) dy dx (hint: reverse the order of integration)


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33. Evaluate the following double integrals.

(1)

Rcos(x+ 2y) dA, R= {(x, y) | 0 x , 0 y /2}

(2)

Rey2 dA, R = {(x, y) | 0 y 1, 0 x y}

(3)

Rx

y2 x2 dA, R = {(x, y) | 0 y 1, 0 x y}

34. Find the volume.

(1) The solid under the surface z= 4 +x2 y2 and above the rectangle

R= {(x, y) | 1 x 1, 0 y 2}

(2) The solid under the surface z = 2x + y2 and above the region bounded bycurves x y2 = 0 and x y3 = 0.

Spring 2009 Exam

35. Let f(x, y) =x2y y2 2y x2.(a) Find all of the critical points offand classify them as either local maxi-

mum, local minimum, or saddle points.

(b) Find the linearization L(x, y) of f at the point (1, 2) and use it to ap-

proximatef(0.9, 2.1).

36. Consider the functionf(x, y) =x2 2xy+ 3y+y2.(a) Find the gradientf(x, y).(b) Find the directional derivative of f at the point (1, 0) in the direction

3, 4.(c) Compute all second partial derivatives off.

(d) Supposex = st2 and y = est. Find fs and ft at s = 2 and t = 1.

37. Consider the functionf(x, y) =exy over the region D given by x2 + 4y2

2.

(a) Find the critical points off.

(b) Find the extreme values on the boundary of D.

(c) What is the absolute maximum value and absolute minimum value off(x, y) on D?

38. (a) Evaluate the following iterated integral.

21

10

(x2y xy) dy dx

(b) Find the volume of the region below z = x2

2xy + 3 and above therectangle R= [0, 1] [1, 1].39. Consider the surface Sgiven by the equation x2 +y3 +z2 = 0.


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6 PRACTICE PROBLEMS FOR FINAL EXAM. 47

(a) Give an equation for the tangent plane ofSat the point (2,2, 2).(b) Give an equation for the normal line toSat the point (2,2, 2).

Fall 2009 Exam

40. (a) Letf(x, y) = sin(xy)+cos(x+y). Compute an equation for the tangentplane to the graph offat the point where x = /4, y = /4.

(b) Let g(x,y ,z) = x2y+y2z+xz2. Compute the directional derivative atthe point (1,1, 1) in the direction of the vector 3i + 4k.

41. Suppose z= ex2+y + sin(x+y2), and x = st, y = s/t. Use the Chain Rule to

find z/t and z/s when s = t = 1.

(a) Use the chain rule to write expressions forz/t and z/s, but do notevaluate the partial derivatives.

(b) Compute all the partial derivatives you wrote in (a). Your answers mayinvolvex, y as well as s, t.

(c) Now use the partial derivatives you computed in (b) together with theformulas in (a) to compute z/t and z/s. Your answer should onlyinvolve the variables s, t.

42. Let f(x, y) =x3/3 +xy2 2xy 3x.(a) Compute the gradient off.

(b) Find all critical points off.

(c) For each critical point you found above, classify it as a local maximum,local minimum, or a saddle point.

43. Find the absolute maximum and minimum values attained by f(x, y) =x2 2x + y2 4y + 2 on the closed square with vertices (0, 0), (4, 0), (0, 4), (4, 4) (inother words, the domain{(x, y) | 0 x 4, 0 y 4}).

44. Use the method of Lagrange multipliers to find the maximum and minimumvalues attained by the function f(x,y ,z) = x+y + z on the ellipsoid 2x2 +3y2 + 6z2 = 1.

6 Practice problems for Final Exam.

Fall 2008 Exam

1. (a) Consider the points

A = (1, 0, 0), B = (2, 1, 0) and C = (1, 2, 3). Find the parametricequations for the line L passing through the points A and C.

(b) Find an equation of the plane in R3 which contains the points A, B, C.

(c) Find the area of the triangle T with vertices A, B and C.

2. Find the volume under the graph off(x, y) =x+ 2xy and over the boundedregion in the first quadrant{(x, y)| x 0, y 0} bounded by the curvey= x2 + 1 and the x and y -axes.


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6 PRACTICE PROBLEMS FOR FINAL EXAM. 48

3. Let

I=

10

22x

sin(y2) dy dx.

(a) Sketch the region of integration.

(b) Write the integral I with the order of integration reversed.

(c) Evaluate the integralI. Show your work.

4. Consider the functionF(x,y ,z) =x2 +xy2 +z.

(a) What is the gradientF(x,y ,z) ofF at the point (1, 2,1)?(b) Calculate the directional derivative of F at the point (1, 2,1) in the

direction of the vector1, 1, 1?(c) What is the maximal rate of change ofF at the point (1, 2,

1)?

(d) Find the equation of the tangent plane to the level surfaceF(x,y ,z) = 4at the point (1, 2,1).

5. Find the volume V of the solid under the surface z = 1 x2 y2 and abovethe xy -plane.

6. (a) Determine whether the following vector fields are conservative or not.Find a potential functionfor those which are indeed conservative.

i. F(x, y) = x2 +ex +xy, xy sin(y).ii. G(x, y) = 3x2y+ex +y2, x3 + 2xy+ 3y2.

7. Evaluate the line integralC

yz dx+xz dy+xy dz,

where C is the curve starting at (0, 0, 0), traveling along a line segment to(1, 0, 0) and then traveling along a second line segment to (1, 2, 1).

8. (a) Use Greens Theorem to show that i

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