Math 243 Exercise Hints and Partial AnswersThere may be some mistakes here, so please email me or ask in class if you can’t get the right

answer.

1.1.1a.

x

y

component form is h4p3/2, 2i ⇡ h3.46, 2i

1.1.1b.

x

y

h0,�8i

1.1.1c.

x

y

h1, 0i

1.1.1d.

x

y

h0, 0i

1.1.1e.

x

y

h16 cos(10�), 16 sin(10�)i ⇡ h15.76, 2.78i

1.1.1f.

x

y

h3.5,�1i

1.1.2a.

x

y

✓ = 45� = ⇡

2

rad, kvk =p22 + 22 =

p8

1.1.2b.

x

y

✓ = 0� = 0rad, kwk = 4

1.1.2c.

x

y

✓ = 270� = 3⇡

2

rad, kuk = 4

1.1.2d.

x

y

✓ = 180� = ⇡rad, krk = 3

1.1.2e.

x

y

✓ = ⇡ + tan�1(�4/3) ⇡ 126.9�, k � 3i+ 4jk = 5

1.1.2f.

x

y

✓ = 225�, k ~ABk =p8

1.1.3

x

y

v

w

v +w

v �w

✓ = 225�, k ~ABk =p8

1.1.4. The resultant has magnitudep

(8 cos 45� + 14 cos 80�)2 + (8 sin 45� + 14 sin 80�)2 ⇡p8.092 + 19.42 ⇡

21.0 knots, and angle ✓ ⇡ tan�1

�19.4

8.09

�⇡ 67.4�.

1.1.5. Magnitude isp

(200 + 160 cos 15�)2 + (sin 15�)2 ⇡ 357lbs.

1.2.1.

1.2.2a. yz-plane, 2D1.2.2b. xz-plane, 2D1.2.2c. xy-plane, 2D1.2.2d. plane parallel to yz-plane, 6 units out, 2D1.2.2e. x = 6 or y = 2. 2 planes, perpendicular to each other, 2D1.2.2f. x = 0 or z = 0. 2 planes forming a cross whose intersection is the y-axis, 2D1.2.2g. The origin, 0D1.2.2h. The z-axis, 1D1.2.2i. A line parallel to the y-axis, 1D1.2.2j. The unit sphere, 2D1.2.2k. The unit ball, 3D1.2.2l. The ball of radius 4 centered at (0,�1, 3), 3D1.2.2m. The first octant, the region where all coordinates are non-negative, 3D1.2.3a Logic “and”, sometimes denoted by ^ or \, is an operator on 2 statements that means

both statements must be satisfied. Logic “or,” denoted _ or [, means either statement is satisfied(or both, to exclude both use “xor”). The truth tables for these logic operators are

T ^ T = TT ^ F = FF ^ T = FF ^ F = FT _ T = TT _ F = TF _ T = TF _ F = F

1.2.3b. A means y = 0 OR z = 0. B means y = 0 AND z = 0. A is a 2D set of crossing planes.B is a 1D line, the x-axis.

1.2.4. answers will vary1.2.5. answers will vary

1.2.6a. h2, 0,�8i1.2.6b. h4, 0,�16i1.2.6c. h3, 0,�16i1.2.6d.

D�2p68

, 0, 8p68

E

1.2.7. If you do it right, part a and part d should be parallel, because they are the same vector.If you are lazy, part d is already done in part a, so you can skip it;)

1.3.1a. �71.3.1b. ⇡ 106�. Does that look right?

1.3.1c.

⌧7

19,�21

19,�21

19

�

1.3.1d.

⌧�1

5,�3

5, 1

�

1.3.1e. answers will vary. the dot product of the answer and u must be 0.1.3.1f. ”1.3.1g. hard until you learn the next section ;)1.3.2a. ↵� �1.3.2b. look it up, write it down

1.3.2c. cos(↵) =adj

hyp=

v1

|v|1.3.2d. if you understand c you can do this1.3.2e. show work. get cos(↵� �) = u·v

|u||v|

1.3.3a. answers will vary.

its about 4 m high. h2.5, 4i1.3.3b. 2.51.3.3c. shadow

1.3.4a.

u·u = hu1

, u2

, u3

i·hu1

, u2

, u3

i (def. of u)

= u2

1

+ u2

2

+ u2

3

(definition of dot product)

=

✓qu2

1

+ u2

2

+ u2

3

◆2

(algebra)

= kuk2 (def. of length)

PAU

1.4.1a. h�24,�2,�6i1.4.1b. h24, 2, 6i1.4.1c. 01.4.1d. 0

1.4.2a. Yes, vector1.4.2b. No1.4.2c. yes, scalar1.4.2d. yes, vector1.4.2e. yes, vector

1.4.2f. no1.4.2g. no1.4.2h. yes, scalar1.4.2i yes, vector1.4.2j yes, vector

1.4.3. The dot product indicates how much two vectors are to parallel, so if its 0 they areperpendicular. The cross product has length indicating how perpendicular 2 vectors are, and if its0 they are parallel. The dot product is a scalar. The cross product is a vector.

1.4.4a. any multiple of h0, 2,�1i works1.4.4b. 9

p5

2

1.4.5. 11

1.4.6a.

u·(u⇥ v) = hu1

, u2

, u3

i·(hu1

, u2

, u3

i ⇥ hv1

, v2

, v3

i) (component forms of u,v)

= hu1

, u2

, u3

i·hu2

v3

� u3

v2

,�u1

v3

+ u3

v1

, u1

v2

� u2

v1

i (def. of cross prod.)

= u1

u2

v3

� u1

u3

v2

� u2

u1

v3

+ u2

u3

v1

+ u3

u1

v2

� u3

u2

v1

(def. of dot prod.)

= 0 (alg.)

= RHS.

PAU1.4.6b. hint: easier than a, but similar

1.4.7. False. Let u = v = w = i. Then

LHS = h1, 0, 0i·(h1, 0, 0i ⇥ h1, 0, 0i)= h1, 0, 0i·h0, 0, 0i= 0,

but

RHS = (h1, 0, 0i·h1, 0, 0i)⇥ (h1, 0, 0i·h1, 0, 0i)= 1⇥ 1

= 1,

so

RHS 6= LHS.

1.5.1a. a line is parametrizedx = x

0

+ aty = y

0

+ btz = z

0

+ ctwhere (x

0

, y0

, z0

) is a point on the line and the vector v = ha, b, ci goes in the direction of the line

1.5.1b. a plane has equation

a(x� x0

) + b(y � y0

) + c(z � z0

) = 0

where (x0

, y0

, z0

) is a point on the plane and n = ha, b, ci is the normal vector, perpendicular to theplane

1.5.2a) A vector has magnitude and a line does not, because lines are infinite. A line has locationand a vector does not. They both have direction. The drawing of a vector is an arrow where thelength matters, and which side the arrowhead is on matters. The drawing of a line is a double-sidedarrow where the arrowheads indicate infinite extension, so the length doesn’t matter.

1.5.2b) A line is 1D, and a plane is 2D. The equations of a line indicate a direction vector, thatgoes parallel to the line. The equation of a plane indicates a normal vector, which is perpendicularto the plane.

1.5.3. Note the points (2, 0, 1) and (3, 0,�1) are on the line, among infinitely many others.9/19/2019 Screenshot 2019-09-19 at 8.29.32 AM.png

chrome-extension://nlkncpkkdoccmpiclbokaimcnedabhhm/gallery.html 1/1

1.5.4. The parametric equations arex = 1 + t

y = 0z = 3� 2t

1.5.5. Yes, the lines are parallel. And yes, they are also identical.

1.5.6. 6(x� 1) + 3y + 2z = 0

1.5.7. z � x� y = �1

1.5.8. x = 2� ty = 3� tz = 4 + t

1.5.9. Note: as number 5 illustrates, there are many di↵erent parametric representations of thesame line. Your answer is correct if its direction is parallel and contains any point with the onebelow.

solution: Since the line of intersection is perpendicular to the normal vectors of both planes, wecross them to get its direction.

��������

i j k

1 �1 �1

2 0 1

��������= h�1,�(1 + 2), 2i

= h�1,�3, 2i

A point on the line can be found by solving the system of equations of the two planes. Theeasiest thing to do is pick a value for one of the variables and plug in. Guess and check. Forexample, say z = 0, so x� y = 2 and 2(x� 1) = 0, which means x = 1 and y = �3.

Answer:x = 1� ty = �3� 3tz = 2t

1.5.10. Find the distance between the point (1, 2, 3) and the plane 2x+ y � z = 12.Solution: Find any point on the plane, say (6, 0, 0), and use the scalar projection

~PQ · nknk =

h5,�2,�3i · h2, 1,�1ip22 + 12 + 12

=11p6

1.5.11. Find the distance between the skew linesx� 1

3=

y � 2

2=

z

4and

x

2=

y � 4

2= (z � 4)

Solution: The path of shortest distance from one line to the other will be perpendicular to bothlines, so its direction is the cross product

��������

i j k

3 2 4

2 2 1

��������= h2� 8,�(3� 8), 6� 4i

= h�6, 5, 2i.

Then find any vector between the two lines and use the scalar projection again. For P = (1, 2, 0)and Q = (0, 4, 4) we get

~PQ · nknk =

h�1, 2, 4i · h�6, 5, 2ip62 + 52 + 22

=24p65

1.6 Use desmos and geogebra3D to check graphs. Hand-drawn answers are available, just ask.

2.1.1a.

x

y

(3, 0)

(0, 4)

2.1.1b.

x

y

2.1.1c.

x

y

(�2, 0)

2.1.1d.

x

y

(1, 0)

2.1.1e.

x

y

(1, 0)

2.1.1f.

x

y

(3, 9)

2.1.1g.

x

y

(4,�2)

(9, 3)

2.1.1h.

2.1.2a.x = 0y = t0 t 4Note: For all these types of problems, there are many correct answers. Another, for example, isx = 0y = 4t0 t 1

2.1.2b.x = 3� 4ty = 1� 4t0 t 1

2.1.2c.x = 2� 4ty = 2no bounds needed

2.1.2d.x = cos ty = sin t0 t 2⇡ (this gives one rotation)

2.1.2e.x = cos ty = � sin t�⇡ t 0

This illustrates how to go backwards. Replace t by �t and adjust bounds. Recall from precal-culus that cos(�t) = cos t because cosine is an even function. Sine is an odd function.

2.1.2f.x = 3 + 5 cos ty = 4 sin t

2.1.2g.x = ty = 2 sin

⇥⇡

2

(x+ 3)⇤+ 1

2.1.2h.x = ty = t3

�1 t 2.

2.1.2i.x = t2 + ty = �t�1 t 0

2.1.2j. Split into 3 pieces, C1

, C2

, and C3

.C

1

:x = ty = 10 t 1

C2

:x = 1y = t1 t e

C3

:x = �ty = e�t

�1 t 0

2.2.1a.cos(⇡/2)

� sin(⇡/2) + 1= undefined

2.2.1b. 1

2.2.2. 8

2.2.3.d2y

dx2

=�2 cos t cos(2t)� 4 sin(2t) sin t

8 cos3(2t)

ccu whend2y

dx2

> 0. The sign of d

2y

dx

2 is the sign ofcos t

cos3(2t), which would be positive when

0 < t < ⇡/4 and 5⇡/4 < t < 7⇡/4 (except at t = 3⇡/2 the second derivative is 0).

2.2.4a.p

(�1� 3)2 + (2� 4)2 =p20

2.2.4b. x = �1 + 4t; y = 2 + 2t; 0 t 12.2.4c.

R1

0

p42 + 22 dt =

p20

2.2.4d. wow

2.2.5. The arc length formula s =R p

dx2 + dy2 or ds =pdx2 + dy2 comes from the Pythagorean

theorem. When you have a curve, you can split it into a bunch of little pieces of length �s, andwhen you add them all up and use infinitesimals, you get an integral.P

�s =Pp

(�x)2 + (�y)2 ⇠R p

dx2 + dy2

2.3.1. Use a grapher like https://www.desmos.com/calculator to check answers. Click on “GraphSettings,” the little wrench in the upper right corner, to see the polar grid option.

2.3.2a. x2 + y2 = 42.3.2b. x = 22.3.2c. x2 + y2 = 2x or (x� 1)2 + y2 = 12.3.2d. x2 + y2 = 2y or x2 + (y � 1)2 = 12.3.2e. y = 02.3.2f. y = 1p

3

x2.3.2g. x = 0

2.3.3a.2.3.3b. Answers will vary. Examples are (2, ⇡/2), (2, 5⇡/2), and (�2,�⇡/2).2.3.3c. (0, 2)2.3.3d. No. Polar coordinates are not unique, but rectangular ones are.

2.3.4a.

2.3.4b. Since r = sin(⇡/4) = 1p2

= cos(⇡/4) = r0,dy

dx=

r0 sin ✓ + r cos ✓

r0 cos ✓ � r sin ✓=

1

2

+ 1

2

1

2

� 1

2

= undefined.

The tangent line at ⇡/4 is vertical.

2.3.5a.2.3.5b. 6⇡

2.3.6.⇡

4� 1

2⇡ 0.2854

2.3.7b. 2⇡2.3.7d. The most common mistake is to get the wrong bounds on the integral. Plot very carefully

all the (r, ✓) points where ✓ = 0, ⇡6

, ⇡3

, ⇡2

, 2⇡3

, 5⇡6

, ⇡. The full circle gets traced out. So the bounds are0 to ⇡.

2.3.8a. (fg)0 = f 0g + g0f

2.3.8b.dy

dx=

d

d✓

r sin ✓d

d✓

r cos ✓. r sin ✓ is a product, so by the product rule the numerator becomes

r0 sin ✓ + cos ✓ · r. Similarly for the denominator.

2.3.9a. 1

12

⇡(30cm)2 ⇡ 236cm2 (note, this was edited. refresh browser for update)2.3.9b.

��✓

2⇡

�⇡r2 = 1

2

r2�✓2.3.9c. Summing sectors turns into an integral

P1

2

r2�✓ ⇠R

1

2

r2 d✓.

2.3.10. Note that for a function f = f(✓), the di↵erential is df = f 0(✓) d✓. So for example, thefunction r(✓) cos ✓, a function of ✓, has di↵erential d(r cos ✓) = (r0 cos ✓ � r sin ✓) d✓. Then

pdx2 + dy2 =

p(r0 cos ✓ � r sin ✓)2 (d✓)2 + (r0 sin ✓ + r cos ✓)2 (d✓)2

=pr02 cos2 ✓ � 2r0r cos ✓ sin ✓ + r2 sin2 ✓ + r02 sin2 ✓ + 2r0r cos ✓ sin ✓ + r2 cos2 ✓ d✓

=pr02 cos2 ✓ + r2 sin2 ✓ + r02 sin2 ✓ + r2 cos2 ✓ d✓

=pr02 + r2 d✓

2.4.1-8 graphs. Find e so you know what type of conic it is (ellipse e < 1, parabola e = 1, orhyperbola e > 1), then plot 4 points. Use desmos to check answers. Just type in the equation in rand ✓. For a polar grid, adjust the settings.

2.4.1-4. e = 1

2.4.5. e = 1

2

2.4.6. e = 22.4.7. e = 2

3

2.4.8. e = 3

2

2.4.9. r =2

1 + 1

2

cos ✓

2.4.10. r =6

1� sin ✓2.4.11. e = 5

3

and r = 16/3

1+

53 cos ✓

3.1.1a.

x

y

(3, 0)

(0, 4)

3.1.1b.

x

y

3.1.1c.

x

y

(�2, 0)

3.1.1d.

x

y

(1, 0)

3.1.1e.

x

y

(3, 9)

3.1.1f.

x

y

(4,�2)

(9, 3)

3.1.2a.

3.1.2b.

3.1.2c.

3.1.2d.

3.1.2e.

3.1.3a. Form 1. x = 3� 3t, y = 4t, 0 t 1. Form 2. r = h3� 3t, 4ti, 0 t 1

3.1.3b. Form 1. x = 2 cos t, y = 2 sin t, 0 t ⇡ Form 2. r = h2 cos t, 2 sin ti, 0 t ⇡.

3.1.3c. Form 1. x = t, y = ln t, 1 t e2. Form 2. r = ht, ln ti, 1 t e2.

3.1.4a. Form 1. x = 2+2t, y = 3, z = 4�2t, 0 t 1. Form 2. r = h2+2t, 3, 4�2ti, 0 t 1.

3.1.4b. Form 1. x = cos t, y = sin t, z = 3. Form 2. r = hcos t, sin t, 3i.

3.1.4c. Review the first chapter. Cross the normal vectors of the planes to get the line direction,and solve the system to find a point on the line. Answers will vary. One is r = h1 + 2t, 5

2

� 7t, 4ti.

3.1.4d. Play with geogebra3d. Answer r = h2 cos t, 2 sin t, ti.

3.1.5a. 03.1.5b. d.n.e.3.1.5c. 13.1.5d. �53.1.5e. 33.1.5f. 1p

2

3.1.5g. 1

3.1.6.

x

y

3.1.6a. 13.1.6b. 03.1.7a. h1, 1, 0i3.1.7b. hsin(1), 1p

2

, 1e

i3.1.7c. h0, 0, 0i

3.2.1a. r0 is blue. r00 is red.

x

y

3.2.1b.

x

y

3.2.1c.

x

y

3.2.1d.

3.2.2. r0(t) = h�2te�t

2, 0, 1

1+t

i

3.2.3. review integration by parts for the first component.Rr dt = hte�t, 2 arctan t, 2eti+C

3.2.4a. h13

, 23

, 23

i

3.2.4b. r(t) = h1, 0, 0i

3.2.5. (1, 0, 4). about 54.7�

3.2.6. r(t) = h4t, 2 + 5t� 4.9t2i

3.3.1a. v = h�1, 0, 4i; a = h0, 0, 0i; T =D

�1p17

, 0, 4p17

E; N is undefined (although many vectors

perpendicular to T exist); B is undefined (since N is), and = 0.

3.3.1b. v = h�a sin t, a cos ti; a = h�a cos t,�a sin ti; T = h� sin t, cos ti; N = � cos t,� sin ti;B = h0, 0, 1i; = 1

a

3.3.1c. Note there was a typo in the original handout. The helix is r = hcos t, sin t, ti.

answers: v = h� sin t, cos t, 1i; a = h� cos t,� sin t, 0i; T =h� sin t, cos t, 1ip

2; N = h� cos t,� sin t, 0i;

B =

2

664

i j k� sin tp

2

cos tp2

1p2

� cos t � sin t 0

3

775

=

⌧sin tp

2,�cos tp

2,1p2

�

= 1

2

3.3.1d. a tad gnarly. use the product rule, and know how to show the work to simplifyp(et cos t+ et sin t)2 + (et cos t� et sin t)2 to

p2et.

answers: v = het cos t+et sin t, 0, et cos t�et sin ti; a = h2et cos t, 0,�2et sin ti; T =hcos t+ sin t, 0, cos t� sin tip

2;

N =hcos t� sin t, 0,� sin t� cos tip

2; B = h0,�1, 0i; = 1p

2e

t

3.3.1e. did in class. know the quotient rule and the trick to simplify T0 with a multiple of(1 + 4t2)1/2 on top and bottom.

v = h1, 2ti; a = h0, 2i, T =

⌧1p

1 + t2,

2tp1 + t2

�; see quiz 13 answers

3.3.2. useful info on the way: T = h �apa

2+b

2 cos t,ap

a

2+b

2 sin t,bp

a

2+b

2 i. Answer: = a

a

2+b

2 ,

⌧ = b

a

2+b

2

3.3.3a. Note s =R

t

0

kr0(u)k du = 3t, so t = s/3 and r(s) = (cos(s))i+ (sin(s))j

3.3.3b. r(s) = 2 cos( s2

)i+ 2 sin( s2

)j

3.3.4. Recall the equation of a plane from the first chapter we covered. The normal plane hasnormal vector T or r0.

answer: x+ y = 1

3.3.5. The rectifying plane.

3.4.1a. r = h9.40t, 3.42t� 4.9t2i3.4.1b. 0.59m3.4.1c. 6.54m3.4.1d. r = h9.40t, 3.42t� 4.9t2 + 2i

3.4.2a.circle radius 33.4.2b. ⇡ seconds (do part a carefully to get that)3.4.2c. 1

⇡

rot/sec3.4.2d. 2 rad/sec3.4.2e. 6 m/sec3.4.2f. kvk = 63.4.2g. wow3.4.2h. aT = 0; aN = 18m/s2

3.4.3. aT = 0; aN = !r2; centripetal

3.4.4. bonus

4.1.1. Because the domain is not numbers, it’s points. Yes, it can be the answer for the range.

4.1.2a. domain R2, both open and closed. Range is (0,1), open.4.1.2b. Domain {(x, y) : x > y}, open. Range is (�1,1), open and closed.4.1.2c. Domain {(x, y) : y � �2x� 1}, closed. Range is [0,1), closed.4.1.2d. Domain {(x, y) : x2 + y2 4}, closed. Range is [0, 2], closed.4.1.2e. Domain {(x, y) : y > 0} open. Range is (0,1), open.

4.1.3a.

x

y

1 20

4.1.3b.

x

y

1 20

4.1.3c.

x

y

0

�1

�1

�2

�2

1 2120

4.1.3d. Note: I included z = 3, 4 in addition because for z 0 they don’t exist, and its hard to

see what is happening with just two curves.

x

y

1

234

4.1.3e. like c but rotated 45 degrees

4.1.4. Play around with geogebra3d. It’s too easy if I just tell you the answers here.

4.2.1 a) As x approaches 1, the y values approach 4. So limx!1

f(x) = 4.

b)

limx!1

x4 � 1

x� 1= lim

x!1

(x� 1)(x3 + x2 + x+ 1)

x� 1= lim

x!1

(x3 + x2 + x+ 1)

= 1 + 1 + 1 + 1 = 4

c) wow

4.2.2a. note this is missing the center row and column

4.2.2b. The limit looks like 3 in the table, excluding the center and diagonal.4.2.2c. Factor by grouping.

lim(x,y)!(1,1)

x3 + x2 � yx2 � y2

x� y= lim

(x,y)!(1,1)

x3 � yx2 + x2 � y2

x� y

= lim(x,y)!(1,1)

x2(x� y) + (x+ y)(x� y)

x� y

= lim(x,y)!(1,1)

(x� y)(x2 + x+ y)

x� y

= lim(x,y)!(1,1)

(x2 + x+ y)

= 1 + 1 + 1 = 3.

4.2.3. did in class

4.2.4a. 5� 8 = �34.2.4b. ln(1/1) = 04.2.4c. lim

r!0

r

2cos ✓ sin ✓

r

= 04.2.4d. DNE4.2.4e. 04.2.4f. Approach along the line x = y = z and get lim

x!0

2x

2

3x

2 = 2

3

. Approach along x = y = 0and get lim

z!0

0

z

2 = 0. The limits from two directions do not agree, so the limit DNE.

4.2.5a. R2

4.2.5b. {(x, y) : x > y}4.2.5c. {(x, y) : 2x+ y + 1 > 0}4.2.5d. {(x, y) : x2 + y2 4}4.2.5e. {(x, y) : y > 0}4.2.6. “If you can plug in, do it.” Since 0+0

1+0

= 0, the first limit is 0 because the rational functionis continuous on its domain and (0, 0) is in the domain. But when we plug in to the second limitand get 0

0

, the fraction is undefined and we need to do more work. Approaching along x = 0, we

get limy!0

0

y

2 = 0. Approaching along x = y, we get limx!0

x

2

2x

2 = 1

2

, and the limits do not agree, so

lim(x,y)!(0,0)

xy

x2 + y2= 0.

4.3.1.

f 0(x) = limh!0

f(x+ h)� f(x)

h

= limh!0

(x+ h)2 � x2

h

= limh!0

x2 + 2xh+ h2 � x2

h

= limh!0

2xh+ h2

h= lim

h!0

(2x+ h) = 2x+ 0 = 2x.

4.3.2a.

@f

@x= lim

h!0

f(x+ h, y)� f(x, y)

h

= limh!0

(x+ h)2 + y � (x2 + y)

h

= limh!0

2xh+ h2

h= 2x (like previous exercise)

4.3.2b.

@f

@y= lim

h!0

f(x, y + h)� f(x, y)

h

= limh!0

x2 + (y + h)� (x2 + y)

h

= limh!0

h

h= 1

4.3.3. There is no such thing as a regular derivative for a real multivariable function. There are

only directional derivatives, the two most important of which are the partial derivatives@f

@xand

@f

@y.

4.3.4a. fxx

= 2y3, fyy

= 6x2y, fxy

= 6xy2 = fyx

4.3.4b. fxx

= �y2 cos(xy), fyy

= �x2cos(xy), fxy

= � sin(xy)� xy cos(xy) = fyx

4.3.4c. fxx

= yex, fyy

= 0, fxy

ex = fyx

4.3.4d. fxx

=y2 � x2

(x2 + y2)2, f

yy

=x2 � y2

(x2 + y2)2, f

xy

=�2xy

(x2 + y2)2= f

yx

4.3.4e. fxy

=y2 � x2

(x2 + y2)2= f

yx

, fyy

=�2xy

(x2 + y2)2= f

xx

4.3.4f. fxx

= 2, fyy

= 0 = fxy

= fyx

4.3.4g. fxx

= y(y � 1)xy�2, fyy

= (ln x)2xy, fxy

= xy�1 � y(ln x)xy�1, fyx

= 1

x

xy + (ln x)yxy�1 =xy�1 + yxy�1 ln x = f

xy

= wow

4.3.5a. @x

@r

= cos ✓4.3.5b. @x

@✓

= �r sin ✓4.3.5c. @L

@v

= mv4.3.5d. @L

@x

= �kx

4.3.5e.@2✓

@t2= �g✓

0

`cos

✓rg

`t

◆

4.4.1a. The formula from Calc 1 is L(x) = f(x0

) + f 0(x0

)(x � x0

) where x0

= 1, so for thisproblem

L(x) = 1 + 100(x� 1),

and L(1.002) = 1 + 100(1.002 � 1) = 1.2. (This type of calculation can be done in your head,whereas actually multiplying a number by itself 100 times cannot, so this is a trick that rare NewYork street “geniuses,” who perform math for money, use.)

4.4.1b. 1.22115882719. |1.22115882719�1.2|1.22115882719

⇡ 1.7%.

4.4.2. Many equivalent forms exist for a plane. Since the answer is an equation, you can multiplyboth sides by anything nonzero to get another form.

4.4.2a. y + z = 04.4.2b. 6(x� 3)� (y � 4)� (z � 5) = 04.4.2c. (x� 2)� y � 2(z � ln 2) = 04.4.2d. x� y + 4(z � ⇡

4

) = 0

4.4.3. did in class

4.4.4. The “level surface” concept involving rF in the Key concepts on the homework assign-ment is a really powerful tool. In OpenStax we have not done gradients yet, but they are a bigdeal, and its never too early to start thinking about them. So this exercise can be done 2 ways,using either hf

x

, fy

,�1i or rF . You have to solve for z if you use the first way, so the rF way isway better. You appreciate that more if you do it both ways;)

4.4.4a. y + z =p2

4.4.4b. x+ y +p2z = 4

4.4.4c. 6x+ 3y + 2z = 6

4.4.5a. L(x, y) = 6(x� 3)� (y � 4) + 54.4.5b. 6.14.4.5c. 6.144.4.5d. ⇡ 0.65%

4.5.1. done in class

4.5.2a.@z

@s=

@z

@x

@x

@s+

@z

@y

@y

@s

4.5.2b.@z

@s= 2xy3 cos t+ 3x2y2 sin t

4.5.3a.z

x y

s t ts

u v u v uvuv

4.5.3b.@z

@u=

@z

@x

✓@x

@s

@s

@u+

@x

@t

@t

@u

◆+

@z

@y

✓@y

@s

@s

@u+

@y

@t

@t

@u

◆

4.5.4a.z

x y

t t

u v uv

4.5.4b.@z

@u=

@z

@x

dx

dt

@t

@u+

@z

@y

dy

dt

@t

@u

4.5.5a. in class

4.6.1a. hyexy, xexyi4.6.1b. h7x6y3, 3x7y2i

4.6.1c.

⌧1 +

z2

x� z

x2

, 0, 2z ln x+1

x

�

4.6.2. Make v a unit vector first. So use u = h 2p5

, 1p5

i. rf·u = 4x+2yp5(x

2+y

2)

.

4.6.3a and c.

x

y

4.6.3b. r = h4, 2i4.6.3d. The gradient at a point is normal (a.k.a. perpendicular) to the level curve through the

point.