Answers to Exercises

Appendix C Answers to Exercises

1 Curves
1.1 Derivatives, Velocity, Etc.

Exercises

1.1.1.

Answer.

(a)

r (y) = \sqrt{a^{2} - y^{2}} \hat{ı ı} + y \hat{ȷ ȷ},

0 \leq y \leq a

(b)

(x (ϕ), y (ϕ)) = (a \sin ϕ, - a \cos ϕ),

\frac{π}{2} \leq ϕ \leq π

(c)

(x (s), y (s)) = (a \cos (\frac{π}{2} - \frac{s}{a}), a \sin (\frac{π}{2} - \frac{s}{a})),

0 \leq s \leq \frac{π}{2} a

1.1.2.

Answer.

(1, 25),

(- 1 / \sqrt{2}, 0),

(0, 25) .

1.1.3.

Answer.

The curve crosses itself at all points

(0, (π n)^{2})

where

n

is an integer. It passes such a point twice,

2 π n

time units apart.

1.1.4.

Answer.

(a)

(a + a θ, a)

(b)

(a + a θ + a \sin θ, a + a \cos θ)

1.1.5.

Answer.

z = - \frac{1}{2} \sqrt{1 - \frac{y^{2}}{2}} - \frac{y}{4}

1.1.6.

Answer.

The particle is moving upwards from

t = 1

t = 2,

and from

t = 3

onwards. The particle is moving downwards from

t = 0

t = 1,

and from

t = 2

t = 3 .

The particle is moving faster when

t = 1

than when

t = 3 .

1.1.7.

Answer.

The red vector is

r (t + h) - r (t) .

The arclength of the segment indicated by the blue line is the (scalar)

s (t + h) - s (t) .

Remark: as

h

approaches 0, the curve (if it’s differentiable at

t

) starts to resemble a straight line, with the length of the vector

r (t + h) - r (t)

approaching the scalar

s (t + h) - s (t) .

This step is crucial to understanding Lemma 1.1.4.

1.1.8.

Answer.

Velocity is a vector-valued quantity, so it has both a magnitude and a direction. Speed is a scalar — the magnitude of the velocity. It does not include a direction.

1.1.9. (✳).

Answer.

(c)

1.1.10.

Answer.

See the solution.

1.1.11. (✳).

Answer.

(d)

1.1.12.

Answer.

velocity = - a \sin t \hat{ı ı} + a \cos t \hat{ȷ ȷ} + c \hat{k}

speed = \sqrt{a^{2} + c^{2}}

acceleration = - a \cos t \hat{ı ı} - a \sin t \hat{ȷ ȷ}

The path is a helix with radius

a

and with each turn having height

2 π c .

1.1.13. (✳).

Answer.

(a)

\hat{T} (1) = \frac{(2, 0, 1)}{\sqrt{5}}

(b)

\frac{1}{3} [5^{3 / 2} - 8]

1.1.14.

Answer.

1.1.15.

Answer.

length = \sqrt{a^{2} + b^{2}} T

1.1.16.

Answer.

1.1.17. (✳).

Answer.

(a)

\frac{20}{3}

(b)

x (t) = - 2 π - 2 t, y (t) = - 2 π t, z (t) = \frac{π^{3}}{3} + π^{2} t

1.1.18. (✳).

Answer.

(a)

r^{'} (t) = (- 3 \sin t, 3 \cos t, 4)

(b) 5

1.1.19.

Answer.

(a)

x (θ) = 3 \cos θ, y (θ) = 3 \sin θ, z (θ) = 6 \cos θ + 9 \sin θ, 0 \leq θ \leq 2 π

(b)

s = \int_{0}^{2 π} \sqrt{45 + 45 \cos^{2} θ - 108 \sin θ \cos θ} d θ

1.1.20. (✳).

Answer.

(a)

\frac{1}{27} (10 \sqrt{10} - 1)

(b)

\frac{2}{27} (10 \sqrt{10} - 1)

1.1.21. (✳).

Answer.

s (t) = \frac{t^{3}}{3} + \frac{t}{2}

1.1.22. (✳).

Answer.

\frac{8}{27} [(2 + \frac{9}{4} b^{m})^{3 / 2} - (2 + \frac{9}{4} a^{m})^{3 / 2}]

1.1.23.

Answer.

(a)

r (x) = x \hat{ı ı} + \sqrt{x} \hat{ȷ ȷ} + \frac{2}{3} x^{3 / 2} \hat{k}

(b)

21

(c)

6 \hat{ı ı} + 3 \hat{ȷ ȷ} + 6 \hat{k}

(d)

- 6 \hat{ı ı} - 12 \hat{ȷ ȷ} + 12 \hat{k}

1.1.24.

Answer.

| t |

1.1.25. (✳).

Answer.

(a)

r (u) = u^{3} \hat{ı ı} + 3 u^{2} \hat{ȷ ȷ} + 6 u \hat{k}

(b)

7

(c)

2

(d)

1

1.1.26. (✳).

Answer.

(a)

r (t) = (\frac{π^{2} t}{2} - \frac{t^{3}}{2}) \hat{ı ı} + (t - \sin t) \hat{ȷ ȷ} + (\frac{1}{2} e^{2 t} - t) \hat{k}

(b)

t = π

(c)

- π^{2} \hat{ı ı} + 2 \hat{ȷ ȷ} + (e^{2 π} - 1) \hat{k}

1.1.27. (✳).

Answer.

(a)

21

(b)

6

(c)

2 \hat{ı ı} + 4 \hat{ȷ ȷ} + 4 \hat{k}

(d)

- \frac{8}{3} (2 \hat{ı ı} + \hat{ȷ ȷ} - 2 \hat{k})

1.1.28.

Answer.

\frac{x (t) y^{'} (t) - y (t) x^{'} (t)}{x^{2} + y^{2}}

1.1.29.

Answer.

Volume:

540 π

Surface area:

360 π

1.1.30.

Answer.

\frac{50}{π \sqrt{9 + \frac{1}{400 π^{2}}}} \approx 5.3 cm

1.1.31.

Answer.

r (t) = r_{0} - \frac{e^{- α t} - 1}{α} v_{0} + g \frac{1 - α t - e^{- α t}}{α^{2}} \hat{k}

1.2 Reparametrization

Exercises

1.2.1.

Answer.

t - 1

1.2.2.

Answer.

(\sin (1 / 2), \cos (1 / 2), \sqrt{3} / 2)

1.2.3.

Answer.

1.2.4. (✳).

Answer.

(a)

(\frac{3}{4}, - \frac{\sqrt{3}}{4}, - \frac{1}{2})

(b)

R (s) = (2 \sin^{3} (\frac{s}{3}), 2 \cos^{3} (\frac{s}{3}), 3 \sin (\frac{s}{3}) \cos (\frac{s}{3}))

1.2.5. (✳).

Answer.

(a)

\sqrt{2}

(b)

\frac{s}{\sqrt{2}} (\cos (\ln (\frac{s}{\sqrt{2}})), \sin (\ln (\frac{s}{\sqrt{2}})))

with

s > 0

1.2.6.

Answer.

(\cos z, z \sin z, z)

for

0 \leq z < π / 2 .

The curve is (the first quarter-turn of) a spiral, with width in the

x

-direction 2, and increasing width in the

y

-direction. The parameter

z

is the height, as well as a radian measure for the spiral.

1.2.7.

Answer.

When

s \leq \frac{1}{3} (2 \sqrt{2} - 1),

R (s) = (\frac{1}{2} [(2 \sqrt{2} - 3 s)^{2 / 3} - 1], - \frac{1}{3} {[(2 \sqrt{2} - 3 s)^{2 / 3} - 1]}^{3 / 2})

and when

s > \frac{1}{3} (2 \sqrt{2} - 1),

R (s) = (\frac{1}{2} [(3 s + 2 - 2 \sqrt{2})^{2 / 3} - 1], \frac{1}{3} {[(3 s + 2 - 2 \sqrt{2})^{2 / 3} - 1]}^{3 / 2})

1.3 Curvature

Exercises

1.3.1.

Answer.

ρ = 3,

κ = \frac{1}{3}

1.3.2.

Answer.

\hat{T} (t) = (\cos t, - \sin t),

\hat{T} (s) = (\cos (s / 3), - \sin (s / 3)),

\hat{N} (t) = (- \sin t, - \cos t),

\hat{N} (s) = (- \sin (s / 3), - \cos (s / 3))

1.3.3.

Answer.

lim_{t \to \infty} κ (t) = 0

1.3.4.

Answer.

\frac{d s}{d t} = \sqrt{e^{2 t} + 9 + \cos^{2} t}

1.3.5.

Answer.

\begin{aligned} \frac{d \hat{T}}{d t} & = \frac{1}{\sqrt{2}} (- \sin t - \cos t, - \sin t + \cos t) \\ \frac{d \hat{T}}{d s} & = \frac{1}{\sqrt{2} s} (- \sin (\ln (s / \sqrt{2})) - \cos (\ln (s / \sqrt{2}))), \\ - \sin (\ln (s / \sqrt{2})) + \cos (\ln (s / \sqrt{2}))) \end{aligned}

1.3.6.

Answer.

See the solution.

1.3.7.

Answer.

$v (t) = (e^{t}, 2 t + 1)$
$a (t) = (e^{t}, 2)$
$\frac{d s}{d t} = \sqrt{e^{2 t} + (2 t + 1)^{2}}$
$\hat{T} (t) = (\frac{e^{t}}{\sqrt{e^{2 t} + (2 t + 1)^{2}}}, \frac{2 t + 1}{\sqrt{e^{2 t} + (2 t + 1)^{2}}})$
$κ (t) = \frac{e^{t} | 1 - 2 t |}{(e^{2 t} + (2 t + 1)^{2})^{3 / 2}}$

1.3.8.

Answer.

κ (t) = \frac{1}{\sqrt{2}}

1.3.9.

Answer.

κ_{m a x} = \frac{a}{b^{2}},

κ_{m i n} = \frac{b}{a^{2}} .

1.3.10. (✳).

Answer.

(a)

κ (0) = 2^{- 3 / 2}

(b)

(x + 2)^{2} + (y - 3)^{2} = 8

1.3.11. (✳).

Answer.

(a)

(b)

κ (t) = \frac{1}{2^{3 / 2} \sqrt{1 - \cos t}}

(c)

4

(d)

(x - π)^{2} + (y + 2)^{2} = 16

1.3.12.

Answer.

κ (s) = π s

1.3.13. (✳).

Answer.

The maximum values occur at

(x, y) = \pm (1 / \sqrt[4]{5}, \frac{1}{3} 5^{- 3 / 4}) .

The limits

lim_{x \to \pm \infty} κ (x) = 0 .

1.4 Curves in Three Dimensions

Exercises

1.4.1.

Answer.

\hat{B}

points out of the page (towards the reader).

1.4.2.

Answer.

arclength

1.4.3.

Answer.

a (t)

and

b (t)

have negative torsion,

c (t)

has zero torsion.

1.4.4.

Answer.

See solution.

1.4.5. (✳).

Answer.

(a), (b)

1.4.6. (✳).

Answer.

(a)

r^{'} (t) = (e^{t} + e^{- t}) \hat{ı ı} + (e^{t} - e^{- t}) \hat{ȷ ȷ} + 2 \hat{k},

r^{″} (t) = (e^{t} - e^{- t}) \hat{ı ı} + (e^{t} + e^{- t}) \hat{ȷ ȷ},

κ (t) = \frac{1}{2 + e^{2 t} + e^{- 2 t}}

(b)

\sqrt{2} [e - \frac{1}{e}]

1.4.7.

Answer.

\frac{3}{181}

1.4.8.

Answer.

\hat{T} (t) = \frac{\hat{ı ı} + t \hat{ȷ ȷ} + t^{2} \hat{k}}{\sqrt{1 + t^{2} + t^{4}}}

\hat{B} (t) = \frac{t^{2} \hat{ı ı} - 2 t \hat{ȷ ȷ} + \hat{k}}{\sqrt{1 + 4 t^{2} + t^{4}}}

\hat{N} (t) = \frac{- (t + 2 t^{3}) \hat{ı ı} + (1 - t^{4}) \hat{ȷ ȷ} + (2 t + t^{3}) \hat{k}}{\sqrt{1 + t^{2} + t^{4}} \sqrt{1 + 4 t^{2} + t^{4}}}

κ (t) = \frac{\sqrt{1 + 4 t^{2} + t^{4}}}{[1 + t^{2} + t^{4}]^{3 / 2}}

τ (t) = \frac{2}{1 + 4 t^{2} + t^{4}}

1.4.9.

Answer.

When

c = 0,

the plane is

z = 1 .

When

c = 1 / 5,

the plane is

(1 / 25) x + 3 y - (30 / e) z = - 10 .

1.4.10. (✳).

Answer.

(a)

2 x + y + 3 z = 6

(b)

κ (t) = \frac{2 \sqrt{1 + 9 t^{2} + 9 t^{4}}}{[1 + 4 t^{2} + 9 t^{4}]^{3 / 2}}

1.4.11. (✳).

Answer.

(a)

2

(b)

- \frac{\sqrt{3}}{2} \hat{ı ı} - \frac{1}{2} \hat{ȷ ȷ} + \frac{π}{6} \hat{k}

(c)

\hat{B} = \frac{1}{2 \sqrt{2}} \hat{ı ı} - \frac{\sqrt{3}}{2 \sqrt{2}} \hat{ȷ ȷ} + \frac{1}{\sqrt{2}} \hat{k}

1.4.12. (✳).

Answer.

(a)

R (t) = (- 1, 0, π^{2}) + t (0, - 1, 2 π)

(b)

a_{T} (t) = \frac{4 t}{\sqrt{1 + 4 t^{2}}}

1.4.13. (✳).

Answer.

(a)

\sqrt{5} t

(b)

a_{T} (t) = \sin t \hat{ı ı} + \cos t \hat{ȷ ȷ} + 2 \hat{k}

(c)

a_{N} (t) = t \cos t \hat{ı ı} - t \sin t \hat{ȷ ȷ}

(d)

κ (t) = \frac{1}{5 t}

1.4.14. (✳).

Answer.

(a)

r (θ) = [- 1 + 3 \cos θ] \hat{ı ı} + 3 \sin θ \hat{ȷ ȷ} + [10 - 6 \cos θ] \hat{k},

0 \leq θ < 2 π

(b) At

(2, 0, 4),

\hat{T} = \hat{ȷ ȷ},

\hat{N} = \frac{- \hat{ı ı} + 2 \hat{k}}{\sqrt{5}},

\hat{B} = \frac{2 \hat{ı ı} + \hat{k}}{\sqrt{5}},

κ (0) = \frac{\sqrt{5}}{3}

1.4.15. (✳).

Answer.

(a)

\hat{T} (t) = \frac{t^{2} \hat{ı ı} + \sqrt{2} t \hat{ȷ ȷ} + \hat{k}}{t^{2} + 1}

(b)

\frac{\sqrt{2}}{{(t^{2} + 1)}^{2}}

(c)

\frac{4 \hat{ı ı} - 3 \sqrt{2} \hat{ȷ ȷ} - 4 \hat{k}}{\sqrt{50}}

1.4.16. (✳).

Answer.

(a) One possible parametrization is

r (θ) = \cos θ \hat{ı ı} + \sin θ \hat{ȷ ȷ} + (1 - \cos θ - \sin θ) \hat{k}

with

0 \leq θ \leq 2 π .

(b)

κ (θ) = \frac{\sqrt{3}}{[2 - \sin (2 θ)]^{3 / 2}}

(c)

maximum curvature = \sqrt{3}

\frac{\hat{ı ı}}{\sqrt{2}} + \frac{\hat{ȷ ȷ}}{\sqrt{2}} + (1 - \sqrt{2}) \hat{k}

and

- \frac{\hat{ı ı}}{\sqrt{2}} - \frac{\hat{ȷ ȷ}}{\sqrt{2}} + (1 + \sqrt{2}) \hat{k}

and

minimum curvature = \frac{1}{3}

- \frac{\hat{ı ı}}{\sqrt{2}} + \frac{\hat{ȷ ȷ}}{\sqrt{2}} + \hat{k}

and

\frac{\hat{ı ı}}{\sqrt{2}} - \frac{\hat{ȷ ȷ}}{\sqrt{2}} + \hat{k}

1.4.17. (✳).

Answer.

\hat{T} (t) = \frac{2 t^{2} \hat{ı ı} + 2 t \hat{ȷ ȷ} + \hat{k}}{2 t^{2} + 1}

\hat{N} (t) = \frac{2 t \hat{ı ı} - (2 t^{2} - 1) \hat{ȷ ȷ} - 2 t \hat{k}}{2 t^{2} + 1}

κ (t) = \frac{2 t}{{(2 t^{2} + 1)}^{2}}

1.4.18. (✳).

Answer.

(a)

\frac{1}{3} [2^{3 / 2} - 1]

(b)

\frac{- \hat{ı ı} - \hat{ȷ ȷ}}{\sqrt{2}}

(c)

\frac{1}{2}

1.4.19. (✳).

Answer.

(a)

v (t) = (1, - 1, t)

(b)

\frac{d s}{d t} (t) = \sqrt{2 + t^{2}}

(c)

a (t) = (0, 0, 1)

(d)

κ (t) = \frac{\sqrt{2}}{{[2 + t^{2}]}^{3 / 2}}

(e)

\hat{N} (t) = \frac{(- t, t, 2)}{\sqrt{2 (2 + t^{2})}}

(f)

x + y = 3

(g)

(2, 1, 2)

1.4.20. (✳).

Answer.

(a)

\hat{T} (t) = \frac{t^{2} \hat{ı ı} + \sqrt{2} t \hat{ȷ ȷ} + \hat{k}}{t^{2} + 1}

(b)

κ (t) = \frac{\sqrt{2}}{{(t^{2} + 1)}^{2}}

(c)

κ (0) = \sqrt{2}

(d)

\hat{N} (0) = \hat{ȷ ȷ}

(e)

\hat{B} (0) = - \hat{ı ı}

1.4.21. (✳).

Answer.

(a)

x = 1 - 2 t,

y = - 1 + t,

z = - 1 + 3 t

(b)

3 x - 3 y - z = - 1

1.4.22. (✳).

Answer.

(a)

\frac{\sqrt{5} π^{2}}{2}

(b)

κ (t) = \frac{1}{5 t}

1.4.23. (✳).

Answer.

(a)

8

(b)

\hat{T} (1) = \frac{1}{\sqrt{2}} (1, 1, 0),

\hat{N} (1) = (0, 0, - 1)

(c)

κ (1) = \frac{1}{8}

1.4.24. (✳).

Answer.

(a)

\frac{5}{2}

(b)

\hat{T} (\frac{π}{6}) = \frac{1}{5} (- \frac{3}{2}, \frac{3 \sqrt{3}}{2}, 4),

\hat{N} (\frac{π}{6}) = \frac{1}{2} (\sqrt{3}, 1, 0),

\hat{B} (\frac{π}{6}) = \frac{1}{5} (- 2, 2 \sqrt{3}, - 3)

1.4.25.

Answer.

(a)

r (θ) = \cos θ \hat{ı ı} + \sin θ \hat{ȷ ȷ} + \cos (2 θ) \hat{k}

0 \leq θ < 2 π

(b)

\frac{1}{5}

(c)

z = \sqrt{2} x - \sqrt{2} y

(d) radius

1 / κ (π / 4) = 5

and centre

(- 2 \sqrt{2}, - 2 \sqrt{2}, 0)

1.4.26. (✳).

Answer.

\frac{4}{9} (\hat{ı ı} - 4 \hat{ȷ ȷ} + \hat{k})

1.4.27. (✳).

Answer.

(a)

\hat{T} (t) = \frac{\hat{ı ı} + t \hat{ȷ ȷ} + \sqrt{3} t \hat{k}}{\sqrt{1 + 4 t^{2}}}

(b)

\hat{N} (t) = \frac{- 4 t \hat{ı ı} + \hat{ȷ ȷ} + \sqrt{3} \hat{k}}{2 \sqrt{1 + 4 t^{2}}}

(d)

- \sqrt{3} y + z = 0

(e)

κ (t) = {(1 + 4 t^{2})}^{- 3 / 2}

(f) The curvature

κ (t)

achieves its maximum value at

r (0) = (0, 0, 0) .

(g) The curvature never achieves a minimum.

(h)

\hat{ı ı} = \frac{u}{2},

\hat{ȷ ȷ} = \frac{v - \sqrt{3} w}{4},

\hat{k} = \frac{\sqrt{3} v + w}{4},

r (t) = t u + t^{2} v

The curve

(a (t), b (t)) = (t, t^{2})

is the curve

y = x^{2} .

It is “curviest” at the origin, which is consistent with part (f). It becomes flatter and flatter as

| t |

increases, but never achieves “perfect flatness”, which is consistent with (g).

1.4.28. (✳).

Answer.

See the solution.

1.4.29. (✳).

Answer.

(a)

\hat{T} = \frac{1}{\sqrt{6}} (0, 2, - \sqrt{2}),

\hat{N} = - \frac{1}{\sqrt{39}} (6, 1, \sqrt{2}),

\hat{B} = \frac{1}{\sqrt{13}} (- 1, 2, 2 \sqrt{2}),

κ = \frac{\sqrt{13}}{3 \sqrt{3}} = \frac{\sqrt{39}}{9}

(b) (i)

\frac{d v}{d t} = \frac{5 \sqrt{2}}{\sqrt{3}}

(ii)

v = (0, \sqrt{2}, - 1) .

1.4.30. (✳).

Answer.

(a)

v (t) = (- \sin t, \cos t, c \cos t),

a (t) = (- \cos t, - \sin t, - c \sin t)

(b)

v (t) = \sqrt{1 + c^{2} \cos^{2} t}

(c)

\frac{- c^{2} \sin t \cos t}{\sqrt{1 + c^{2} \cos^{2} t}}

(d) The curve lies on the plane

z = c y .

1.4.31. (✳).

Answer.

(a)

\frac{\sqrt{17}}{4}

(b)

\frac{4}{\sqrt{17}}

4 \sqrt{π}

(ii)

(16 π, - 4, - 4 π)

(iii)

4 \sqrt{17} π

1.6 Integrating Along a Curve

Exercises

1.6.1.

Answer.

\int_{C} d s

1.6.2.

Answer.

(a) See the solution.

(b)

8

1.6.3.

Answer.

\frac{4}{21 \sqrt{3}} (2^{7} - 1) + \frac{2}{5 \sqrt{3}} (2^{5} - 1)

1.6.4.

Answer.

π

1.6.5.

Answer.

26

1.6.6.

Answer.

(a)

\frac{5^{3 / 2} - 1}{12}

(b)

\frac{8 - 3^{3 / 2}}{3 / 2}

1.6.7.

Answer.

\frac{1}{2} \ln 2

1.6.8. (✳).

Answer.

2 e^{3}

1.6.9. (✳).

Answer.

(a)

\frac{1}{\sqrt{1 + 5 π^{2}}} (- \hat{ı ı} - π \hat{ȷ ȷ} + 2 π \hat{k})

(b)

\frac{1}{15} [(1 + 5 π^{2})^{3 / 2} - 1]

(c)

z = x^{2} + y^{2}

(d)

1.6.10.

Answer.

(\frac{412}{55}, - \frac{92}{55}, \frac{4736}{693})

1.7 Sliding on a Curve
1.7.4 Exercises

1.7.4.1.

Answer.

1.7.4.2.

Answer.

time

1.7.4.3.

Answer.

positive

1.7.4.4.

Answer.

y = \frac{E}{m g}

— just like a circular culvert (if the culvert is high enough).

1.7.4.5.

Answer.

2940 J

1.7.4.6.

Answer.

at least

5 \sqrt{9.8}

m/s

1.7.4.7.

Answer.

(- \frac{3}{\sqrt{2}} + 2.352, - \frac{5}{\sqrt{2}} + 3.92, - 2 \sqrt{2} + 3.136)

1.7.4.8.

Answer.

\sqrt{\frac{9.8}{\sqrt{6}} (100 + \frac{1}{\sqrt{2}})} \approx 20

m/s

1.7.4.9.

Answer.

| v | > 504

kph

1.7.4.10.

Answer.

U = m g \frac{d y}{d s}

1.7.4.11.

Answer.

(a)

M = m g \hat{ȷ ȷ} \cdot \hat{T}

(b) negative

(c)

- \frac{1960}{\sqrt{3}} \approx - 1131.6

1.7.4.12.

Answer.

(a)

y_{S} = \frac{E}{m g}

(b)

\frac{24 (E - m g y_{A})}{{(9 + 7 {(\frac{y_{A} - 3}{3})}^{2})}^{3 / 2}} = 4 m g (\frac{\frac{y_{A} - 3}{3}}{\sqrt{9 + 7 {(\frac{y_{A} - 3}{3})}^{2}}})

(or equivalent)

1.7.4.13.

Answer.

(a), (b) See the solution.

(c)

2 {[\frac{a^{2} + b^{2}}{g b} π]}^{1 / 2}

1.8 Optional — Polar Coordinates

Exercises

1.8.1.

Answer.

The upper sketch below contains the points,

(x_{1}, y_{1}),

(x_{3}, y_{3}),

(x_{5}, y_{5}),

that are on the axes. The lower sketch below contains the points,

(x_{2}, y_{2}),

(x_{4}, y_{4}),

that are not on the axes.

r_{1} = 3,

θ_{1} = 0

r_{2} = \sqrt{2},

θ_{2} = \frac{π}{4}

r_{3} = 1,

θ_{3} = \frac{π}{2}

r_{4} = \sqrt{2},

θ_{4} = \frac{3 π}{4}

r_{5} = 2,

θ_{5} = π

1.8.2.

Answer.

(a)

(r = 2, θ = n π, n odd integer)

(r = - 2, θ = n π, n even integer)

(b)

(r = \sqrt{2}, θ = \frac{π}{4} + 2 n π)

(r = - \sqrt{2}, θ = \frac{5 π}{4} + 2 n π),

with

n

integer.

(c)

(r = \sqrt{2}, θ = \frac{5 π}{4} + 2 n π)

(r = - \sqrt{2}, θ = \frac{π}{4} + 2 n π),

with

n

integer.

1.8.3.

Answer.

(a) Both

{\hat{e}}_{r} (θ)

and

{\hat{e}}_{θ} (θ)

have length 1. The angle between them is

\frac{π}{2} .

The cross product is

{\hat{e}}_{r} (θ) \times {\hat{e}}_{θ} (θ) = \hat{k} .

(b) Here is a sketch of

(x_{i}, y_{i}),

{\hat{e}}_{r} (θ_{i}),

{\hat{e}}_{θ} (θ_{i})

for

i = 1, 3, 5

(the points on the axes)

and here is a sketch (to a different scale) of

(x_{i}, y_{i}),

{\hat{e}}_{r} (θ_{i}),

{\hat{e}}_{θ} (θ_{i})

for

i = 2, 4

(the points off the axes).

1.8.4. (✳).

Answer.

(a)

\leftrightarrow

(E)

(b)

\leftrightarrow

(B)

(c)

\leftrightarrow

(F)

(d)

\leftrightarrow

(C)

(e)

\leftrightarrow

(A)

(f)

\leftrightarrow

(D)

1.8.5.

Answer.

κ (θ) = \frac{| f (θ)^{2} + 2 f^{'} (θ)^{2} - f (θ) f^{″} (θ) |}{{[f (θ)^{2} + f^{'} (θ)^{2}]}^{3 / 2}}

1.8.6.

Answer.

κ (θ) = \frac{3}{2^{3 / 2} a \sqrt{1 - \cos θ}} = \frac{3}{2 \sqrt{2 a r (θ)}}

1.9 Optional — Central Forces

Exercises

1.9.1. (✳).

Answer.

(a) See the solutions.

(b)

f (r) = 0

for all

r \geq 0 .

f (r)

which is a positive constant times

- \frac{1}{r^{3}}

works.

1.9.2. (✳).

Answer.

(a) See the solution.

(b)

| a (t) | = \frac{h^{2}}{r (t)^{3}}

2 Vector Fields
2.1 Definitions and First Examples

Exercises

2.1.1.

Answer.

v (x, y) \cdot \hat{ı ı} {\begin{cases} > 0 & when x > 0 \\ = 0 & when x = 0 \\ < 0 & when x < 0 \end{cases}}

and

v (x, y) \cdot \hat{ȷ ȷ} {\begin{cases} > 0 & when - 2 < x < 2 \\ = 0 & when x \in {- 2, 2} \\ < 0 & when x < - 2 or x > 2 \end{cases}}

at least for

(x, y)

shown in the sketch.

2.1.2.

Answer.

v (x, y) \cdot \hat{ı ı} {\begin{cases} > 0 & when y > - x \\ = 0 & when y = - x \\ < 0 & when y < - x \end{cases}}

and

v (x, y) \cdot \hat{ȷ ȷ} {\begin{cases} > 0 & when y < x \\ = 0 & when y = x \\ < 0 & when y > x \end{cases}}

at least for

(x, y)

shown in the sketch.

2.1.3.

Answer.

v (x, y) = \frac{- | y |}{\sqrt{x^{2} + y^{2}}} (x, y)

2.1.4.

Answer.

P > 0

Q > 0

\frac{\partial Q}{\partial x} < 0

\frac{\partial Q}{\partial y} > 0

2.1.5.

Answer.

(a)

(1.01, 1.01)

(b)

(0, 0)

(c)

(0, 0)

2.1.6.

Answer.

(0, - 10)

2.1.7.

Answer.

If your face is at the origin, then

v (x, y, z) = - \frac{α}{x^{2} + y^{2} + z^{2}} (x, y, z)

for some positive constant

α .

2.1.8.

Answer.

2.1.9.

Answer.

2.1.10.

Answer.

2.1.11.

Answer.

2.1.12.

Answer.

(a)

(b)

(c)

2.1.13.

Answer.

f (x, y) = \frac{- 5 G (x, y)}{(x^{2} + y^{2})^{3 / 2}} + \frac{3 G (2 - x, 3 - y)}{((x - 2)^{2} + (y - 3)^{2})^{3 / 2}} + \frac{7 G (4 - x, - y)}{((x - 4)^{2} + y^{2})^{3 / 2}}

2.1.14.

Answer.

v (p) = ((1 - \frac{p}{2}) \frac{1}{2 \sqrt{3}}, - \frac{p}{4})

V (x, y, z) = (- \frac{x}{6}, - \frac{y}{6}, \frac{z}{2})

or equivalent

2.2 Optional — Field Lines
2.2.2 Exercises

2.2.2.1.

Answer.

2.2.2.2.

Answer.

v (x, y) = (- x - y, x - y)

2.2.2.3. (✳).

Answer.

(a)

\frac{x^{2}}{2} = \frac{y^{2}}{2} + C

(b)

2.2.2.4. (✳).

Answer.

x = y^{2},

z = e^{y}

2.2.2.5. (✳).

Answer.

The field lines are

y = C^{'} x^{3}

with

C^{'}

a nonzero constant, as well as

x = 0

and

y = 0 .

2.3 Conservative Vector Fields

Exercises

2.3.1.

Answer.

In general, false.

2.3.2.

Answer.

a. C

b. B

c. C

d. B

2.3.3.

Answer.

Let

φ

be a potential for

F .

Define

ϕ = φ + a x + b y + c z .

Then

\nabla \nabla ϕ = \nabla \nabla φ + (a, b, c) = F + (a, b, c) .

2.3.4.

Answer.

If $F + G$ is conservative for any particular $F$ and $G,$ then by definition, there exists a potential $φ$ with $F + G = \nabla \nabla φ .$

Since $F$ is conservative, there also exists a potential $ψ$ with $F = \nabla \nabla ψ .$

But now $G = (F + G) - F = \nabla \nabla φ - \nabla \nabla ψ = \nabla \nabla (φ - ψ) .$ That means the function $(φ - ψ)$ is a potential for $G .$ However, this is impossible: since $G$ is non-conservative, no function with this property exists.

So it is not possible that $F + G$ is conservative. It must be non-conservative.
Counterexample: if $F = - G,$ then $F + G = 0 = \nabla \nabla c$ for any constant $c .$
Since both fields are conservative, they both have potentials, say $F = \nabla \nabla φ$ and $G = \nabla \nabla ψ .$ Then $F + G = \nabla \nabla φ + \nabla \nabla ψ = \nabla \nabla (φ + ψ) .$ That is, $(φ + ψ)$ is a potential for $F + G,$ so $F + G$ is conservative.

2.3.5. (✳).

Answer.

Yes,

F

is conservative on

D .

A potential is

φ (x, y) = \arctan \frac{y}{x} .

2.3.6.

Answer.

φ = \frac{1}{2} x^{2} + x y - \frac{1}{2} y^{2}

2.3.7.

Answer.

φ = \ln | x | - \frac{x}{y}

2.3.8.

Answer.

None exists:

\frac{\partial F_{2}}{\partial z} = \frac{1}{3} x^{3},

while

\frac{\partial F_{3}}{\partial y} = \frac{1}{3} x^{3} + 1,

F

fails the screening test, Theorem 2.3.9.

2.3.9.

Answer.

φ (x, y, z) = \frac{1}{2} \ln (x^{2} + y^{2} + z^{2})

2.3.10.

Answer.

(a)

F

is conservative with potential

ϕ (x, y, z) = \frac{1}{2} x^{2} - y^{2} + \frac{3}{2} z^{2} + C

for any constant

C .

(b)

F

is not conservative.

2.3.11.

Answer.

(a)

A = 2,

B

is arbitrary.

(b)

φ (x, y, z) = x e^{(z^{2})} + B y^{2} z^{3} + C

for any constant

C .

2.3.12.

Answer.

v = m \frac{x \hat{ı ı} + y \hat{ȷ ȷ}}{x^{2} + y^{2}}

φ = \frac{1}{2} m \ln (x^{2} + y^{2}) + C

for any constant

C

2.3.13.

Answer.

It can never escape the sphere centred at the origin with radius

\sqrt{20} .

2.3.14.

Answer.

\sqrt{14}

2.3.15.

Answer.

φ = f^{2} (x) + g (y) h (z)

is a potential for

F,

F

is conservative.

2.3.16.

Answer.

The line through the origin in the direction of the vector

(2, 1, 2) .

2.4 Line Integrals
2.4.2 Exercises

2.4.2.1.

Answer.

\frac{1}{6}

2.4.2.2.

Answer.

a. A

b. B

c. A

d. B

2.4.2.3.

Answer.

2.4.2.4.

Answer.

2.4.2.5. (✳).

Answer.

a = 1,

b = c = 0

2.4.2.6.

Answer.

(a) Not conservative

(b) Not conservative

(d) Conservative

2.4.2.7. (✳).

Answer.

(a) The (largest possible) domain is

D = {(x, y, z) | x^{2} + y^{2} \neq 0} .

(b)

\nabla \nabla \times F = 0

D

(c)

\int_{C} F \cdot d r = 4 π

(d)

F

is not conservative.

2.4.2.8.

Answer.

9 \frac{1}{2}

for all paths from

(1, 0, - 1)

(0, - 2, 3)

2.4.2.9.

Answer.

2 (e - 1) + \frac{π^{2}}{2} + 3 π

2.4.2.10.

Answer.

(a)

- \frac{1}{4}

(b)

- 1

2.4.2.11. (✳).

Answer.

- \frac{40}{3}

2.4.2.12.

Answer.

(a)

λ = - 1

(b)

ϕ (x, y, z) = 2 x^{3} y z^{2} - x y z + y^{2} + K,

for any constant

K

(c)

e^{2} + 2 e - 2

2.4.2.13. (✳).

Answer.

\frac{7}{3}

2.4.2.14. (✳).

Answer.

\frac{1}{3} [1 - \frac{1}{2^{3 / 2}}] \approx 0.2155

2.4.2.15. (✳).

Answer.

\frac{π^{3}}{8} + \frac{π^{2}}{4} - 1

2.4.2.16. (✳).

Answer.

- \frac{2}{3}

2.4.2.17. (✳).

Answer.

The line integral is independent of path because it is of the form

\int_{C} F \cdot d r

with

F

being a conservative field. The value of the integral is

1 + \frac{π}{2} .

2.4.2.18. (✳).

Answer.

\frac{1}{2}

2.4.2.19. (✳).

Answer.

π e^{π}

2.4.2.20. (✳).

Answer.

(a)

α = 1,

β = γ

(b)

e^{e} - β (e + 1)

2.4.2.21. (✳).

Answer.

(a)

0

(b) Yes. In fact

F = \nabla \nabla f

with

f = \sin x + 2 y - \cos y + e^{z} .

(c)

- 4

2.4.2.22. (✳).

Answer.

(a)

\nabla \nabla \times F = 0 .

F

is conservative.

(b)

\int_{C} F \cdot d r = 2 π^{2}

2.4.2.23. (✳).

Answer.

(a)

a = - 1,

b = 3

(b)

f (x, y, z) = x y e^{x} + y z^{3} + C

works for any constant

C

(c)

π e^{π} - 2

(d)

π e^{π} - \frac{32}{15}

2.4.2.24. (✳).

Answer.

(a)

A = 2,

B = 3

(b)

φ (x, y, z) = x y^{2} e^{3 z} + x^{2} y^{3}

is one allowed scalar potential.

(c)

6 + e - 2 [e - 1] = 8 - e \approx 5.2817

2.4.2.25. (✳).

Answer.

(a)

a = π, b = 3

(b)

φ (x, y, z) = x^{2} \sin (π y) - x e^{z} - 3 y e^{z} + C

for any constant

C

(c)

- 8

(d)

- \frac{13}{2}

2.4.2.26. (✳).

Answer.

(a)

f (x, y, z) = y e^{y z} + y \cos^{2} x + C

works for any constant

C

(b)

2 e^{π} - e^{- π^{2}} - 1

2.4.2.27. (✳).

Answer.

(a)

0 .

(b)

F

is conservative with potential

φ (x, y, z) = x^{2} + y^{2} + z^{2} .

So the integral is

φ (a_{1}, a_{2}, a_{3}) - φ (0, 0, 0) = a \cdot a .

2.4.2.28. (✳).

Answer.

(a)

\nabla \nabla \times F = 0

(b)

\frac{π e}{2} - 1

2.4.2.29. (✳).

Answer.

(a), (b)

f (x, y) = y \sin (x^{2}) + \cos (y) + C

is a potential for any constant

C .

Because

F

has a potential, it is conservative.

(c)

- 1 - \frac{π}{2} \sin (1)

2.4.2.30. (✳).

Answer.

(a)

p = 2,

m = 2,

n = 2,

but

q \in R

is completely free

(b)

4 q

2.4.2.31. (✳).

Answer.

(a)

r (t) = t \hat{ı ı} + (1 + \frac{t^{2}}{2}) \hat{ȷ ȷ} + \sin t \hat{k}

(b)

r (π / 2) = \frac{π}{2} \hat{ı ı} + (1 + \frac{π^{2}}{8}) \hat{ȷ ȷ} + \hat{k}

(c)

\frac{π^{2}}{8} - \frac{1}{2}

2.4.2.32.

Answer.

(a)

\frac{2}{3} [14^{3 / 2} - 1] \approx 34.26

(b)

\sin 1 + \frac{3}{2} \approx 2.3415

2.4.2.33. (✳).

Answer.

2 π + \frac{1}{3}

2.4.2.34. (✳).

Answer.

(a)

18

(b)

3 - e

2.4.2.35. (✳).

Answer.

(a)

r (t) = t \hat{ı ı} + (1 + \frac{t^{2}}{2}) \hat{ȷ ȷ} + \sin t \hat{k}

(b)

r_{1} = \frac{π}{2} \hat{ı ı} + (1 + \frac{π^{2}}{8}) \hat{ȷ ȷ} + \hat{k}

(c)

\frac{π^{2}}{8} - \frac{1}{2}

2.4.2.36. (✳).

Answer.

(a) -5

(b) One possibility is the path consisting of the line segment from

(2, 2)

(2, - 3),

followed by the line segment from

(2, - 3)

(1, - 3),

followed by the line segment from

(1, - 3)

(1, 1) .

Another possibility is the path from

(2, 2)

(1, 1)

along the parabola

27 x^{2} - 80 x + 54 .

2.4.2.37. (✳).

Answer.

One possibility is the path consisting of the line segment from

(0, 0)

(0, 1),

followed by the line segment from

(0, 1)

(2, 1),

followed by the line segment from

(2, 1)

(2, 0) .

Another possibility is the path tracing out the half ellipse

(\cos t + 1, \frac{4}{π} \sin t),

with

t

running from

π

0 .

2.4.2.38. (✳).

Answer.

See the solution.

2.4.2.39. (✳).

Answer.

a = 4

2.4.2.40. (✳).

Answer.

(a)

\nabla \nabla \times F = [- (b + 2) x \cos (x^{2} z) + (b + 2) x^{3} z \sin (x^{2} z)] \hat{ȷ ȷ} + (6 - a) x^{2} e^{3 x^{2}} \hat{k}

(b)

a = 6,

b = - 2

(c)

f (x, y, z) = x y e^{3 x^{2}} + \sin (x^{2} z) + C

for any constant

C

(d)

\frac{1}{3} e^{3} + \sin 1 - \frac{1}{3}

2.4.2.41. (✳).

Answer.

(a)

\frac{23}{15} = 1.5 \dot{3}

(b)

\frac{2}{3} [14^{3 / 2} - 1] \approx 34.26

(c)

\sin 1 + \frac{3}{2} \approx 2.3415

2.4.2.42. (✳).

Answer.

(a)

A = - 4,

B = - 2

(b)

φ (x, y, z) = - x^{4} y^{2} z + y z^{3} + C

with

C

being an arbitrary constant.

(c)

- 2

(d)

- \frac{37}{24} \approx - 1.5417

(e)

\frac{1}{2}

2.4.2.43. (✳).

Answer.

(a)

v (t) = (t^{2}, t^{3}, - t^{2})

(b)

r (t) = (\frac{t^{3}}{3} + 1, \frac{t^{4}}{4} + 2, - \frac{t^{3}}{3} + 3)

(c)

κ (t) = \frac{\sqrt{2}}{t^{2} {(2 + t^{2})}^{3 / 2}}

(d)

2 T^{4} + T^{6}

2.4.2.44. (✳).

Answer.

(a)

8

(b)

\frac{1}{8}

(c)

- \frac{16}{5} (3^{5} - 1) \approx - 774.4

(d)

(0, 0, - \frac{9}{8})

3 Surface Integrals
3.1 Parametrized Surfaces

Exercises

3.1.1.

Answer.

r (x, y) = x \hat{ı ı} + y \hat{ȷ ȷ} + (e^{x + 1} + x y) \hat{k}

3.1.2. (✳).

Answer.

parabolic bowl

3.1.3. (✳).

Answer.

(a) No

(b) Yes

(d) Yes

(e) No

3.1.4. (✳).

Answer.

(a) No.

(b) Yes.

(d) Yes.

(e) Yes.

3.1.5. (✳).

Answer.

(a) No

(b) Yes

3.1.6. (✳).

Answer.

(a) A, F

(b) B, E

(d) H, L

3.1.7.

Answer.

(a)

(x, y, z) = (2 + \frac{1}{\sqrt{2}} \cos θ, 2 + \frac{1}{\sqrt{2}} \cos θ, 4 + \sin θ),

0 \leq θ \leq 2 π .

3.2 Tangent Planes

Exercises

3.2.1.

Answer.

Yes. The plane

z = 0

is the tangent plane to both surfaces at

(0, 0, 0) .

3.2.2.

Answer.

See the solution.

3.2.3.

Answer.

\begin{aligned} (x - x_{0}, y - y_{0}, z - z_{0}) = t (- f_{x} (x_{0}, y_{0}), - f_{y} (x_{0}, y_{0}), 1) or \\ x = x_{0} - t f_{x} (x_{0}, y_{0}) y = y_{0} - t f_{y} (x_{0}, y_{0}) z = f (x_{0}, y_{0}) + t \end{aligned}

3.2.4.

Answer.

The normal plane is

n \cdot (x - x_{0}, y - y_{0}, z - z_{0}) = 0,

where the normal vector

n = \nabla \nabla F (x_{0}, y_{0}, z_{0}) \times \nabla \nabla G (x_{0}, y_{0}, z_{0}) .

3.2.5.

Answer.

Tangent line is

\begin{aligned} x & = x_{0} + t [g_{y} (x_{0}, y_{0}) - f_{y} (x_{0}, y_{0})] \\ y & = y_{0} + t [f_{x} (x_{0}, y_{0}) - g_{x} (x_{0}, y_{0})] \\ z & = z_{0} + t [f_{x} (x_{0}, y_{0}) g_{y} (x_{0}, y_{0}) - f_{y} (x_{0}, y_{0}) g_{x} (x_{0}, y_{0})] \end{aligned}

3.2.6. (✳).

Answer.

2 x + y + 9 z = 2

3.2.7. (✳).

Answer.

2 x + y + z = 6

3.2.8. (✳).

Answer.

z = - \frac{3}{4} x - \frac{3}{2} y + \frac{11}{4}

3.2.9. (✳).

Answer.

(a)

2 a x - 2 a y + z = - a^{2}

(b)

a = \frac{1}{2} .

3.2.10. (✳).

Answer.

x + 3 y - 2 z = 1

3.2.11. (✳).

Answer.

y = 2 x - 2

3.2.12. (✳).

Answer.

The tangent plane is

\frac{8}{25} x - \frac{6}{25} y - z = - \frac{8}{5} .

The normal line is

(x, y, z) = (- 1, 2, \frac{4}{5}) + t (\frac{8}{25}, - \frac{6}{25}, - 1) .

3.2.13. (✳).

Answer.

\pm (1, 0, - 2)

3.2.14. (✳).

Answer.

(\frac{1}{\sqrt{2}}, - 1, - \frac{1}{2})

and

(- \frac{1}{\sqrt{2}}, - 1, - \frac{1}{2})

3.2.15. (✳).

Answer.

\pm (\frac{1}{2}, - 1, - 1)

3.2.16. (✳).

Answer.

(a)

(1, 0, 3)

(b)

(3, 3, - 1)

(c)

r (t) = (1, 1, 3) + t (3, 3, - 1)

3.2.17. (✳).

Answer.

{49.11}^{\circ}

(to two decimal places)

3.2.18.

Answer.

The horizontal tangent planes are

z = 0,

z = e^{- 1}

and

z = - e^{- 1} .

The largest and smallest values of

z

are

e^{- 1}

and

- e^{- 1},

respectively.

3.3 Surface Integrals
3.3.6 Exercises

3.3.6.1.

Answer.

a b \sqrt{1 + \tan^{2} θ} = a b \sec θ

3.3.6.2.

Answer.

(a)

\frac{1}{2} \sqrt{a^{2} b^{2} + a^{2} c^{2} + b^{2} c^{2}}

(b) See the solution.

3.3.6.3.

Answer.

\frac{π a h}{2}

3.3.6.4.

Answer.

\frac{116}{15} π

3.3.6.5. (✳).

Answer.

\frac{π}{6} [{(1 + 4 a^{2})}^{3 / 2} - 1]

3.3.6.6. (✳).

Answer.

5 \sqrt{2} π

3.3.6.7. (✳).

Answer.

\frac{4}{15} [9 \sqrt{3} - 8 \sqrt{2} + 1]

3.3.6.8. (✳).

Answer.

(a)

F (x, y) = \sqrt{1 + f_{x} (x, y)^{2} + f_{y} (x, y)^{2}}

(b) (i)

\int_{0}^{2 π} d θ \int_{0}^{1} d r \frac{2 r}{\sqrt{4 - r^{2}}}

(b) (ii)

16 π

3.3.6.9.

Answer.

\sqrt{2} π

3.3.6.10. (✳).

Answer.

a^{2} [π - 2]

3.3.6.11.

Answer.

\sqrt{2} π

3.3.6.12.

Answer.

(2 π)^{2} R r

3.3.6.13.

Answer.

(\frac{a}{2}, \frac{a}{2}, \frac{a}{2})

3.3.6.14.

Answer.

16 a^{2}

3.3.6.15. (✳).

Answer.

\frac{π}{2} a^{3} \sqrt{a^{2} + b^{2}}

3.3.6.16.

Answer.

(a)

4 π a^{2 n + 3}

(b)

3 a b c

(c)

\frac{π}{3}

3.3.6.17. (✳).

Answer.

(a)

\frac{8}{3}

(b)

\frac{16}{3}

3.3.6.18. (✳).

Answer.

(a)

\frac{1}{6}

(b)

\frac{1}{2}

3.3.6.19. (✳).

Answer.

\frac{8}{27} [{(\frac{13}{4})}^{3 / 2} - 1]

3.3.6.20. (✳).

Answer.

9 π

3.3.6.21. (✳).

Answer.

(a)

\begin{aligned} r (θ, z) = \frac{2}{3} (3 - z) \cos θ \hat{ı ı} + \frac{2}{3} (3 - z) \sin θ \hat{ȷ ȷ} + z \hat{k} \\ 0 \leq θ < 2 π, 0 \leq z \leq 3 \end{aligned}

(b)

1

3.3.6.22. (✳).

Answer.

(0, 0, \frac{a}{3})

3.3.6.23. (✳).

Answer.

2 π

3.3.6.24. (✳).

Answer.

- \frac{14}{3}

3.3.6.25. (✳).

Answer.

\frac{\sqrt{2} π}{4}

3.3.6.26. (✳).

Answer.

\frac{16}{3} π

3.3.6.27. (✳).

Answer.

(0, 0, \frac{2}{3})

3.3.6.28. (✳).

Answer.

4

3.3.6.29. (✳).

Answer.

\frac{81}{16}

3.3.6.30. (✳).

Answer.

(a)

r (Y, θ) = e^{Y} \sin θ \hat{ı ı} + Y \hat{ȷ ȷ} + e^{Y} \cos θ \hat{k} 0 \leq Y \leq 1, 0 \leq θ \leq 2 π

(b)

\frac{2 π}{3} [(1 + e^{2})^{3 / 2} - 2^{3 / 2}]

(c)

π (1 - e^{2})

3.3.6.31. (✳).

Answer.

12 π

3.3.6.32. (✳).

Answer.

\frac{\sqrt{5} π}{8}

3.3.6.33. (✳).

Answer.

- 20 π

3.3.6.34. (✳).

Answer.

3

3.3.6.35. (✳).

Answer.

2 π

3.3.6.36. (✳).

Answer.

2 π

3.3.6.37.

Answer.

192 π

3.3.6.38. (✳).

Answer.

(a)

x + y + z = 1 + π / 4

(b)

\frac{2 π}{3} [2 \sqrt{2} - 1]

3.3.6.39. (✳).

Answer.

(a) Yes. See the solution for the explanation.

(b) See the solution for the proof.

3.3.6.40. (✳).

Answer.

(a) (i)

r (u, v) = (u, v, \frac{1}{3} (16 - 2 u - 4 v)) \hat{k} u \geq 0, v \geq 0, u + 2 v \leq 8

(a) (ii)

r (u, v) = (4 \cos u \sin v, 4 \sin u \sin v, 4 \cos v) 0 \leq u \leq 2 π, 0 \leq v \leq \frac{π}{4}

(a) (iii)

r (u, v) = (u, v, \sqrt{1 + u^{2} + v^{2}}) u^{2} + v^{2} \leq 99

r (u, v) = (u \cos v, u \sin v, \sqrt{1 + u^{2}}) 0 \leq v \leq 2 π, 0 \leq u \leq \sqrt{99}

(b)

32 π [1 - \frac{1}{\sqrt{2}}]

3.3.6.41. (✳).

Answer.

(a)

\frac{π}{2}

(b)

\frac{π}{4} + \frac{2}{3}

3.3.6.42. (✳).

Answer.

- \frac{5}{6}

4 Integral Theorems
4.1 Gradient, Divergence and Curl
4.1.6 Exercises

4.1.6.1. (✳).

Answer.

(a) A

(b) B

(d) A

(e) B

4.1.6.2.

Answer.

No.

4.1.6.3.

Answer.

See solution.

4.1.6.4.

Answer.

(a)

\nabla \nabla \cdot F = 3,

\nabla \nabla \times F = 0

(b)

\nabla \nabla \cdot F = y^{2} - z^{2} + x^{2},

\nabla \nabla \times F = 2 y z \hat{ı ı} - 2 x z \hat{ȷ ȷ} - 2 x y \hat{k}

(c)

\nabla \nabla \cdot F = \frac{1}{\sqrt{x^{2} + y^{2}}},

\nabla \nabla \times F = 0

(d)

\nabla \nabla \cdot F = 0,

\nabla \nabla \times F = \frac{\hat{k}}{\sqrt{x^{2} + y^{2}}}

4.1.6.5. (✳).

Answer.

(a)

\frac{2}{r}

(b)

(x e^{x y} - 2 x) \hat{ı ı} + y (1 - e^{x y}) \hat{ȷ ȷ} + z \hat{k}

4.1.6.6. (✳).

Answer.

(a)

k = - 3

(b)

k = 2

(c)

k = - 2

4.1.6.7. (✳).

Answer.

(a)

3

(b)

2 r

(c)

- 2 a

(d)

\frac{2}{r}

4.1.6.8. (✳).

Answer.

(a)

a = - 3

(b)

a = 4

(c)

a = 12

4.1.6.9.

Answer.

(a)

F

cannot have a vector potential.

(b) Two solutions are

A = \frac{1}{2} (z^{2} - y^{2}) x \hat{ı ı} - \frac{1}{2} y z^{2} \hat{ȷ ȷ}

and

A = \frac{1}{2} x z^{2} \hat{ı ı} + \frac{1}{2} (x^{2} - z^{2}) y \hat{ȷ ȷ} .

4.1.6.10. (✳).

Answer.

(a)

D = {(x, y, z) | x^{2} + z^{2} \neq 0}

(b)

\nabla \nabla \times F = 0

D

(c)

\nabla \nabla \cdot F = 1

D

(d)

F

is not conservative on the domain

D

of part (a).

4.1.6.11. (✳).

Answer.

(a)

α = β = - 1

(b) Any function of the form

g (x, y, z) = x y z + w (z)

will work.

4.1.6.12.

Answer.

(a) See the solution

(b)

\nabla \nabla \times (Ω Ω \times r) = 2 Ω Ω \nabla \nabla \cdot (Ω Ω \times r) = 0

(c)

1095

km/hr

4.1.6.13.

Answer.

See the solution.

4.2 The Divergence Theorem
4.2.6 Exercises

4.2.6.1.

Answer.

See the solution.

4.2.6.2.

Answer.

See the solution.

4.2.6.3.

Answer.

(a), (b)

\frac{8 π}{3}

4.2.6.4.

Answer.

(a), (b)

\frac{4}{3} π a^{3}

4.2.6.5.

Answer.

(a)

- \frac{81}{4} π

(b)

2 | V |

(c)

2 | V | + \frac{81}{4} π

4.2.6.6.

Answer.

(a), (b)

2 π

4.2.6.7. (✳).

Answer.

(a)

z

(b)

0

4.2.6.8.

Answer.

π

4.2.6.9.

Answer.

[3 + 3 x_{0} - y_{0}] V

4.2.6.10. (✳).

Answer.

- π

4.2.6.11. (✳).

Answer.

\frac{16}{3} π

4.2.6.12. (✳).

Answer.

(a)

\nabla \nabla \cdot F (x, y, z) = 0

except at

(x, y, z) = (0, 0, 0),

where

F

is not defined.

(b)

4 π

(d)

4 π

(e)

0

4.2.6.13. (✳).

Answer.

(a)

\begin{aligned} r (θ, φ) = \sin φ \cos θ \hat{ı ı} + 2 \sin φ \sin θ \hat{ȷ ȷ} + 2 \cos φ \hat{k} \\ 0 \leq θ < 2 π, 0 \leq φ \leq π \end{aligned}

(b)

16 π

(c)

16 π,

again

4.2.6.14. (✳).

Answer.

40 π

4.2.6.15. (✳).

Answer.

24 π

4.2.6.16. (✳).

Answer.

\frac{π}{2}

4.2.6.17. (✳).

Answer.

\frac{3}{2} π

4.2.6.18. (✳).

Answer.

\frac{32}{3} π

4.2.6.19. (✳).

Answer.

3 π

4.2.6.20. (✳).

Answer.

\frac{26}{3}

4.2.6.21. (✳).

Answer.

2 π

4.2.6.22.

Answer.

(a)

\frac{64}{3} π

(b)

\frac{128}{3} π

4.2.6.23. (✳).

Answer.

(a)

\nabla \nabla \cdot F = 0

(x, y, z) \neq 0

and is not defined if

(x, y, z) = 0 .

(b)

4 π

(c)

0

(d) The flux integrals

\iint_{S_{1}} F \cdot \hat{n} d S

and

\iint_{S_{2}} F \cdot \hat{n} d S

are different, because the one point,

(0, 0, 0),

where

\nabla \nabla \cdot F

fails to be well-defined and zero, is contained inside

S_{1}

but is not contained inside

S_{2} .

4.2.6.24. (✳).

Answer.

72

4.2.6.25. (✳).

Answer.

(a)

3

(b)

- 14 π

4.2.6.26. (✳).

Answer.

- \frac{4 π}{5} 3^{5 / 2}

4.2.6.27. (✳).

Answer.

(a)

0

(b)

\frac{625}{2} π

4.2.6.28. (✳).

Answer.

4 π

4.2.6.29. (✳).

Answer.

π

4.2.6.30. (✳).

Answer.

12 π

4.2.6.31. (✳).

Answer.

\frac{188}{15} π \approx 39.37

4.2.6.32. (✳).

Answer.

5 π

4.2.6.33.

Answer.

[\frac{π}{6} - \frac{1}{3}] a^{3}

4.2.6.34.

Answer.

(a)

- 8 \sqrt{2} π

(b)

8 \sqrt{2} π

(c)

16 \sqrt{2} π

4.2.6.35.

Answer.

See the solution.

4.2.6.36.

Answer.

See the solution.

4.2.6.37. (✳).

Answer.

(a), (b)

36 π

4.2.6.38. (✳).

Answer.

\frac{e}{4}

4.2.6.39. (✳).

Answer.

(a)

\frac{1}{\sqrt{3}} (- 1, - 1, 1)

(b)

- \frac{27 π}{2}

(c)

- \frac{81 π}{2}

4.2.6.40. (✳).

Answer.

(a)

\nabla \nabla \cdot F = 2 + 2 z

(b)

π \frac{23}{6} 5^{3} = \frac{2875}{6} π

S

be an oriented surface that encloses a solid

V

and has outward pointing normal. If

\bar{z} = - \frac{9}{2 | V |} - 1,

where

| V |

is the volume of

V

and

\bar{z}

is the

z

-component of the centroid (i.e. centre of mass with constant density) of

V,

then

\iint_{S} F \cdot \hat{n} d S = - 9 .

One surface which obeys this condition is the unit cube (with outward normal) centred on

(0, 0, - \frac{11}{2}) .

4.2.6.41. (✳).

Answer.

(a)

\frac{\sqrt{5} π}{8}

(b)

20

(c)

18

4.2.6.42. (✳).

Answer.

(a)

\frac{π}{4} + \frac{π (b + d)}{6}

(b)

\iint_{σ_{1} \cup σ_{3}} F \cdot \hat{n} d S

is zero if and only if

d = - b .

(c)

\iint_{σ_{1} \cup σ_{3}} F \cdot \hat{n} d S

is zero for all

a,

b,

c,

d .

4.2.6.43. (✳).

Answer.

(a)

4 π

(b)

0 .

4.2.6.44. (✳).

Answer.

See the solution.

4.2.6.45. (✳).

Answer.

See the solution.

4.2.6.46. (✳).

Answer.

\frac{3}{4}

4.2.6.47. (✳).

Answer.

9 π a^{3} + 9 π a^{2}

4.2.6.48. (✳).

Answer.

See the solution.

4.2.6.49. (✳).

Answer.

(a)

0

(b)

\frac{15}{2} π

4.2.6.50. (✳).

Answer.

30 + 24 π

4.2.6.51. (✳).

Answer.

(a)

\frac{64}{3} π

(b)

\frac{128}{3} π

4.3 Green’s Theorem

Exercises

4.3.1.

Answer.

See the solution.

4.3.2.

Answer.

See the solution.

4.3.3.

Answer.

(a)

1

(b)

1

(c)

0

4.3.4.

Answer.

See the solution.

4.3.5.

Answer.

- 54

4.3.6.

Answer.

9

4.3.7. (✳).

Answer.

\frac{32}{3} π

4.3.8. (✳).

Answer.

- 6

4.3.9. (✳).

Answer.

(a)

(b)

- \frac{8}{3}

4.3.10. (✳).

Answer.

- \frac{1}{3}

4.3.11. (✳).

Answer.

54

4.3.12. (✳).

Answer.

\frac{10}{3}

4.3.13. (✳).

Answer.

(a)

- \frac{π}{2}

(b)

\frac{3 π}{2}

4.3.14. (✳).

Answer.

54

4.3.15. (✳).

Answer.

(a)

I_{2} = 0

(b)

I_{3} = π

(c)

I_{4} = π

4.3.16. (✳).

Answer.

(a)

Q_{x} - P_{y} = 0

except at

(0, 0)

where it is not defined.

(b)

- 2 π

(d)

0

(e)

- 2 π

4.3.17. (✳).

Answer.

(a)

\frac{π}{2}

(b)

\frac{π}{4} + \frac{2}{3}

4.3.18. (✳).

Answer.

\frac{3 π}{2}

4.3.19. (✳).

Answer.

\frac{3 π}{8}

4.3.20. (✳).

Answer.

A = - 2

4.3.21. (✳).

Answer.

- π

4.3.22. (✳).

Answer.

\oint_{C_{1}} F \cdot d r = 0

and

\oint_{C_{2}} F \cdot d r = 2 π

4.3.23. (✳).

Answer.

(a)

\frac{1}{2} + \frac{1}{12} [5^{3 / 2} - 1] \approx 1.3484

(b)

\frac{3}{4}

4.3.24. (✳).

Answer.

(a) The projection of the curve on the

x y

-plane (i.e. the top view of the curve) is a circle. See the solution for more details.

(b) (i)

0

(b) (ii)

0

4.3.25.

Answer.

6 x^{2} + 3 y^{2} = 1

4.4 Stokes’ Theorem
4.4.3 Exercises

4.4.3.1.

Answer.

(a)

(b)

(c)

4.4.3.2.

Answer.

See the solution.

4.4.3.3.

Answer.

See the solution.

4.4.3.4.

Answer.

(a)

2 π

(b)

2 π

4.4.3.5.

Answer.

π

4.4.3.6. (✳).

Answer.

8 π

4.4.3.7. (✳).

Answer.

12 π

4.4.3.8. (✳).

Answer.

π

4.4.3.9.

Answer.

8

4.4.3.10.

Answer.

1

4.4.3.11. (✳).

Answer.

4 π

4.4.3.12. (✳).

Answer.

(a)

8

(b)

4 \sqrt{3}

4.4.3.13. (✳).

Answer.

(a)

(b)

S = {(x, y, z) | x^{2} + y^{2} + z^{2} = 4, x \geq 0, y \geq 0, z \geq 0}

with

\begin{aligned} r (θ, φ) = 2 \cos θ \sin φ \hat{ı ı} + 2 \sin θ \sin φ \hat{ȷ ȷ} + 2 \cos φ \hat{k} \\ 0 \leq θ \leq \frac{π}{2}, 0 \leq φ \leq \frac{π}{2} \end{aligned}

and

\hat{n} = \cos θ \sin φ \hat{ı ı} + \sin θ \sin φ \hat{ȷ ȷ} + \cos φ \hat{k} = \frac{1}{2} r (θ, φ)

(c)

- 4 π

4.4.3.14. (✳).

Answer.

(a)

- 128 π,

(b)

- 126 π

4.4.3.15. (✳).

Answer.

4 π

4.4.3.16. (✳).

Answer.

(a) 8

(b)

4 \sqrt{3}

4.4.3.17. (✳).

Answer.

5 π / 4

4.4.3.18. (✳).

Answer.

- \frac{10}{\sqrt{3}}

4.4.3.19. (✳).

Answer.

- 2

4.4.3.20. (✳).

Answer.

- π

4.4.3.21. (✳).

Answer.

\frac{3 π}{4}

4.4.3.22. (✳).

Answer.

\frac{π}{3}

4.4.3.23. (✳).

Answer.

(a)

(b)

\int_{C} F \cdot d r = 10

4.4.3.24. (✳).

Answer.

- π

4.4.3.25. (✳).

Answer.

24 π

4.4.3.26. (✳).

Answer.

2 \sqrt{3} π R^{2}

4.4.3.27.

Answer.

\frac{4}{3}

4.4.3.28.

Answer.

- 2 π

4.4.3.29.

Answer.

24 π

4.4.3.30. (✳).

Answer.

(a)

\nabla \nabla \times F = (1 - 2 x z) \hat{ȷ ȷ}

(b)

\frac{20}{3}

4.4.3.31. (✳).

Answer.

(a)

- 18 π

(b)

- 18 π

4.4.3.32. (✳).

Answer.

(a)

D = {(x, y, z) | x > 0, y > 0, z > 0}

(b) The domain

D

is both connected and simply connected.

(c)

\nabla \nabla \times F = (2 x - \frac{1}{x}) \hat{k}

(d)

2 \ln 2 - 24

(e) No.

F

is not conservative.

4.4.3.33. (✳).

Answer.

(a)

a = 2,

b = - 1

(b)

\frac{π}{4}

4.4.3.34. (✳).

Answer.

- 15

4.4.3.35. (✳).

Answer.

12 π

4.4.3.36. (✳).

Answer.

Rewrite

\oint_{C} E \cdot d r

as a surface integral. For the details, see the solution.

4.4.3.37. (✳).

Answer.

\sqrt{2} π

4.4.3.38. (✳).

Answer.

\frac{2 π}{3 \sqrt{3}}

4.4.3.39. (✳).

Answer.

(a) One possible parametrization is

r (r, θ) = r \cos θ \hat{ı ı} + r \sin θ \hat{ȷ ȷ} + r \hat{k}

with

0 \leq r \leq 1,

0 \leq θ \leq π .

(b)

π

4.4.3.40.

Answer.

- \frac{4}{\sqrt{3}} π

5 True/False and Other Short Questions
5.2 Exercises

5.2.1. (✳).

Answer.

(a) True

(b) True

(d) False

(e) True

(f) That depends. If

κ = 0,

the curve is part of a straight line. If

κ > 0

it is part of a circle of radius

\frac{1}{κ} .

(g) False.

(h) False.

(i) False.

5.2.2. (✳).

Answer.

(a) False

(b) False

(d) False

(e) True

(f) True

(g) False

(h) False

(i) False

(j) True

5.2.3. (✳).

Answer.

(a) False.

(b)

\hat{N} (t),

\hat{B} (t)

(d) False.

(e) False.

(f) True.

5.2.4. (✳).

Answer.

(a) decreasing

(b)

f (x)

D

(c)

r (θ) = \cos θ \hat{ı ı} + \sin θ \hat{k} + \sin θ \cos θ \hat{ȷ ȷ},

0 \leq θ < 2 π

(d) We want parametrisation (d) with domain

| u | \geq 2,

0 \leq v \leq 5 .

(e) One possible answer is

r (t) = t \hat{ı ı},

0 \leq t \leq 1 .

(f)

C = 6

(g)

{(a, b, c, d) | a, b, c, d all real and b = c}

(h) 2

(i) (1) True

(2) False

(3) False

(4) False

(5) False

(j) Any vector field whose divergence is

1

everywhere will work. One such vector field is

F = x \hat{ı ı} .

(k) negative

5.2.5. (✳).

Answer.

(a) false

(b) false

(d) false

(e) true, assuming that the second derivatives of the vector field exist and are continuous.

(f) silly, but true

(g) true

(h) false

(i) false

(j) false

5.2.6. (✳).

Answer.

(a) False

(b) False

(d) True

(e) True

(f) True

(g) True

(h) False

5.2.7. (✳).

Answer.

(a)

P_{y} < 0

(b)

Q_{x} > 0

(c)

\nabla \nabla \times F

is in the direction of

+ \hat{k}

A

(d)

\int_{C_{1}} F \cdot d r > 0

(e)

\int_{C_{2}} F \cdot d r < 0

(f)

F

is not conservative

5.2.8. (✳).

Answer.

(a) False

(b) True

(d) False

(e) True

(f) False

(g) False

(h) False

(i) True

(j) True

5.2.9. (✳).

Answer.

(a) True

(b) False

(d) False

(e) False

(f) True

(g) True

(h) False

(i) False

(j) True

5.2.10. (✳).

Answer.

(a) False.

(b) False.

(d) False.

(e) True.

(f) False.

(g) False.

(h) True.

(i) False.

5.2.11. (✳).

Answer.

(a) True

(b) False

(d) False

(e) True

(f) True

(g) True

(h) False

(i) False

(j) True

5.2.12. (✳).

Answer.

(b)

5.2.13. (✳).

Answer.

(a) False.

(b) False.

(d) True.

(e) False.

(f) True.

(g) False.

(h) False.

(i) True.

(j) False.

5.2.14. (✳).

Answer.

(a) True

(b) False

r (t)

is not indentically

0 .

(d) False

(e) False

5.2.15. (✳).

Answer.

(a)

2 x y + e^{y} \sin x + x e^{x z}

(b)

y^{3} \hat{ı ı} - z \hat{ȷ ȷ}

(d) False.

5.2.16. (✳).

Answer.

(a) True

(b) True

(d) False

(e) True

(f) True

(g) True

5.2.17. (✳).

Answer.

(a) True

(b) False

5.2.18. (✳).

Answer.

(a) True

(b) False

5.2.19. (✳).

Answer.

(a), (b), (c) See the solution.

(d) Yes

(d) No

5.2.20. (✳).

Answer.

(a) yes

(b) no

(d) yes

5.2.21. (✳).

Answer.

(a) True

(b) True

5.2.22. (✳).

Answer.

(a), (c) See the solution.

(b)

8 π b^{2}

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Appendix C Answers to Exercises

1 Curves1.1 Derivatives, Velocity, Etc.

Exercises

1.1.1.

1.1.2.

1.1.3.

1.1.4.

1.1.5.

1.1.6.

1.1.7.

1.1.8.

1.1.9. (✳).

1.1.10.

1.1.11. (✳).

1.1.12.

1.1.13. (✳).

1.1.14.

1.1.15.

1.1.16.

1.1.17. (✳).

1.1.18. (✳).

1.1.19.

1.1.20. (✳).

1.1.21. (✳).

1.1.22. (✳).

1.1.23.

1.1.24.

1.1.25. (✳).

1.1.26. (✳).

1.1.27. (✳).

1.1.28.

1.1.29.

1.1.30.

1.1.31.

1.2 Reparametrization

Exercises

1.2.1.

1.2.2.

1.2.3.

1.2.4. (✳).

1.2.5. (✳).

1.2.6.

1.2.7.

1.3 Curvature

Exercises

1.3.1.

1.3.2.

1.3.3.

1.3.4.

1.3.5.

1.3.6.

1.3.7.

1.3.8.

1.3.9.

1.3.10. (✳).

1.3.11. (✳).

1.3.12.

1.3.13. (✳).

1.4 Curves in Three Dimensions

Exercises

1.4.1.

1.4.2.

1.4.3.

1.4.4.

1.4.5. (✳).

1.4.6. (✳).

1.4.7.

1.4.8.

1.4.9.

1.4.10. (✳).

1.4.11. (✳).

1.4.12. (✳).

1.4.13. (✳).

1.4.14. (✳).

1.4.15. (✳).

1.4.16. (✳).

1.4.17. (✳).

1.4.18. (✳).

1.4.19. (✳).

1.4.20. (✳).

1 Curves
1.1 Derivatives, Velocity, Etc.

1.7 Sliding on a Curve
1.7.4 Exercises

2 Vector Fields
2.1 Definitions and First Examples

2.2 Optional — Field Lines
2.2.2 Exercises