Calculus one and several variables 10E Salas solutions manual ch14

P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU

JWDD027-14 JWDD027-Salas-v1 November 30, 2006 13:43

724 SECTION 14.1

CHAPTER 14

SECTION 14.1

1. f ′(t) = 2 i − j + 3k 2. f ′(t) = sin tk

3. f ′(t) = − 12√

1 − ti

12√

1 + tj +

1(1 − t)2

k 4. f ′(t) =1

1 + t2(i − j)

5. f ′(t) = cos t i − sin t j + sec2 tk 6. f ′(t) = et[i + (1 + t) j + (2t + t2)k

]7. f ′(t) =

−11 − t

i − sin t j + 2tk 8. f ′(t) = et(i − j) − 2e−2t(j − k)

9. f ′(t) = 4 i + 6t2 j + (2t + 2)k; f ′′(t) = 12t j + 2k

10. f ′(t) = (sin t + t cos t) i + (cos t− t sin t)k

f ′′(t) = (2 cos t− t sin t) i + (−2 sin t− t cos t)k

11. f ′(t) = −2 sin 2t i + 2 cos 2t j + 4tk; f ′′(t) = −4 cos 2t i − 4 sin 2t j

12. f ′(t) =12t−1/2 i +

32t1/2 j +

1tk

f ′′(t) = −14t−3/2 i +

34t−1/2 j − 1

t2k

13. (a) r′(t) = −2te−t2 i − e−t j; r′(0) = −j

(b) r′(t) = cot t i − tan t j + (2 cos t + 3 sin t)k; r′(π/4) = i − j +5√2

k

14. (a) r′′(t) =(2e−t − 4te−t + t2e−t

)i + (−2e−t + te−t) j; r′′(0) = 2 i − 2 j

(b) r′′(t) =1ti +

2 − 2 ln t

t2j − 1 + 2 ln t

4t2(ln t)3/2k; r′′(e) = e−1 i − 3

4e−2 k

15.∫ 2

1

(i + 2t j) dt =[t i + t2 j

]21

= i + 3 j

16.∫ π

0

(sin t i + cos t j + tk) dt =[− cos t i + sin t j +

t2

2k]π0

= 2 i +12π2 k

17.∫ 1

0

(et i + e−t k

)dt =

[et i − e−t k

]10

= (e− 1) i +(

1 − 1e

)k

18.∫ 1

0

(te−t i + 4e2t j + e−t k) dt =[(−te−t − e−t) i + 2e2t j − e−t k

]10

= 2e−1 i + 2e2 j − e−1 k

19.∫ 1

0

(1

1 + t2i + sec2 t j

)dt =

[tan−1 t i + tan t j

]10

=π

4i + tan(1) j



SECTION 14.1 725

20.∫ 3

1

(1ti +

ln t

tj + e−2t k

)dt =

[ln t i +

12(ln t)2 j − 1

2e−2t k

]31

= ln 3 i +12(ln 3)2 j +

12(e−2 − e−6)k

21. limt→0

f (t) =(

limt→0

sin t

2t

)i +(limt→0

e2t)j +(

limt→0

t2

et

)k =

12

i + j

22. Does not exist ( because oft

|t| k)

23. limt→0

f (t) =(limt→0

t2)

i +(

limt→0

1 − cos t3t

)j +(

limt→0

t

t + 1

)k = 0 i +

13

(limt→0

1 − cos tt

)j + 0k = 0

24. limt→0

f (t) = j + k

25. (a)∫ 1

0

(tet i + tet

2j)dt = i +

e− 12

j

(b)∫ 8

3

(t

t + 1i +

t

(t + 1)2j +

t

(t + 1)3k)

dt =[5 + ln

(49

)]i +[− 5

36 + ln(

49

)]j + 295

2592 k

26. (a) 34 i + 1

4 j + k (b) 2e2 i + 2e−4 j + 2k

27. 28. 29.

30. 31. 32.

33. 34. 35.



726 SECTION 14.1

36. 37. 38.

2 π 4 π 6 πx

1

2

y

39. (a) f (t) = 3 cos t i + 2 sin t j (b) f (t) = 3 cos t i − 2 sin t j

40. (a) f (t) = (1 + cos t) i + sin t j (b) f (t) = (1 + cos t) i − sin t j

41. (a) f (t) = t i + t2 j (b) f (t) = −t i + t2 j

42. (a) f (t) = t i + t3 j (b) f (t) = −t i − t3 j

43. f (t) = (1 + 2t) i + (4 + 5t) j + (−2 + 8t)k, 0 ≤ t ≤ 1

44. f (t) = (3 + 4t) i + 2 j + (−5 + 14t)k, 0 ≤ t ≤ 1

45. f ′(t0) = i + m j, ∫ b

a

f (t) dt =[12t2 i]ba

+[∫ b

a

f(t) dt]j =

12(b2 − a2

)i + A j,

∫ b

a

f ′(t) dt = [t i + f(t) j]ba = (b− a) i + (d− c) j

46. f (t) =(

12t2 + 1

)i + (√

1 + t2 + 1) j + (tet − et + 4)k

47. f ′(t) = i + t2 j

f (t) = (t + C1) i +(

13 t

3 + C2

)j + C3 k

f (0) = j − k =⇒ C1 = 0, C2 = 1, C3 = −1

f (t) = t i +(

13 t

3 + 1)

j − k

48. f (t) = e2t i − e2t k = e2t(i − k)

49. f ′(t) = α f (t) =⇒ f (t) = eαt f (0) = eαt c

50. For each ε > 0 there exists δ > 0 such that

if 0 < |t− t0| < δ, then ||f (t) − L|| < ε.

51. (a) If f ′ (t) = 0 on an interval, then the derivative of each component is 0 on that interval, each

component is constant on that interval, and therefore f itself is constant on that interval.

(b) Set h (t) = f (t) − g (t) and apply part (a).



SECTION 14.1 727

52. ||[f (t) · g(t)] − [L · M]|| = ||[f (t) · g(t)] − [L · g(t)] + [L · g(t)] − [L · M]||

= ||[(f (t) − L) · g(t)] + [L · (g(t) − M)]||

≤ ||(f (t) − L) · g(t)|| + ||L · (g(t) − M)||

≤ ||f (t) − L|| ||g(t)|| + ||L|| ||g(t) − M||

by Schwarz’s inequality

As t → t0, the right side tends to (0)||M|| + ||L||(0) = 0.

53. If f is differentiable at t, then each component is differentiable at t, each component is continuous at

t, and therefore f is continuous at t.

54. Set f (t) = f1(t) i + f2(t) j + f3(t)k and apply the fundamental Theorem of Calculus to f1, f2, f3.

55. no; as a counter-example, set f (t) = i = g (t).

56.∫ b

a

[c · f (t)] dt =∫ b

a

[c1f1(t) + c2f2(t) + c3f3(t)] dt

= c1

∫ b

a

f1(t) dt + c2

∫ b

a

f2(t) dt + c3

∫ b

a

f3(t) dt

= c ·∫ b

a

f (t) dt∫ b

a

[c× f (t)] dt can be written

∫ b

a

{[c2f3(t) − c3f2(t)] i − [c1f3(t) − c3f1(t)] j + [c1f2(t) − c2f1(t)]k} dt.

This gives [c2

∫ b

a

f3(t) dt− c3

∫ b

a

f2(t) dt]

i −[c1

∫ b

a

f3(t) dt− c3

∫ b

a

f1(t) dt]j

+[c1

∫ b

a

f2(t) dt− c2

∫ b

a

f1(t) dt]k

which is c ×∫ b

a

f (t) dt

57. Suppose f (t) = f1(t) i + f2(t) j + f3(t)k. Then ‖ f (t) ‖ =√

f21 (t) + f2

2 (t) + f23 (t) and

d

dt(‖ f ‖) =

12[f21 + f2

2 + f23

]−1/2 (2f1 · f ′1 + 2f2 · f ′

2 + 2f3 · f ′3) =

f (t) · f ′ (t)‖ f (t) ‖

The Answer Section of the text gives an alternative approach.

58.d

dt

(f (t)‖f (t)‖

)=

d

dt

(1

‖f (t)‖ f (t))

=1

‖f (t)‖ f ′(t) + f (t) · −1‖f (t)‖2

· d

dt[‖f (t)‖]

=f ′(t)‖f (t)‖ − f (t) · f ′(t)

‖f (t)‖3f (t), using the result of Exercise 57.



728 SECTION 14.2

SECTION 14.2

1. f ′t = b, f ′′(t) = 0

2. f ′t = b + 2t c, f ′′(t) = 2 c

3. f ′ (t) = 2e2t i − cos t j, f ′′(t) = 4e2t i + sin t j

4. f (t) = 2t2 i =⇒ f ′t = 4t i, f ′′(t) = 4 i

5. f ′t = [(t2 i − 2t j) · (i + 3t2 j) + (2t i − 2 j) · (t i + t3 j] j = [3t2 − 8t3] j

f ′′(t) = (6t− 24t2) j

6. f (t) = (t− t5)k =⇒ f ′(t) = (1 − 5t4)k, f ′′(t) = −20t3 k

7. f ′(t) =[(et i + tk

)× d

dt

(t j + e−t k

)]+[d

dt

(et i + tk

)×(t j + e−t k

)]=[(et i + tk

)×(j − e−t k

)]+[(et i + k

)×(t j + e−t k

)]=(−t i + j + et k

)+(−t i − j + tet k

)= −2t i + et(t + 1)k

f ′′(t) = −2 i + et(t + 2)k

8. f ′(t) = a × b + 2tb, f ′′(t) = 2b

9. f ′(t) = (a × tb) × 2tb + (a × b) × (a + t2 b), f ′′(t) = (a × tb) × 2b + 4t(a × b) × b

10. f ′(t) = g(t2) + tg′(t2)2t = g(t2) + 2t2g′(t2)

f ′′(t) = g′(t2)2t + 4tg′(t2) + 2t2g′′(t2)2t = 6tg′(t2) + 4t3g′′(t2)

11. f ′(t) =12

√tg′(√

t)

+ g(√

t), f ′′(t) =

14

g′′(√

t)

+3

4√tg′ (

√t )

12. f (t) = 2e−2t i − 2k =⇒ f ′(t) = −4e−2t i, f ′′(t) = 8e−2t i.

13. −(sin t) ecos t i + (cos t) esin t j

14.d2

dt2[et cos t i + et sin t j] =

d

dt[et(cos t− sin t) i + et(sin t + cos t) j]

= −2et sin t i + 2et cos t j

15. (et i + e−t j) · (et i − e−t j) = e2t − e−2t; therefore

d2

dt2[(et i + e−t j) · (et i − e−t j)

]=

d2

dt2[e2t − e−2t

]=

d

dt

[2e2t + 2e−2t

]= 4e2t − 4e−2t



SECTION 14.2 729

16. (ln t i + t j) × (2t j − k) +(

1ti + j)×(t2 j − tk

)

17.d

dt[(a + tb)× (c + td)] = [(a + tb) ×d ] + [b× (c + td)] = (a × d) + (b × c) + 2t (b × d)

18. b× (a + tb + t2 c) + (a + tb) × (b + 2t c) = 2t(a× c) + 3t2(b× c).

19.d

dt[(a + tb) · (c + td)] = [(a + tb) · d ] + [b · (c + td)] = (a · d) + (b · c) + 2t (b · d)

20. b · (a + tb + t2 c) + (a + tb) · (b + 2t c) = 2(a · b) + 2t(a · c) + 2t ‖ b ‖2 +3t2(b · c)

21. r(t) = a + tb 22. r(t) = a + tb +12t2 c

23. r(t) =12t2 a +

16t3 b + t c + d 24. r(t) =

(1 + 2t− 1

4cos 2t

)i +(

1 − 14

sin 2t)

j

25. r(t) = sin t i + cos t j, r′(t) = cos t i − sin t j, r′′(t) = − sin t i − cos t j = − r(t).

Thus r(t) and r′′(t) are parallel, and they always point in opposite directions.

26. r′′(t) = k2ekt i + k2e−kt j = k2 r(t), so r′′(t) and r(t) are parallel.

27. r (t) · r′(t) = (cos t i + sin t j) · (− sin t i + cos t j) = 0

r (t)× r′(t) = (cos t i + sin t j) × (− sin t i + cos t j)

= cos2 tk + sin2 tk =(cos2 t + sin2 t

)k = k

28. (g × f )′(t) = [g(t)× f ′(t)] + [g′(t)× f (t)]

= −[f ′(t)×g(t)] − [f (t)×g′(t)]

= −{[f (t)×g′(t)] + [f ′(t)×g(t)]} = −(f ×g)′(t)

29.d

dt[ f (t) × f ′(t) ] = [ f (t) × f ′′(t) ] + [f ′(t)× f ′(t) ]︸︷︷︸

0

= f (t) × f ′′(t).

30.d

dt[u1(t)r1(t) ×u2(t)r2(t)] =

d

dt[(u1(t)u2(t))(r1(t)× r2(t))]

= u1(t)u2(t)d

dt[r1(t) × r2(t)] + [r1(t) × r2(t)]

d

dt[u1(t)u2(t)]

31. [f · g ×h]′ = f ′ · (g ×h) + f · (g ×h)′ = f ′ · (g ×h) + f · [g′ ×h + g ×h′]

and the result follows.

32.d

dt(f × f ′) = f (t)× f ′′(t) by Exercise 29. If f (t) and f ′′(t) are parallel, their cross product is zero,

sod

dt(f × f ′) = 0, hence f × f ′ is constant.



730 SECTION 14.3

33. ‖ r (t) ‖ is constant ⇐⇒ ‖ r (t) ‖2 = r (t) · r (t) is constant

⇐⇒ d

dt[r (t) · r (t) ] = 2 [r (t) · r′(t) ] = 0 identically

⇐⇒ r (t) · r′(t) = 0 identically

34. (a) Routine

(b) Write

[f (t + h) · g(t + h)] − [f (t) · g(t)]h

so (f (t + h) ·

[g(t + h) − g(t)

h

])+([

f (t + h) − f (t)h

]· g(t)

)and take the limit as h → 0. (Appeal to Theorem 13.1.3)

35. Write[ f (t + h) × g (t + h) ] − [ f (t) × g (t) ]

h

as (f (t + h) ×

[g (t + h) − g (t)

h

])+([

f (t + h) − f (t)h

]×g (t)

)and take the limit as h → 0. (Appeal to Theorem 13.1.3.)

36. (a) and (b) can derived routinely by using components. An ε, δ derivation of (a) is also simple:

Let ε > 0. Since f is continuous at u(t0), there exists δ1 > 0 such that

if |z − u(t0)| < δ1, then ‖ f (z) − f (u(t0)) ‖< ε.

Since u is continuous at t0, there exists δ > 0 such that

if |t− t0| < δ, then |u(t) − u(t0)| < δ1.

Thus

|t− t0| < δ =⇒ |u(t) − u(t0)| < δ1 =⇒ ‖ f (u(t)) − f (u(t0)) ‖< ε.

SECTION 14.3

1. r′ (t) = −π sinπt i + π cosπt j + k, r′(2) = π j + k

R (u) = (i + 2k) + u(π j + k)

2. r′(t) = et i − e−t j − 1tk, r′(1) = e i − e−1 j − k; R(u) = (e i + e−1 j) + u(e i − e−1 j − k)

3. r′(t) = b + 2t c, r′(−1) = b − 2 c, R (u) = (a − b + c) + u(b − 2 c)

4. r′(0) = i; R(u) = (i + j + k) + u i.



SECTION 14.3 731

5. r′(t) = 4t i − j + 4tk, P is tip of r (1), r′(1) = 4 i − j + 4k

R (u) = (2 i + 5k) + u (4 i − j + 4k)

6. r′(2) = 3a − 4 c; R(u) = (6a + b − 4 c) + u(3a − 4 c)

7. r′(t) = −2 sin t i + 3 cos t j + k, r′(π/4) = −√

2 i + 32

√2 j + k

R (u) =(√

2 i + 32

√2 j + π

4 k)

+ u(−√

2 i + 32

√2 j + k

)8. r′(t) = (sin t + t cos t) i + (cos t− t sin t) j + 2k

r′(π

2

)= i − π

2j + 2k; R(u) =

(π2

i + π k)

+ u(i − π

2j + 2k

)9. The scalar components x(t) = at and y(t) = bt2 satisfy the equation

a2y(t) = a2(bt2) = b (a2t2) = b [x(t) ]2

and generate the parabola a2y = bx2.

10. x(t)2 − y(t)2 =a2

4[(ewt + e−wt)2 − (ewt − e−wt)2

]= a2, with x(t) > 0; the right branch of the

hyperbola x2 − y2 = a2.

11. r (t) = t i +(1 + t2

)j, r′(t) = i + 2t j

(a) r (t) ⊥ r′(t) =⇒ r (t) · r′(t) =[t i +

(1 + t2

)j]

· (i + 2t j)

= t(2t2 + 3

)= 0 =⇒ t = 0

r (t) and r′(t) are perpendicular at (0, 1).

(b) and (c) r (t) = α r′(t) with α = 0 =⇒ t = α and 1 + t2 = 2tα =⇒ t = ±1.

If α > 0, then t = 1. r (t) and r′(t) have the same direction at (1, 2).

If α < 0, then t = −1. r (t) and r′(t) have opposite directions at (−1, 2).

12. r(t) = eαt(i + 2 j + 3k)

13. The tangent line at t = t0 has the form R (u) = r (t0) + u r′(t0). If r′(t0) = α r (t0), then

R (u) = r (t0) + uαr (t0) = (1 + uα) r (t0).

The tangent line passes through the origin at u = −1/α.

14. r′1(0) = i, r′2(0) = 2 i + j + k

θ = cos−1 r′1(0) · r′2(0)‖ r′1(0) ‖ ‖ r′2(0) ‖ = cos−1 2√

6∼= 0.62 radian, or 35.3◦

15. r1(t) passes through P (0, 0, 0) at t = 0; r2(u) passes through P (0, 0, 0) at u = −1.

r′1(t) = et i + 2 cos t j +1

t + 1k; r′1(0) = i + 2 j + k

r′2(u) = i + 2u j + 3u2 k; r′2(−1) = i − 2 j + 3k

cos θ =r′1(0) · r′2(1)

‖r′1(0)‖ ‖r′2(1)‖ = 0; θ =π

2∼= 1.57, or 90◦.



732 SECTION 14.3

16. r′1(0) = −i, r′2(−1) = i − 4 j − 8k

θ = cos−1 r′1(0) · r′2(−1)‖ r′1(0) ‖ ‖ r′2(−1) ‖ = cos−1

(−19

)∼= 1.68 radians, or 96.4◦

17. r1(t) = r2(u) implies ⎧⎪⎪⎨⎪⎪⎩

et = u

2 sin(t + 1

2π)

= 2

t2 − 2 = u2 − 3

⎫⎪⎪⎬⎪⎪⎭ so that t = 0, u = 1.

The point of intersection is (1, 2,−2).

r′1(t) = et i + 2 cos(t +

π

2

)j + 2tk, r′1(0) = i

r′2(u) = i + 2uk, r′2(1) = i + 2k

cos θ =r′1(0) · r′2(1)

‖r′1(0)‖ ‖r′2(1)‖ =15

√5 ∼= 0.447, θ ∼= 1.11 radians

18. (a) R(u) = r(t0) + ur′(t0) = (t0 i + f(t0) j) + u(i + f ′(t0) j)

=⇒ x(u) = t0 + u, y(u) = f(t0) + uf ′(t0)

(b) From (a), we get u = x(u) − t0 and y(u) − f(t0) = f ′(t0)u. so

y − f(t0) = f ′(t0)(x− t0), as expected.

19. (a) r (t) = a cos t i + b sin t j (b) r (t) = a cos t i − b sin t j

(c) r (t) = a cos 2t i + b sin 2t j (d) r (t) = a cos 3t i − b sin 3t j

20. (a) r(t) = −a sin t i + b cos t j (b) r(t) = a sin t i + b cos t j

(c) r(t) = −a sin 2t i + b cos 2t j (d) r(t) = a sin 3t i + b cos 3t j

21. r′(t) = t3 i + 2t j 22. r′(t) = 2 i + 2t j 23. r′(t) = 2e2t i − 4e−4t j

24. r′(t) = cos t i + 2 sin t j 25. r′(t) = −2 sin t i + 3 cos t j 26. r′(t) = sec t tan t i + sec2 t j



SECTION 14.3 733

27. r (t) = (t2 + 1) i + t j, t ≥ 1; or, r (t) = sec2 t i + tan t j, t ∈[14π,

12π)

28. r(t) = cos t(1 − cos t) i + sin t(1 − cos t) j, t ∈ [0, 2π]

29. r (t) = cos t sin 3t i + sin t sin 3t j, t ∈ [0, π]

30. r(t) = t4 i + t3 j, t ≤ 0

31. y3 = x2

There is no tangent vector at the origin.

32. (a) r(0) = i + 2 j = r(1)

(b) T(0) =r′(0)

‖ r′(0) ‖ =1√

π2 + 2(−i − j + π k), T(1) =

r′(1)‖ r′(1) ‖ =

1√π2 + 5

(i + 2 j − π k)

33. We substitute x = t, y = t2, z = t3 in the plane equation to obtain

4t + 2t2 + t3 = 24, (t− 2)(t2 + 4t + 12

)= 0, t = 2.

The twisted cubic intersects the plane at the tip of r(2), the point (2, 4, 8).

The angle between the curve and the normal line at the point of intersection is the angle between the

tangent vector r′(2) = i + 4 j + 12k and the normal N = 4 i + 2 j + k :

cos θ =(i + 4 j + 12k) · (4 i + 2 j + k)‖ i + 4 j + 12k‖ ‖4 i + 2 j + k‖ =

24√161

√21

∼= 0.412, θ ∼= 1.15 radians.

34. (a) T (t) =1√

a2 sin2 t + b2 cos2 t(−a sin t i + b cos t j)

N(t) =1√

a2 sin2 t + b2 cos2 t(−b cos t i + a sin t j)

(b) r(u) =√

22

(a i + b j) + u(−a i + b j), R(u) =√

22

(a i + b j) + u(−b i − a j)

35. r′(t) = 2 j + 2tk, ‖r′(t)‖ = 2√

1 + t2

T (t) =r′(t)‖r′(t)‖ =

1√1 + t2

(j + tk) ,

T′(t) =1

(1 + t2)3/2[−t j + k]

at t = 1: tip of r = (1, 2, 1), T = T(1) =1√2

j +1√2

k;



734 SECTION 14.3

T′(1) = − 12√

2j +

12√

2k; ‖T′(1)‖ =

12; N = N(1) =

T′(1)‖T′(1)‖ = − 1√

2j +

1√2

k

normal for osculating plane:

T×N =(

1√2

j +1√2

k)

×(− 1√

2j +

1√2

k)

=12

i

equation for osculating plane:12(x− 1) + 0(y − 2) + 0(z − 1) = 0, which gives x− 1 = 0

36. T(t) =r′(t)

‖ r′(t) ‖ =i + 2t j + 4tk√

20t2 + 1T(1) =

1√21

(i + 2 j + 4k)

N(t) =T′(t)

‖ T′(t) ‖ =−20t i + 2 j + 4k√

400t2 + 20N(1) =

1√420

(−20 i + 2 j + 4k) =1√105

(−10 i + j + 2k)

T(1)×N(1) =1√5(−2 j + k).

Osculating plane at (1, 1, 2) : −2(y − 1) + (z − 2) = 0 =⇒ −2y + z = 0

37. r′(t) = −2 sin 2t i + 2 cos 2t j + k, ‖r′(t)‖ =√

5

T (t) =r′(t)

‖r′(t)‖ =15

√5 (−2 sin 2t i + 2 cos 2t j + k)

T′(t) = −45

√5 (cos 2t i + sin 2t j), ‖T′(t)‖ =

45

√5

N (t) =T′(t)

‖T′(t)‖ = −(cos 2t i + sin 2t j)

at t = π/4: tip of r = (0, 1, π/4), T =15

√5 (−2 i + k), N = −j


T×N =15

√5 (−2 i + k)× (−j) =

15

√5 i +

25

√5k

equation for osculating plane:15

√5 (x− 0) +

25

√5(z − π

4

)= 0, which gives x + 2z =

π

2

38. T(t) =r′(t)

‖ r′(t) ‖ =i + 2 j + 2tk√

4t2 + 5T(2) =

1√21

(i + 2 j + 4k)

N(t) =T′(t)

‖ T′(t) ‖ =−2t i − 4t j + 5k√

20t2 + 25N(2) =

1√105

(−4 i − 8 j + 5k)

T(2)×N(2) =1√5(2 i − j)

Osculating plane at (2, 4, 2) : 2(x− 2) − (y − 4) = 0 =⇒ 2x− y = 0



SECTION 14.3 735

39. r′(t) = cosh t i + sinh t j + k, ‖r′(t)‖ =√

cosh2 t + sinh2 t + 1 =√

2 cosh t

T (t) =r′(t)

‖r′(t)‖ =1√2

(i + tanh t j + sech tk) ,

T′(t) =1√2

(sech2t j − sech t tanh tk

)at t = 0: tip of r = (0, 1, 0), T =

1√2

(i + k), T′(0) =1√2

j; N = j


T×N =(

1√2

(−i + k))× j =

1√2

(−i + k) or i − k

equation for osculating plane: x− z = 0

40. T(t) =r′(t)

‖ r′(t) ‖ =−3 sin 3t i + j − 3 cos 3tk√

10T(π

3

)=

1√10

(j + 3k)

N(t) =T′(t)

‖ T′(t) ‖ =−9 cos 3t i + 9 sin 3tk

9N(π

3

)= i

T(π

3

)×N

(π3

)= 3 j − k

Osculating plane at (−1, π3 , 0) : 3(y − π

3 ) − z = 0 or 3y − z − π = 0

41. r′(t) = et [(sin t + cos t) i + (cos t− sin t) j + k] , ‖r′(t)‖ = et√

3

T (t) =r′(t)

‖r′(t)‖ =1√3

[(sin t + cos t) i + (cos t− sin t) j + k] ,

T′(t) =1√3

[(cos t− sin t) i − (sin t + cos t) j]

at t = 0: tip of r = (0, 1, 1), T = T(0) =1√3

(i + j + k);

T′(0) =1√3

(i − j); ‖T′(0)‖ =√

2√3; N = N(0) =

T′(0)‖T′(0)‖ =

1√2

(i − j)


T×N =1√3(i + j + k)× 1√

2(i − j) =

1√6(i + j − 2k)

equation for osculating plane:1√6(x− 0) +

1√6(y − 1) − 2√

6(z − 1) = 0, or x + y − 2z + 1 = 0



736 SECTION 14.4

42. T(t) =r′(t)

‖ r′(t) ‖ =t cos t i + t sin t j

t= cos t i + sin t j T

(π4

)=

√2

2i +

√2

2j

N(t) =T′(t)

‖ T′(t) ‖ = − sin t i + cos t j N(π

4

)= −

√2

2i +

√2

2j

T(π

4

)×N

(π4

)= k

Osculating plane at(√

22 [1 + π

4 ],√

22 [1 − π

4 ], 2)

: z − 2 = 0 or z = 2

43. T1 =R′(u)

‖R′(u) ‖ = − r′(a + b− u)‖ r′(a + b− u) ‖ = −T.

Therefore T1′(u) = T′(a + b− u) and N1 = N.

44. (a) r′(t) = −√

2 sin t i +√

2 cos t j + k

tangent line (t = π/4) : x = 1 − t, y = 1 + t, z = 14π + t

(c) Since 14π + t > 0, t ∈ [0, 2π], the tangent line is never parallel to the x, y-plane.

45. (a) r′(t) = −√

2 sin t i +√

2 cos t j + 5 cos tk

tangent line (t = π/4) : x = 1 − t, y = 1 + t, z = −√

22

− 5√

22

t

(c) The tangent line is parallel to the x, y-plane at the points where t =(2n + 1)π

10, n = 0, 1, 2, . . . , 9.

46. (a) r′(t) = −√

2 sin t i +√

2 cos t j +1tk

tangent line (t = π/4) : x = 1 − t, y = 1 + t, z = ln(π/4) +4π

+ t

SECTION 14.4

1. r′(t) = i + t1/2 j, ‖r′(t)‖ =√

1 + t 2. r′(t) = (t2 − 1)i + 2t j; ‖ r′(t) ‖= t2 + 1

L =∫ 8

0

√1 + t dt =

[23

(1 + t)3/2]80

=523

L =∫ 2

0

(t2 + 1) dt =143

3. r′(t) = −a sin t i + a cos t j + bk, ‖r′(t)‖ =√a2 + b2

L =∫ 2π

0

√a2 + b2 dt = 2π

√a2 + b2

4. r′(t) = i +√

2t1/2 j + tk; ‖ r′(t) ‖= t + 1

L =∫ 2

0

(t + 1) dt = 4

5. r′(t) = i + tan t j, ‖r′(t)‖ =√

1 + tan2 t = | sec t|

L =∫ π/4

0

| sec t | dt =∫ π/4

0

sec t dt = [ ln | sec t + tan t| ]π/40 = ln (1 +√

2 )



SECTION 14.4 737

6. r′(t) =1

1 + t2i +

t

1 + t2j; ‖ r′(t) ‖= 1√

1 + t2

L =∫ 1

0

dt√1 + t2

=[ln |t +

√t2 + 1|

]10

= ln(1 +√

2)

7. r′(t) = 3t2 i + 2t j, ‖r′(t)‖ =√

9t4 + 4t2 = |t|√

4 + 9t2

L =∫ 1

0

∣∣∣ t√4 + 9t2∣∣∣ dt =

∫ 1

0

t√

4 + 9t2 dt =[

127(4 + 9t2

)3/2]10

=127

(13√

13 − 8)

8. r′(t) = i +(

12t2 − 1

2t−2

)k; ‖ r′(t) ‖= 1

2t2 +

12t−2

L =∫ 3

1

(12t2 +

12t−2

)dt =

143

9. r′(t) = (cos t− sin t)et i + (sin t + cos t)et j, ‖r′(t)‖ =√

2 et

L =∫ π

0

√2 et dt =

√2 (eπ − 1)

10. r′(t) = 2t i + 2t j − 2tk; ‖ r′(t) ‖= 2t√

3 =⇒ L =∫ 2

0

2t√

3 dt = 4√

3

11. r′(t) =1ti + 2 j + 2tk, ‖r′(t)‖ =

√1t2

+ 4 + 4t2

L =∫ e

1

√1t2

+ 4 + 4t2 dt =∫ e

1

(1t

+ 2t)

dt =[ln |t| + t2

]e1

= e2

12. r′(t) = t cos t i − t sin t j; ‖ r′(t) ‖= t =⇒ L =∫ 2

0

t dt = 2

13. r′(t) = t cos t i + t sin t j +√

3 tk, ‖r′(t)‖ =√t2 cos2 t + t2 sin2 t + 3t2 =

√4t2 = 2t

L =∫ 2π

0

2t dt =[t2]2π0

= 4π2

14. r′(t) = (1 + t)1/2 i + (1 − t)1/2 j +√

2k, ‖r′(t)‖ =√

(1 + t) + (1 − t) + 2 =√

4 = 2

L =∫ 1/2

−1/2

2 dt = 2

15. r′(t) = 2 i + 2t j − 2tk, ‖r′(t)‖ = 2√

1 + 2t2

L =∫ 2

0

2√

1 + 2t2 dt =√

2∫ tan−1 (2

√2)

0

sec3 u du

∧(t√

2 = tanu)

= 12

√2[secu tanu + ln | secu + tanu|

]tan−1 (2√

2)

0= 6 + 1

2

√2 ln (3 + 2

√2 )



738 SECTION 14.4

16. r′(t) = (3 cos t− 3t sin t)i + (3 sin t + 3t cos t)j + 4k; ‖ r′(t) ‖=√

9t2 + 25

L =∫ 4

0

3

√t2 +

259

dt =

[32t

√t2 +

259

+32· 25

9ln

∣∣∣∣∣t +

√t2 +

259

∣∣∣∣∣]4

0

= 26 +256

ln 5.

17. s = s(t) =∫ t

a

‖r′(u)‖ du

s′(t) = ‖r′(t)‖ = ‖x′(t) i + y′(t) j + z′(t)k‖

=√

[x′(t)]2 + [y′(t)]2 + [z′(t)]2.

In the Leibniz notation this translates to

ds

dt=

√(dx

dt

)2

+(dy

dt

)2

+(dz

dt

)2

.

18. Parameterize the graph of f by setting r(x) = x i + f(x) j, x ∈ [a, b].

19. s = s(x) =∫ x

a

√1 + [f ′(t)]2 dt

s′(x) =√

1 + [f ′(x)]2.

In the Leibniz notation this translates to

ds

dx=

√1 +(dy

dx

)2

.

20. L1 =∫ e

1

‖ r′(t) ‖ dt =∫ e

1

√2 +

2t2

dt; L2 =∫ 1

0

√1 + e2x dx

Setting t = ex, we get L1 =√

2L2

21. r′(t) = − sin t i + cos t j. Since ‖r′‖ ≡ 1, the parametrization is by arc length.

22. r′(t) = −a + b; ‖r′‖ = ‖b − a‖.

s =∫ t

0

‖b − a‖ du = ‖b − a‖ t; t =s

‖b − a‖ .

R(s) =(

1 − s

‖b − a‖

)a +

s

‖b − a‖ b, 0 ≤ s ≤ ‖b − a‖.

23. r′(t) = t sin t i + t cos t j + tk; ‖r′‖ = t√

2.

s =∫ t

0

u√

2 du =√

22

t2; t = 21/4√s.

R(s) =(sin 21/4

√s− 21/4

√s cos 21/4

√s)i +(cos 21/4

√s + 21/4

√s sin 21/4

√s)j +

1√2sk.

24. r′(t) = (et cos t− et sin t) i + (et sin t + et cos t) j; ‖r′‖ =√

2 et.

s =∫ t

0

√2 eu du =

√2(et − 1); t = ln (1 + s/

√2).

R(s) = ev[cos v i + sin v j] where v = ln (1 + s/√

2), 0 ≤ s ≤√

2 (eπ − 1).



SECTION 14.4 739

25. r′(t) = t3/2 j + k, ‖r′(t)‖ =√(

t3/2)2 + 1 =

√t3 + 1

s =∫ 1/2

0

√t3 + 1 dt ∼= 0.5077

26. r′(t) = i + t2 j; ‖ r′(t) ‖=√

1 + t4 L =∫ 2

0

√1 + t4 ∼= 3.6535

27. r′(t) = −3 sin t i + 4 cos t j, ‖r′(t)‖ =√

9 sin2 t + 16 cos2 t dt

s =∫ 2π

0

√9 sin2 t + 16 cos2 t dt ∼= 22.0939

28. r′(t) = i + 2tj +1tk; ‖ r′(t) ‖=

√1 + 4t2 +

1t2

L =∫ 4

1

√1 + 4t2 +

1t2

dt ∼= 15.4480

29. (a) (b) s =∫ 2π

0

√1 + 16 cos2 4t dt ∼= 17.6286

30. (a) (b) s =∫ 2π

0

√1 +(

11 + t

)2

dt ∼= 6.6818

PROJECT 14.4

1. Given the differentiable curve r = r(t), t ∈ I. Let t = φ(u) be a continuously differentiable one-to-one

function that maps the interval J onto the interval I, and let R(u) = r(φ(u)), u ∈ J . Suppose that

φ′(u) > 0 on J . Then, as u increases across J , t = φ(u) increases across I. As a result, R(u) takes



740 SECTION 14.5

on exactly the same values in exactly the same order as r. If φ′(u) < 0 on J , then R(u) takes on

exactly the same values as r but in the reverse order.

2. Let r = r(t), t ∈ I be a differentiable curve. Let t = φ(u) be a continuously differentiable one-to-one

function that maps the interval J onto the interval I with φ′(u) > 0. Set R(u) = r(φ(u)), u ∈ J .

R′(u) = [r(φ(u))]′ = r′(φ(u))φ′(u);R′(u)

||R′(u)|| =r′(φ(u))φ′(u)

||r′(φ(u))φ′(u)|| =r′(φ(u))φ′(u)

φ′(u)||r′(φ(u))|| =r′(t)

||r′(t)|| .

since φ′(u) > 0∧

Therefore, the unit tangent is left unchanged by a sense-preserving change of parameter. The invariance

of the principal normal and the osculating plane follows from the invariance of the unit tangent.

3. Suppose that φ′(u) < 0 on J . Then

R′(u) = [r(φ(u))]′ = r′(φ(u))φ′(u);R′(u)R′(u)

=r′(φ(u))φ′(u)

||r′(φ(u))φ′(u)|| = − r′(φ(u))φ′(u)φ′(u)||r′(φ(u))|| = − r′(t)

||r′(t)|| .

since φ′(u) < 0∧

Thus, the unit tangent is reversed by a sense-reversing change of parameter. That is, if TR and Tr

are the respective unit tangents, then TR = −Tr.

Now consider the principal normals:

T ′R

||T ′R|| = − T ′

r(φ(u)φ′(u)||T ′

r(φ(u)φ′(u)|| =T ′r(φ(u)φ′(u)

φ′(u)||T ′r(φ(u)φ′(u)|| =

T ′r(t)

||T ′r(t)||

.

since φ′(u) < 0∧

Thus the principal normal is unchanged by a sense-reversing change of parameter. The osculating plane

is also unchanged since T × N and −T × N are each normal to the osculating plane.

4. s =∫ t

0

{r(v)|| dv = ψ(t). ψ is an continuously differentiable increasing function on I; t = ψ−1(s).

5. Let L be the length as computed from r and L∗ the length as computed from R. Then

L∗ =∫ d

c

‖R′(u) ‖ du =∫ d

c

‖ r′(φ (u)) ‖φ′(u) du =∫ b

a

‖ r′(t) ‖ dt = L.

t = φ (u)∧

SECTION 14.5

1. r (t) = a[cos θ(t) i + sin θ(t) j], r′(t) = a[− sin θ(t) i + cos θ(t) j)]θ′(t)

‖r′(t)‖ = v =⇒ a|θ′(t)| = v =⇒ |θ′(t)| = v/a

r′′(t) = a[− cos θ(t) i − sin θ(t) j] [θ′(t)]2, ‖r′′(t)‖ = a[θ′(t)]2 = v2/a



SECTION 14.5 741

2. r′(t) = (−πa sinπt + 2bt)i + (πa cosπt− 2bt)j

r′′(t) = (−π2a cosπt + 2b)i + (−π2a sinπt− 2b)j

At t = 1, v = 2b i − (aπ + 2b)j, v =‖ v ‖=√

4b2 + (aπ2 + 2b)2

a = (aπ2 + 2b)i − 2b j, ‖ a ‖=√

4b2 + (aπ2 + 2b)2

3. r (t) = at i + b sin at j, = a i + ab cos at j

r′(t) = −a2b sin at j, ‖r′(t)‖ = a2|b sin at| = a2|y(t)|

4. r′(t) = 2t j + 2(t− 1)k; speed is minimum when ‖ r′(t) ‖2 is minimum

4t2 + 4(t− 1)2 = 8t2 − 8t + 4 is minimum at t =12

5. y = cosπx, 0 ≤ x ≤ 2 6. x = y3, all real x

7. x =√

1 + y2 , y ≥ −1 8. x = sinπy, 0 ≤ y ≤ 2

9. (a) initial position is tip of r (0) = x0 i + y0 j + z0 k

(b) r′(t) = (α cos θ) j + (α sin θ − 32t)k, r′(0) = (α cos θ) j + (α sin θ)k

(c) |r′(0)| = |α| (d) r′′(t) = −32k

(e) a parabolic arc from the parabola

z = z0 + (tan θ ) (y − y0) − 16(y − y0)2

α2 cos2 θin the plane x = x0



742 SECTION 14.5

10. (a) x(t) = 2 cos 2t = 4 cos2 t− 2 and y(t) = 3 cos t =⇒ x =49y2 − 2. Since

−2 ≤ x(t) ≤ 2, −3 ≤ y(t) ≤ 3,

the path consists only of the bounded arc

x =49y2 − 2, −3 ≤ y ≤ 3.

The motion traces out this arc twice on every t-interval of length 2π.

(b) included in part (d)

(c) r(t) = 2 cos 2t i + 3 cos t j, r′(t) = −4 sin 2t i − 3 sin t j, r′′(t) = −8 cos 2t i − 3 cos t j

The velocity r′(t) is 0 when t = nπ. At such points

the acceleration =

{−8 i − 3 j, if n is even

−8 i + 3 j, if n is odd

(b) and (d)

11. ‖r(t)‖ = C iff ‖r(t)‖2 = r(t) · r(t) = C

iffd

dt‖r(t)‖ = 2r(t) · r′(t) = 0

iff r(t) ⊥ r′(t)

12. Problem 11 with v in place of r.

13. κ =e−x

(1 + e−2x)3/2

14. y′ = 3x2, y′′ = 6x; κ =6|x|

(1 + 9x4)3/2

15. y′ =1

2x1/2; y′′ =

−14x3/2

κ =

∣∣−1/4x3/2∣∣[

1 +(1/2x1/2

)2]3/2 =2

(1 + 4x)3/2

16. y′ = 1 − 2x, y′′ = −2; κ =2

[1 + (1 − 2x)2]3/2=

√2

2(1 − 2x + 2x2)3/2

17. κ =sec2 x

(1 + tan2 x)3/2= | cosx |

18. y′ = sec2 x, y′′ = 2 sec2 x tanx; κ =|2 sec2 x tanx|(1 + sec4 x)3/2



SECTION 14.5 743

19. κ =| sinx|

(1 + cos2 x)3/2

20. 2x− 2yy′ = 0 =⇒ y′ =x

y, y′′ =

y − x(xy )

y2= −a2

y3; κ =

∣∣∣∣∣ a2/y3

[1 + (xy )2]3/2

∣∣∣∣∣ = a2

(x2 + y2)3/2

21. κ =|x|

(1 + x4/4)3/2; at

(2,

43

), κ =

25√

5

22. y′ = x, y′′ = 1, κ =1

(1 + x2)3/2; at (0, 0), κ = 1

23. κ =

∣∣− 1/y3∣∣

(1 + 1/y2)3/2=

1

(1 + y2)3/2; at (2, 2), κ =

15√

5

24. y′ = 4 cos 2x, y′′ = −8 sin 2x; at x =π

4, y′ = 0, y′′ = −8 =⇒ κ = 8

25. y′(x) =1

x + 1, y′(2) =

13

; y′′(x) =−1

(x + 1)2, y′′(2) = − 1

9.

At x = 2, κ =

∣∣∣∣− 19

∣∣∣∣[1 +(

13

)2]3/2 =

310√

10

26. At x =π

4, y′ = secx tanx =

√2, y′′ = secx tan2 x + sec3 x =

√2 + 23/2 = 3

√2

=⇒ κ =3√

2(1 + 2)3/2

=

√23

27. κ(x) =

∣∣− 1/x2∣∣

(1 + 1/x2)3/2=

x

(x2 + 1)3/2, x > 0

κ′(x) =

(1 − 2x2

)(x2 + 1)5/2

, κ′(x) = 0 =⇒ x =12

√2

Since κ increases on(0, 1

2

√2]

and decreases on[12

√2, ∞

), κ is maximal at

(12

√2, 1

2 ln 12

).

28. y′ = 3 − 3x2, y′′ = −6x. local max at x = 1 : y′ = 0, y′′ = −6, κ =6

(1 + 0)3/2= 6

29. x(t) = t, x′(t) = 1, x′′(t) = 0; y(t) = 12 t

2, y′(t) = t, y′′(t) = 1 κ =1

(1 + t2)3/2

30. x′ = et, x′′ = et, y′ = −e−t, y′′ = e−t

κ =|x′y′′ − x′′y′|

[(x′)2 + (y′)2]3/2=

2(e2t + e−2t)3/2



744 SECTION 14.5

31. x(t) = 2t, x′(t) = 2, x′′(t) = 0; y(t) = t3, y′(t) = 3t2, y′′(t) = 6t; κ =12| t |

(4 + 9t4)3/2

32. x′ = 2t, x′′ = 2, y′ = 3t2, y′′ = 6t; κ =|12t2 − 6t2|

(4t2 + 9t4)3/2=

6|t|(4 + 9t2)3/2

33. x(t) = et cos t, x′(t) = et(cos t− sin t), x′′(t) = −2 et sin t

y(t) = et sin t, y′(t) = et(sin t + cos t), y′′(t) = 2 et cos t

κ =

∣∣2e2t cos t (cos t− sin t) + 2e2t sin t (cos t + sin t)∣∣

[ e2t(cos t− sin t)2 + e2t(cos t + sin t)2 ]3/2=

2e2t

(2e2t)3/2=

12

√2 e−t

34. x′ = −2 sin t, x′′ = −2 cos t, y′ = 3 cos t, y′′ = −3 sin t; κ =6

(4 sin2 t + 9 cos2 t)3/2

35. x(t) = t cos t, x′(t) = cos t− t sin t, x′′(t) = −2 sin t− t cos t

y(t) = t sin t, y′(t) = sin t + t cos t, y′′(t) = 2 cos t− t sin t

κ =

∣∣(cos t− t sin t)(2 cos t− t sin t) − (sin t + t cos t)(−2 sin t− t cos t)∣∣

[ (cos t− t sin t)2 + (sin t + t cos t)2 ]3/2=

2 + t2

[1 + t2]3/2

36. x′ = t cos t, x′′ = cos t− t sin t, y′ = t sin t, y′′ = sin t + t cos t

κ =∣∣∣∣ t cos t(sin t + t cos t) − t sin t(cos t− t sin t)

(t2 cos2 t + t2 sin2 t)3/2

∣∣∣∣ = t2

t3=

1t. (t > 0)

37. κ =

∣∣2/x3∣∣

[1 + 1/x4]3/2=

2∣∣x3∣∣

(x4 + 1)3/2; at x = ±1, κ =

√2

2

38. From Exercise 20, κ =1

(x2 + y2)3/2. At (±1, 0), κ = 1

39. We use (14.5.3) and the hint to obtain

κ =

∣∣ab sinh2 t− ab cosh2 t∣∣[

a2 sinh2 t + b2 cosh2 t]3/2 =

∣∣∣aby2 − b

ax2∣∣∣[(ay

b

)2

+(bx

a

)2]3/2

=a3b3∣∣∣aby2 − b

ax2∣∣∣[

a4y2 + b4x2]3/2 =

a4b4

[a4y2 + b4x2]3/2.

40. x′ = r(1 − cos t), x′′ = r sin t, y′ = r sin t, y′′ = r cos t

Highest point when t = π =⇒ x′ = 2r, x′′ = 0, y′ = 0, y′′ = −r

κ =2r2

(4r2)3/2=

14r



SECTION 14.5 745

41. r′(t) = et(cos t− sin t) i + et(sin t + cos t) j + et k

ds

dt= ‖r′(t)‖ =

√3 et,

d2s

dt2=

√3 et

T (t) =r′(t)‖r′(t)‖ =

1√3

[(cos t− sin t) i + (sin t + cos t) j + k]

T′(t) =1√3

[(− sin t− cos t) i + (cos t− sin t) j]; ‖T′(t)‖ =√

2/3

κ =‖T′(t)‖ds/dt

=

√2/3√3 et

=13

√2 e−t; aT =

d2s

dt2=

√3 et, aN = κ

(ds

dt

)2

=√

2 et

42. r′(t) = cosh t i + sinh t j + k;ds

dt= ||r′|| =

√2 cosh t;

d2s

dt2=

√2 sinh t.

T(t) =r′(t)

||r′(t)|| =1√2

(i + tanh t j + sech tk).

T′(t) =1√2

(sech2 t j − sech t tanh tk

); ||T′(t)|| =

1√2

sech t

κ =||T′(t)||ds/dt

=12

sech2 t; aT =d2s

dt2=

√2 sinh t, aN = κ

(ds

dt

)2

= 1

43. r′(t) = −2 sin 2t i + 2 cos 2t j;ds

dt= ||r′(t)|| = 2,

d2s

dt2= 0

T (t) =r′(t)

||r′(t)|| = − sin 2t i + cos 2t j

T′(t) = −2 (cos 2t i + sin 2t j) ; ||T′(t)|| = 2


=22

= 1; aT =d2s

dt2= 0, aN = κ

(ds

dt

)2

= 1 · 4 = 4.

44. r′(t) = v(t) = i + 2t j +1tk, a(t) = 2 j − 1

t2k

ds

dt=‖ v ‖=

√1 + 4t2 + 1/t2 =

√4t4 + t2 + 1

t; v ×a = − 4

ti +

1t2

j + 2k

κ =||v ×a||(ds/dt)3

=t√

4t4 + 16t2 + 1(4t4 + t2 + 1)3/2

aT =d2s

dt2=

4t4 − 1t2√

4t4 + t2 + 1; aN = κ

(ds

dt

)2

=1t

√4t4 + 16t2 + 14t4 + t2 + 1

45. r′(t) = −3 sin 3t i + 4 j − 3 cos 3tk;ds

dt= ||r′|| = 5;

d2s

dt2= 0.

T(t) =r′(t)

||r′(t)|| = −35

sin 3t i +45

j − 35

cos 3tk

T′(t) = −95

cos 3t i +95

sin 3tk; ||T′(t)|| =95


=9/55

=925

; aT =d2s

dt2= 0, aN = κ

(ds

dt

)2

=925

· 25 = 9.



746 SECTION 14.5

46. r′(t) = t cos t i + t sin t j + t√

3k,ds

dt= ||r′(t)|| =

√4t2 = 2t,

d2s

dt2= 2

T (t) =r′(t)||r′(t)| = 1

2 cos t i + 12 sin t j + 1

2

√3k

T′(t) = − 12 sin t i + 1

2 cos t j, ‖T′(t)‖ =12

Then, κ =‖T′(t)‖ds/dt

=1/22t

=14t

aT =d2s

dt2= 2, aN = κ

(ds

dt

)2

=14t

(4t2) = t.

47. r′(t) =√

1 + t i −√

1 − t j +√

2k,ds

dt= ‖r′(t)‖ =

√(1 + t) + (1 − t) + 2 = 2,

d2s

dt2= 0

T (t) =r′(t)‖r′(t)| =

√1 + t

2i −

√1 − t

2j +

√2

2k

T′(t) =1

4√

1 + ti +

14√

1 − tj.

‖T′(t)‖ =

√1

16(1 + t)+

116(1 − t)

=14

√2

1 − t2

Then, κ =‖T′(t)‖ds/dt

=18

√2

1 − t2

aT =d2s

dt2= 0, aN = κ

(ds

dt

)2

=12

√2

1 − t2.

48. (b) κ(t) =√

1 + 4t2 + t4

6 (1 + t2 + t4)3/2maximum curvature occurs at x ∼= ±0.2715

49. tangential component: aT =6t + 12t3√1 + t2 + t4

; normal component: aN = 6

√1 + 4t2 + t4

1 + t2 + t4

50. Set r(θ) = cos θf(θ)i + sin θf(θ)j

51. By Exercise 50

κ =

∣∣ (eaθ)2 + 2(aeaθ)2 − (eaθ) (a2eaθ

) ∣∣[(eaθ)2 + (aeaθ)2

]3/2 =e−aθ

√1 + a2

.

52. f(θ) = aθ, f ′(θ) = a, f ′′(θ) = 0 =⇒ κ =a2θ2 + 2a2

(a2θ2 + a2)3/2=

θ2 + 2|a|(θ2 + 1)3/2

53. By Exercise 50,

κ =

∣∣a2(1 − cos θ)2 + 2a2 sin2 θ − a2(1 − cos θ)(cos θ)∣∣[

a2(1 − cos θ)2 + a2 sin2 θ]3/2 =

3a2(1 − cos θ)

[2a2(1 − cos θ)]3/2=

3ar[2ar]3/2

=3

2√

2ar.



SECTION 14.4 747

54. By Exercise 50,

κ =

∣∣(a sin 2θ)2 + 2(2a cos 2θ)2 − (a sin 2θ)(2a cos 2θ)∣∣

[(a sin 2θ)2 + (2a cos 2θ)2]32

=

∣∣a2 sin2 2θ + 8a2 cos2 2θ − 3a2 sin 2θ cos 2θ∣∣

(a2 sin2 2θ + 4a2 cos2 2θ)32

=

∣∣r2 + 8(a2 − r2) + 3r√a2 − r2

∣∣[r2 + 4(a2 − r2)]

32

=

∣∣8a2 − 7r2 + 3r√a2 − r2

∣∣(4a2 − 3r2)

32

.

PROJECT 14.5A

1. The system of equations generated by the specified conditions is:a + b + c + d = 3 27a + 9b + 3c + d = 7

6a + 2b = 0 27α + 9β + 3γ + δ = 7

729α + 81β + 9γ + δ = −2 54α + 2β = 0

27a + 6b + c = 27α + 6β + γ 18a + 2b = 18α + 2β

a ∼= −0.1094 b ∼= 0.3281 c ∼= 2.1094 d ∼= 0.6719

α ∼= 0.0365 β ∼= −0.9844 γ ∼= 6.0469 δ ∼= −3.2656

2. Clearly p and q are continuous on their

respective intervals. The conditions p(3) = q(3),

p′(3) = q′(3) and p′′(3) = q′′(3) imply that

F, F ′, and F ′′ are continuous on [1, 9].

3. (a), (b) The system of equations generated by the specified conditions (and the derivative conditions of

Problem 1) is:27α + 9b + 3c + d = 10 64α + 16b + 4c + d = 15

18a + 2b = 0 64α + 64β + 4γ + δ = 15

216α + 36β + 6γ + δ = 35 36α + 2β = 0

48α + 8b + c = 48α + 8β + γ 24α + 2b = 24α + 2β

(c) a = b = 0, c = 5, d = −5; α = 1.25, β = −15, γ = 65, δ = −85

4. κ =|y′′|

[1 + (y′)2]3/2=

Cn(n− 1)xn−2

[1 + (Cnxn−1)2]3/2



748 SECTION 14.6

At x = 0, κ = 0 as desired. At x = 1, want κ =Cn(n− 1)

(1 + C2n2)3/2=

96125

y =14x3 works.

PROJECT 14.5B

1.dTdt

=dTds

ds

dt=⇒ dT

ds=

dT/dt

ds/dt=

T′(t)‖T′(t)‖

‖T/(t)‖ds/dt

= κN.

2.dB

ds=

d

ds(T×N) =

(T× dN

ds

)+(dTds

×N)

= T× dNds

. ThereforedBds

⊥ T.

Since B has constant length,dBds

⊥ B. Being perpendicular to both T and B,dBds

is parallel to N and is therefore a scalar multiple of N.

3.dNds

=d

ds(B×T) =

(B× dT

ds

)+(dBds

×T)

= (B×κN) + τ(N×T)

= −κ(N×B) − τ(T×N) = −κT − τB

4. We know that dB/ds = τN. It follows that ‖ dB/ds ‖= |τ | ‖ N ‖= |τ |. Thus |τ | is the magnitude

of the change in direction of B per unit arc length, or equivalently, the rate per unit arc length at which

the curve tends away from the osculating plane.

SECTION 14.6

1. (a) r′(t) =aω

2(eωt − e−ωt

)i +

bω

2(eωt + e−ωt

)j, r′(0) = bωj

(b) r′′(t) =aω2

2(eωt + e−ωt

)i +

bω2

2(eωt − e−ωt

)j = ω2r (t)

(c) The torque τ is 0 : τ (t) = r (t) × ma (t) = r (t) × mω2r (t) = 0.

The angular momentum L (t) is constant since L′(t) = τ (t) = 0.

2. (a) F(t) = mr′′(t) = mb2r(t), mb2 > 0

(b) F(t) = −mr(t), −m < 0

(c) L(t) = r(t)×mv(t) = m(i + j − k)

3. We begin with the force equation F(t) = αk. In general, F (t) = ma (t), so that here

a (t) =α

mk.

Integration gives

v (t) = C1i + C2 j +( αm

t + C3

)k.

Since v (0) = 2 j, we can conclude that C1 = 0, C2 = 2, C3 = 0. Thus

v (t) = 2 j +α

mtk.



SECTION 14.6 749

Another integration gives

r (t) = D1i + (2t + D2) j +( α

2mt2 + D3

)k.

Since r (0) = y0 j + z0 k, we have D1 = 0, D2 = y0, D3 = z0, and therefore

r (t) = (2t + y0) j +( α

2mt2 + z0

)k.

The conditions of the problem require that t be restricted to nonnegative values.

To obtain an equation for the path in Cartesian coordinates, we write out the components

x(t) = 0, y(t) = 2t + y0, z(t) =α

2mt2 + z0. (t ≥ 0)

From the second equation we have

t = 12 [y(t) − y0]. (y(t) ≥ y0)

Substituting this into the third equation, we get

z(t) =α

8m[y(t) − y0]

2 + z0. (y(t) ≥ y0)

Eliminating t altogether, we have

z =α

8m(y − y0)

2 + z0. (y ≥ y0)

Since x = 0, the path of the object is a parabolic arc in the yz-plane.

Answers to (a) through (d):

(a) velocity: v(t) = 2 j +α

mtk. (b) speed: v(t) =

1m

√4m2 + α2t2.

(c) momentum: p(t) = 2m j + α tk.

(d) path in vector form: r (t) = (2t + y0) j +( α

2mt2 + z0

)k, t ≥ 0.

path in Cartesian coordinates: z =α

8m(y − y0)2 + z0, y ≥ y0, x = 0.

4. F(t) · v(t) = 0 for all t

=⇒ a(t) · v(t) = v′(t) · v(t) =12d

dt[v(t) · v(t)] = 0 for all t

=⇒ v(t) · v(t) = [v(t)]2 is constant =⇒ v(t) is constant

5. F (t) = ma (t) = m r′′(t) = 2mk

6. If v′(t) = 0, then L′(t) = r′(t)×mv(t) + r(t)×mv′(t)

= v(t)×mv(t) + r(t)×0 = 0

7. From F (t) = ma (t) we obtain

a (t) = π2[a cosπt i + b sinπt j].

By direct calculation using v (0) = −πbj + k and r (0) = bj we obtain

v (t) = aπ sinπt i − bπ cosπt j + k

r (t) = a(1 − cosπt) i + b(1 − sinπt) j + tk.



750 SECTION 14.6

(a) v (1) = bπj + k (b) ‖v (1)‖ =√π2b2 + 1

(c) a (1) = −π2ai (d) mv(1) = m (πbj + k)

(e) L (1) = r (1) × mv(1) = [2ai + bj + k] × [m (bπj + k)]

= m [b(1 − π)i − 2aj + 2abπk]

(f) τ (1) = r (1) × F (1) = [2ai + bj + k] ×[−mπ2ai

]= −mπ2a [j − bk]

8. d

dt

(12m[v(t)]2

)=

12m

d

dt[v(t) · v(t)] =

12m[2v′(t) · v(t)]

= ma(t) · v(t) = F(t) · v(t)

9. We have mv = mv1 + mv2 and 12mv2 = 1

2mv12 + 1

2mv22.

Therefore v = v1 + v2 and v2 = v12 + v2

2.

Since v2 = v · v = (v1 + v2) · (v1 + v2) = v12 + v2

2 + 2(v1 · v2),

we have v1 · v2 = 0 and v1 ⊥ v2.

10. That the path is of the form r(t) = cosωtA + sinωtB can be seen by applying the hint to each

component of the equation F(t) = −mω2r(t).

A = r(0) = the initial position

B = r′(0)/ω = the initial velocity divided by ω

The path is circular if ‖ A ‖=‖ B ‖ and A ⊥ B.

11. r′′(t) = a, r′(t) = v (0) + ta, r (t) = r (0) + tv(0) + 12 t

2 a.

If neither v (0) nor a is zero, the displacement r (t) − r (0) is a linear combination of v (0) and a and

thus remains on the plane determined by these vectors. The equation of this plane can be written

[a × v (0)] · [r − r (0)] = 0.

(If either v (0) or a is zero, the motion is restricted to a straight line; if both of these vectors are zero,

the particle remains at its initial position r (0). )

12. Clearly we can take φ1 ∈ [0, 2π). With ω = −1, the path takes the form

r(t) = [A1 cos(−t + φ1) + D1]i − [A1 sin(−t + φ1) + D2]j + [Ct + D3]k.

Differentiation gives

v(t) = A1 sin(−t + φ1)i + A1 cos(−t + φ1)j + Ck.

r(0) = a i =⇒ A1 cosφ1 + D1 = a, A sinφ1 + D2 = 0, D3 = 0

v(0) = a j + bk =⇒ A1 sinφ1 = 0, A1 cosφ1 = a, C = b



SECTION 14.7 751

From these equations we have

D1 = D2 = D3 = 0, c = b, A1 sinφ1 = 0, A1 cosφ1 = a.

The last two equations give

φ1 = 0 and A1 = a or φ1 = π and A1 = −a.

The first possibility gives

r(t) = a cos(−t) i − a sin(−t) j + btk

= a cos t i + a sin t j + btk.

The second possibility gives the same path:

r(t) = −a cos(−t + π) i − a sin(−t + π) j + btk

= a cos t i + a sin t j + btk

13. r(t) = i + t j +(qE0

2m

)t2k

14. Since v has magnitude w ‖ r ‖, lies in the plane of the wheel, and makes an angle of 90◦

counterclockwise with r, we have v = ω× r

15. r(t) =(

1 +t3

6m

)i +

t4

12mj + tk 16. r(t) = sin

(ωt +

12π

)k

17. d

dt

(12mv2

)= mv

dv

dt= m

(v · dv

dt

)= m

dvdt

· v = F · drdt

= 4r2

(r · dr

dt

)= 4r2

(rdr

dt

)= 4r3 dr

dt=

d

dt

(r4).

Therefore d/dt(

12mv2 − r4

)= 0 and 1

2mv2 − r4 is a constant E. Evaluating E from t = 0, we

find that E = 2m.

Thus 12mv2 − r4 = 2m and v =

√4 + (2/m) r4 .

SECTION 14.7

1. On Earth: year of length T , average distance from sun d.

On Venus: year of length αT, average distance from sun 0.72d.

Therefore(αT )2

T 2=

(0.72d)3

d3.

This gives α2 = (0.72)3 ∼= 0.372 and α ∼= 0.615. Answer: about 61.5% of an Earth year.



752 REVIEW EXERCISES

2.dE

dt=

12m

d

dt(v2) −mρ

d

dt(r−1)

d

dt(v2) =

d

dt(v · v) = 2(a · v)

d

dt(r−1) = − 1

r2

dr

dt= − 1

r3

(rdr

dt

)= − 1

r3(r · v) (using 13.2.3)

dE

dt= m(a · v) +

mρ

r3(r · v) = (a · mv) +

(ρrr3

· mv)

=(a +

ρrr3

)· mv = 0

3.(dx

dt

)2

+(dy

dt

)2

=[d

dt(r cos θ )

]2+[d

dt(r sin θ)

]2

=[r (− sin θ )

dθ

dt+

dr

dtcos θ

]2+[r cos θ

dθ

dt+

dr

dtsin θ

]2

= r2 sin2 θ

(dθ

dt

)2

+(dr

dt

)2

cos2 θ + r2 cos2 θ

(dθ

dt

)2

+(dr

dt

)2

sin2 θ

=(dr

dt

)2

+ r2

(dθ

dt

)2

4. Using r = drdθ θ and θ = L

mr2

we have

E +mρ

r=

12m(r2 + r2θ2) =

12

[(dr

dθ

)2

θ2 + r2θ2

]

=12m

[(dr

dθ

)2L2

m2r4+

L2

m2r2

]=

L2

2m

[1r4

(dr

dθ

)2

+1r2

]

and therefore

E =L2

2m

[1r2

+1r4

(dr

dθ

)2]− mρ

r

5. Substitute

r =a

1 + e cos θ,

(dr

dθ

)2

=[ −a

(1 + e cos θ)2· (−e sin θ)

]2=

(ae sin θ)2

(1 + e cos θ)4

into the right side of the equation and you will see that, with a and e2 as given, the expression reduces

to E.

REVIEW EXERCISES

1. f ′(t) = 6t i − 15t2 j, f ′′(t) = 6 i − 30t j

2. f ′(t) = 2e2t i +2t

1 + t2j, f ′′(t) = 4e2t i +

2 − 2t2

(1 + t2)2j

3. f ′(t) = (et cos t− et sin t) i + 2 sin 2t j, f ′′(t) = −2et sin t i + 4 cos 2t j



REVIEW EXERCISES 753

4. f ′(t) = cosh t i − (2t− t2)e−t j + sinh tk, f ′′(t) = sinh t i + (t2 − 4t + 2)e−t j + cosh tk

5.∫ 2

0

[2t i + (t2 − 1) j

]dt =

[t2 i +

(13t3 − t

)j]20

= 4 i +23

j

6.∫ π

0

[sin 2t i + 2 cos t j +√tk] dt =

[−1

2cos 2t i + 2 sin t j +

23t3/2 k

]π0

=23π3/2 k

7.

50 100 150 200x

−5

−10

y8.

1 2 3 4x

6

12

y

9. 10.

11. (a) r(t) = 2 cos(t +

π

2

)i + 4 sin

(t +

π

2

)j (b) r(t) = −2 cos 2t i + 4 sin 2t j

12. direction vector: d = (2, 4, 6); r(t) = (1 + 2t) i + (1 + 4t) j + (−2 + 6t)k, 0 ≤ t ≤ 1

13. f (t) = 13 t

3 i + ( 12e

2t + t) j + 13 (2t + 1)3/2k + C.

f (0) = i − 3 j + 3k =⇒ C = i − 72 j + 8

3k; f (t) =(

13 t

3 + 1)i +(

12e

2t + t− 72

)j +(

13 (2t + 1)3/2 + 8

3

)k

14. f ′(t) = −f (t) =⇒ f (t) = f 0e−t

f (0) = i + 2k =⇒ f 0 = i + 2k and so f (t) = e−t i + 2e−tk

15. f ′(t) = (6 i + 12t3 j) + (8t i − 12k) = (6 + 8t) i + 12t3 j − 12k

16. Note: f is not a vector function. f(t) = et + 1 =⇒ f ′(t) = et




17. f (t) = (t2 + 2t3) i −(

2t2 +1t2

)j + (t4 − t)k, f ′(t) = (2t + 6t2) i −

(4t− 2

t3

)j + (4t3 − 1)k

18. Note: f is not a vector function. f(t) = t3 cos t + t2 sin t + 3t cos t = (t3 + 3t) cos t + t2 sin t

f ′(t) = −(t3 + 3t) sin t + (3t2 + 3) cos t + 2t sin t + t2 cos t = (4t2 + 3) cos t− (t3 + t) sin t

19. r′(t) = 2r(t) =⇒ r(t) = r0e2t

r(0) = (1, 2, 1) =⇒ r0 = (1, 2, 1) and r(t) = (e2t, 2e2t, e2t)

20. F(t) = e2t i + e−2t j, F ′(t) = 2e2t i − 2e−2t j, F ′′(t) = 4e2t i + 4e−2t j

Since F ′′(t) = 4F(t) for all t, F and F ′′ are parallel.

F and F ′ will have the same direction for some value of t iff there is a number k > 0 such that

e2t i + e−2t j = k(2e2t i − 2e−2t j). No such value of k exists.

21. The tip of r(t) is P (1, 1, 1) when t = 0.

r′(t) = (2t + 2) i + 3 j + (3t2 + 1)k, r′(0) = 2 i + 3 j + k

Scalar parametric equations for the tangent line are: x = 1 + 2t, y = 1 + 3t, z = 1 + t.

22. r(π/3) = (√

3/2,−1/2, π/3)

r′(t) = (2 cos 2t,−2 sin 2t, 1); r′(π/3) = (−1,−√

3, 1)

Scalar parametric equations for the tangent line are: x =√

32 − t, y = − 1

2 −√

3 t, z = π3 + t

23. r1(t) = (2, 1, 1) at t = 1; r2(u) = (2, 1, 1) at u = −1. Therefore the curves intersect at the point

(2, 1, 1).

r′1(t) = (2 i + 2t j + k, r1(1) = 2 i + 2 j + k; r2(u) = −i − 2u j + 2uk, r′2(−1) = −i + 2 j − 2k.

Since r′1(1) · r′2(−1) = 0, the angle of intersection is π/2 radians

24. r(t) = 2t i + (1 − t2) j − t2 k, r′(t) = 2t i − 2t j − 2tk.

(2t i + (1 − t2) j − t2 k) · (2t i − 2t wj − 2tk) = 4t3; 4t3 = 0 =⇒ t = 0

The curve and the tangent line meet at right angles at the point where t = 0; (0, 1, 0).

25. r(t) = t i + e2t j, r(0) = j;

r′(t) = i + 2e2t j, r′(0) = i + 2 j

−1 1x

2

4

6

8

y

26. r(t) = 2 sin t i − 3 cos t j, r(π/6) = i − 32

√3 j

r′(t) = 2 cos t i + 3 sin t j r′(π/6) =√

3 i + 32 j

−2 2x

−3

3y




27. r′(t) = t cos t i + t sin t j +√

3tk; ||r′|| =ds

dt= 2t

unit tangent vector: T =r′(t)

||r′(t)|| =12(cos t i + sin t j +

√3k).

T′(t) = −12

sin t i +12

cos t j; ||T′(t)|| =12.

principal normal vector: N =T′(t)

||T′(t)|| = − sin t i + cos t j

28. r′(t) = (−a sin t i + a cos t j + bk.

The cosine of the angle θ between r′(t) and k is:r′ · k

||r′|| ||k|| =|b|√

a2 + b2= constant.

Therefore θ is a constant.

29. r′(t) = 2 i +1tj − 2tk; ||r′(t)|| =

1 + 2t2

t.

T(t) =r′(t)

||r′(t)|| =2t

2t2 + 1i +

12t2 + 1

j − 2t2

2t2 + 1k; T(1) =

23

i +13

j − 23 k

T′(t) =2 − 4t2

(2t2 + 1)2i − 4t

(2t2 + 1)2j − 4t

(2t2 + 1)2k; T′(1) = − 2

9 i − 49 j − 4

9 k, ||T′(1)|| = 23

N(1) = T′(1)||T′(1)|| = − 1

3 i − 23 j − 2

3 k

A normal vector for the osculating plane is: (2 i + j − 2k) × (i + 2 j + 2k) = 6 i − 6 j + 3k.

Since r(1) = 2 i − k, an equation for the osculating plane is

6(x− 2) − 6y + 3(z + 1) = 0 or 2x− 2y + z = 3.

30. r′(t) = − sin t i − sin t j −√

2 cos tk; ||r′(t)|| =√

2.

T(t) =r′(t)

||r′(t)|| =1√2

(− sin t i − sin t j −√

2 cos tk); T(π/4) = −12

i − 12

j − 1√2

k

T′(t) =1√2

(− cos t i − cos t j +√

2 sin tk); T′(π/4) = −12

i − 12

j +1√2

k, ||T′(π/4)|| = 1

N(π/4) =T′(π/4)

||T′(π/4)|| = −12

i − 12

j +1√2

k

A normal vector for the osculating plane is: (i + j +√

2k) × (i + j −√

2k) = −2√

2 i + 2√

2 j.

Since r(π/4) =√

22

(i + j − 2k, an equation for the osculating plane is

−2√

2

(x−

√2

2

)+ 2

√2

(y −

√2

2

)= 0 or x− y = 0.

31. r′(t) = 2 i + t1/2 j; L =∫ 5

0

||r′(t)||dt =∫ 5

0

√4 + tdt =

383

32. r′(t) = et i − e−t j −√

2k; ||r′(t)|| = et + e−t; L =∫ ln 3

0

(et + e−t)dt =[et − e−t

]ln 3

0=

83




33. r′(t) = cosh t i + sinh t j + k; ||r′(t)|| =√

cosh2 t + sinh2 t + 1 =√

2 cosh t;

L =∫ 1

0

√2 cosh t dt =

[√2 sinh t

]10

=√

2 sinh 1.

34. r′(t) = − sin t i − cos t j + sinh tk; ||r′|| =√

1 + sinh2 t = cosh t;∫ ln 2

0

cosh t dt =[sinh t

]ln 2

0=

34

35. (a) r′(t) = − sin t i + cos t j + t1/2 k; ||r′(t)|| =√

1 + t.

s =∫ t

0

||r′(u)|| du =∫ t

0

√1 + t dt =

[23(1 + u)3/2

]t0

=23(1 + t)3/2 − 2

3

(b) t =(

32s + 1

)2/3

− 1 = φ(s); R(s) = cos φ(s) i + sin φ(s) j +23[φ(s)]3/2 k

(c) R′(s) =[− sin φ(s) i + cos φ(s) j + φ(s)1/2 k

]φ′(s)

||R′(s)|| = φ′(s)√

1 + φ(s) =23

[32s + 1

]−1/3(32

)√(32s + 1

)2/3

= 1

36. velocity: v = r′(t) = − sin t i + cos t j + 2 sin 2tk speed: s = ||v|| =√

1 + 4 sin2 2t

acceleration: a = r′′(t) = − cos t i − sin t j + 4 cos 2tk; ||a|| =√

1 + 16 cos2 2t

37. r′′(t) = − cos t i − sin t j and r′(0) = k =⇒ r′(t) = − sin t i + (cos t− 1) j + k.

Thus: velocity v = − sin t i + (cos t− 1) j + k and speed ||v|| =√

3 − 2 cos t.

r′(t) = − sin t i + (cos t− 1) j + k and r(0) = i =⇒ r(t) = cos t i + (sin t− t) j + tk.

38. The acceleration vector remains perpendicular to the path means:

r′′ · r′ = 0.

r′(t) = f ′(t) i + 2f(t)f ′(t) j, r′′(t) = f ′′(t) i + [2f ′(t)2 + 2f(t)f ′′(t)] j

r′′ · r′ = 0 =⇒ f ′(t)f ′′(t) + 4f(t)[f ′(t)]3 + 4f2(t)f ′(t)f ′′(t) = 0

39. y′ =32x1/2, y′′ =

34x−1/2;

κ =|y′′|

[1 + (y′)2]3/2=

34x

−1/2

[1 + 94x]3/2

=6√

x(4 + 9x)3/2

40. y′ = −2 sin 2x, y′′ = −4 cos 2x; κ =4| cos 2x|

[1 + 4 sin2 2x]3/2

41. x(t) = 2e−t, y(t) = e−2t =⇒ x′(t) = −2e−t, y′(t) = −2e−2t =⇒ x′′(t) = 2e−t, y′′(t) = 4e−2t

κ =|(−2e−t)(4e−2t) − (−2e−2t)(2e−t)|

[4e−2t + 4e−4t]3/2=

12(1 + e−2t)3/2

=e3t

2(e2t + 1)3/2




42. x(t) = 13 t

3, y(t) = 12 t

2 =⇒ x′(t) = t2, y′(t) = t =⇒ x′′(t) = 2t, y′′(t) = 1

κ =|t2 − 2t2|(t2 + t4)

32

=1

|t|(1 + t2)32

43. r′(t) = −3 sin 3t i − 4 j + 3 cos 3tk,ds

dt= |r′(t)| = 5

T(t) = −35

sin 3t i − 45

j +35

cos 3tk, T′(t) = −95

cos 3t i − 95

sin 3tk; ||T′(t)|| = 9/5


=925

44. r′(t) = 1 i +√

2t1/2 j + tk,ds

dt= ||r′(t)|| = 1 + t

T =1

1 + ti +

√2t

1 + tj +

t

1 + tk T′(t) =

−1(1 + t2)

i +

√2

2 (1/√t−

√t)

(1 + t)2j +

1(1 + t)2

k

||T′(t)|| =1√

2t(1 + t); κ =

||T′(t)||ds/dt

=1√

2t(1 + t)2

45. y′ = sinh (x/a), y′′ =1a

cosh (x/a); κ =|y′′|

[1 + (y′)2]32

=1

a cosh2(x/a)=

a

y2

46. (a) Suppose ||v|| = c constant. Then ||v||2 = c2 =⇒ r′ · r′ = c2 =⇒ 2r′r′′ = 0 =⇒ a ⊥ v.

(b) T =v

||v|| , T′ =v′

||v|| =a

||v|| since ||v|| is constant. κ =|T′||v| =

|a||v|2

47. r′(t) = −45

sin t i +35

sin t j + cos tk;ds

dt= ||r′(t)|| = 1

T = r′(t) = −45

sin t i +35

sin t j + cos tk, T′(t) = −45

cos t i +35

cos t j − sin tk; ||T′(t)|| = 1

κ = 1; aT =d2s

dt2= 0, aN = κ

(ds

dt

)2

= 1

48. r′(t) = 2 i + 2t j + t2 k;ds

dt= ||r′(t)|| = 2 + t2

T(t) =1

2 + t2(2 i + 2t j + t2 k), T′(t) =

1(2 + t2)2

(−4t i + (4 − 2t2) j + 4tk); ||T′(t)|| =2

2 + t2

κ =2

(2 + t2)2; aT =

d2s

dt2= 2t, aN = κ

(ds

dt

)2

= 2.

Calculus one and several variables 10E Salas solutions manual ch14

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Transcript of Calculus one and several variables 10E Salas solutions manual ch14