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534
CHAPTER 13
Vector-Valued Functions
EXERCISE SET 13.
- (−∞, +∞); r(π) = −i − 3 πj 2. [− 1 / 3 , +∞); r(1) = 〈 2 , 1 〉
- [2, +∞); r(3) = −i − ln 3j + k 4. [− 1 , 1); r(0) = 〈 2 , 0 , 0 〉
- r = 3 cos ti + (t + sin t)j 6. r = (t 2 + 1)i + e − 2 t j
- r = 2ti + 2 sin 3tj + 5 cos 3tk 8. r = t sin ti + ln tj + cos 2 tk
- x = 3t 2 , y = − 2 10. x = sin 2 t, y = 1 − cos 2t
- x = 2t − 1, y = − 3
t, z = sin 3t 12. x = te −t , y = 0, z = − 5 t 2
- the line in 2-space through the point (2, 0) and parallel to the vector − 3 i − 4 j
- the circle of radius 3 in the xy-plane, with center at the origin
- the line in 3-space through the point (0, − 3 , 1) and parallel to the vector 2i + 3k
- the circle of radius 2 in the plane x = 3, with center at (3, 0 , 0)
- an ellipse in the plane z = −1, center at (0, 0 , −1), major axis of length 6 parallel to x-axis, minor
axis of length 4 parallel to y-axis
- a parabola in the plane x = −2, vertex at (− 2 , 0 , −1), opening upward
- (a) The line is parallel to the vector − 2 i + 3j;the slope is − 3 /2.
(b) y = 0 in the xz-plane so 1 − 2 t = 0, t = 1/2 thus x = 2 + 1/2 = 5/2 and z = 3(1/2) = 3/2; the coordinates are (5/ 2 , 0 , 3 /2).
- (a) x = 3 + 2t = 0, t = − 3 /2 so y = 5(− 3 /2) = − 15 / 2
(b) x = t, y = 1 + 2t, z = − 3 t so 3(t) − (1 + 2t) − (− 3 t) = 2, t = 3/4;the point of intersection is (3/ 4 , 5 / 2 , − 9 /4).
- (a)
x
y
(1, 0)
(0, 1)
(b)
x
y
(1, -1)
(1, 1)
Exercise Set 13.1 535
- (a)
y
x
z
(0, 0, 1)
(1, 1, 0)
(b)
y
x
z
(1, 1, 0)
(1, 1, 1)
- r = (1 − t)(3i + 4j), 0 ≤ t ≤ 1 24. r = (1 − t)4k + t(2i + 3j), 0 ≤ t ≤ 1
- x = 2
2
x
y
y = 2x + 10
10
x
y
- (x − 1) 2
1
3
x
y
- x 2 /4 + y 2 /25 = 1
2
5
x
y
- x 2 − y 2 = 1, x ≥ 1
1
2
x
y
- y = 2x 2
1
4
x
y
Exercise Set 13.1 537
- x 2
- y 2 = (t sin t) 2 + (t cos t) 2 = t 2 (sin 2 t + cos 2 t) = t 2 = z
- x − y + z + 1 = t − (1 + t)/t + (1 − t 2 )/t + 1 = [t 2 − (1 + t) + (1 − t 2 ) + t]/t = 0
- x = sin t, y = 2 cos t, z =
3 sin t so x 2
2 t + 4 cos 2 t + 3 sin
2 t = 4 and z =
3 x;it is the curve of intersection of the sphere x 2
- y 2
- z 2 = 4 and the plane z =
3 x, which is a circle with center at (0, 0 , 0) and radius 2.
- x = 3 cos t, y = 3 sin t, z = 3 sin t so x^2 + y^2 = 9 cos^2 t + 9 sin 2 t = 9 and z = y;it is the curve of intersection of the circular cylinder x^2 + y^2 = 9 and the plane z = y, which is an ellipse with major axis of length 6
2 and minor axis of length 6.
- The helix makes one turn as t varies from 0 to 2π so z = c(2π) = 3, c = 3/(2π).
- 2 t = 10, t = 50;the helix has made one revolution when t = 2π so when t = 50 it has made
50 /(2π) = 25/π ≈ 7 .96 revolutions.
- x^2 + y^2 = t^2 cos^2 t + t^2 sin 2 t = t^2 ,
x^2 + y^2 = t = z;a conical helix.
- The curve wraps around an elliptic cylinder with axis along the z-axis;an elliptical helix.
- (a) III, since the curve is a subset of the plane y = −x
(b) IV, since only x is periodic in t, and y, z increase without bound
(c) II, since all three components are periodic in t
(d) I, since the projection onto the yz-plane is a circle and the curve increases without bound in the x-direction
- (a) Let x = 3 cos t and y = 3 sin t, then z = 9 cos 2 t. (b) z
y
x
- The plane is parallel to a line on the surface of
the cone and does not go through the vertex so
the curve of intersection is a parabola. Eliminate
z to get y + 2 =
x^2 + y^2 , (y + 2)^2 = x^2 + y^2 ,
y = x^2 / 4 − 1;let x = t, then y = t^2 / 4 − 1
and z = t^2 /4 + 1.
z
y
x
538 Chapter 13
- (a)
-4 -2 2 4
1
2
x
y (b) In Part (a) set x = 2t;
then y = 2/(1 + (x/2) 2 ) = 8/(4 + x 2 )
EXERCISE SET 13.
- 9 i + 6j 2. 〈
- j 5. 2 i − 3 j + 4k 6. 〈 3 , 1 / 2 , sin 2〉
- (a) continuous, lim t→ 0
r(t) = 0 = r(0) (b) not continuous, lim t→ 0
r(t) does not exist
- (a) not continuous, lim t→ 0
r(t) does not exist. (b) continuous, lim t→ 0
r(t) = 5i − j + k = r(0)
-2 2
2
x
y
x
y
r '( p /4)
r ''( p )
r (2 p ) – r (3 p /2)
- r ′ (t) = 5i + (1 − 2 t)j 12. r ′ (t) = sin tj 13. r ′ (t) = −
t^2
i + sec
2 tj + 2e
2 t k
- r ′ (t) =
1 + t^2
i + (cos t − t sin t)j −
t
k
- r ′ (t) = 〈 1 , 2 t〉,
r ′ (2) = 〈 1 , 4 〉,
r(2) = 〈 2 , 4 〉
2
4
〈1, 4 〉
x
y
- r ′ (t) = 3t 2 i + 2tj,
r ′ (1) = 3i + 2j
r(1) = i + j
1 2 3 4
1
2
3
x
y
540 Chapter 13
- r ′ (t) =
t
i − e −t j + 3t 2 k, r ′ (2) =
i − e − 2 j + 12k,
r(2) = ln 2i + e − 2 j + 8k; x = ln 2 +
t, y = e − 2 − e − 2 t, z = 8 + 12t
- r ′ (t) = 2i +
3 t + 4
j, t = 0 at P 0 so r ′ (0) = 2i +
j,
r(0) = −i + 2j; r = (−i + 2j) + t
2 i +
j
- r ′ (t) = −4 sin ti − 3 j, t = π/3 at P 0 so r ′ (π/3) = − 2
3 i − 3 j,
r(π/3) = 2i − πj; r = (2i − πj) + t(− 2
3 i − 3 j)
- r ′ (t) = 2ti +
(t + 1)^2
j − 2 tk, t = −2 at P 0 so r ′ (−2) = − 4 i + j + 4k,
r(−2) = 4i + j; r = (4i + j) + t(− 4 i + j + 4k)
- r ′ (t) = cos ti + sinh tj +
1 + t^2
k, t = 0 at P 0 so r ′ (0) = i + k, r(0) = j; r = ti + j + tk
- (a) lim t→ 0
(r(t) − r
′ (t)) = i − j + k
(b) lim t→ 0
(r(t) × r ′ (t)) = lim t→ 0
(− cos ti − sin tj + k) = −i + k
(c) lim t→ 0
(r(t) · r ′ (t)) = 0
- r(t) · (r′(t) × r′′(t)) =
t t^2 t^3 1 2 t 3 t^2 0 2 6 t
= 2t^3 , so lim t→ 1
r(t) · (r ′ (t) × r ′′ (t)) = 2
- (a) r′ 1 = 2i + 6tj + 3t^2 k, r′ 2 = 4t^3 k, r 1 · r 2 = t^7 ;
d
dt
(r 1 · r 2 ) = 7t 6 = r 1 · r ′ 2 +^ r
′ 1 ·^ r^2
(b) r 1 × r 2 = 3t 6 i − 2 t 5 j,
d
dt
(r 1 × r 2 ) = 18t 5 i − 10 t 4 j = r 1 × r ′ 2 +^ r
′ 1 ×^ r^2
- (a) r′ 1 = − sin ti+cos tj+k, r′ 2 = k, r 1 · r 2 = cos t+t^2 ;
d
dt
(r 1 · r 2 ) = − sin t+2t = r 1 · r ′ 2 +r
′ 1 ·^ r^2
(b) r 1 × r 2 = t sin ti + t(1 − cos t)j − sin tk,
d
dt
(r 1 × r 2 ) = (sin t + t cos t)i + (1 + t sin t − cos t)j − cos tk = r 1 × r ′ 2 +^ r
′ 1 ×^ r^2
- 3 ti + 2t 2 j + C 36. (sin t)i − (cos t)j + C
- (−t cos t + sin t)i + tj + C 38. 〈(t − 1)et, t(ln t − 1)〉 + C
- (t 3 /3)i − t 2 j + ln |t|k + C 40. 〈−e −t , e t , t 3 〉 + C
sin 3t,
cos 3t
〉]π/ 3
0
t
3 i +
t
4 j
)] 1
0
i +
j
Exercise Set 13.2 541
0
t^2 + t^4 dt =
0
t(1 + t 2 ) 1 / 2 dt =
1 + t 2
] 2
0
(3 − t) 5 / 2 ,
(3 + t) 5 / 2 , t
〉] 3
− 3
t 3 / 2 i + 2t 1 / 2 j
)] 9
1
i + 4j 46.
(e 2 − 1)i + (1 − e − 1 )j +
k
- y(t) =
y ′ (t)dt = 1 3 t
3 i + t 2 j + C, y(0) = C = i + j, y(t) = ( 1 3 t
3
- y(t) =
y ′ (t)dt = (sin t)i − (cos t)j + C,
y(0) = −j + C = i − j so C = i and y(t) = (1 + sin t)i − (cos t)j.
- y ′ (t) =
y ′′ (t)dt = ti + e t j + C 1 , y ′ (0) = j + C 1 = j so C 1 = 0 and y ′ (t) = ti + e t j.
y(t) =
y ′ (t)dt =
t 2 i + e t j + C 2 , y(0) = j + C 2 = 2i so C 2 = 2i − j and
y(t) =
t 2
i + (e t − 1)j
- y ′ (t) =
y ′′ (t)dt = 4t 3 i − t 2 j + C 1 , y′(0) = C 1 = 0 , y′(t) = 4t^3 i − t^2 j
y(t) =
y ′ (t)dt = t 4 i −
t 3 j + C 2 , y(0) = C 2 = 2i − 4 j, y(t) = (t^4 + 2)i − (
t 3
- r′(t) = −4 sin ti + 3 cos tj, r(t) · r′(t) = −7 cos t sin t, so r and r′^ are perpendicular for
t = 0, π/ 2 , π, 3 π/ 2 , 2 π. Since
‖r(t)‖ =
16 cos^2 t + 9 sin 2 t, ‖r′(t)‖ =
16 sin 2 t + 9 cos^2 t,
‖r‖‖r′‖ =
144 + 337 sin 2 t cos^2 t, θ = cos−^1
[
−7 sin t cos t √ 144 + 337 sin 2 t cos^2 t
]
, with the graph
0 o 0
3
From the graph it appears that θ is bounded away from 0 and π, meaning that r and r ′ are never parallel. We can check this by considering them as vectors in 3-space, and then r × r ′ = 12 k = 0 , so they are never parallel.
Exercise Set 13.3 543
- Let c = c 1 i + c 2 j, r(t) = x(t)i + y(t)j, r 1 (t) = x 1 (t)i + y 1 (t)j, r 2 (t) = x 2 (t)i + y 2 (t)j and use
properties of derivatives.
- Let r 1 (t) = x 1 (t)i + y 1 (t)j + z 1 (t)k and r 2 (t) = x 2 (t)i + y 2 (t)j + z 2 (t)k, in both (6) and (7);show
that the left and right members of the equalities are the same.
- (a)
kr(t) dt =
k(x(t)i + y(t)j + z(t)k) dt
= k
x(t) dt i + k
y(t) dt j + k
z(t) dt k = k
r(t) dt
(b) Similar to Part (a) (c) Use Part (a) on Part (b) with k = − 1
EXERCISE SET 13.
- (a) The tangent vector reverses direction at the four cusps.
(b) r ′ (t) = −3 cos 2 t sin ti + 3 sin 2 t cos tj = 0 when t = 0, π/ 2 , π, 3 π/ 2 , 2 π.
- r′(t) = cos ti + 2 sin t cos tj = 0 when t = π/ 2 , 3 π/2. The tangent vector reverses direction at (1, 1) and (− 1 , 1).
- r ′ (t) = 3t 2 i + (6t − 2)j + 2tk;smooth
- r′(t) = − 2 t sin(t^2 )i + 2t cos(t^2 )j − e−tk;smooth
- r ′ (t) = (1 − t)e −t i + (2t − 2)j − π sin(πt)k;not smooth, r ′ (1) = 0
- r ′ (t) = π cos(πt)i + (2 − 1 /t)j + (2t − 1)k;not smooth, r ′ (1/2) = 0
- (dx/dt) 2 + (dy/dt) 2 + (dz/dt) 2 = (−3 cos 2 t sin t) 2 + (3 sin
2 t cos t) 2
2 t cos 2 t,
L =
∫ (^) π/ 2
0
3 sin t cos t dt = 3/ 2
- (dx/dt) 2 + (dy/dt) 2 + (dz/dt) 2 = (−3 sin t) 2 + (3 cos t) 2 + 16 = 25, L =
∫ (^) π
0
5 dt = 5π
- r′(t) = 〈et, −e−t,
2 〉, ‖r′(t)‖ = et^ + e−t, L =
0
(e t
- (dx/dt)^2 + (dy/dt)^2 + (dz/dt)^2 = 1/4 + (1 − t)/4 + (1 + t)/4 = 3/4, L =
− 1
3 /2)dt =
- r′(t) = 3t^2 i + j +
6 tk, ‖r′(t)‖ = 3t^2 + 1, L =
1
(3t 2
- r ′ (t) = 3i − 2 j + k, ‖r ′ (t)‖ =
14, L =
3
14 dt =
- r ′ (t) = −3 sin ti + 3 cos tj + k, ‖r ′ (t)‖ =
10, L =
∫ (^2) π
0
10 dt = 2π
544 Chapter 13
- r ′ (t) = 2ti + t cos tj + t sin tk, ‖r ′ (t)‖ =
5 t, L =
∫ (^) π
0
5 t dt = π 2
- (dr/dt)(dt/dτ ) = (i + 2tj)(4) = 4i + 8tj = 4i + 8(4τ + 1)j;
r(τ ) = (4τ + 1)i + (4τ + 1)^2 j, r′(τ ) = 4i + 2(4)(4τ + 1) j
- (dr/dt)(dt/dτ ) = 〈−3 sin t, 3 cos t〉(π) = 〈− 3 π sin πτ , 3π cos πτ 〉;
r(τ ) = 〈3 cos πτ, 3 sin πτ 〉, r′(τ ) = 〈− 3 π sin πτ, 3 π cos πτ 〉
- (dr/dt)(dt/dτ ) = (e t i − 4 e −t j)(2τ ) = 2τ e τ 2 i − 8 τ e −τ 2 j;
r(τ ) = e τ 2 i + 4e −τ 2 j, r ′ (τ ) = 2τ e τ 2 i − 4(2)τ e −τ 2 j
- (dr/dt)(dt/dτ ) =
t 1 / 2 j + k
(− 1 /τ 2 ) = −
2 τ 5 /^2
j −
τ 2
k;
r(τ ) = i + 3τ − 3 / 2 j +
τ
k, r ′ (τ ) = −
τ − 5 / 2 j −
τ 2
k
- (a) ‖r′(t)‖ =
2 , s =
∫ (^) t
0
2 dt =
2 t; r =
s √ 2
i +
s √ 2
j, x =
s √ 2
, y =
s √ 2
(b) Similar to Part (a), x = y = z =
s √ 3
- (a) x = −
s √ 2
, y = −
s √ 2
(b) x = −
s √ 3
, y = −
s √ 3
, z = −
s √ 3
- (a) r(t) = 〈 1 , 3 , 4 〉 when t = 0,
so s =
∫ (^) t
0
1 + 4 + 4 du = 3t, x = 1 + s/ 3 , y = 3 − 2 s/ 3 , z = 4 + 2s/ 3
(b) r
]
s=
- (a) r(t) = 〈− 5 , 0 , 1 〉 when t = 0, so s =
∫ (^) t
0
9 + 4 + 1 du =
14 t,
x = −5 + 3s/
14 , y = 2s/
14 , z = 5 + s/
(b) r(s)
]
s=
- x = 3 + cos t, y = 2 + sin t, (dx/dt) 2
s =
∫ (^) t
0
du = t so t = s, x = 3 + cos s, y = 2 + sin s for 0 ≤ s ≤ 2 π.
- x = cos^3 t, y = sin 3 t, (dx/dt)^2 + (dy/dt)^2 = 9 sin 2 t cos^2 t,
s =
∫ (^) t
0
3 sin u cos u du =
sin 2 t so sin t = (2s/3)^1 /^2 , cos t = (1 − 2 s/3)^1 /^2 ,
x = (1 − 2 s/3)^3 /^2 , y = (2s/3)^3 /^2 for 0 ≤ s ≤ 3 / 2
546 Chapter 13
- (a) (dr/dt) 2
- r 2 (dθ/dt) 2 + (dz/dt) 2 = 9e 4 t , L =
∫ (^) ln 2
0
3 e 2 t dt =
e 2 t
]ln 2
0
(b) (dr/dt) 2
- r 2 (dθ/dt) 2
- (dz/dt) 2 = 5t 2
- t 4 = t 2 (5 + t 2 ),
L =
1
t(5 + t 2 ) 1 / 2 dt = 9 − 2
dx
dt
= sin φ cos θ
dρ
dt
dφ
dt
− ρ sin φ sin θ
dθ
dt
dy
dt
= sin φ sin θ
dρ
dt
dφ
dt
dθ
dt
dz
dt
= cos φ
dρ
dt
− ρ sin φ
dφ
dt
dx
dt
dy
dt
dz
dt
dρ
dt
dθ
dt
dφ
dt
- (a) (dρ/dt)^2 + ρ^2 sin 2 φ(dθ/dt)^2 + ρ^2 (dφ/dt)^2 = 3e−^2 t, L =
0
3 e −t dt =
3(1 − e − 2 )
(b) (dρ/dt) 2
- ρ 2 sin 2 φ(dθ/dt) 2
- ρ 2 (dφ/dt) 2 = 5, L =
1
5 dt = 4
- (a)
d
dt
r(t) = i + 2tj is never zero, but
d
dτ
r(τ
3 ) =
d
dτ
(τ
3 i + τ
6 j) = 3τ
2 i + 6τ
5 j is zero at τ = 0.
(b)
dr
dτ
dr
dt
dt
dτ
, and since t = τ 3 ,
dt
dτ
= 0 when τ = 0.
- (a) g(τ ) = πτ (b) g(τ ) = π(1 − τ ) 38. t = 1 − τ
- Represent the helix by x = a cos t, y = a sin t, z = ct with a = 6.25 and c = 10/π, so that the
radius of the helix is the distance from the axis of the cylinder to the center of the copper cable, and the helix makes one turn in a distance of 20 in. (t = 2π). From Exercise 29 the length of the
helix is 2π
252 + (10/π)^2 ≈ 44 in.
r(t) = cos ti + sin tj + t^3 /^2 k, r ′ (t) = − sin ti + cos tj +
t 1 / 2 k
(a) ‖r′(t)‖ =
sin 2 t + cos^2 t + 9t/4 =
4 + 9t
(b)
ds
dt
4 + 9t (c)
0
4 + 9t dt =
- r′(t) = (1/t)i + 2j + 2tk
(a) ‖r′(t)‖ =
1 /t^2 + 4 + 4t^2 =
(2t + 1/t)^2 = 2t + 1/t
(b)
ds
dt
= 2t + 1/t (c)
1
(2t + 1/t)dt = 8 + ln 3
- If r(t) = x(t)i + y(t)j + z(t)k is smooth, then ‖r ′ (t)‖ is continuous and nonzero. Thus the angle between r ′ (t) and i, given by cos − 1 (x ′ (t)/‖r ′ (t)‖), is a continuous function of t. Similarly, the angles between r ′ (t) and the vectors j and k are continuous functions of t.
- Let r(t) = x(t)i + y(t)j and use the chain rule.
Exercise Set 13.4 547
EXERCISE SET 13.
- (a)
x
y (b)
x
y
x
y
- r ′ (t) = 2ti + j, ‖r ′ (t)‖ =
4 t^2 + 1, T(t) = (4t 2
− 1 / 2 (2ti + j),
T ′ (t) = (4t 2
− 1 / 2 (2i) − 4 t(4t 2
− 3 / 2 (2ti + j);
T(1) =
i +
j, T ′ (1) =
(i − 2 j), N(1) =
i −
j.
- r′(t) = ti + t^2 j, T(t) = (t^2 + t^4 )−^1 /^2 (ti + t^2 j),
T′(t) = (t^2 + t^4 )−^1 /^2 (i + 2tj) − (t + 2t^3 )(t^2 + t^4 )−^3 /^2 (ti + t^2 j);
T(1) =
i +
j, T ′ (1) =
(−i + j), N(1) = −
i +
j
- r ′ (t) = −5 sin ti + 5 cos tj, ‖r ′ (t)‖ = 5, T(t) = − sin ti + cos tj, T ′ (t) = − cos ti − sin tj;
T(π/3) = −
i +
j, T ′ (π/3) = −
i −
j, N(π/3) = −
i −
j
- r
′ (t) =
t
i + j, ‖r
′ (t)‖ =
1 + t^2
t
, T(t) = (1 + t 2 ) − 1 / 2 (i + tj),
T
′ (t) = (1 + t 2 ) − 1 / 2 (j) − t(1 + t 2 ) − 3 / 2 (i + tj); T(e) =
1 + e^2
i +
e √ 1 + e^2
j,
T
′ (e) =
(1 + e^2 )^3 /^2
(−ei + j), N(e) = −
e √ 1 + e^2
i +
1 + e^2
j
- r ′ (t) = −4 sin ti + 4 cos tj + k, T(t) =
(−4 sin ti + 4 cos tj + k),
T
′ (t) =
(−4 cos ti − 4 sin tj), T(π/2) = −
i +
k
T
′ (π/2) = −
j, N(π/2) = −j
- r′(t) = i + tj + t^2 k, T(t) = (1 + t^2 + t^4 )−^1 /^2 (i + tj + t^2 k),
T′(t) = (1 + t^2 + t^4 )−^1 /^2 (j + 2tk) − (t + 2t^3 )(1 + t^2 + t^4 )−^3 /^2 (i + tj + t^2 k),
T(0) = i, T′(0) = j = N(0)
Exercise Set 13.5 549
- r(π/4) =
i +
j + k, T = − sin ti + cos tj =
(−i + j), N = −(cos ti + sin tj) = −
(i + j),
B = k;the rectifying, osculating, and normal planes are given (respectively) by x + y =
z = 1, −x + y = 0.
- r(0) = i + j, T =
(i + j + k), N =
(−j + k), B =
(2i − j − k);the rectifying, osculating,
and normal planes are given (respectively) by −y + z = − 1 , 2 x − y − z = 1, x + y + z = 2.
- (a) By formulae (1) and (11), N(t) = B(t) × T(t) =
r′(t) × r′′(t)
‖r′(t) × r′′(t)‖
×
r′(t)
‖r′(t)‖
(b) Since r ′ is perpendicular to r ′ × r ′′ it follows from Lagrange’s Identity (Exercise 32 of Section 12.4) that ‖(r ′ (t) × r ′′ (t)) × r ′ (t)‖ = ‖r ′ (t) × r ′′ (t)‖‖r ′ (t)‖, and the result follows.
(c) From Exercise 39 of Section 12.4, (r ′ (t) × r ′′ (t)) × r ′ (t) = ‖r ′ (t)‖ 2 r ′′ (t) − (r ′ (t) · r ′′ (t))r ′ (t) = u(t), so N(t) = u(t)/‖u(t)‖
- (a) r′(t) = 2ti + j, r′(1) = 2i + j, r′′(t) = 2i, u = 2i − 4 j, N =
i −
j
(b) r′(t) = −4 sin ti + 4 cos tj + k, r′(
π
2
) = − 4 i + k, r ′′ (t) = −4 cos ti − 4 sin tj,
r ′′ (
π
2
) = − 4 j, u = 17(− 4 j), N = −j
- r ′ (t) = cos ti−sin tj+k, r ′′ (t) = − sin ti−cos tj, u = −2(sin ti+cos tj), ‖u‖ = 2, N = − sin ti−cos tj
- r′(t) = i + 2tj + 3t^2 k, r′′(t) = 2j + 6tk, u(t) = −(4t + 18t^3 )i + (2 − 18 t^4 )j + (6t + 12t^3 )k,
N =
81 t^8 + 117t^6 + 54t^4 + 13t^2 + 1
−(4t + 18t 3 )i + (2 − 18 t 4 )j + (6t + 12t 3 )k
EXERCISE SET 13.
- κ ≈
= 2 2. κ ≈
- r′(t) = 2ti + 3t^2 j, r′′(t) = 2i + 6tj, κ = ‖r′(t) × r′′(t)‖/‖r′(t)‖^3 =
t(4 + 9t^2 )^3 /^2
- r ′ (t) = −4 sin ti+cos tj, r ′′ (t) = −4 cos ti−sin tj, κ = ‖r ′ (t)×r ′′ (t)‖/‖r ′ (t)‖ 3 =
(16 sin 2 t + cos^2 t)^3 /^2
- r ′ (t) = 3e 3 t i − e −t j, r ′′ (t) = 9e 3 t i + e −t j, κ = ‖r ′ (t) × r ′′ (t)‖/‖r ′ (t)‖ 3 =
12 e 2 t
(9e^6 t^ + e−^2 t)
3 / 2
- r′(t) = − 3 t^2 i + (1 − 2 t)j, r′′(t) = − 6 ti − 2 j, κ = ‖r′(t) × r′′(t)‖/‖r′(t)‖^3 =
6 |t^2 − t|
(9t^4 + 4t^2 − 4 t + 1)^3 /^2
- r′(t) = −4 sin ti + 4 cos tj + k, r′′(t) = −4 cos ti − 4 sin tj,
κ = ‖r ′ (t) × r ′′ (t)‖/‖r ′ (t)‖ 3 = 4/ 17
- r′(t) = i + tj + t^2 k, r′′(t) = j + 2tk, κ = ‖r′(t) × r′′(t)‖/‖r′(t)‖^3 =
t^4 + 4t^2 + 1
(t^4 + t^2 + 1)^3 /^2
550 Chapter 13
- r ′ (t) = sinh ti + cosh tj + k, r ′′ (t) = cosh ti + sinh tj, κ = ‖r ′ (t) × r ′′ (t)‖/‖r ′ (t)‖ 3 =
2 cosh
2 t
- r ′ (t) = j + 2tk, r ′′ (t) = 2k, κ = ‖r ′ (t) × r ′′ (t)‖/‖r ′ (t)‖ 3 =
(4t^2 + 1)^3 /^2
- r ′ (t) = −3 sin ti + 4 cos tj + k, r ′′ (t) = −3 cos ti − 4 sin tj,
r ′ (π/2) = − 3 i + k, r ′′ (π/2) = − 4 j; κ = ‖ 4 i + 12k‖/‖ − 3 i + k‖ 3 = 2/ 5 , ρ = 5/ 2
- r ′ (t) = e t i − e −t j + k, r ′′ (t) = e t i + e −t j,
r ′ (0) = i − j + k, r ′′ (0) = i + j; κ = ‖ − i + j + 2k‖/‖i − j + k‖ 3 =
2 / 3 , ρ = 3/
- r ′ (t) = e t (cos t − sin t)i + e t (cos t + sin t)j + e t k,
r ′′ (t) = − 2 e t sin ti + 2e t cos tj + e t k, r ′ (0) = i + j + k,
r ′′ (0) = 2j + k; κ = ‖ − i − j + 2k‖/‖i + j + k‖
3
2 / 3 , ρ = 3
- r ′ (t) = cos ti − sin tj + tk, r ′′ (t) = − sin ti − cos tj + k,
r′(0) = i, r′′(0) = −j + k; κ = ‖ − j − k‖/‖i‖^3 =
2 , ρ =
- r ′ (s) =
cos
s
2
i −
sin
s
2
j +
k, ‖r ′ (s)‖ = 1, so
dT
ds
sin
s
2
i −
cos
s
2
j, κ =
dT
ds
- r′(s) = −
3 − 2 s
3
si +
2 s
3
j, ‖r′(s)‖ = 1, so
dT
ds
9 − 6 s
i +
6 s
j, κ =
dT
ds
9 − 6 s
6 s
2 s(9 − 6 s)
- (a) r ′ = x ′ i + y ′ j, r ′′ = x ′′ i + y ′′ j, ‖r ′ × r ′′ ‖ = |x ′ y ′′ − x ′′ y ′ |, κ =
|x′y′′^ − y′x′′|
(x′^2 + y′^2 )^3 /^2
(b) Set x = t, y = f (x) = f (t), x′^ = 1, x′′^ = 0, y ′ =
dy
dx
, y ′′ =
d 2 y
dx^2
, κ =
|d 2 y/dx 2 |
(1 + (dy/dx)^2 )^3 /^2
dy
dx
= tan φ, (1 + tan 2 φ) 3 / 2 = (sec 2 φ) 3 / 2 = | sec φ| 3 , κ(x) =
|y′′|
| sec φ|^3
= |y ′′ cos 3 φ|
- κ(x) =
| sin x|
(1 + cos^2 x)^3 /^2
, κ(π/2) = 1 20. κ(x) =
2 |x|
(1 + x^4 )^3 /^2
, κ(0) = 0
- κ(x) =
2 |x|^3
(x^4 + 1)^3 /^2
, κ(1) = 1/
2 22. κ(x) =
e−x
(1 + e−^2 x)^3 /^2
, κ(1) =
e−^1
(1 + e−^2 )^3 /^2
- κ(x) =
2 sec 2 x| tan x|
(1 + sec^4 x)^3 /^2
, κ(π/4) = 4/(
- By implicit differentiation, dy/dx = 4x/y, d 2 y/dx 2 = 36/y 3 so κ =
36 /|y| 3
(1 + 16x^2 /y^2 )^3 /^2
if (x, y) = (2, 5) then κ =
(1 + 64/25)^3 /^2
552 Chapter 13
- (a) II takes the value zero at x = 0, yet the curvature of I is large there;hence I is the curvature
of II.
(b) I has constant zero curvature;II has constant, positive curvature;hence I is the curvature of II.
- (a)
(^0 ) 0
(^1) (b) 1
0
0 5
(a) 4
-1 1
(b) 4
-1 1
- (a) κ =
| 12 x 2 − 4 |
(1 + (4x^3 − 4 x)^2 )
3 / 2
(b)
f ( x )
k
-2 2
8
x
y
(c) f ′(x) = 4x^3 − 4 x = 0 at x = 0, ± 1 , f ′′(x) = 12x^2 − 4, so extrema at x = 0, ±1, and ρ = 1/ 4 for x = 0 and ρ = 1/8 when x = ±1.
- (a)
-30 30
30
x
y (c) κ(t) =
t 2
(t^2 + 1)^3 /^2
(d) lim t→+∞
κ(t) = 0
Exercise Set 13.5 553
- r ′ (θ) =
−r sin θ + cos θ
dr
dθ
i +
r cos θ + sin θ
dr
dθ
j;
r
′′ (θ) =
−r cos θ − 2 sin θ
dr
dθ
d 2 r
dθ^2
i +
−r sin θ + 2 cos θ
dr
dθ
d 2 r
dθ^2
j;
κ =
r^2 + 2
dr
dθ
− r
d 2 r
dθ^2
∣ ∣ ∣ ∣ ∣ [ r
2
dr
dθ
) 2 ]^3 /^2
- Let r = a be the circle, so that dr/dθ = 0, and κ(θ) =
r
a
- κ(θ) =
2(1 + cos θ)^1 /^2
, κ(π/2) =
- κ(θ) =
5 e^2 θ^
, κ(1) =
5 e^2
- κ(θ) =
10 + 8 cos 2 3 θ
(1 + 8 cos^2 θ)^3 /^2
, κ(0) =
- κ(θ) =
θ 2
(θ^2 + 1)^3 /^2
, κ(1) =
- The radius of curvature is zero when θ = π, so there is a cusp there.
dr
dθ
= − sin θ,
d^2 r
dθ^2
= − cos θ, κ(θ) =
23 /^2
1 + cos θ
- Let y = t, then x =
t^2
4 p
and κ(t) =
1 /| 2 p|
[t^2 /(4p^2 ) + 1]^3 /^2
t = 0 when (x, y) = (0, 0) so κ(0) = 1/| 2 p|, ρ = 2|p|.
- κ(x) =
e x
(1 + e^2 x)^3 /^2
, κ ′ (x) =
e x (1 − 2 e 2 x )
(1 + e^2 x)^5 /^2
; κ′(x) = 0 when e^2 x^ = 1/2, x = −(ln 2)/2. By the first
derivative test, κ(−
ln 2) is maximum so the point is (−
ln 2, 1 /
- Let x = 3 cos t, y = 2 sin t for 0 ≤ t < 2 π, κ(t) =
(9 sin 2 t + 4 cos^2 t)^3 /^2
so
ρ(t) =
(9 sin 2 t + 4 cos 2 t) 3 / 2 =
(5 sin 2 t + 4) 3 / 2 which, by inspection, is minimum when
t = 0 or π. The radius of curvature is minimum at (3, 0) and (− 3 , 0).
- κ(x) =
6 x
(1 + 9x^4 )^3 /^2
for x > 0, κ ′ (x) =
6(1 − 45 x 4 )
(1 + 9x^4 )^5 /^2
; κ ′ (x) = 0 when x = 45 − 1 / 4 which, by the
first derivative test, yields the maximum.
- r ′ (t) = − sin ti + cos tj − sin tk, r ′′ (t) = − cos ti − sin tj − cos tk,
‖r ′ (t) × r ′′ (t)‖ = ‖ − i + k‖ =
2, ‖r ′ (t)‖ = (1 + sin 2 t) 1 / 2 ; κ(t) =
2 /(1 + sin 2 t) 3 / 2 ,
ρ(t) = (1 + sin 2 t) 3 / 2 /
- The minimum value of ρ is 1/
2;the maximum value is 2.
