MA261-A Calculus III
2006 Fall
Homework 14 Solutions
Due 12/11/2006 8:00AM
13.1 #12 Match the vector …eld F (x; y) = h1; sin yi with the plot. Give reasons for your choices.
[Solution]
The vector …eld matches VI since the vector h1; sin yi sits on the line x = 1 and when
y changes sin y is between y = 1 and y = 1. These vectors looks like
y
2
1
-1.0 -0.8 -0.6 -0.4 -0.2
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
x
-1
-2
The vector …eld looks like
y
5
4
3
2
1
-5
-4
-3
-2
-1
1
-1
-2
-3
-4
-5
1
2
3
4
5
x
2
13.1 #16 Match the vector …eld F (x; y; z) = i + 2j + zk on R3 with the plot. Give reasons for your
choices.
[Solution]
The vector …eld matches I since the vector i+2j+zk sits on the line fx = 1 and y = 2g.
These vectors looks like the vectors in 13.1#12 but in R3 .
The vector …eld looks like
5
4
3
2
1
-5
-4
-3
z
-2
-1
0
4
-1
-2
02 0
-4
-2
1
y
2
3
4
5
x
-3
-4
-5
x
13.1 #22 Find the gradient vector …eld of f (x; y) = x e
[Solution]
The gradient vector …eld is
rf =
@f @f
;
@x @y
=
x
1
e
x
.
x e
x
;0 .
13.1 #26 Find the gradient vector …eld rf of f (x; y) = 14 (x + y)2 and sketch it.
[Solution]
The gradient vector …eld is
rf =
@f @f
;
@x @y
The gradient vector …eld looks like
=
1
1
(x + y) ; (x + y) .
2
2
3
y
5
4
3
2
1
-5
-4
-3
-2
-1
1
2
3
4
-1
5
x
-2
-3
-4
-5
R
13.2 #2 Evaluate the integral C xy ds, where C : x = t4 ; y = t3 ; 12 t 1.
[Solution]
The line integral is
Z
Z 1 p
Z 1 3 q
y
t
1
2
2
3
2
ds =
125
(4t ) + (3t ) dt =
t 16t2 + 9dt =
4
1
1
t
48
C x
2
2
p
13 13 .
R
13.2 #6 Evaluate the integral C sin xdx + cos ydy, where C consists of the top half of the circle
x2 + y 2 = 1 from (1; 0) to ( 1; 0) and the line segment from ( 1; 0) to ( 2; 3).
[Solution]
C is a piecewise-smooth curve consisted by C1 and C2 . C1 can be described as x = cos t
and y = sin t where 0 t
(because we only have the top half of the circle.) C2 can
be described as
r (t) = ( 1; 0) + t [( 2; 3)
( 1; 0)] = ( 1
where 0 t 1. Thus, the line integral is
Z
sin xdx + cos ydy
C
Z
Z
=
sin xdx + cos ydy +
sin xdx + cos ydy
C1
=
Z
0
=
Z
0
C2
Z
sin (cos t) ( sin tdt) + cos (sin t) (cos tdt) +
Z
[cos t cos (sin t) sin t sin (cos t)] dt + dt +
= 0 + (cos 1
t; 3t) ,
1
sin ( 1
t) ( dt) + cos 3t (3dt)
0
1
(sin (1 + t) + 3 cos 3t) dt
0
cos 2 + sin 3) = cos 1
13.2 #10 Evaluate the integral
[Solution]
R
C
cos 2 + sin 3.
(2x + 9z) ds, where C : x = t; y = t2 ; z = t3 ; 0
t
1.
4
The line integral is
Z
Z 1
Z
q
2
2
2
3
2
2t + 9t
(2x + 9z) ds =
(1) + (2t) + (3t ) dt =
C
0
7p
=
14
3
1
2t + 9t3
p
1 + 4t2 + 9t4 dt
0
1
.
6
R
13.2 #18 Evaluate the integral C F dr, where F (x; y; z) = yzi + xzj + xyk and C is given by
r (t) = ti + t2 j + t3 k, 0 t 2.
[Solution]
The line integral is
Z
Z
Z 2
t2 t3 (dt) + (t) t3 (2tdt) + (t) t2 3t2 dt
F dr =
P dx + Qdy + Rdz =
0
C
C
Z 2
=
6t5 dt = 64.
0
R
13.2 #24(a) Evaluate the integral C F dr, where F (x; y; z) = xi zj + yk and C is given by
r (t) = 2ti + 3tj t2 k, 1 t 1.
[Solution]
The line integral is
Z 1
Z
Z
(2t) (2dt) + t2 (3dt) + (3t) ( 2tdt)
P dx + Qdy + Rdz =
F dr =
C
1
C
Z
=
1
4t
3t2 dt =
2.
1
13.2 #34 Find the work done by the force …eld F (x; y) = x sin yi + yj on a particle that moves
along the parabola y = x2 from ( 1; 1) to (2; 4).
[Solution]
Since the curve is the parabola y = x2 from ( 1; 1) to (2; 4), we can parametrized it
as r (x) = (x; x2 ) where 1 x 2. Thus, the work done is
Z
Z 2
Z 2
0
W =
F Tds =
F (r (x)) r (x) dx =
x sin x2 ; x2 (1; 2x) dx
C
2
=
Z
1
x sin x2 + 2x3 dx =
1
1
1
(cos 1
2
cos 4) +
15
.
2
13.3 #4 Determine whether or not F (x; y) = (x3 + 4xy) i + (4xy y 3 ) j is a conservative vector
…eld. If it is, …nd a function f such that F = rf .
[Solution]
To see if F is a conservative …eld, we check if @P
= @Q
. Since 3x2 + 4y = @P
6= @Q
=
@y
@x
@y
@x
2
3x 3y , F is not a conservative …eld.
5
13.3 #8 Determine whether or not F (x; y) = (1 + 2xy + ln x) i + (x2 ) j is a conservative vector
…eld. If it is, …nd a function f such that F = rf .
[Solution]
To see if F is a conservative …eld, we check if @P
= @Q
. Since @P
= 2x = @Q
, F is a
@y
@x
@y
@x
conservative …eld.
= 1 + 2xy + ln x. Then,
To …nd a function f such that F = rf , we assume that @f
@x
f (x; y) = x + x2 y + x ln x
x + h (y) = x2 y + x ln x + h (y) .
By taking partial derivative of f with respect to y, we have x2 = @f
= x2 + h0 (y). This
@y
implies that h0 (y) = 0. Thus, h (y) = C. So, f (x; y) = x2 y + x ln x + C.
2
y
2
13.3 #14 Let F (x; y) = 1+x
t 1.
2 i + 2y arctan xj and C : r (t) = t i + 2tj, 0
(a) Find a function
f
such
that
F
=
rf
.
R
(b) Evaluate C F dr.
[Solution]
(a) Let f (x; y) = y 2 arctan x. Then, it is eacy to check that F = rf .
(b) Obviously, C is a smooth curve and f is of C 1 . So, by Fundamental Theorem of Line
Integrals, we have
Z
F dr = f (r (1)) f (r (0)) = f (1; 2) f (0; 0) = (2)2 arctan (1) (0)2 arctan (0) = .
C
13.3 #16 Let F (x; y; z) = (2xz + y 2 ) i+2xyj+(x2 + 3z 2 ) k and C : x = t2 ; y = t+1; z = 2t 1; 0
t 1.
(a) Find a function
f such that F = rf .
R
(b) Evaluate C F dr.
[Solution]
@g
= 2xz+y 2 . Then, we have f (x; y; z) = x2 z+xy 2 +g (y; z). So, 2xy+ @y
= @f
=
(a) Let @f
@x
@y
@g
2
2
2xy. This implies that @y = 0, or, g (y; z) = h (z). Thus, f (x; y; z) = x z+xy +h (z).
By taking partial derivative with respect to z, we have x2 + h0 (z) = @f
= x2 + 3z 2 .
@z
Therefore, h0 (z) = 3z 2 , or, h (z) = z 3 + k. Then, we have f (x; y; z) = x2 z + xy 2 + z 3
and F = rf .
(b) Obviously, C is a smooth curve and f is of C 1 . So, by Fundamental Theorem of Line
Integrals, we have
Z
F dr = f (r (1)) f (r (0)) = f (1; 2; 1) f (0; 1; 1)
C
= (1)2 (1) + (1) (2)2 + (1)3
= 7.
(0)2 ( 1) + (0) (1)2 + ( 1)3