Thomas Calculus 13th [Solutions]

Chapter 13 Practice Exercises 963
i j k
5. v 3i 4j and a 5i 15j v a 3 4 0 25k v ×a 25;
2 2 v×a
v 3 4 5
25 1
3 3
v 5 5
5 15 0
= y e x 2
3 2
2
3 2
2
5 2
3
2
1 e x d x x x x x
1 1 2
dx
e e e 2
e e
1+ y
6.
2
3 2
x 2x 3 2
3x 2x 5 2
x 2x 5 2
2x 2x x 2x 5 2
2x
e 1 e 3e 1 e e 1 e 1 e 3e e 1 e 1 2 e ;
d
dx
2x
2x
0 1 2e 0 e 1 2x ln 2 x 1 ln 2 ln 2 y 1 ; therefore is at a
2 2 2
maximum at the point ln 2, 1
2
7.
dx dy
r xi yj v i j and v i y dx y.
Since the particle moves around the unit circle
dt dt
dt
2 2
dy dy dy
x y 1, 2x dx 2y 0 x dx x ( y) x . Since dx
dy
y and x,
dt dt dt y dt dt y
dt
dt
we have v yi xj at 1, 0 , v j and the motion is clockwise.
8.
9y 3 dy 2 dy 1 2
x 9 3 x dx x dx . If r
dt dt dt 3 dt
xi yj , where x and y are differentiable functions of t,
then dx dy
.
dt
dt
v i 4 dx
dt
4
dy 2 2
and v j 1 x dx 1 (3) (4)
dt 3 dt 3
12 at (3, 3).
Also, a
2 2
2
d x d y d y 2
2
i j and 2 1 2
2 2
2 2
dt dt
3 dx d x .
dt
dt 3
2
a i 2 d x
2
dt
dt
2 and
2
d y 2 2 1 2
a j (3)(4) (3) ( 2) 26 at the point ( x, y) (3, 3).
2
dt 3 3
9. dr orthogonal to r 1 1 1
dt
dr r dr r r dr
d
dt 2 dt 2 dt 2 dt
( r r ) r r K , a constant. If r xi yj , where x
and y are differentiable functions of t, then r r
2
x
2
y
2
x
2
y K, which is the equation of a circle
centered at the origin.
10. (a) (b) v cos t i sin t j
v(0)
2 2
a sin t i cos t j;
v(1) 2
v(2)
0 and a(0) 2
j;
i and a(1) 2
j;
0 and a(2) 2
j;
a(3)
2
j
v(3) 2 i ; and
(c) Forward speed at the topmost point is v(1) v (3) 2 ft/sec; since the circle makes 1 revolution per
2
second, the center moves ft parallel to the x -axis each second the forward speed of C is ft/sec.
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