Thomas Calculus 13th [Solutions]

212 Chapter 3 Derivatives
147. Given dx
dy
2 2 2
10 m/sec and 5 m/sec, let D be the distance from the origin D x y
dt
dt
2D dD 2x dx 2y dy D dD dx dy
.
dt dt dt dt
x dt
y dt
When ( x , y ) (3, 4), 2 2
D 3 ( 4) 5 and
5 dD (3)(10) ( 4)(5) dD 10 2. Therefore, the particle is moving away from the origin at 2 m/sec
dt
dt 5
(because the distance D is increasing).
2
148. Let D be the distance from the origin. We are given that dD 11 units/sec. Then D
dt
2 3
x x 2D
dD
2
2 x dx 3 x dx x (2 3 x ) dx ;
2 3
x 3 3 3 6
dt dt dt
equation gives (2)(6)(11) (3)(2 9) dx
dt
dx
dt
2 2 2 3/2 2
x y x ( x )
dt
D and substitution in the derivative
4 units/sec.
149. (a) From the diagram we have 10 4 r 2 h
h r 5 .
2
2
(b) V 1 r h 1 2
3 3 5 h h 3
2
4 h dV 4 h dh , so dV
75 dt 25 dt dt
5 and h 6 dh 125
dt 144
ft/min.
150. From the sketch in the text, s r ds d dr . Also
dt
r dt dt
r 1.2 is constant dr
dt
Therefore, ds 6 ft/sec and r 1.2 ft d 5 rad/sec
dt
dt
0 ds
dt
r d
dt
(1.2) d .
dt
151. (a) From the sketch in the text, d 0.6 rad/sec and x tan . Also x tan dx 2
sec d ; at point A,
dt
dt dt
x 0 0 dx 2
(sec 0)( 0.6) 0.6. Therefore the speed of the light is 0.6 3 km/sec when it
dt
5
reaches point A.
(3/5) rad 1 rev
(b)
60sec 18 revs/min
sec 2 rad min
152. From the figure, a b a b . We are given
r BC r 2 2
b r
that r is constant. Differentiation gives,
1 da
r dt
2 2
b r b
db
( )
b db
dt
2 2 dt
b r
2 2
b
r
. Then, b 2r and
db
dt
0.3r
2 2
(2 r) r ( 0.3 r) (2 r)
da
dt r
2 2
(2 r ) r
2 r ( 0.3 r)
(2 r )
2
r
2
2
3 r ( 0.3 r)
3r
4 r
2
(0.3 r)
3r
2
increasing when OB 2 ,
2 2
(3 r )( 0.3 r) (4 r )(0.3 r)
0.3r
2
3 3r
3 3
r and B is moving toward O at the rate of 0.3r
m/sec.
r m/sec. Since da is positive, the distance OA is
10 3
dt
153. (a) If f ( x) tan x and x
4 , then f ( x)
2
sec x,
f 1 and f 2. The linearization of
4 4
f ( x ) is L( x) 2 x ( 1) 2 x 2 .
4
2
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