Thomas Calculus 13th [Solutions]

332 Chapter 4 Applications of Derivatives
r
r
118. (2 ) dr 2
ln 2 C 119. 3 1
dx 3 1 dx 3 sec x C
2 2 2
2x x 1 x x 1
2
120.
1
d
4
2 2
2
16 1
16
16
d d 1
sin
16
4
C
1
121.
122.
2
2
y x 1 dx (1 x )
2
x
y x 1 1
x
1
dx x x C x 1 C ; y 1 when 1
x
x 1 1 C 1 C 1
1
2
1
2
y x dx 1
2 2
3
( x 2 ) dx ( 2 ) x
1
3
x x dx 2x
x C x 2 x 1 C;
x
2
x
3
3 x
3
y 1 when x 1 1 2 1 C 1 C 1 y x 2x
1 1
3 1
3 3 x 3
123. dr 15 3
1/2 1/2 3/2 1/2
t dt (15t 3 t ) dt 10 6 ; dr
3/2 1/2
t t C 8 when t 1 10(1) 6(1) C 8
dt
t
dt
3/2 1/2
C 8. Thus dr
3/2 1/2
5/2 3/2
10t
6t
8 r (10t 6t 8) dt 4t 4t 8 t C; r 0 when t 1
dt
5/2 3/2
5/2 3/2
4(1) 4(1) 8(1) C1 0 C 1 0. Therefore, r 4t 4t 8t
2
124. d r cos t dt sin t C; r 0 when t 0 sin 0 C 0 C 0. Thus, d r sin t
2 2
dt
dt
dr sin t dt cos t C
dt
1; r 0 when t 0 1 C1 0 C 1 1. Then
dr cos t 1 r (cost 1) dt sin t t
dt
C 2 ; r 1 when t 0 0 0 C2
1 C 2 1. Therefore,
r sin t t 1
2
125. Yes,
sin
1
x and
1
cos x differ by the constant 2
126. Yes, the derivatives of
1
y cos x C and
1
y cos ( x)
C are both
1
1
x
2
.
2 2 2 2
x x x x 2
127. A xy xe dA e ( x)( 2 x) e e (1 2 x ). Solving dA
2
0 1 2x 0 x 1 ; dA 0
dx
dx
2 dx
for x 1 and dA 0 for 0 x 1 absolute maximum of 1 1/2
e 1 at x 1 units long by
2 dx
2
2 2e 2
1/2
y e 1 units high.
e
A xy x ln x ln x dA 1 ln x 1 ln x .
x x dx
Solving dA 0 1 ln x 0 x e ; dA 0
x x x
dx
dx
dA 0 for x < e absolute maximum of ln e 1
dx
e e at x = e units long and y 1 units high.
2
e
128.
2 2 2 2
for x > e and
Copyright
2014 Pearson Education, Inc.