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Thomas Calculus 13th [Solutions]

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138 Chapter 3 Derivatives<br />

74. (a) We use the Quotient rule to derive the Reciprocal Rule (with u 1): d 1<br />

v 0 1 1<br />

dx<br />

dx v 2<br />

v v<br />

(b) Now, using the Reciprocal Rule and the Product Rule, well derive the Quotient Rule:<br />

d u d<br />

u 1 u<br />

d 1 1 du<br />

(Product Rule) u 1 dv 1 du<br />

(Reciprocal Rule)<br />

2<br />

dx v dx v dx v v dx<br />

dv du du dv<br />

d u<br />

u v v u<br />

dx dx dx dx<br />

,<br />

dx v 2 2<br />

v v<br />

the Quotient Rule.<br />

75. (a)<br />

d<br />

( uvw) d<br />

(( uv) w ) ( uv )<br />

dw<br />

w d<br />

( uv ) uv dw w u dv v du uv<br />

dw<br />

wu<br />

dv<br />

wv<br />

du<br />

dx<br />

dx<br />

uvw uv w u vw<br />

dx<br />

dx<br />

v<br />

dx<br />

dx dx dx<br />

d d du d<br />

dx dx dx dx<br />

d<br />

du4 du3<br />

du2 du1<br />

dx 1 2 3 4 1 2 3 dx 4 1 2 dx 3 1 dx 3 2 dx<br />

4<br />

(b) ( u1u 2u3u4 ) u1u 2u3 u4 u1u 2u3 u4 u1u 2u3<br />

v dx<br />

dv<br />

dx dx dx<br />

u u u u u u u u u u u u u u (using (a) above)<br />

d<br />

du4 du3<br />

du2 du1<br />

u1u 2u3u4 u1u 2u3 u<br />

dx dx 1u2u4 u<br />

dx 1u3u4 u<br />

dx 2u3u4<br />

dx<br />

dv<br />

dx<br />

2 2<br />

v<br />

u1u 2u3u4 u1u 2u3u4 u1u 2u3u4 u1u 2u3u4<br />

(c) Generalizing (a) and (b) above,<br />

d<br />

( 1 )<br />

dx u u n u 1 u 2 u n 1 u n u 1 u 2 u n 2 u n 1 u n u 1 u 2 u n<br />

1 dv<br />

.<br />

dx<br />

m<br />

dx dx x<br />

76.<br />

d<br />

( x )<br />

d 1<br />

m<br />

m m 1<br />

x 0 1( m x ) m 1<br />

m x<br />

m 2<br />

x<br />

2m<br />

x<br />

( )<br />

m 1 2m m 1<br />

m x m x<br />

77.<br />

2<br />

nRT<br />

V nb<br />

an2 .<br />

V<br />

2 2<br />

dP ( V nb) 0 ( nRT )(1) V (0) ( an )(2 V ) 2<br />

nRT 2an<br />

dV<br />

2 2 2 2 3<br />

V nb V V nb V<br />

P We are holding T constant, and a, b, n, R are also constant so their derivatives are zero<br />

( ) ( ) ( )<br />

78.<br />

A( q)<br />

km<br />

q<br />

cm<br />

hq<br />

2<br />

1<br />

( km) q cm<br />

h<br />

q<br />

2<br />

dA<br />

dq<br />

2<br />

( km)<br />

q<br />

h km h<br />

2<br />

2<br />

q 2<br />

2<br />

d A<br />

3 2<br />

2<br />

2( km)<br />

q<br />

km<br />

3<br />

dt<br />

q<br />

3.4 THE DERIVATIVE AS A RATE OF CHANGE<br />

1.<br />

2.<br />

2<br />

s t 3t 2, 0 t 2<br />

(a) displacement s s(2) s(0) 0 m 2 m 2 m, v<br />

s 2<br />

av<br />

1 m/sec<br />

t 2<br />

2<br />

(b) v<br />

ds<br />

2t 3 | v (0)| | 3| 3 m/sec and | v(2)| 1 m/sec; a<br />

d s<br />

2 a(0)<br />

dt<br />

2<br />

dt<br />

2<br />

a(2) 2 m/sec<br />

(c) v 0 2t 3 0 t 3 . v is negative in the interval 0 t 3 and v is positive when 3 2<br />

2<br />

2<br />

body changes direction at t 3<br />

2 .<br />

2<br />

s 6 t t , 0 t 6<br />

(a) displacement s s(6) s(0) 0 m,<br />

s 0<br />

ds<br />

dt<br />

v av t<br />

6<br />

0 m/ sec<br />

(b) v 6 2t | v (0)| |6| 6 m/ sec and | v(6)| | 6| 6 m/ sec; a<br />

2<br />

2<br />

d s<br />

2<br />

dt<br />

2<br />

2 m/sec and<br />

t 2 the<br />

2 a(0) 2 m/ sec and<br />

a(6) 2 m/ sec<br />

(c) v 0 6 2t 0 t 3. v is positive in the interval 0 t 3 and v is negative when 3 t 6 the<br />

body changes direction at t 3.<br />

2<br />

Copyright<br />

2014 Pearson Education, Inc.

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