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Thomas Calculus 13th [Solutions]

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Chapter 7 Additional and Advanced Exercises 541<br />

6. (a)<br />

g g(0) g(0) 2 3 3<br />

(0 0) 1 (0) (0) 2 (0) (0) (0) 2 (0) (0) (0) 0<br />

1 g(0) g(0)<br />

g g g g g g g g<br />

2<br />

g(0) g (0) 1 0 g(0) 0<br />

(b)<br />

2<br />

g( x) g ( h)<br />

1 g( x) g ( h)<br />

( )<br />

2<br />

g( x h) g( x) g x<br />

g( x) g( h) g( x) g ( x) g( h)<br />

g ( x) lim lim lim<br />

h 0<br />

h<br />

h 0<br />

h<br />

h 0 h 1 g( x) g( h)<br />

g( h) 1 g ( x) 2 2<br />

2<br />

lim 1 1 g ( x) 1 g ( x) 1 g( x)<br />

h 0<br />

h 1 g( x) g( h)<br />

dy 2 dy<br />

1 1 1<br />

(c) 1 y dx tan y x C tan g( x) x C; g(0) 0 tan 0 0 C<br />

dx<br />

2<br />

1 y<br />

1<br />

C 0 tan g( x) x g( x) tan x<br />

1 1<br />

1<br />

7. M 2 dx 2 tan x and<br />

0<br />

2<br />

1 x<br />

0 2<br />

M 1 1<br />

2x<br />

2<br />

M<br />

ln 1 ln 2<br />

y ln 2 ln 4<br />

y<br />

;<br />

0 dx x x y<br />

2<br />

1 x<br />

0<br />

M<br />

0 by<br />

symmetry<br />

2<br />

8. (a)<br />

(b)<br />

V 4 2<br />

1<br />

4<br />

1 4 1<br />

4<br />

1/4 dx ln | | ln 4 ln ln16 ln 2 ln 2<br />

2<br />

4 1/4 x<br />

dx 4 x<br />

x<br />

1/4 4 4 4 4<br />

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

4<br />

M 1 1 1 8 1 64 1 63<br />

y x dx x dx x<br />

;<br />

1/4 2 x 2 1/4<br />

3 1/4 3 24 24 24<br />

M 4<br />

1 1 1 1<br />

4<br />

1 1<br />

4<br />

1 1<br />

x<br />

1/4 2 dx ln | | ln16 ln 2;<br />

2 2 8 1/4 x<br />

dx x x<br />

8 x 1/4 8 2<br />

M 4 4 1/2 1/2<br />

4<br />

1 1 1 3<br />

1/4 dx 2 1/4 2 x dx x 2 ;<br />

x<br />

1/4 2 2<br />

therefore, M<br />

x<br />

y 63 2 21 7 and<br />

M 24 3 12 4<br />

M<br />

y x 1 ln 2 2 ln 2<br />

M 2 3 3<br />

rt<br />

rt rt<br />

9. A( t) A0<br />

e ; A( t) 2A ln 2 .7 70 70<br />

0 2A0 A0e e 2 rt ln 2 t t<br />

r r 100 r ( r%)<br />

10. In order to maximize the amount of sunlight, we need to maximize the angle formed by extending the two red<br />

line segments to their vertex. The angle between the two lines is given by 1 ( 2) . From trig we<br />

1<br />

have tan 350 350<br />

1 450 1 tan<br />

x 450 x and 200 1<br />

tan<br />

200<br />

2 2 tan<br />

x<br />

x<br />

1 350 1<br />

( 200<br />

1 ( 2)) tan tan<br />

450 x<br />

x<br />

d 1 350 1 200 350 200<br />

dx 350<br />

2 2<br />

200<br />

2 2 2 2<br />

1 (450 x) 1<br />

x (450 x) 122500 x 40000<br />

450 x<br />

x<br />

d<br />

2 2<br />

0 350 200 0 200 (450 x) 122500 350( x 40000)<br />

dx 2 2<br />

(450 x) 122500 x 40000<br />

2<br />

3x 3600x 1020000 0 x 600 100 70. Since x 0, consider only x 600 100 70. Using the<br />

first derivative test, d<br />

9 0 and d<br />

9 0 local max when<br />

dx x 100 3500 dx x 400 5000<br />

x 600 100 70 236.67 ft.<br />

Copyright<br />

2014 Pearson Education, Inc.

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