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Thomas Calculus 13th [Solutions]

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Chapter 4 Additional and Advanced Exercises 337<br />

11. From (ii), f ( 1) 1 a<br />

b c 2<br />

0 a 1; from (iii), either 1 lim f ( x ) or 1 lim f ( x ). In either case,<br />

x<br />

x<br />

1<br />

1<br />

lim ( ) lim x 1<br />

1<br />

1<br />

x<br />

x<br />

f x lim 1 b 0 and c 1. For if b 1, then lim 0 and if<br />

2 2<br />

x x bx cx 2 x bx c<br />

x c<br />

2<br />

x<br />

c 0, then<br />

lim<br />

x<br />

1<br />

bx<br />

1<br />

x<br />

2<br />

x<br />

1<br />

x<br />

2<br />

x<br />

1<br />

lim .<br />

x<br />

x<br />

Thus a 1, b 0, and c 1.<br />

x<br />

12.<br />

dy<br />

dx<br />

2<br />

3x<br />

2kx<br />

3<br />

0<br />

x<br />

2<br />

2k<br />

4k<br />

36<br />

6<br />

2 2<br />

x has only one value when 4k 36 0 k 9 or k 3.<br />

2<br />

13. The area of the ABC is A( x) 1 (2) 1 x<br />

2<br />

2 1/2<br />

(1 x ) , where 0 x 1. Thus A ( x) x<br />

2<br />

1 x<br />

0 and 1 are critical points. Also A ( 1) 0 so<br />

A (0) 1 is the maximum. When x 0 the ABC is<br />

isosceles since AC BC 2.<br />

14.<br />

lim f ( c h) f ( c)<br />

f ( c ) for 1<br />

h 0<br />

h<br />

2 | f ( c ) | 0 there exists a 0 such that 0 | h |<br />

f ( c h) f ( c) f ( c) 1 | f ( c) | . Then 1<br />

f ( c h) f ( c) 0 | f ( c) |<br />

f ( c) 1 | f ( c) |<br />

h<br />

2<br />

2<br />

h<br />

2<br />

1<br />

f ( c h) f ( c) | f ( c) | f ( c) 1 | f ( c) | . If f ( c ) 0, then | f ( c) | f ( c)<br />

2<br />

h<br />

2<br />

3 f ( c h)<br />

f ( c) 1 f ( c) 0; likewise if f ( c ) 0, then 1 f ( c h) 0 f ( c) 3 f ( c).<br />

2 h 2<br />

2 h 2<br />

(a) If f ( c ) 0, then h 0 f ( c h ) 0 and 0 h f ( c h ) 0. Therefore, f ( c ) is a local<br />

maximum.<br />

(b) If f ( c ) 0, then h 0 f ( c h ) 0 and 0 h f ( c h ) 0. Therefore, f ( c ) is a local<br />

minimum.<br />

2 y<br />

15. The time it would take the water to hit the ground from height y is<br />

g , where g is the acceleration of gravity.<br />

The product of time and exit velocity (rate) yields the distance the water travels:<br />

D ( )<br />

2 y<br />

2 2 1/2<br />

64( ) 8 ( ) ,<br />

g<br />

g<br />

2 2 1/2<br />

0 y h D ( y) 4 ( hy y ) ( h 2 y) 0, and h<br />

g<br />

2<br />

are critical points. Now D(0) 0, D 2<br />

1/2<br />

h 8 2 h h 4 2<br />

2 g<br />

2 2<br />

g<br />

and D ( h ) 0 the best place to<br />

drill the hole is at y h<br />

2 .<br />

16. From the figure in the text, tan( ) b a<br />

tan tan<br />

; tan( ) ; and tan a . These equations give<br />

n<br />

1 tan tan<br />

h<br />

tan<br />

a<br />

b a<br />

h h tan a<br />

. Solving for tan gives tan bh<br />

h a<br />

1 tan h a tan<br />

or 2<br />

( h a( b a)) tan bh.<br />

2<br />

h a( b a)<br />

h<br />

2<br />

2 d<br />

Differentiating both sides with respect to h gives 2h tan ( h a( b a))<br />

sec b . Then<br />

dh<br />

d<br />

0 2h tan b 2 bh<br />

2 2<br />

2<br />

h<br />

b 2 bh bh ab( b a)<br />

h a( b a ) h a( a b).<br />

dh<br />

2<br />

h a( b a)<br />

Copyright<br />

2014 Pearson Education, Inc.

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