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Thomas Calculus 13th [Solutions]

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Section 3.1 Tangents and the Derivative at a Point 117<br />

20. At x 2, y 3 m<br />

[1 (2 h) ] ( 3)<br />

lim<br />

h 0<br />

h<br />

2<br />

(1 4 4 h h ) 3<br />

lim<br />

h 0<br />

h<br />

2<br />

lim<br />

h 0<br />

h(4 h)<br />

h<br />

4, slope<br />

21. At x 3, y 1 m<br />

2<br />

lim<br />

h 0<br />

1 1<br />

(3 h ) 1 2<br />

h<br />

2 (2 h)<br />

lim<br />

h 0<br />

2 h(2 h)<br />

lim h 1 , slope<br />

h 0<br />

2 h(2 h) 4<br />

22. At x 0, y 1 m<br />

lim<br />

h 0<br />

h<br />

h<br />

1<br />

1 ( 1)<br />

h<br />

( 1) ( 1)<br />

lim h h<br />

h 0<br />

h( h 1)<br />

lim 2h<br />

0<br />

h( h 1)<br />

2<br />

h<br />

23. (a) It is the rate of change of the number of cells when t 5. The units are the number of cells per hour.<br />

(b) P (3) because the slope of the curve is greater there.<br />

2 2 2<br />

6.10(5 h) 9.28(5 h) 16.43 [6.10(5) 9.28(5) 16.43] 61.0h 6.10h 9.28h<br />

(c) P (5) lim lim<br />

h 0 h<br />

h 0 h<br />

lim 51.72 6.10h<br />

51.72 52 cells/hr.<br />

h<br />

0<br />

24. (a) From t 0 to t 3, the derivative is positive.<br />

(b) At t 3, the derivative appears to be 0. From t 2 to t 3, the derivative is positive but decreasing.<br />

25. At a horizontal tangent the slope m<br />

2 2 2 2<br />

2 2<br />

[( ) 4( ) 1] ( 4 1)<br />

0 0 m lim x h x h x x<br />

h 0<br />

h<br />

lim ( x 2xh h 4x 4h 1) ( x 4x 1) lim (2xh h 4 h)<br />

lim (2 4) 2 4;<br />

h 0<br />

h<br />

h 0<br />

h<br />

h 0<br />

x 2. Then ( 2) 4 8 1 5 ( 2, 5)<br />

x h x 2x<br />

4 0<br />

f is the point on the graph where there is<br />

a horizontal tangent.<br />

26.<br />

27.<br />

3 3<br />

[( ) 3( )] ( 3 )<br />

h 0<br />

h<br />

2 2<br />

0 m lim x h x h x x<br />

lim (3x 3xh h 3)<br />

h<br />

0<br />

3 2 2 3 3<br />

( 3 3 3 3 ) ( 3 )<br />

lim x x h xh h x h x x<br />

h 0<br />

h<br />

2 2 3<br />

lim<br />

3 x h 3 xh h 3 h<br />

h 0<br />

h<br />

2 2<br />

3x 3; 3x 3 0 x 1 or x 1. Then f ( 1) 2 and f (1) 2 ( 1, 2)<br />

and (1, 2) are the points on the graph where a horizontal tangent exists.<br />

1 1<br />

( ) 1 1<br />

0<br />

h<br />

1 m lim x h x<br />

h<br />

x( x 2) 0 0<br />

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

0<br />

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

lim x<br />

h<br />

lim<br />

h<br />

h<br />

1<br />

0<br />

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

2<br />

1)<br />

2 2<br />

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

0<br />

x or x 2. If x 0, then y 1 and m 1 y 1 ( x 0) ( x 1). If x 2,<br />

then y 1 and m 1 y 1 ( x 2) ( x 3).<br />

28.<br />

1<br />

4<br />

m<br />

Thus, 1 1<br />

4 2 x<br />

lim x h x<br />

h 0<br />

h<br />

lim x h x x h x<br />

h 0<br />

h x h x<br />

( )<br />

lim x h x<br />

h 0 h x h x<br />

h<br />

h 0 h x h x<br />

x x<br />

4<br />

4<br />

1<br />

2 x<br />

lim .<br />

x 2 x 4 y 2. The tangent line is y 2<br />

1<br />

( 4) 1.<br />

29.<br />

(2 ) (2)<br />

lim f h f<br />

h 0<br />

h<br />

lim<br />

h<br />

0<br />

2 2<br />

(100 4.9(2 ) ) (100 4.9(2) )<br />

h<br />

h<br />

2<br />

4.9(4 4 h h ) 4.9(4)<br />

lim<br />

h 0<br />

h<br />

h 0<br />

The minus sign indicates the object is falling downward at a speed of 19.6 m/sec.<br />

lim ( 19.6 4.9 h) 19.6.<br />

30.<br />

(10 ) (10)<br />

lim f h f<br />

h 0<br />

h<br />

lim<br />

h<br />

0<br />

2 2<br />

3(10 ) 3(10)<br />

h<br />

h<br />

lim<br />

h<br />

0<br />

2<br />

3(20 )<br />

h h<br />

h<br />

60 ft/sec.<br />

31.<br />

(3 ) (3)<br />

lim f h f<br />

h 0<br />

h<br />

lim<br />

h<br />

0<br />

2 2<br />

(3 ) (3) )<br />

h<br />

h<br />

2<br />

[9 6h h 9]<br />

lim<br />

h 0<br />

h<br />

h 0<br />

lim (6 h)<br />

6<br />

Copyright<br />

2014 Pearson Education, Inc.

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