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Thomas Calculus 13th [Solutions]

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Chapter 3 Practice Exercises 209<br />

112.<br />

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

2 dy 0 dy x<br />

dy<br />

0<br />

e y d e y d e y e m<br />

e 1<br />

dx dx dx dx 2 y tan dx<br />

(0, 1)<br />

2(1) 2<br />

;<br />

m 1 2; tangent line: y 1 1 ( x 0) y 1 x ; normal line: y 1 = 2(x 0) y = 2x + 1<br />

m<br />

2 2<br />

tan<br />

113. xy 2x 5y 2 x dy<br />

dy<br />

2 5<br />

dx<br />

y dx<br />

0 dy<br />

dx ( x 5) 2 dy y 2 dy<br />

y<br />

dx x 5 dx<br />

2<br />

(3, 2)<br />

line is y 2 2( x 3) 2x 4 and the normal line is y 2 1 ( x 3) 1 7<br />

2<br />

2 x 2 .<br />

114.<br />

2<br />

dy<br />

( y x)<br />

2x 4 2( y x) 1 2 ( ) dy<br />

dy 1 y x dy<br />

y x 1 ( y x)<br />

3<br />

dx<br />

dx<br />

dx y x dx<br />

(6, 2)<br />

4<br />

line is y 2 3 ( x 6) 3 x 5 and the normal line is y 2 4 ( x 6) 4 x 10.<br />

4 4 2<br />

3 3<br />

the tangent<br />

the tangent<br />

115. x xy 6 1 dy<br />

dy<br />

dy 2 xy y dy<br />

1 x y 0 x y 2 xy<br />

5<br />

2 xy dx<br />

dx<br />

dx x dx<br />

(4,1)<br />

4<br />

is y 1 5 ( x 4) 5 x 6 and the normal line is y 1 4 ( x 4) 4 x 11.<br />

4 4<br />

5 5 5<br />

the tangent line<br />

3/2 3/2<br />

1/2 1/2.<br />

116. x 2y<br />

3<br />

dy dy 1/2<br />

17 x 3y 0<br />

x dy 1<br />

2<br />

dx dx 1/2<br />

2 y dx<br />

(1, 4)<br />

4<br />

y 4 1 ( x 1) 1 x 17 and the normal line is y 4 4( x 1) 4 x.<br />

4<br />

4 4<br />

the tangent line is<br />

3 3 2<br />

3 2 dy 3 2 dy dy 3 2 dy dy dy 2 3<br />

117. x y y x y x 3 y y (3 x ) 2y<br />

1 3x y 2y 1 3x y<br />

dx<br />

dx dx dx dx dx<br />

2 3<br />

dy 3 2<br />

2 3 dy 1 3x y dy<br />

(3x y 2y<br />

1) 1 3x y<br />

2 , but dy is undefined. Therefore, the<br />

dx<br />

dx 3 2<br />

3x y 2y<br />

1 dx<br />

(1,1)<br />

4 dx<br />

(1, 1)<br />

curve has slope 1 at (1,1) but the slope is undefined at (1, 1).<br />

2<br />

dy<br />

118. y sin( x sin x ) [cos( x sin x)](1 cos x);<br />

y 0 sin( x sin x ) 0 x sin x k , k 2, 1, 0, 1, 2<br />

dx<br />

dy<br />

(for our interval) cos( x sin x) cos( k ) 1. Therefore, 0 and y 0 when 1 cos x 0 and x k .<br />

dx<br />

For 2 x 2 , these equations hold when k 2, 0, and 2(since cos( ) cos 1.) Thus the curve has<br />

horizontal tangents at the x -axis for the x -values 2 , 0, and 2 (which are even integer multiples of ) the<br />

curve has an infinite number of horizontal tangents.<br />

119. B graph of f , A graph of f . Curve B cannot be the derivative of A because A has only negative slopes<br />

while some of Bs values are positive.<br />

120. A graph of f , B graph of f . Curve A cannot be the derivative of B because B has only negative slopes<br />

while A has positive values for x 0.<br />

121. 122.<br />

123. (a) 0, 0 (b) largest 1700, smallest about 1400<br />

Copyright<br />

2014 Pearson Education, Inc.

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