math1a - quiz 5 - solns

math1a - quiz 5 - solns - x = 2 , we get f ( 2 ) = e 2 1 2...

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Math 1A Quiz 5 Solutions March 6th, 2008 1. A particle moves along the curve y = 1 + x 3 . As it reaches the point (2 , 3), the y -coordinate is increasing at a rate of 4 cm/s. How fast is the x -coordinate of the point changing at that instant? Solution: Differentiate both sides with respect to t . We have dy dt = 1 2 (1 + x 3 ) - 1 / 2 · 3 x 2 · dx dt . Plugging in x = 2 , y = 3 , dy dt = 4 and solving for dx dt gives dx dt = 2. 2. Let f ( x ) = e x · (sin x ) x . Find f 0 ( π 2 ). Solution 1: We have (sin x ) x = e x ln(sin x ) , so e x (sin x ) x = e x + x ln(sin x ) . Differentiating using chain rule gives: f 0 ( x ) = e 1+ x ln(sin x ) · d dx ( x + x ln(sin x )) = e 1+ x ln(sin x ) · (1 + ln(sin x ) + x cos x sin x ) = e x (sin x ) x · (1 + ln(sin x ) + x cos x sin x ) Plugging in
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Unformatted text preview: x = 2 , we get f ( 2 ) = e 2 1 2 (1 + ln(1) + 2 1 ) = e 2 . Solution 2: Take logarithms of both sides of y = e x (sin x ) x , to get ln y = ln( e x (sin x ) x ) = ln( e x ) + ln((sin x ) x ) = x + x ln(sin x ) . Now take derivatives of both sides to get 1 y y = 1 + ln(sin x ) + x cos x sin x . Multiplying both sides by y = e x (sin x ) x gives exactly the same derivative formula as in solution 1, and the rest follows indenti-cally. 1...
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