Oct 5, 2010 - Then dx dt. = 20. Let y be the distance between the boat and spectator, so we are looking for dy dt . By Pythagorean theorem, y2 = 1202 + x2, ...
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1. A spectator watches a rowing race from the edge of a reiver bank. The lead boat is moving in a straight line that is 120 ft from the river bank. I the boat is moving at a constant speed of 20 f t/sec, how fast is the boat moving away from the spectator when it is 50 ft past her ? Credits will be given to each step. 3 points Answer Let x be the distance past by the boat from the spectator. Then dx dt = 20. Let y be the distance between the boat and spectator, so we are looking for dy dt . dy dx 2 2 2 By Pythagorean theorem, y = 120 + x , so 2y dt = 2x dt . √ When x = 50, we have y = 1202 + 502 = 130. 2x dx 2(50).(20) 2000 dt Therefore dy dt = 2y = 2(130) = 260 = 7.69 f t/sec.
2. Given x2 y 2 + xy = 2, find
Answer
dy dx
=
2xy 2 +y 2x2 y+x
dy dx
using implicit differentiation. 2 points
3. Compute the second derivative of f (x) = √
Answer f 0 (x) = f 00 (x) =
√ x . 2−x2
2−x2 (1)−x( 21 )(2−x2 )−1/2 (−2x) √ ( 2−x2 )2
=
2 points
2 (2−x2 )3/2
6x (2−x2 )5/2
4. Let f (x) = 12 x4 + 23 x3 − 2x2 + 10, find the intervals of increasing and decreasing and classify the relative extrema. 3 points
Answer f 0 (x) = 2x3 + 2x2 − 4x = 2x(x − 1)(x + 2) −2, 0 and 1 are the critical points. If x < −2 then f’(x) is negative, so f is decreasing. If −2 < x < 0 then f’(x) is positive, so f is increasing. If 0 < x < 1 then f’(x) is negative, so f is decreasing. If 1 < x then f’(x) is positive, so f is increasing. Therefore, −2 and 1 are relative minimum and 0 is a relative maximum.
