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HW08-solutions

# HW08-solutions - hinojosa(jah4698 HW08 seckin(57195 This...

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hinojosa (jah4698) – HW08 – seckin – (57195) 2 Explanation: The area, A , of a circle having radius r is given by A = πr 2 . Differentiating implicitly with respect to t we thus see that dA dt = 2 πr dr dt . When r = 2 , dr dt = 6 , therefore, the speed at which the area of the ripple is increasing is given by speed = 24 π sq. ft/sec . 003 (part 1 of 2) 10.0 points A point is moving on the graph of 5 x 3 + 6 y 3 = xy. When the point is at P = parenleftBig 1 11 , 1 11 parenrightBig , its x -coordinate is decreasing at a speed of 5 units per second. (i) What is the speed of the y -coordinate at that time? 1. speed y -coord = 20 7 units/sec correct 2. speed y -coord = - 27 7 units/sec 3. speed y -coord = 13 7 units/sec 4. speed y -coord = 27 7 units/sec 5. speed y -coord = - 20 7 units/sec Explanation: Differentiating 5 x 3 + 6 y 3 = xy implicitly with respect to t we see that 15 x 2 dx dt + 18 y 2 dy dt = y dx dt + x dy dt . Thus dy dt = parenleftBig 15 x 2 - y x - 18 y 2 parenrightBig dx dt . Now at P , 15 x 2 - y = parenleftBig 15 (11) 2 - 1 11 parenrightBig = 1 (11) 2 (4) , while x - 18 y 2 = parenleftBig 1 11 - 18 (11) 2 parenrightBig = - 1 (11) 2 (7) . Hence, at P , dy dt = - 4 7 dx dt .
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