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

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hinojosa (jah4698) – HW08 – seckin – (57195) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A 5 foot ladder is leaning against a wall. If the foot of the ladder is sliding away from the wall at a rate of 9 ft/sec, at what speed is the top of the ladder falling when the foot of the ladder is 3 feet away from the base of the wall? 1. speed = 13 2 ft/sec 2. speed = 27 4 ft/sec correct 3. speed = 7 ft/sec 4. speed = 25 4 ft/sec 5. speed = 6 ft/sec Explanation: Let y be the height of the ladder when the foot of the ladder is x feet from the base of the wall as shown in figure x ft. 5 ft. We have to express dy/dt in terms of x, y and dx/dt . But by Pythagoras’ theorem, x 2 + y 2 = 25 , so by implicit differentiation, 2 x dx dt + 2 y dy dt = 0 . In this case dy dt = - x y dx dt . But again by Pythagoras, if x = 3, then y = 4. Thus, if the foot of the ladder is moving away from the wall at a speed of dx dt = 9 ft/sec , and x = 3, then the velocity of the top of the ladder is given by dy dt = - 3 4 dx dt . Consequently, the speed at which the top of the ladder is falling is speed = vextendsingle vextendsingle vextendsingle dy dt vextendsingle vextendsingle vextendsingle = 27 4 ft/sec . keywords: speed, ladder, related rates 002 10.0 points A rock is thrown into a still pond and causes a circular ripple. If the radius of the ripple is increasing at a rate of 6 ft/sec, at what speed is the area of the ripple increasing when its radius is 2 feet? 1. speed = 25 sq. ft/sec 2. speed = 24 π sq. ft/sec correct 3. speed = 27 π sq. ft/sec 4. speed = 26 π sq. ft/sec 5. speed = 24 sq. ft/sec 6. speed = 28 sq. ft/sec 7. speed = 25 π sq. ft/sec 8. speed = 27 sq. ft/sec
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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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