Solutions HW 8

# Solutions HW 8 - mao(tm23477 – Hw 08 – gualdani...

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Unformatted text preview: mao (tm23477) – Hw 08 – gualdani – (57180) 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 10 foot ladder is leaning against a wall. If the foot of the ladder is sliding away from the wall at a rate of 10 ft/sec, at what speed is the top of the ladder falling when the foot of the ladder is 8 feet away from the base of the wall? 1. speed = 40 3 ft/sec correct 2. speed = 41 3 ft/sec 3. speed = 38 3 ft/sec 4. speed = 14 ft/sec 5. speed = 13 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. 10 ft. We have to express dy/dt in terms of x, y and dx/dt . But by Pythagoras’ theorem, x 2 + y 2 = 100 , 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 = 8, then y = 6. Thus, if the foot of the ladder is moving away from the wall at a speed of dx dt = 10 ft/sec , and x = 8, then the velocity of the top of the ladder is given by dy dt =- 4 3 dx dt . Consequently, the speed at which the top of the ladder is falling is speed = vextendsingle vextendsingle vextendsingle dy dt vextendsingle vextendsingle vextendsingle = 40 3 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 5 feet? 1. speed = 63 sq. ft/sec 2. speed = 60 π sq. ft/sec correct 3. speed = 62 π sq. ft/sec 4. speed = 63 π sq. ft/sec 5. speed = 61 sq. ft/sec 6. speed = 61 π sq. ft/sec 7. speed = 60 sq. ft/sec 8. speed = 64 sq. ft/sec mao (tm23477) – Hw 08 – gualdani – (57180) 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 = 5 , dr dt = 6 , therefore, the speed at which the area of the ripple is increasing is given by speed = 60 π sq. ft/sec . 003 (part 1 of 2) 10.0 points A point is moving on the graph of 5 x 3 + 3 y 3 = xy. When the point is at P = parenleftBig 1 8 , 1 8 parenrightBig , its x-coordinate is decreasing at a speed of 9 units per second. (i) What is the speed of the y-coordinate at that time? 1. speed y-coord = 64 units/sec 2. speed y-coord = 63 units/sec correct 3. speed y-coord =- 63 units/sec 4. speed y-coord = 62 units/sec 5. speed y-coord =- 64 units/sec Explanation: Differentiating 5 x 3 + 3 y 3 = xy implicitly with respect to t we see that 15 x 2 dx dt + 9 y 2 dy dt = y dx dt + x dy dt ....
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## This note was uploaded on 05/12/2010 for the course M 408k taught by Professor Schultz during the Fall '08 term at University of Texas.

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Solutions HW 8 - mao(tm23477 – Hw 08 – gualdani...

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