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HW10-solutions

# HW10-solutions - wei(jw35975 HW10 milburn(54685 This...

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wei (jw35975) – HW10 – milburn – (54685) 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 10.0points A circle of radius r has area A and circum- ference C are given respectively by A = πr 2 , C = 2 πr . If r varies with time t , for what value of r is the rate of change of A with respect to t half the rate of change of C with respect to t ? 1. r = 1 2. r = 1 2 correct 3. r = π 4. r = π 2 5. r = 2 6. r = 2 π Explanation: Differentiating A = πr 2 , C = 2 πr implicitly with respect to t we see that dA dt = 2 πr dr dt , dC dt = 2 π dr dt . Thus the rate of change, dA/dt , of area is half the rate of change, dC/dt , of circumference when dA dt = 1 2 dC dt , i.e. , when 2 πr dr dt = 1 2 parenleftBig 2 π dr dt parenrightBig . This happens when r = 1 2 . 002 10.0points A point is moving on the graph of 3 x 3 + 5 y 3 = xy. When the point is at P = parenleftBig 1 8 , 1 8 parenrightBig , its y -coordinate is increasing at a speed of 7 units per second. What is the speed of the x -coordinate at that time and in which direction is the x - coordinate moving? 1. speed = 48 units/sec , decreasing x 2. speed = 51 units/sec , increasing x 3. speed = 48 units/sec , increasing x 4. speed = 49 units/sec , decreasing x correct 5. speed = 50 units/sec , increasing x 6. speed = 51 units/sec , decreasing x 7. speed = 49 units/sec , increasing x 8. speed = 50 units/sec , decreasing x Explanation: Differentiating 3 x 3 + 5 y 3 = xy implicitly with respect to t we see that 9 x 2 dx dt + 15 y 2 dy dt = y dx dt + x dy dt . Thus dx dt = parenleftBig x - 15 y 2 9 x 2 - y parenrightBig dy dt .

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wei (jw35975) – HW10 – milburn – (54685) 2 Now at P , x - 15 y 2 = - 7 64 , while 9 x 2 - y = 1 64 . Hence, at P , dx dt = - 7 dy dt . When the y -coordinate at P is increasing at a rate of 7 units per second, therefore, dx dt = - 49 . Consequently, the x -coordinate is moving at speed = 49 units/sec , and the negative sign indicates that it is mov- ing in the direction of decreasing x . 003 10.0points If the radius of a melting snowball decreases at a rate of 2 ins/min, find the rate at which the volume is decreasing when the snowball has diameter 2 inches. 1. rate = 10 π cu.ins/min 2. rate = 7 π cu.ins/min 3. rate = 8 π cu.ins/min correct 4. rate = 9 π cu.ins/min 5. rate = 11 π cu.ins/min Explanation: The volume, V , of a sphere of radius r is given by V = 4 3 πr 3 . Thus by implicit differentiation, dV dt = 4 πr 2 dr dt = - 8 πr 2 , since dr/dt = - 2 ins/min. When the snow- ball has diameter 2 inches, therefore, its ra- dius r = 1 and dV dt = - 8 (1) π . Consequently, when the snowball has diam- eter 2 inches, the volume of the snowball is decreasing at a rate = 8 π cu.ins/min . 004 10.0points A street light is on top of a 12 foot pole. Joe, who is 3 feet tall, walks away from the pole at a rate of 3 feet per second. At what speed is the tip of Joe’s shadow moving from the base of the pole when he is 12 feet from the pole?
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