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Unformatted text preview: wei (jw35975) HW10 milburn (54685) 1 This printout should have 20 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points A circle of radius r has area A and circum ference C are given respectively by A = r 2 , C = 2 r . 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 = 4. r = 2 5. r = 2 6. r = 2 Explanation: Differentiating A = r 2 , C = 2 r implicitly with respect to t we see that dA dt = 2 r dr dt , dC dt = 2 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 r dr dt = 1 2 parenleftBig 2 dr dt parenrightBig . This happens when r = 1 2 . 002 10.0 points 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 ycoordinate is increasing at a speed of 7 units per second. What is the speed of the xcoordinate 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 . 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 ycoordinate at P is increasing at a rate of 7 units per second, therefore, dx dt = 49 . Consequently, the xcoordinate is moving at speed = 49 units/sec , and the negative sign indicates that it is mov ing in the direction of decreasing x . 003 10.0 points 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 cu.ins/min 2. rate = 7 cu.ins/min 3. rate = 8 cu.ins/min correct 4. rate = 9 cu.ins/min 5. rate = 11 cu.ins/min Explanation: The volume, V , of a sphere of radius r is given by V = 4 3 r 3 . Thus by implicit differentiation, dV dt = 4 r 2 dr dt = 8 r 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) . Consequently, when the snowball has diam eter 2 inches, the volume of the snowball is decreasing at a rate = 8 cu.ins/min ....
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This note was uploaded on 10/25/2011 for the course MATHEMATIC 408K taught by Professor Milburn during the Fall '11 term at University of Texas at Austin.
 Fall '11
 MILBURN
 Calculus

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