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Unformatted text preview: pokharel (yp624) – HW08 – Radin – (56520) 1 This printout should have 24 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 = 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 equal to the rate of change of C with respect to t ? 1. r = π 2. r = 1 2 3. r = π 2 4. r = 2 π 5. r = 2 6. r = 1 correct 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 equal to the rate of change, dC/dt , of circum ference when dA dt = dC dt , i.e. , when 2 πr dr dt = 2 π dr dt . This happens when r = 1 . 002 10.0 points A point is moving on the graph of xy = 5. When the point is at (3 , 5 3 ), its xcoordinate is increasing at a rate of 4 units per second. What is the speed of the ycoordinate at that moment and in which direction is it mov ing? 1. speed = − 29 9 units/sec, decreasing y 2. speed = 38 9 units/sec, increasing y 3. speed = 29 9 units/sec, increasing y 4. speed = − 20 9 units/sec, decreasing y 5. speed = 20 9 units/sec, decreasing y correct 6. speed = − 38 9 units/sec, increasing y Explanation: Provided x, y negationslash = 0, the equation xy = 5 can be written as y = 5 /x . Differentiating implicitly with respect to t we thus see that dy dt = − 5 x 2 dx dt . whenever x negationslash = 0. When x = 3 , dx dt = 4 , therefore, the corresponding rate of change of the ycoordinate is given by dy dt vextendsingle vextendsingle vextendsingle x =3 = − 4 parenleftBig 5 x 2 parenrightBigvextendsingle vextendsingle vextendsingle x =3 = − 20 9 . Consequently, the speed of the ycoordintate is 20 9 units per second and the negative sign indicates that the point is moving in the di rection of decreasing y . pokharel (yp624) – HW08 – Radin – (56520) 2 003 10.0 points The dimensions of a cylinder are changing, but the height is always equal to the diameter of the base of the cylinder. If the height is increasing at a speed of 4 inches per second, determine the speed at which the volume, V , is increasing (in cubic inches per second) when the height is 2 inches. 1. dV dt = 12 π cub. ins./sec correct 2. dV dt = 9 π cub. ins./sec 3. dV dt = 10 π cub. ins./sec 4. dV dt = 13 π cub. ins./sec 5. dV dt = 11 π cub. ins./sec Explanation: Since the height h of the cylinder is equal to its diameter D , the radius of the cylinder is r = 1 2 h . Thus, as a function of h , the volume of the cylinder is given by V ( h ) = πr 2 h = π 4 h 3 ....
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 Spring '10
 TSOI
 Critical Point, per capita, dt, Inch

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