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Unformatted text preview: vasquez (mpv244) HW08 Schultz (56445) 1 This print-out should have 24 questions. Multiple-choice 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 twice the rate of change of C with respect to t ? 1. r = 2 2. r = 2 correct 3. r = 2 4. r = 5. r = 1 6. r = 1 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 twice the rate of change, dC/dt , of circumfer- ence when dA dt = 2 dC dt , i.e. , when 2 r dr dt = 2 parenleftBig 2 dr dt parenrightBig . This happens when r = 2 . 002 10.0 points A point is moving on the graph of xy = 6. When the point is at (3 , 2), its x-coordinate is increasing at a rate of 4 units per second. What is the speed of the y-coordinate at that moment and in which direction is it mov- ing? 1. speed = 8 3 units/sec, decreasing y 2. speed = 11 3 units/sec, increasing y 3. speed = 11 3 units/sec, decreasing y 4. speed = 14 3 units/sec, increasing y 5. speed = 8 3 units/sec, decreasing y correct 6. speed = 14 3 units/sec, increasing y Explanation: Provided x, y negationslash = 0, the equation xy = 6 can be written as y = 6 /x . Differentiating implicitly with respect to t we thus see that dy dt = 6 x 2 dx dt . whenever x negationslash = 0. When x = 3 , dx dt = 4 , therefore, the corresponding rate of change of the y-coordinate is given by dy dt vextendsingle vextendsingle vextendsingle x =3 = 4 parenleftBig 6 x 2 parenrightBigvextendsingle vextendsingle vextendsingle x =3 = 8 3 . Consequently, the speed of the y-coordintate is 8 3 units per second and the negative sign indicates that the point is moving in the di- rection of decreasing y . 003 10.0 points vasquez (mpv244) HW08 Schultz (56445) 2 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 = 13 cub. ins./sec 2. dV dt = 9 cub. ins./sec 3. dV dt = 10 cub. ins./sec 4. dV dt = 12 cub. ins./sec correct 5. dV dt = 11 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 ) = r 2 h = 4 h 3 ....
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This note was uploaded on 04/18/2009 for the course CH 52375 taught by Professor Ruth during the Spring '09 term at University of Texas at Austin.
- Spring '09