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Unformatted text preview: myers (nmm698) HW08 Hamrick (54868) 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 The radius of a circle is increasing at a constant rate of 8 ft/sec. Express the rate at which the area of the circle is changing in terms of the circumfer ence, C of the circle. 1. rate = 16 C sq. ft./sec 2. rate = 8 C sq. ft./sec correct 3. rate = 4 C sq. ft./sec 4. rate = 4 C sq. ft./sec 5. rate = 16 C sq. ft./sec 6. rate = 8 C sq. ft./sec Explanation: The area and circumference of a circle of radius r are given by A = r 2 . C = 2 r respectively. Thus dA dt = 2 r dr dt = C dr dt . When dr/dt = 8 ft/sec, therefore, dA dt = 8 C sq. ft./sec . 002 10.0 points A point is moving on the graph of 4 x 3 + 3 y 3 = xy. When the point is at P = parenleftBig 1 7 , 1 7 parenrightBig , its ycoordinate is increasing at a speed of 4 units per second. What is the speed of the xcoordinate at that time and in which direction is the x coordinate moving? 1. speed = 2 units/sec , decreasing x 2. speed = 7 5 units/sec , increasing x 3. speed = 2 units/sec , increasing x 4. speed = 9 5 units/sec , decreasing x 5. speed = 8 5 units/sec , decreasing x correct 6. speed = 9 5 units/sec , increasing x 7. speed = 7 5 units/sec , decreasing x 8. speed = 8 5 units/sec , increasing x Explanation: Differentiating 4 x 3 + 3 y 3 = xy implicitly with respect to t we see that 12 x 2 dx dt + 9 y 2 dy dt = y dx dt + x dy dt . Thus dx dt = parenleftBig x 9 y 2 12 x 2 y parenrightBig dy dt . Now at P , x 9 y 2 = 2 49 , while 12 x 2 y = 5 49 . Hence, at P , dx dt = 2 5 dy dt . myers (nmm698) HW08 Hamrick (54868) 2 When the ycoordinate at P is increasing at a rate of 4 units per second, therefore, dx dt = 8 5 . Consequently, the xcoordinate is moving at speed = 8 5 units/sec , and the negative sign indicates that it is mov ing in the direction of decreasing x . 003 10.0 points If a snowball melts so that its surface area decreases at a rate of 1 cm 2 / min, find the rate at which the diameter decreases when the diameter is 14 cm. 1. 1 42 cm / min 2. 1 56 cm / min 3. 1 7 cm / min 4. 1 28 cm / min correct 5. 1 14 cm / min Explanation: If the radius is r and the diameter is x = 2 r then r = 1 2 x and S = 4 r 2 = 4 parenleftbigg 1 2 x parenrightbigg 2 = x 2 dS dt = dS dx dx dt = 2 x dx dt dS dt = 1 2 x dx dt = 1 dx dt = 1 2 x When x = 14 , dx dt = 1 28 , so the rate of decrease is 1 28 cm / min. 004 10.0 points 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 6 feet per second. At what speed is the tip of Joes shadow moving from the base of the pole when he is 14 feet from the pole?...
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This note was uploaded on 03/10/2011 for the course M 408k taught by Professor Schultz during the Fall '08 term at University of Texas at Austin.
 Fall '08
 schultz

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