HW09-solutions - cadena(jc59484 HW09 ben-zvi(55400 This print-out should have 23 questions Multiple-choice questions may continue on the next column or

HW09-solutions - cadena(jc59484 HW09 ben-zvi(55400 This...

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cadena (jc59484) – HW09 – ben-zvi – (55400) 1 This print-out should have 23 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points A 15 foot ladder is leaning against a wall. If the foot of the ladder is sliding away from the wall at a rate of 6 ft/sec, at what speed is the top of the ladder falling when the foot of the ladder is 9 feet away from the base of the wall? 1. speed = 4 ft/sec 2. speed = 9 2 ft/sec correct 3. speed = 15 4 ft/sec 4. speed = 17 4 ft/sec 5. speed = 19 4 ft/sec Explanation: Let y be the height of the ladder when the foot of the ladder is x feet from the base of the wall as shown in figure x ft. 15 ft. We have to express dy/dt in terms of x, y and dx/dt . But by Pythagoras’ theorem, x 2 + y 2 = 225 , so by implicit differentiation, 2 x dx dt + 2 y dy dt = 0 . In this case dy dt = - x y dx dt . But again by Pythagoras, if x = 9, then y = 12. Thus, if the foot of the ladder is moving away from the wall at a speed of dx dt = 6 ft/sec , and x = 9, then the velocity of the top of the ladder is given by dy dt = - 3 4 dx dt . Consequently, the speed at which the top of the ladder is falling is speed = vextendsingle vextendsingle vextendsingle dy dt vextendsingle vextendsingle vextendsingle = 9 2 ft/sec . keywords: speed, ladder, related rates 002 10.0points The radius of a circle is increasing at a constant rate of 5 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 = 5 πC sq. ft./sec 2. rate = 5 2 C sq. ft./sec 3. rate = 10 πC sq. ft./sec 4. rate = 10 C sq. ft./sec 5. rate = 5 C sq. ft./sec correct 6. rate = 5 2 πC sq. ft./sec Explanation:
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cadena (jc59484) – HW09 – ben-zvi – (55400) 2 The area and circumference of a circle of radius r are given by A = πr 2 . C = 2 πr respectively. Thus dA dt = 2 πr dr dt = C dr dt . When dr/dt = 5 ft/sec, therefore, dA dt = 5 C sq. ft./sec . 003 10.0points Determine the value of dy/dt at x = 2 when y = x 2 - 5 x and dx/dt = 3. 1. dy dt vextendsingle vextendsingle vextendsingle x =2 = 5 2. dy dt vextendsingle vextendsingle vextendsingle x =2 = - 1 3. dy dt vextendsingle vextendsingle vextendsingle x =2 = 1 4. dy dt vextendsingle vextendsingle vextendsingle x =2 = - 3 correct 5. dy dt vextendsingle vextendsingle vextendsingle x =2 = 3 Explanation: Differentiating implicitly with respect to t we see that dy dt = (2 x - 5) dx dt = 3 (2 x - 5) . At x = 2, therefore, dy dt = 3( - 1) = - 3 . 004(part1of2)10.0points A point is moving on the graph of 3 x 3 + 2 y 3 = xy . When the point is at P = parenleftBig 1 5 , 1 5 parenrightBig , its x -coordinate is decreasing at a speed of 4 units per second. What is the speed of the y -coordinate at that time? 1. speed y -coord = 15 units/sec 2. speed y -coord = 16 units/sec correct 3. speed y -coord = - 17 units/sec 4. speed y -coord = 17 units/sec 5. speed y -coord = - 16 units/sec Explanation: Differentiating 3 x 3 + 2 y 3 = xy implicitly with respect to t we see that 9 x 2 dx dt + 6 y 2 dy dt = y dx dt + x dy dt .
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