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Unformatted text preview: suleimenov (bs26835) HW 06 rusin (55565) 1 This printout should have 20 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. Here are problems for the rest of chapter 11, plus section 13.1. I have made them due the day before Spring break, which in particular means they arent due until after the next test, but please try most of them because I will ask similar material on the exam. 001 10.0 points Find d 2 y dx 2 for the curve given parametrically by x ( t ) = 3 + 2 t 2 , y ( t ) = 2 t 2 + t 3 . 1. d 2 y dx 2 = 3 4 t 2. d 2 y dx 2 = 3 16 t correct 3. d 2 y dx 2 = 10 t 3 4. d 2 y dx 2 = 8 t 3 5. d 2 y dx 2 = 10 3 t 6. d 2 y dx 2 = 3 t 16 Explanation: Differentiating with respect to t we see that x ( t ) = 4 t, y ( t ) = 4 t + 3 t 2 . Thus dy dx = y ( t ) x ( t ) = 4 t + 3 t 2 4 t = 1 + 3 4 t. On the other hand, by the Chain Rule, d dt parenleftBig dy dx parenrightBig = dx dt braceleftBig d dx parenleftBig dy dx parenrightBigbracerightBig = parenleftBig dx dt parenrightBig d 2 y dx 2 . Consequently, d 2 y dx 2 = d dt parenleftBig dy dx parenrightBigslashBig dx dt = 3 16 t . 002 10.0 points Determine all values of t for which the curve given parametrically by x = 2 t 3 3 t 2 + 2 t, y = 3 t 3 + 2 t 2 3 has a horizontal tangent? 1. t = 4 9 2. t = 1 3. t = 0 , 4 9 4. t = 0 , 1 5. t = 0 , 4 9 correct 6. t = 1 Explanation: After differentiation with respect to t we see that y ( t ) = 9 t 2 + 4 t, x ( t ) = 6 t 2 6 t + 2 . Now dy dx = y ( t ) x ( t ) = 9 t 2 + 4 t 6 t 2 6 t + 2 , so the tangent line to the curve will be hori zontal at the solutions of y ( t ) = t (9 t + 4) = 0 , hence at t = 0 , 4 9 . 003 10.0 points The region A enclosed by the line y = 5 and the graph of the curve given parametrically by x ( t ) = t 1 t , y ( t ) = 4 t + 1 t suleimenov (bs26835) HW 06 rusin (55565) 2 is similar to the shaded region in x y Determine the area of A . 1. area( A ) = 83 8 5 ln4 2. area( A ) = 75 8 5 ln4 correct 3. area( A ) = 75 8 + 5 ln4 4. area( A ) = 83 8 + 5 ln4 5. area( A ) = 67 8 5 ln4 6. area( A ) = 67 8 + 5 ln4 Explanation: If the graph of x ( t ) = t 1 t , y ( t ) = 4 t + 1 t intersects y = 5 when t = t and t = t 1 , then A is similar to the shaded region in 5 x y a b with a = x ( t ) and b = x ( t 1 ). In this case area( A ) = integraldisplay b a (5 y ) dx = integraldisplay x ( t 1 ) x ( t ) parenleftBig 5 4 t 1 t parenrightBig dx, where x = x ( t ) = t 1 t , dx = parenleftBig 1 + 1 t 2 parenrightBig dt. Thus after a change of variable from x to t we reduce the integral describing area( A ) to integraldisplay t 1 t parenleftBig 5 4 t 1 t parenrightBigparenleftBig 1 + 1 t 2 parenrightBig dt = integraldisplay t 1 t parenleftBig 5 4 t 5 t + 5 t 2 1 t 3 parenrightBig dt = bracketleftBig 5 t 2 t 2 5 ln t 5 t + 1 2 t 2 bracketrightBig t 1 t ....
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 Spring '07
 Treisman
 Calculus

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