solution_pd6f - suleimenov(bs26835 HW 06 rusin(55565 This...

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suleimenov (bs26835) – HW 06 – rusin – (55565) 1 This print-out should have 20 questions. Multiple-choice 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 aren’t due until after the next test, but please try most of them because I will ask similar material on the exam. 001 10.0points 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 10.0points 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 10.0points 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
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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 ln 4 2. area( A ) = 75 8 - 5 ln 4 correct 3. area( A ) = 75 8 + 5 ln 4 4. area( A ) = 83 8 + 5 ln 4 5. area( A ) = 67 8 - 5 ln 4 6. area( A ) = 67 8 + 5 ln 4 Explanation: If the graph of x ( t ) = t - 1 t , y ( t ) = 4 t + 1 t intersects y = 5 when t = t 0 and t = t 1 , then A is similar to the shaded region in 5 x y a b with a = x ( t 0 ) and b = x ( t 1 ). In this case area( A ) = integraldisplay b a (5 - y ) dx = integraldisplay x ( t 1 ) x ( t 0 ) 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 0 parenleftBig 5 - 4 t - 1 t parenrightBigparenleftBig 1 + 1 t 2 parenrightBig dt = integraldisplay t 1 t 0 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 0 . But the graph of x ( t ) = t - 1 t , y ( t ) = 4 t + 1 t intersects y = 5 when 4 t + 1 t = 5 , i.e. , when 4 t 2 - 5 t + 1 = (4 t - 1)( t - 1) = 0 . Thus t 0 = 1 / 4 while t 1 = 1, so area( A ) = parenleftBig - 3 2 - 5 ln 1 parenrightBig - parenleftBig - 87 8 - 5 ln parenleftBig 1 4 parenrightBigparenrightBig .
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