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University of Texas M 408D Rusin HW 06

University of Texas M 408D Rusin HW 06 - byrne(cmb3744 HW...

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byrne (cmb3744) – 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 16 t correct 2. d 2 y dx 2 = 10 t 3 3. d 2 y dx 2 = 3 4 t 4. d 2 y dx 2 = 10 3 t 5. d 2 y dx 2 = 3 t 16 6. d 2 y dx 2 = 8 t 3 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 = 4 t 3 - t 2 + 2 t , y = t 3 + 4 t 2 - 3 has a horizontal tangent? 1. t = 0 , 8 3 2. t = 0 , 1 6 3. t = - 8 3 4. t = 0 , - 8 3 correct 5. t = 1 6 6. t = - 1 6 Explanation: After differentiation with respect to t we see that y ( t ) = 3 t 2 + 8 t , x ( t ) = 12 t 2 - 2 t + 2 . Now dy dx = y ( t ) x ( t ) = 3 t 2 + 8 t 12 t 2 - 2 t + 2 , so the tangent line to the curve will be hori- zontal at the solutions of y ( t ) = t (3 t + 8) = 0 , hence at t = 0 , - 8 3 . 003 10.0points The region A enclosed by the line y = 9 and the graph of the curve given parametrically by x ( t ) = t - 1 t , y ( t ) = 4 t + 2 t
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byrne (cmb3744) – HW 06 – rusin – (55565) 2 is similar to the shaded region in x y Determine the area of A . 1. area( A ) = 181 8 + 6 ln 8 2. area( A ) = 189 8 + 6 ln 8 3. area( A ) = 197 8 + 6 ln 8 4. area( A ) = 189 8 - 6 ln 8 correct 5. area( A ) = 181 8 - 6 ln 8 6. area( A ) = 197 8 - 6 ln 8 Explanation: If the graph of x ( t ) = t - 1 t , y ( t ) = 4 t + 2 t intersects y = 9 when t = t 0 and t = t 1 , then A is similar to the shaded region in 9 x y a b with a = x ( t 0 ) and b = x ( t 1 ). In this case area( A ) = integraldisplay b a (9 - y ) dx = integraldisplay x ( t 1 ) x ( t 0 ) parenleftBig 9 - 4 t - 2 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 9 - 4 t - 2 t parenrightBigparenleftBig 1 + 1 t 2 parenrightBig dt = integraldisplay t 1 t 0 parenleftBig 9 - 4 t - 6 t + 9 t 2 - 2 t 3 parenrightBig dt = bracketleftBig 9 t - 2 t 2 - 6 ln t - 9 t + 1 t 2 bracketrightBig t 1 t 0 . But the graph of x ( t ) = t - 1 t , y ( t ) = 4 t + 2 t intersects y = 9 when 4 t + 2 t = 9 , i.e. , when 4 t 2 - 9 t + 2 = (4 t - 1)( t - 2) = 0 . Thus t 0 = 1 / 4 while t 1 = 2, so area( A ) = parenleftBig 23 4 - 6 ln 2 parenrightBig - parenleftBig - 143 8 - 6 ln parenleftBig 1 4 parenrightBigparenrightBig .
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