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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 aren’t due until after the next test, but please try most of them because I will ask similar material on the exam. 001 10.0 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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This note was uploaded on 03/20/2011 for the course M 408 C taught by Professor Treisman during the Spring '07 term at University of Texas.
 Spring '07
 Treisman
 Calculus

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