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Unformatted text preview: huang (th22624) HW03 Gilbert (55015) 1 This printout should have 21 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points Find the yintercept of the tangent line at P (2 , 1) to the graph of the curve defined para metrically by x ( t ) = 4 cos 2 t, y ( t ) = 2 sin t. 1. yintercept = 3 2. yintercept = 3 3. yintercept = 3 2 correct 4. yintercept = 3 2 5. yintercept = 3 4 6. yintercept = 3 4 Explanation: The point P (2 , 1) on the graph of x = 4 cos 2 t, y = 2 sin t is the point at which t = / 6. On the other hand, x ( t ) = 8 sin 2 t, y ( t ) = 2 cos t, and so the slope of the tangent line at P is given by y parenleftBig 6 parenrightBigslashBig x parenleftBig 6 parenrightBig = 3 4 3 = 1 4 . But then by the pointslope formula, y 1 = 1 4 ( x 2) is an equation for the tangent line at P , which after simplification becomes y + 1 4 x = 3 2 . Consequently, this tangent line has yintercept = 3 2 . 002 10.0 points Find dy dx for the curve given parametically by x ( t ) = 2 + 3 t 2 , y ( t ) = 2 t 2 + t 3 . 1. dy dx = 6 1 + 6 t 2. dy dx = 1 3 + t 3. dy dx = 1 3 + 2 t 4. dy dx = 2 3 + t 5. dy dx = 2 3 + 1 2 t correct 6. dy dx = 3 4 + 3 t 7. dy dx = 6 4 + 3 t 8. dy dx = 3 1 + 6 t Explanation: Differentiating with respect to t we see that x ( t ) = 6 t, y ( t ) = 4 t + 3 t 2 . Consequently, dy dx = y ( t ) x ( t ) = 4 t + 3 t 2 6 t = 2 3 + 1 2 t . 003 10.0 points Find d 2 y dx 2 when x ( t ) = 5 + t 2 , y ( t ) = t ln( t 4 ) . huang (th22624) HW03 Gilbert (55015) 2 1. d 2 y dx 2 = 2 ln t t 3 2. d 2 y dx 2 = ln t t 3 3. d 2 y dx 2 = ln t t 2 4. d 2 y dx 2 = ln t t 3 correct 5. d 2 y dx 2 = 2 ln t t 2 6. d 2 y dx 2 = 2 ln t t 2 Explanation: First notice that y ( t ) = 4 t ln t . After differentiation with respect to t , therefore, we see that x ( t ) = 2 t, y ( t ) = 4(1 + ln t ) . Thus dy dx = 2 parenleftBig 1 + ln t t parenrightBig . On the other hand, by the Chain Rule, d dt parenleftBig dy dx parenrightBig = dx dt braceleftBig d dx parenleftBig dy dx parenrightBigbracerightBig = dx dt d 2 y dx 2 , in which case d 2 y dx 2 = d dt parenleftBig dy dx parenrightBigslashBig dx dt . But d dt parenleftBig dy dx parenrightBig = 2 parenleftBig t ( 1 /t ) (1 + ln t ) t 2 parenrightBig = 2 ln t t 2 . Consequently, d 2 y dx 2 = ln t t 3 . 004 10.0 points Determine all values of t for which the curve given parametrically by x = 3 t 3 2 t 2 + 1 , y = 2 t 3 + 3 t 2 1 has a vertical tangent? 1. t = 0 , 1 2. t = 0 , 1 3. t = 0 , 4 9 4. t = 1 5. t = 0 , 4 9 correct 6. t = 1 7. t = 4 9 8. t = 4 9 Explanation: After differentiation with respect to t we see that y ( t ) = 6 t 2 + 6 t, x ( t ) = 9 t 2 4 t....
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This note was uploaded on 10/20/2011 for the course M 408M taught by Professor Gilbert during the Spring '07 term at University of Texas at Austin.
 Spring '07
 Gilbert

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