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HW03-solutions-1

# HW03-solutions-1 - huang(th22624 HW03 Gilbert(55015 This...

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huang (th22624) – HW03 – Gilbert – (55015) 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Find the y -intercept 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. y -intercept = 3 2. y -intercept = 3 3. y -intercept = 3 2 correct 4. y -intercept = 3 2 5. y -intercept = 3 4 6. y -intercept = 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 = π/ 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 π 6 parenrightBigslashBig x parenleftBig π 6 parenrightBig = 3 4 3 = 1 4 . But then by the point-slope 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 y -intercept = 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 ) .

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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 . Now dy dx = y ( t ) x ( t ) , so the tangent line to the curve will be vertical at the solutions of x ( t ) = t (9 t 4) = 0 , hence at t = 0 , 4 9 . 005 10.0 points Determine all points P at which the tangent line to the curve given parametrically by x ( t ) = t 3 6 t , y = 3 t 2 is parallel to the line ( t, 2 t ).
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