# Homework 07-solutions.pdf - park(jp53827 Homework 07...

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park (jp53827) – Homework 07 – staron – (53735) 1 This print-out should have 31 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points Find dy dx when x ( t ) = t ln t , y ( t ) = sin 8 t . 1. dy dx = 7 sin 7 t cos t 1 + ln t 2. dy dx = 8 cos 7 t sin t 1 + ln t 3. dy dx = 1 + ln t 8 sin 7 t cos t 4. dy dx = 7 cos 7 t sin t 1 + ln t 5. dy dx = 1 + ln t 7 sin 7 t cos t 6. dy dx = 8 sin 7 t cos t 1 + ln t correct 7. dy dx = 1 + ln t 7 cos 7 t sin t 8. dy dx = 1 + ln t 8 cos 7 t sin t Explanation: Differentiating with respect to t we see that x ( t ) = t t + ln t , y ( t ) = 8 sin 7 t cos t . Consequently, dy dx = y ( t ) x ( t ) = 8 sin 7 t cos t 1 + ln t . 002 10.0points Find dy dx when x ( t ) = 2 te t , y ( t ) = t - e t . 1. dy dx = 2 e t (1 - t ) 1 + e t 2. dy dx = 1 - e t 2 e t (1 + t ) 3. dy dx = 2 e t (1 + t ) 1 + e t 4. dy dx = 1 - e t 2 e t (1 - t ) correct 5. dy dx = 2 e t (1 - t ) 1 - e t 6. dy dx = 1 + e t 2 e t (1 + t ) Explanation: Differentiating with respect to t we see that x ( t ) = 2( e t - te t ) , y ( t ) = 1 - e t . Consequently, dy dx = y ( t ) x ( t ) = 1 - e t 2 e t (1 - t ) . 003 10.0points Find dy/dx when 2 x 3 + y 3 - 9 xy + 1 = 0 . 1. dy dx = 3 y + 2 x 2 y 2 + 3 x 2. dy dx = 3 y - 2 x 2 y 2 - 3 x correct 3. dy dx = 2 x 2 - 3 y y 2 - 3 x 4. dy dx = 2 x 2 - 3 y y 2 + 3 x 5. dy dx = 2 x 2 + 3 y y 2 - 3 x Explanation:
park (jp53827) – Homework 07 – staron – (53735) 2 We use implicit differentiation. For then 6 x 2 + 3 y 2 dy dx - 9 y - 9 x dy dx = 0 , which after solving for dy/dx and taking out the common factor 3 gives 3 parenleftBig (2 x 2 - 3 y ) + dy dx ( y 2 - 3 x ) parenrightBig = 0 . Consequently, dy dx = 3 y - 2 x 2 y 2 - 3 x . keywords: implicit differentiation, Folium of Descartes, derivative, 004 10.0points Find an equation for the tangent line to the curve given parametrically by x ( t ) = 4 e t , y ( t ) = t - ln t 4 at the point P = ( x (1) , y (1)). 1. y = 3 4 e x + 7 2. y + 3 2 e x = 7 correct 3. y + 3 4 e x = 4 4. y = 3 2 e x - 5 5. y = 3 4 e x - 3 6. y + 3 2 e x = 5 Explanation: The point slope formula with t = 1 can be used to find an equation for the tangent line at P . We first need to find the slope at P . By the Chain Rule, x ( t ) = 4 2 t e t , y ( t ) = 1 - 4 t 3 t 4 , and so at t = 1, x (1) = 2 e , y ( t ) = 1 - 4 = - 3 . Thus at P , slope = y (1) x (1) = - 3 2 e . Consequently, by the point slope formula, the tangent line at P has equation y - 1 = - 3 2 e parenleftBig x - 4 e parenrightBig , which after simplification becomes y + 3 2 e x = 7 . keywords: derivative, parametric curve, tan- gent line, exp function, log function 005 10.0points Find an equation for the tangent line to the curve given parametrically by x ( t ) = e 2 t , y ( t ) = t 2 - 4 t + 3 at the point P (1 , 3). 1. y = 2 x + 5 2. y = - 2 x + 5 correct 3. y = - 2 x + 1 4. y = - 4 x + 5 5. y = - 4 x + 1 6. y = 2 x + 1 Explanation: Notice first that P (1 , 3) is the point corre- sponding to the choice t = 0. We can thus use
park (jp53827) – Homework 07 – staron – (53735) 3 the point slope formula with t = 0 to find an equation for the tangent line at P .