Postclass 10.2-solutions

# Postclass 10.2-solutions - im(gi768 Postclass 10.2...

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im (gi768) – Postclass 10.2 – sadun – (53088) 1 This print-out should have 9 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points Which one of the following could be the graph of the curve given parametrically by ( x ( t ) , y ( t )) when the graphs of x ( t ) and y ( t ) are shown in 1 1 t x ( t ) : y ( t ) : 1. 1 1 x y 2. 1 1 x y correct 3. 1 1 x y 4. 1 1 x y 5. 1 1 x y 6. 1 1 x y Explanation:

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im (gi768) – Postclass 10.2 – sadun – (53088) 2 Since x (1) = y (1) = 0 , x (0) = 1 , y (0) = 0 , the graph passes through the origin and the point (1 , 0). This already eliminates three of the graphs. There are two ways of deciding which of the remaining three it is: (i) since x parenleftBig 1 2 parenrightBig = 3 4 , y parenleftBig 1 2 parenrightBig = 1 , thegraphmustpassalsothrough (3 / 4 , 1), (ii) thegraphof y ( t ) isincreasingnear t = 0 atthesamerateasitisdecreasingnear t = 1 , while the graph of x ( t ) is decreasing faster near t = 1 than itis near t = 0 . So thegraph of ( x ( t ) , y ( t )) is decreasing faster near t = 1 thanitisincreasingnear t = 0. Consequently, the graph of ( x ( t ) , y ( t )) is 1 1 x y keywords: parametric curve, graph, 002 10.0points Find an equation for the tangent line to the curve given parametrically by x ( t ) = e t , y ( t ) = t - ln t 2 at the point P = ( x (1) , y (1)). 1. y + 1 e x = 2 2. y + 2 e x = 3 correct 3. y = 1 e x + 3 4. y + 2 e x = 1 5. y = 2 e x - 1 6. y = 1 e x - 1 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 ) = 1 2 t e t , y ( t ) = 1 - 2 t t 2 , and so at t = 1, x (1) = 1 2 e , y ( t ) = 1 - 2 = - 1 .
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