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HW01-solutions

# HW01-solutions - hayes(tsh489 HW01 milburn(55800 This...

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hayes (tsh489) – HW01 – milburn – (55800) 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points Determine a Cartesian equation For the curve given in parametric Form by x ( t ) = 4 ln(9 t ) , y ( t ) = t . 1. y = 1 3 e x/ 8 correct 2. y = 1 4 e x/ 6 3. y = 1 3 e 8 /x 4. y = 1 4 e 6 /x 5. y = 1 4 e x/ 3 6. y = 1 3 e x/ 4 Explanation: We have to eliminate the parameter t From the equations For x and y . Now From the equation For x it Follows that t = 1 9 e x/ 4 . But then y = p 1 9 e x/ 4 P 1 / 2 = 1 3 e x/ 8 . 002 10.0 points Which one oF the Following could be the graph oF the curve given parametrically by x ( t ) = 3 - t 2 , y ( t ) = 2 t - t 3 , where the arrows indicate the direction oF increasing t ? 1. x y 2. x y cor- rect 3. x y 4. x y

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hayes (tsh489) – HW01 – milburn – (55800) 2 5. x y 6. x y Explanation: All the graphs are symmetric either about the y -axis or the x -axis. Let’s check which it is for the graph of ( x ( t ) , y ( t )) = (3 - t 2 , 2 t - t 3 ) . Now x ( - t ) = 3 - ( - t ) 2 = 3 - t 2 = x ( t ) , and y ( - t ) = 2( - t ) - ( - t ) 3 = - (2 t - t 3 ) = - y ( t ) , so ( x ( - t ) , y ( - t )) = ( x ( t ) , - y ( t )) . Thus the graph is symmetric about the x -axis, eliminating three choices. To decide which one of the remaining three it is, we can check the path traced out as t ranges from -∞ to + by looking at sign charts for x ( t ) and y ( t ) because this will tell us in which quadrant the graph lies: 1. x ( t ) = 3 - t 2 : -∞ - 3 0 - + 3 0 - + 2. y ( t ) = t (2 - t 2 ): -∞ - 2 0 + 0 0 - + 2 0 - + Thus the graph starts in quadrant II, crosses the y -axis into quadrant I, then crosses the x -axis into quadrant IV, crosses back into quadrant I and so on. Consequently, the graph is x y keywords: parametric curve, graph, direction, 003 10.0 points 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 ) :
hayes (tsh489) – HW01 – milburn – (55800) 3 1. 1 1 x y 2. 1 1 x y 3. 1 1 x y correct 4. 1 1 x y 5. 1 1 x y 6. 1 1 x y Explanation: Since x (0) = y (0) = 0 , x (1) = 1 , y (1) = 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 p 1 2 P = 1 4 , y p 1 2 P = 1 , the graph must pass also through (1 / 4 , 1), (ii) the graph of y ( t ) is increasing near t = 0 at the same rate as it is decreasing near t = 1 , while the graph of x ( t ) is increasing faster near t = 0 than it is near t = 1 . So the graph of ( x ( t ) , y ( t )) is increasing faster near t = 0 than it is decreasing near t = 1. Consequently, the graph of ( x ( t ) , y ( t )) is

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hayes (tsh489) – HW01 – milburn – (55800) 4 1 1 x y keywords: parametric curve, graph, 004 10.0 points When a mortar shell is fred with an initial velocity oF v 0 Ft/sec at an angle α above the horizontal, then its position aFter t seconds is given by the parametric equations x = ( v 0 cos α ) t , y = ( v 0 sin α ) t - 16 t 2 .
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HW01-solutions - hayes(tsh489 HW01 milburn(55800 This...

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