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HW01-solutions

# HW01-solutions - qazi(kaq87 HW01 berg(55290 This print-out...

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qazi (kaq87) – HW01 – berg – (55290) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points If the constant C is chosen so that the curve given parametrically by parenleftBig Ct, C 2 16 t 2 parenrightBig , 0 t 3 , is the arc of the parabola 16 y = x 2 from (0 , 0) to (8 , 4), find the coordinates of the point P on this arc corresponding to t = 2. 1. P = parenleftBig 4 9 , 8 3 parenrightBig 2. P = parenleftBig 16 9 , 16 3 parenrightBig 3. P = parenleftBig 8 3 , 4 9 parenrightBig 4. P = parenleftBig 8 3 , 16 9 parenrightBig 5. P = parenleftBig 16 3 , 16 9 parenrightBig correct 6. P = parenleftBig 4 9 , 16 3 parenrightBig Explanation: The point P has coordinates parenleftBig Ct vextendsingle vextendsingle vextendsingle t =2 , C 2 16 t 2 vextendsingle vextendsingle vextendsingle t =2 parenrightBig = parenleftBig 2 C, C 2 4 parenrightBig , so we need to find C . But we are told that the graph of parenleftBig Ct, C 2 16 t 2 parenrightBig passes through (8 , 4) when t = 3. Thus 3 C = 8 , i . e ., C = 8 3 . Consequently, P = parenleftBig 2 C, C 2 4 parenrightBig = parenleftBig 16 3 , 16 9 parenrightBig . keywords: parametric curve, parabola 002 10.0points Find a Cartesian equation for the curve given in parametric form by x ( t ) = 4 t 2 , y ( t ) = 8 t 3 . 1. x = 2 y 2 / 3 2. x = y 2 / 3 correct 3. x = y 3 / 2 4. x = 2 y 3 / 2 5. x = 2 y 4 / 3 6. x = y 4 / 3 Explanation: We have to eliminate the parameter t from the equations for x and y . But from the equation for y , it follows that t = 1 2 y 1 / 3 , in which case x = 4 parenleftbigg 1 2 y 1 / 3 parenrightbigg 2 = y 2 / 3 . 003 10.0points Determine a Cartesian equation for the curve given in parametric form by x ( t ) = 4 e t , y ( t ) = 3 e 2 t . 1. x 2 y = 48 correct 2. x y 2 = 12

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qazi (kaq87) – HW01 – berg – (55290) 2 3. x 2 y = 48 4. x 2 y = 36 5. x 2 y = 36 6. xy 2 = 12 Explanation: We have to eliminate the parameter t from the equations for x and y . Now from the equation for x it follows that e t = x 4 , from which in turn it follows that y = 3 parenleftBig 4 x parenrightBig 2 . Consequently, x 2 y = 48 . 004 10.0points 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 3. x y 4. x y 5. x y
qazi (kaq87) – HW01 – berg – (55290) 3 6. x y cor- rect 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, 005 10.0points

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