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Unformatted text preview: hayes (tsh489) HW01 milburn (55800) 1 This printout should have 19 questions. Multiplechoice questions may continue on the next column or page find 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 = parenleftBig 1 9 e x/ 4 parenrightBig 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 hayes (tsh489) HW01 milburn (55800) 2 5. x y 6. x y Explanation: All the graphs are symmetric either about the yaxis or the xaxis. Lets 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 xaxis, 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 + 3 + 2. y ( t ) = t (2 t 2 ): 2 + + 2 + Thus the graph starts in quadrant II, crosses the yaxis into quadrant I, then crosses the xaxis 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 parenleftBig 1 2 parenrightBig = 1 4 , y parenleftBig 1 2 parenrightBig = 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....
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This note was uploaded on 06/09/2011 for the course CH 310M taught by Professor Iverson during the Spring '05 term at University of Texas at Austin.
 Spring '05
 Iverson

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