Quiz 1 Solution

# Quiz 1 Solution - x = e t-t, y = 4 e t 2 ,-8 t 3 Sol (1)....

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Quiz1 ( Solution ) 1. Sketch the curve by using the following parametric equations to plot points. Indicate with an arrow the direction in which the curve is traced as t increases. Eliminate the parameter to ﬁnd a Cartesian equation of the curve. x = 1 + t,y = 5 - 2 t , - 2 t 3 Sol : (1). You may calculate values of x and y at diﬀerent t to get the sketch of the graph; Or it’s much easier if you know it’s a straight line -1 -0.5 0 0.5 1 1.5 2 2.5 3 3.5 4 -1 0 1 2 3 4 5 6 7 8 9 -1 x 4 Figure 1: Graph of the parametric curve x = 1 + t,y = 5 - 2 t , - 2 t 3. (2). Solving t from x = 1+ t gives us t = x - 1 , and after plugging this into y , we are led to y = 5 - 2( x - 1) = 5 - 2 x +2 = 7 - 2 x ,i.e y = 7 - 2 x , - 1 x 4

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2. Considering the parametric curve:
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Unformatted text preview: x = e t-t, y = 4 e t 2 ,-8 t 3 Sol (1). Find dy/dx dy/dx = dy/dt dx/dt = 2 e t 2 e t-1 (2). Find the length of the curve. Recall the formula for the length of the parametric curve : Length L = R p ( x ) 2 + ( y ) 2 dt where x ,y are the derivatives of x and y with respect to t . By what we got in (1),we have L= R 3-8 q ( e t-1) 2 + (2 e t 2 ) 2 dt = Z 3-8 p e 2 t-2 e t + 1 + 4 e t dt = Z 3-8 p ( e t + 1) 2 dt = Z 3-8 ( e t + 1) dt = ( e t + t ) | 8-3 = e 8-e-3 + 11 (1)...
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## This note was uploaded on 02/23/2010 for the course MATH 221 taught by Professor Staff during the Spring '08 term at Tulane.

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Quiz 1 Solution - x = e t-t, y = 4 e t 2 ,-8 t 3 Sol (1)....

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