Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 10 - Parametric Equations and Polar Coordinates - 10.1 Curves Defined by Parametric Equations - 10.1 - Page 685: 5

Answer

a. See the graph b. $y=\frac{1}{4}x+\frac{5}{4}$

Work Step by Step

a. Using the given equations, $x=2t-1$ and $y=\frac{1}{2}t+1$ create a table of values in terms of x and y, using values of t Then, plot the calculated points from the lowest t value to the highest t value. This will give the direction of the curve. b. $x=2t-1$ $2t=x+1$ $t=\frac{1}{2}x+\frac{1}{2}$.......(1) $y=\frac{1}{2}t+1$ .......(2) by substituting (1) into (2) we get: $y=\frac{1}{2} (\frac{1}{2}x+\frac{1}{2})+1$ $y=\frac{1}{4}x+\frac{1}{4}+1$ $y=\frac{1}{4}x+\frac{1}{4}+\frac{4}{4}$ $y=\frac{1}{4}x+\frac{5}{4}$
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