Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 10 - Section 10.1 - Curves Defined by Parametric Equations - 10.1 Exercises - Page 645: 5

Answer

$y=\displaystyle \frac{1}{4}x+\frac{5}{4}$ See image:

Work Step by Step

Build a table of coordinates (x,y) $x=f(t)=2t-1,\displaystyle \quad \qquad y=g(t)=\frac{1}{2}t+1$, Plot the points and join with a smooth curve. Taking the initial t to be the first point in the table, track the direction in which the points "travel" as t increases. From $ x=2t-1,$ $x+1=2t$ $\displaystyle \frac{x+1}{2}=t,\qquad $which we insert into the parametric equation for y $\displaystyle \quad y=\frac{1}{2}t+1$ $\displaystyle \quad y=\frac{1}{2}$($\displaystyle \frac{x+1}{2}$)$+1$ $\displaystyle \quad y=\frac{1}{4}x+\frac{5}{4}$
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