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: 6

Answer

$y=\displaystyle \frac{2}{3}x+\frac{5}{3}$ See image:

Work Step by Step

Build a table of coordinates (x,y) $x=f(t)=3t+2,\quad \qquad y=g(t)=2t+3$ 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=3t+2,$ $x-2=3t$ $\displaystyle \frac{x-2}{3}=t,\qquad $which we insert into the parametric equation for y $y=2t+3\quad $ $y=2\displaystyle \cdot\frac{x-2}{3}+3$ $y=\displaystyle \frac{2}{3}x+\frac{5}{3}$
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