Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 12 - Parametric Equations, Polar Coordinates, and Conic Sections - 12.1 Parametric Equations - Exercises - Page 603: 40


The answer is $c(t)=(t-1,3t-7)$.

Work Step by Step

At $t=3$, we have $(x(3),y(3))=(2,2)$. We may take $x(t)=t-1$. So, $y(t)=3(t-1)-4=3t-7$. Thus, the parametrization of $y=3 x-4$ is $c(t)=(t-1,3t-7)$ for $-\infty
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