Answer
The answer is $c(t)=(t+3,t^2+6t+9)$.
Work Step by Step
At $t=0$, we have $(x(0),y(0))=(3,9)$.
We may take $x(t)=t+3$. So, $y(t)=(t+3)^2=t^2+6t+9$.
Thus, the parametrization of $y=x^2$ is $c(t)=(t+3,t^2+6t+9)$ for $-\infty
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
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: 41
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