Calculus (3rd Edition)

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

Chapter 14 - Calculus of Vector-Valued Functions - 14.1 Vector-Valued Functions - Exercises - Page 711: 40

Answer

$$r(t)=\lt 9+2\cos t, -4+3\sin t,0\gt.$$

Work Step by Step

We know that the equation of the given ellipse centered at $(9,-4,0)$ is $$ \left(\frac{x-9}{2}\right)^{2}+\left(\frac{y+4}{3}\right)^{2}=1 $$ To parametrize the ellipse, we put $\frac{x-9}{2}=\cos t$ and $\frac{y+4}{3}=\sin t$. Thus, we have $x=9+2\cos t$ and $y=-4+3\sin t.$ That is, $$r(t)=\lt 9+2\cos t, -4+3\sin t,0\gt.$$
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