Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 14 - Partial Derivatives - 14.3 Partial Derivatives - 14.3 Exercises - Page 967: 99


$x=1+t\\y=2\\ z=2-2t$

Work Step by Step

Set $x=1$ in the given equation. Thus, we have $4x^2+2(2)^2+z^2=16 \implies 4x^2+z^2=8$ This gives: $\dfrac{x^2}{2}+\dfrac{z^2}{8}=1$ The parametric equations are given as: $x=r \cos t$ and $y=r \sin t$ $r=\sqrt 2$ Thus, $x=r \cos t$ and $y=r \sin t \implies x=\sqrt 2 \cos t; y=2; z=2 \sqrt 2 \sin t$ $x'=-\sqrt 2 \sin t$ and $ y=0; z'=2 \sqrt 2 \cos t$ For the point $(1,2,2)$, and $t=\dfrac{\pi}{4}$ we get $x=1+t\\y=2\\z=2-2t$
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