Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 14 - Section 14.3 - Partial Derivatives - 14.3 Exercise - Page 927: 99


$x=1+t; y=2, ; z=2-2t$

Work Step by Step

Given: $4x^2+2y^2+z^2=16$ intersects with the plane $x=1$ in a parabola. This gives: $4x^2+2(2)^2+z^2=16$ or, $4x^2+z^2=8$ Re-arrange as: $\dfrac{x^2}{2}+\dfrac{z^2}{8}=1$ The parametric equations are given as: $x=r \cos t; y=r \sin t$ Here, we have $r=\sqrt 2$ Thus, $x=\sqrt 2 \cos t; y=2; z=2 \sqrt 2 \sin t$ and $x'=-\sqrt 2 \sin t; y=0; z'=2 \sqrt 2 \cos t$ At the point $(1,2,2)$, and $t=\dfrac{\pi}{4}$ we have $x=1+t; y=2, ; z=2-2t$
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