Calculus 8th Edition

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

Chapter 14 - Partial Derivatives - Review - Exercises - Page 1023: 30

Answer

$2x+4y-z=4$ and $\frac{x-1}{2}=\frac{y-1}{4}=\frac{z-2}{-1}=t$ or $x=1+2t$, $y=1+4t$, $z=2-t$ See the attached graphs 1 and 2.

Work Step by Step

Given:$z=x^2+y^4$ Equation for a tangent plane is given as: $z-2=2(x-1)+4(y-1)$ $z-2=2x-2+4y-4$ $z=2x+4y-4$ $2x+4y-z=4$ Equation of normal line is: $\frac{x-1}{2}=\frac{y-1}{4}=\frac{z-2}{-1}=t$ or $x=1+2t$, $y=1+4t$, $z=2-t$ See the attached graphs 1 and 2.
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