Calculus 8th Edition

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

Chapter 14 - Partial Derivatives - Review - Exercises - Page 1023: 25


(a) $z=8x+4y+1$ (b) $\frac{x-1}{8}=\frac{y+2}{4}=\frac{z-1}{-1}$

Work Step by Step

$z=3x^2-y^2+2x$ and $(1,-2,1)$ (a) Equation for a tangent plane is given as: $z-1=8(x-1)+4(y+2)$ $z-1=8x+4y$ $z=8x+4y+1$ (b) $8x+4y-z=-1$ Equation of normal line is: $\frac{x-1}{8}=\frac{y+2}{4}=\frac{z-1}{-1}$
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