Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 14: Partial Derivatives - Section 14.6 - Tangent Planes and Differentials - Exercises 14.6 - Page 834: 13


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

Work Step by Step

The vector equation is given by: $r(x,y,z)=r_0+t \nabla f(r_0)$ Given: $f(x,y,z)=x+y^2+z-4$ The equation of tangent line $v=0 i +2j-2k$ is given as follows: Now, we have the parametric equations for $\nabla f(1,1,1)=\lt 1,2,2 \gt$ are: $x=1; y=1+2t; z=1-2t$
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