Thomas' Calculus 13th Edition

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

Chapter 14: Partial Derivatives - Practice Exercises - Page 865: 47


$4x-y-5z=4$ and $x=2+4t; y=-1-t; z=1-5t$

Work Step by Step

Formula to calculate the vector equation is:$\nabla f(r_0) \cdot (r-r_0)=0$ As we know that the equation of tangent for $\nabla f( 2,-1,1)=\lt 4,-1,-5 \gt$ is given by $4(x-2)-1(y+1)-5(z-1)=0 \implies 4x-8-y-1-5z+5 =0$ Thus, $4x-y-5z=4$ The parametric equations $r-r_0+\nabla f(r_0) t$ for $\nabla f( 2,-1,1)=\lt 4,-1,-5 \gt$ are given as follows: Thus, $x=2+4t; y=-1-t; z=1-5t$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.