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: 15


$x=1-2t; y=1; z=\dfrac{1}{2}+2t$

Work Step by Step

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