University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 13 - Section 13.6 - Tangent Planes and Differentials - Exercises - Page 727: 10

Answer

$z=1$

Work Step by Step

As we are given that $e^{-x^2-y^2}-z=0$ Since, we have the vector equation $r(x,y,z)=r_0+t \nabla f(r_0)$ The equation of the tangent line for $\nabla f(0,0,1)=\lt 0,0,-1 \gt$ is $0(x-0)+0(y-0)-1(z-1)=0$ or, $-z+1=0$ or, $z=1$
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.