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 833: 10



Work Step by Step

The vector equation is given by: $r(x,y,z)=r_0+t \nabla f(r_0)$ Given: $e^{-x^2-y^2}-z=0$ The equation of tangent line for $\nabla f(0,0,1)=\lt 0,0,-1 \gt$ is given a follows: $(0)(x-0)+(0)(y-0)-(1)(z-1)=0 \implies -z+1=0$ Thus, $z=1$
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