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: 9


$ 2x-z=2$

Work Step by Step

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