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: 3


a) $-2x+z=-2$ and b) $x=2-4t,y=0; z=2+2t$

Work Step by Step

a. The vector equation is given by: $r(x,y,z)=r_0+t \nabla f(r_0)$ The equation of tangent line is given as: $\nabla f(2,0,2)=\lt -4,0,2 \gt$ Now, $-4(x-2)+0(y-0)+2(z-2)=0$ or, $-4x+2z=-4$ This implies that $-2x+z=-2$ b. The vector equation is given by: $r(x,y,z)=r_0+t \nabla f(r_0)$ The parametric equations for $\nabla f(2,0,2)=\lt -4,0,2 \gt$ are: $x=2-4t,y=0; z=2+2t$
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