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: 6


a) $x-3y-z=-1$ b) $x=1+t,y=1-3t; z=-1-t$

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: $\nabla f(1,1,-1)=\lt 1,-3,-1 \gt$ Now, $(1)(x-1)-(3)(y-1)-(1)(z+1)=0 \implies x-3y-z=-1$ b. The vector equation is given by: $r(x,y,z)=r_0+t \nabla f(r_0)$ we have the parametric equations as follows: $x=1+t,y=1-3t; z=-1-t$
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