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: 1


a. $x+y+z=3$ b. $x=1+2t,y=1+2t; z=1+2t$

Work Step by Step

a. As we know that 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: $\nabla f(1,1,1)=\lt 2,2,2 \gt$ Now, $2(x-1)+2(y-1)+2(z-1)=0$ or, $2x+2y+2z=6 $ This implies that $x+y+z=3$ b. As we know that the vector equation is given by: $r(x,y,z)=r_0+t \nabla f(r_0)$ The parametric equations for $\nabla f(1,1,1)=\lt 2,2,2 \gt$ are as follows: $x=1+2t,y=1+2t; z=1+2t$
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