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: 7


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

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: $\nabla f(0,1,0)=\lt 1,1,1 \gt$ Now, $(1)(x-0)+(1)(y-1)-(1)(z-0)=0 \implies x+y+z=1$ 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: $x=0+t=t,y=1+t; z=0+t=t$ Thus, $x=t,y=1+t; z=t$
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