Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 14: Partial Derivatives - Practice Exercises - Page 865: 48


$2x+2y+z=6$ and $x=1+2t; y=1+2t; z=2+t$

Work Step by Step

Formula to calculate the vector equation is:$\nabla f(r_0) \cdot (r-r_0)=0$ As we know that the equation of tangent for $f( 1,1,2)=\lt 2,2,1 \gt$ is given by $2(x-1)+2(y-1)+1(z-2)=0 \implies 2x-2+2y-2+z-2 =0 $ Thus, $2x+2y+z=6$ Now, the parametric equations can be written as: $r-r_0+\nabla f(r_0) t$ for $\nabla f( 1,1,2)=\lt 2,2,1 \gt$: So, $x=1+2t; y=1+2t; z=2+t$
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