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 834: 16


$x=\dfrac{1}{2}-t; y=1; z=\dfrac{1}{2}+t$

Work Step by Step

The vector equation is given by: $r(x,y,z)=r_0+t \nabla f(r_0)$ Given: $f(x,y,z)=x+y^2+z-2=0$ The equation of tangent line is $v=- i +k$ or, $v=\lt -1,0,1 \gt$ Now, we have the parametric equations for $\nabla f(\dfrac{1}{2},1,\dfrac{1}{2})=\lt -1,0,1 \gt$ as follows: $x=\dfrac{1}{2}-t; y=1+0t=1; z=\dfrac{1}{2}+t$ Thus, $x=\dfrac{1}{2}-t; y=1; z=\dfrac{1}{2}+t$
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