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: 54


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

Work Step by Step

Formula to calculate the vector equation is:$\nabla f(r_0) \cdot (r-r_0)=0$ $\nabla f \times \nabla g=-i+k$ The parametric equations for $\nabla f( \dfrac{1}{2},1,\dfrac{1}{2})=\lt -1,0,1 \gt$ are given as follows: This implies that $x=(\dfrac{1}{2})-t; y=1+(0)t=1; z=(\dfrac{1}{2})+t$ Hence, $x=(\dfrac{1}{2})-t; y=1; z=(\dfrac{1}{2})+t$
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