University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 13 - Practice Exercises - Page 750: 54

Answer

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

Work Step by Step

The vector equation can be calculated as: $\nabla f(r_0) \cdot (r-r_0)=0$ and $\nabla f \times \nabla g=-i+k$ Now, the parametric equations can be written as: $r-r_0+\nabla f(r_0) t$ for $\nabla f( \dfrac{1}{2},1,\dfrac{1}{2})=\lt -1,0,1 \gt$: Thus, $x=\dfrac{1}{2}-t; y=1+0t=1; z=\dfrac{1}{2}+t$ or, $x=\dfrac{1}{2}-t; y=1; z=\dfrac{1}{2}+t$
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