## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 14: Partial Derivatives - Practice Exercises - Page 865: 53

#### Answer

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

#### 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 the tangent vector becomes: $\nabla f \times \nabla g=((2x)i+2j+2k ) \times (y-1)=-2i+2k$ The the parametric equations for $\nabla f( 1,1,\dfrac{1}{2})=\lt 0,-2,2 \gt$ are given by: We have, $x=1-(2)t; y=1+(0)t=1; z=(\dfrac{1}{2})+(2)t$ Thus, $x=1-2t; y=1; z=(\dfrac{1}{2})+2t$

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