## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 14: Partial Derivatives - Section 14.6 - Tangent Planes and Differentials - Exercises 14.6 - Page 834: 17

#### Answer

$x=1+90t; y=1-90t; z=3$

#### Work Step by Step

The vector equation is given by: $r(x,y,z)=r_0+t \nabla f(r_0)$ Given: $f(x,y,z)=x^3+3x^2+y^3+4xy-z^3=0$ Since, we have the vector equation $r(x,y,z)=r_0+t \nabla f(r_0)$ The equation of tangent line is $v=90i -90j+0k$ or, $v=\lt 90,-90,0\gt$ Now, we have the parametric equations for $\nabla f(1,1,3)=\lt 90,-90,0\gt$ as follows: $x=1+90t; y=1-90t; z=3+0t=3$ Thus, $x=1+90t; y=1-90t; z=3$

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