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: 18


$x=\sqrt 2-2\sqrt 2t; y=\sqrt 2 +2\sqrt 2t; z=4$

Work Step by Step

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