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: 14


$x=1+2t; y=1-4t; z=1+2t$

Work Step by Step

The vector equation is given by: $r(x,y,z)=r_0+t \nabla f(r_0)$ Given: $f(x,y,z)=xyz; x^2+2y^2+3z^2=6$ The equation of tangent line is $v=2 i -4j+2k$ or, $v=\lt 2,-4,2 \gt$ Now, the parametric equations are as follows: $x=1+2t; y=1+0t=1; z=1+2t$ Thus, $x=1+2t; y=1-4t; z=1+2t$
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