University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 13 - Section 13.6 - Tangent Planes and Differentials - Exercises - Page 727: 6


a) $x-3y-z=-1$ b) $x=1+t,y=1-3t; z=-1-t$

Work Step by Step

a) Since, we have the vector equation $r(x,y,z)=r_0+t \nabla f(r_0)$ The equation of the tangent line is: $\nabla f(1,1,-1)=\lt 1,-3,-1 \gt$ Thus, $1(x-1)-3(y-1)-1(z+1)=0$ or, $x-3y-z=-1$ b) Since, we have the vector equation $r=r_0+t \nabla f(r_0)$ Now, the parametric equations are: $x=1+t,y=1-3t; z=-1-t$
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