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 833: 4


a) $2y+3z=7$ b) $x=1,y=-1+4t; z=3+6t$

Work Step by Step

a. The vector equation is given as: $r(x,y,z)=r_0+t \nabla f(r_0)$ The equation of tangent line is given as:$\nabla f(1,-1,3)=\lt 0,4,6 \gt$ Now, $(0)(x-1)+(4)(y+1)+(6)(z-3)=0 \implies 4y+6z=14$ Thus, $2y+3z=7$ b. The vector equation is given as: $r(x,y,z)=r_0+t \nabla f(r_0)$ we have the parametric equations as follows: $x=1+0t=1,y=-1+4t; z=3+6t$ Thus, $x=1,y=-1+4t; z=3+6t$
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