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: 12



Work Step by Step

The vector equation is given by: $r(x,y,z)=r_0+t \nabla f(r_0)$ Given: $4x^2+y^2-z=0$ The equation of tangent line for $\nabla f(1,1,5)=\lt 8,2,-1 \gt$ is given as follows: Now, $(8)(x-1)+(2)(y-1)-(1)(z-5)=0$ or, $8x-8+2y-2-z+5=0$ Thus, $8x+2y-z=5$
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