Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 14: Partial Derivatives - Practice Exercises - Page 865: 50



Work Step by Step

Formula to calculate the vector equation is:$\nabla f(r_0) \cdot (r-r_0)=0$ As we know that the equation of tangent for $\nabla f( 1,1,\dfrac{1}{2})=\lt -\dfrac{1}{2},-\dfrac{1}{2},-1 \gt$ is given by $-\dfrac{1}{2}(x-1)-\dfrac{1}{2}(y-1)-1(z-\dfrac{1}{2})=0$ or, $-(\dfrac{1}{2})x+\dfrac{1}{2}-(\dfrac{1}{2})y+\dfrac{1}{2}-z+(\dfrac{1}{2})=0$ Thus, $x+y+2z=3$
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