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 14.3 - Partial Derivatives - Exercises 14.3 - Page 808: 61

Answer

(a) $3$ and (b) $2$

Work Step by Step

(a) We know that the slope of the line tangent to a surface $f(x,y)$ at the point $(p,q)$ and lying in the plane $x=p$ is equal to $f_y(p,q)$. Thus, $f_y(x,y)=3$ This means that $f_y(-2,1)=3$ (b) We know that the slope of the line tangent to a surface $f(x,y)$ at the point $(p,q)$ and lying in the plane $y=q$ is equal to $f_x(p,q)$. Thus, $f_x(x,y)=2$ This means that $f_x(-2,1)=2$
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