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 14.5 - Directional Derivatives and Gradient Vectors - Exercises 14.5 - Page 826: 39


$[(x-x_0) i+(y-y_0) j] \cdot (Ai+Bj) =A(x-x_0) +B(y-y_0)=0$; Thus, we have verified the given formula.

Work Step by Step

The equation for a line vector can be expressed as: $v=(x-x_0) i+(y-y_0) j$ Next, the line is passing through the points $(x_0,y_0)$ and $(x,y)$ . So, the vector $n = Ai+Bj$ is normal to it. $\implies v \cdot n =0$ $\implies [(x-x_0) i+(y-y_0) j] \cdot (Ai+Bj) =A(x-x_0) +B(y-y_0)=0$ Verified.
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