Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 14 - Partial Derivatives - 14.6 Directional Derivatives and the Gradient Vector - 14.6 Exercises - Page 999: 58

Answer

Every normal line to the sphere passes through the center of the sphere.

Work Step by Step

Our aim is to determine the normal line equation. The general form is: $\dfrac{(x_2-x_1)}{f_x(x_1,y_1,z_1)}=\dfrac{(y_2-y_1)}{f_y(x_1,y_1,z_1)}=\dfrac{(z_2-z_1)}{f_x(x_1,y_1,z_1)}$ ...(1) From the given data, we have $f(x,y,z)=(p,q,r)$ Equation (1), becomes: Thus, $\dfrac{(x-p)}{2p}=\dfrac{(y-q)}{2q}=\dfrac{(z-r)}{2r}$ After simplifications, we get:$\dfrac{x}{p}=\dfrac{y}{q}=\dfrac{z}{r}$ Hence, $x=y=z=0$ This has been proved that every normal line to the sphere passes through the center of the sphere.
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