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 998: 52

Answer

$\dfrac{xx_0}{a^2}+\dfrac{yy_0}{b^2}-\dfrac{zz_0}{c^2}=1$

Work Step by Step

Our aim is to determine the tangent plane equation. The general form is: $(x_2-x_1)f_x(x_1,y_1,z_1)+(y_2-y_1)f_y(x_1,y_1,z_1)+(z_2-z_1)f_x(x_1,y_1,z_1)=0$ ...(1) From the given data, we have $f(x,y,z)=(x_0,y_0,z_0)$ Equation (1), becomes: Thus, $(x-x_0)(\dfrac{2x_0}{a^2})+(y-y_0)(\dfrac{2y_0}{b^2})+(z-z_0)(\dfrac{-2z_0}{c^2})=0$ After simplifications, we get $\dfrac{2x_0}{a^2}(x)+\dfrac{2y_0}{b^2}(y)+\dfrac{2z_0}{c^2}(z)-[\dfrac{2x_0^2}{a^2}+\dfrac{2y_0^2}{b^2}+\dfrac{2z_0^2}{c^2}]=0$ As we are given $\dfrac{x_0^2}{a^2}+\dfrac{y_0^2}{b^2}+\dfrac{z_0^2}{c^2}=1$ The desired answer is:$\dfrac{xx_0}{a^2}+\dfrac{yy_0}{b^2}-\dfrac{zz_0}{c^2}=1$
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