Calculus 8th Edition

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

Chapter 14 - Partial Derivatives - 14.4 Tangent Planes and Linear Approximations - 14.4 Exercises - Page 975: 21


$L(x,y,z)=\frac{3}{7}x+\frac{2}{7}y+\frac{6}{7}z$. The function evaluated at the point $(3.02,1.97,5.99)$: $L(3.02,1.97,5.99)=6.9914$

Work Step by Step

The linear approximation to a three-variable function at a point $(x_0,y_0,z_0)$ is given by the formula: $L(x,y,z)=f_x(x-x_0)+f_y(y-y_0)+f_z(z-z_0)+w_0$ where $w_0$ is the output of the function evaluated at the given point. The fucntion $f(x,y,z)=\sqrt{x^2+y^2+z^2}$ is provided and the linear approximation must be found to at the point $(3,2,6)$. Finding the partial derivatives w.r.t $x$, $y$, and $z$: $f_x=\frac{1}{2}(x^2+y^2+z^2)^{-\frac{1}{2}} \cdot 2x=\frac{x}{\sqrt{x^2+y^2+z^2}}$ Because of the symmetry of the function, all the derivatives will assume the same general form: $f_y=\frac{1}{2}(x^2+y^2+z^2)^{-\frac{1}{2}} \cdot 2y=\frac{y}{\sqrt{x^2+y^2+z^2}}$ $f_x=\frac{1}{2}(x^2+y^2+z^2)^{-\frac{1}{2}} \cdot 2z=\frac{z}{\sqrt{x^2+y^2+z^2}}$ We find the value of the function and its partial derivatives at the point $(3,2,6)$: $w_0=f(3,2,6)=\sqrt{3^2+2^2+6^2}=7$ $f_x(3,2,6)=\frac{3}{7}$ $f_y(3,2,6)=\frac{2}{7}$ $f_z(3,2,6)=\frac{6}{7}$ Combining all the results above, the linear approximation to $f$ is as follows: $L(x,y,z)=\frac{3}{7}(x-3)+\frac{2}{7}(y-2)+\frac{6}{7}(z-6)+7=\frac{3}{7}x+\frac{2}{7}y+\frac{6}{7}z$. Using the linear approximation, the function can be evaluated at the point $(3.02,1.97,5.99)$: $L(3.02,1.97,5.99)=\frac{3}{7}(3.02)+\frac{2}{7}(1.97)+\frac{6}{7}(5.99)=6.9914$ The actual value of the function at that point is $6.9915$. Thus, the linear approximation proves to be useful in simplifying computations.
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