Answer
$L(x,y,z)=\frac{3}{7}x+\frac{2}{7}y+\frac{6}{7}z$.
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.
