Q. 54 - Calculus

Question

Find all points where the first-order partial derivatives of the functions in Exercises 43–54 are continuous. Then use Theorems 12.28 and 12.31 to determine the sets in which the functions are differentiable.$f(x,y,z)=\frac{x}{y^2+z^2}$.

Step-by-Step Solution

Answer

Answer for the equation is$f(x,y,z)=\frac{x}{y^2+z^2}$ is $x-4 z-4 w=-8$.

Step1: Given.

Given $f(x,y,z)=\frac{x}{y^2+z^2}$ at point  

The equation of line of tangent is

$f_x\left(x_0,y_0,z_0\right)\left(x-x_0\right)+f_y\left(x_0,y_0,z_0\right)\left(y-y_0\right)+f_z\left(x_0,y_0,z_0\right)\left(z-z_0\right)=w-f\left(x_0,y_0,z_0\right)$

$f_x(4,0,2)(x-4)+f_y(4,0,2)(y-0)+f_z(4,0,2)(z-2)=w-f(4,0,2)$                                   $\left(1\right)$

Step2: Consider.

$f_x\left(x_0,y_0,z_0\right)={\left.\frac{d}{dx}f(x,y,z)\right|}_{\left(x_0,y_0,z_0\right)}={\left.\frac{d}{dx}\frac{x}{y^2+z^2}\right|}_{\left(x_0,y_0,z_0\right)}$

$f_x(4,0,2)={\left.\frac{1}{y^2+z^2}\frac{d}{dx}\left(x\right)\right|}_{(4,0,2)}$

$f_x(4,0,2)={\left.\frac{1}{y^2+z^2}·1\right|}_{(4,0,2)}$

$f_x(4,0,2)={\left.\frac{1}{0^2+2^2}\right|}_{(4,0,2)}$

$f_x(4,0,2)=\frac{1}{4}$                                                              $\left(2\right)$

Step3: Equation 3 .

$f_y\left(x_0,y_0,z_0\right)={\left.\frac{d}{dy}f(x,y,z)\right|}_{\left(x_0,y_0,z_0\right)}={\left.\frac{d}{dy}\frac{x}{y^2+z^2}\right|}_{\left(x_0,y_0,z_0\right)}$

$f_y\left(x_0,y_0,z_0\right)={\left.x\frac{d}{dy}\left(\frac{1}{y^2+z^2}\right)\right|}_{\left(x_0,y_0,z_0\right)}={\left.x\frac{d}{dy}\left({\left(y^2+z^2\right)}^{-1}\right)\right|}_{\left(x_0,y_0,z_0\right)}$

$f_y\left(x_0,y_0,z_0\right)={\left.x\frac{d}{dy}\left({\left(y^2+z^2\right)}^{-1}\right)\frac{d}{dy}\left(y^2+z^2\right)\right|}_{\left(x_0,y_0,z_0\right)}$

$f_y\left(x_0,y_0,z_0\right)={\left.x\left(-\frac{1}{{\left(y^2+z^2\right)}^2}\right)·2y\right|}_{\left(x_0,y_0,z_0\right)}$

$f_y\left(x_0,y_0,z_0\right)=-{\left.\frac{2xy}{{\left(y^2+z^2\right)}^2}\right|}_{\left(x_0,y_0,z_0\right)}$

$f_y(4,0,2)=-{\left.\frac{2xy}{{\left(y^2+z^2\right)}^2}\right|}_{(4,0,2)}=-\frac{2·4·0}{{\left(0^2+2^2\right)}^2}$

$f_y(4,0,2)=0$                                                                                         $\left(3\right)$

Step4: Equation 4 .

$f_z\left(x_0,y_0,z_0\right)={\left.\frac{d}{dz}f(x,y,z)\right|}_{\left(x_0,y_0,z_0\right)}={\left.\frac{d}{dz}\frac{x}{y^2+z^2}\right|}_{\left(x_0,y_0,z_0\right)}$

$f_z\left(x_0,y_0,z_0\right)={\left.x\frac{d}{dz}\left(\frac{1}{y^2+z^2}\right)\right|}_{\left(x_0,y_0,z_0\right)}={\left.x\frac{d}{dz}\left({\left(y^2+z^2\right)}^{-1}\right)\right|}_{\left(x_0,y_0,z_0\right)}$

$f_z\left(x_0,y_0,z_0\right)={\left.x\frac{d}{dz}\left({\left(y^2+z^2\right)}^{-1}\right)\frac{d}{dz}\left(y^2+z^2\right)\right|}_{\left(x_0,y_0,z_0\right)}$

$f_z\left(x_0,y_0,z_0\right)={\left.x\left(-\frac{1}{{\left(y^2+z^2\right)}^2}\right)·2z\right|}_{\left(x_0,y_0,z_0\right)}=-{\left.\frac{2xz}{{\left(y^2+z^2\right)}^2}\right|}_{\left(x_0,y_0,z_0\right)}$

$f_y(4,0,2)=-{\left.\frac{2xy}{{\left(y^2+z^2\right)}^2}\right|}_{(4,0,2)}=-\frac{2·4·2}{{\left(0^2+2^2\right)}^2}$

$f_z(4,0,2)=-\frac{16}{16}$

$f_z(4,0,2)=-1$                                                                                    $\left(4\right)$

Step5: Equation 5 .

$f\left(x_0,y_0,z_0\right)={\left.\frac{x}{y^2+z^2}\right|}_{\left(x_0,y_0,z_0\right)}$

$f\left(x_0,y_0,z_0\right)=f(4,0,2)=\frac{4}{0^2+2^2}$

$f\left(x_0,y_0,z_0\right)=1$                                                 $\left(5\right)$

Step6: Substituting equation ( 2 ) , ( 3 ) , ( 4 ) a n d ( 5 ) in equation ( 1 ) .

$\frac{1}{4}(x-4)+0(y-0)-1(z-2)=w-1$

$\frac{1}{4}x-1+0-z+2=w-1$

$\frac{1}{4}x-z-w=-1+1-2$

$\frac{1}{4}x-z-w=-2$

Multiply by 4 on both sides, get

$x-4 z-4 w=-8$