Q. 51 - 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. $f(x,y)=\sqrt{\frac{y}{x}}$

Step-by-Step Solution

Answer

points where the first-order partial derivatives of the functions in  $f(x,y)=\sqrt{\frac{y}{x}}$ is $9x-y+6z=18$

Step1: Given.

Given $f(x,y)=\sqrt{\frac{y}{x}}$ at point $P=\left(x_0,y_0\right)=(1,9)$

The equation of line of tangent is

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

$f_x(1,9)(x-1)+f_y(1,9)(y-9)=z-f(1,9)$                                 $\left(1\right)$

Step2: Considering.

$f_x\left(x_0,y_0\right)={\left.\frac{d}{dx}f(x,y)\right|}_{\left(x_0,y_0\right)}={\left.\frac{d}{dx}\sqrt{\frac{y}{x}}\right|}_{\left(x_0,y_0\right)}$

$f_x\left(x_0,y_0\right)=\frac{d}{dx}\left({\left.\sqrt{\left.\frac{y}{x}\right)}\frac{d}{dx}\left(\frac{y}{x}\right)\right|}_1\right.$

$f_x\left(x_0,y_0\right)={\left.\frac{1}{2\sqrt{\frac{y}{x}}}\left(-\frac{y}{x^2}\right)\right|}_{\left(x_0,y_0\right)}$

$f_x(1,9)={\left.\frac{1}{2\sqrt{\frac{y}{x}}}\left(-\frac{y}{x^2}\right)\right|}_{(1,9)}$

$f_x(1,9)=\frac{1}{2\sqrt{\frac{9}{1}}}\left(-\frac{9}{1^2}\right)$

$f_x(1,9)=-\frac{9}{6}$                                                    $\left(2\right)$

Step3: Considering.

$f_y\left(x_0,y_0\right)={\left.\frac{d}{dy}f(x,y)\right|}_{\left(x_0,y_0\right)}={\left.\frac{d}{dy}\sqrt{\frac{y}{x}}\right|}_{\left(x_0,y_0\right)}$

$f_y\left(x_0,y_0\right)={\left.\frac{d}{dy}\left(\sqrt{\frac{y}{x}}\right)\frac{d}{dy}\left(\frac{y}{x}\right)\right|}_{\left(x_0,y_0\right)}$

$f_y\left(x_0,y_0\right)={\left.\frac{1}{2\sqrt{\frac{y}{x}}}·\frac{1}{x}\right|}_{\left(x_0,y_0\right)}$

$f_y(1,9)={\left.\frac{1}{2x\sqrt{\frac{y}{x}}}\right|}_{(1,9)}$

$f_y(1,9)=\frac{1}{2\times 1\times \sqrt{\frac{9}{1}}}$

$f_y(1,9)=\frac{1}{6}$                                              $\left(3\right)$

Step4: Equation ( 4 ) .

$f\left(x_0,y_0\right)=f(1,9)=\sqrt{\frac{9}{1}}$

$\$f(1,9)=3\$$                                                      $\left(4\right)$

Step5: Substituting ( 2 ) , ( 3 )   a n d   ( 4 )   in equation ( 1 ) to get.

$$\begin{gathered} -\frac{9}{6}\left(x-1\right)+\frac{1}{6}\left(y-9\right)=z-3 \\ -\frac{9}{6}x+\frac{9}{6}+\frac{1}{6}y-\frac{9}{6}=z-3 \\ -\frac{9}{6}x+\frac{1}{6}y-z=-3 \end{gathered}$$

Multiply by 6 on both sides, get

$$\begin{gathered} -9x+y-6z=-18 \\ 9x-y+6z=18 \end{gathered}$$