Q. 63 - Calculus

Question

$f(x,y)=\sqrt{\frac{y}{x}},P=(1,9)$

Step-by-Step Solution

Answer

The equation's solution $f(x,y)=\sqrt{\frac{y}{x}},P=(1,9)$is $9x-y+6z=18$

Step 1:Given data

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

The line of tangent equation is

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

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

Step 2:Find fx

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

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

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

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

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

Step 3:Find fy

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

$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}}}\cdot \frac{1}{x}\right|}_{\left(x_0,y_0\right)}$

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

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

Step4: Find f ( 1 , 9 )

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

$f(1,9)=3$$\left(4\right)$       

Step 5:Solution

Substituting equation $f_x(1,9)=-\frac{9}{6}$,$f_y(1,9)=\frac{1}{6}$and $f(1,9)=3$in$f_x(1,9)(x-1)+f_y(1,9)(y-9)=z-f(1,9)$ get

$-\frac{9}{6}(x-1)+\frac{1}{6}(y-9)=z-3$

$-\frac{9}{6}x+\frac{9}{6}+\frac{1}{6}y-\frac{9}{6}=z-3$

$-\frac{9}{6}x+\frac{1}{6}y-z=-3$

Multiple$6$ on both side

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