Q. 64 - Calculus

Question

$f(x,y)=\ln \left(x{y}^2\right),P=(1,-3)$

Step-by-Step Solution

Answer

The solution's response $f(x,y)=\ln \left(x{y}^2\right),P=(1,-3)$ is$3x-2y-3z=2.4084$

Step1: Given data

The equation of line of segment is

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

$f_x(1,-3)(x-1)+f_y(1,-3)(y+3)=z-f(1,-3)$  $\left(1\right)$

Step2: Find fx

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

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

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

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

$f_x(1,-3)={\left.\frac{1}{1}\right|}_{(1,-3)}$

$f_x(1,-3)=1$  $\left(2\right)$

Step3: Find fy

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

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

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

$f_y(1,-3)={\left.\frac{2}{y}\right|}_{(1,-3)}$

$f_y(1,-3)=-\frac{2}{3}$  $\left(3\right)$

Step4: Find f ( 1 , - 3 )

$$\begin{gathered} f\left(x_0,y_0\right)=f(1,-3) \\ =\ln \left(1\times (-3{)}^2\right) \end{gathered}$$

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

$f(1,-3)=2.1972$  $\left(4\right)$

Step 5:Solution

Subsututing Equation  $f_x(1,-3)=1$,$f_y(1,-3)=-\frac{2}{3}$and $f(1,-3)=2.1972$ in$f_x(1,-3)(x-1)+f_y(1,-3)(y+3)=z-f(1,-3)$ get

$1(x-1)-\frac{2}{3}(y+3)=z-2.1972$

$x-1-\frac{2}{3}y-2-z=-2.1972$

$x-\frac{2}{3}y-z=-2.1972+1+2$

$x-\frac{2}{3}y-z=0.8028$

multiple $3$on both side ,get

$3x-2y-3z=2.4084$