Q. 52 - 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 $f(x,y)=\ln \left(x{y}^2\right)$.

Step-by-Step Solution

Answer

The answer for the equation$f(x,y)=\ln \left(x{y}^2\right)$ is $3 x-2 y-3 z=2.4084$.

Step1: Given.

Given  $f(x,y)=\ln \left(x{y}^2\right)$ at point $P=\left(x_0,y_0\right)=(1,-3)$

The equation of line of tangent is

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

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

Step2: Considering.

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

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

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

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

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

$f_x(1,-3)=1$                                                      $\left(2\right)$

Step3: Equation 3 .

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

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

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

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

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

Step4: Equation 4 .

$f\left(x_0,y_0\right)=f(1,-3)=\ln \left(1·(-3{)}^2\right)$

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

$f(1,-3)=2.1972$                                          $\left(4\right)$

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

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

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

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

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

Multiply by 3 on both sides, get

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