Q. 53 - 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. $f(x, y,z) = x 2 + y 2 - z 3$.

Step-by-Step Solution

Answer

Answer for the equation is$f(x, y,z) = x 2 + y 2 - z 3$ is $2 x-10 y-27 z-w=-28$

Step1: Given.

Given $f(x,y,z)=x^2+y^2-z^3$ at point $P=\left(x_0,y_0,z_0\right)=(1,-5,3)$

The equation to the hyperplane tangent is

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

$f_x(1,-5,3)(x-1)+f_y(1,-5,3)(y+5)+f_z(1,-5,3)(z-3)=w-f(1,-5,3)$                     $\left(1\right)$

Step2: Consider.

$f_x\left(x_0,y_0,z_0\right)={\left.\frac{d}{dx}f(x,y,z)\right|}_{\left(x_0,y_0,z_0\right)}={\left.\frac{d}{dx}\left(x^2+y^2-z^3\right)\right|}_{\left(x_0,y_0,z_0\right)}$

$f_x(1,-5,3)={\left.2x\right|}_{(1,-5,3)}$

$f_x(1,-5,3)=2$                                                                                                       $\left(2\right)$

Step3: Equation 3 .

$f_y\left(x_0,y_0,z_0\right)={\left.\frac{d}{dy}f(x,y,z)\right|}_{\left(x_0,y_0,z_0\right)}={\left.\frac{d}{dy}\left(x^2+y^2-z^3\right)\right|}_{\left(x_0,y_0,z_0\right)}$

$f_y(1,-5,3)={\left.2y\right|}_{(1,-5,3)}$

$f_y(1,-5,3)=-10$                                                                    $\left(3\right)$                                   

Step4: Equation 4 .

$f_z\left(x_0,y_0,z_0\right)={\left.\frac{d}{dz}f(x,y,z)\right|}_{\left(x_0,y_0,z_0\right)}={\left.\frac{d}{dz}\left(x^2+y^2-z^3\right)\right|}_{\left(x_0,y_0,z_0\right)}$

$f_z(1,-5,3)=-{\left.3z^2\right|}_{(1,-5,3)}$

$f_z(1,-5,3)=-27$                                                                                            $\left(4\right)$

Step5: Equation 5 .

$f\left(x_0,y_0,z_0\right)=f(1,-5,3)=x^2+y^2-{\left.z^3\right|}_{\left(x_0,y_0,z_0\right)}$

$f\left(x_0,y_0,z_0\right)=1^2+(-5{)}^2-3^3$

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

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

$2(x-1)-10(y+5)-27(z-3)=w+1$

$2 x-2-10 y-50-27 z+81=w+1$

$2 x-10 y-27 z-w=1+2+50-81$

$2 x-10 y-27 z-w=-28$