Q. 65 - Calculus

Question

Use the first-order partial derivatives of the functions in Exercises$65$ and $66$to find the equation of the hyperplane tangent to the graph of the function at the indicated point P. Note that these are the same functions as in Exercises $53$ and $54 .$

$f(x,y,z)=x^2+y^2-z^3,P=(1,-5,3)$

Step-by-Step Solution

Answer

The solution's equation is$f(x,y,z)=x^2+y^2-z^3,P=(1,-5,3)$ is$2x-10y-27z-w=-28$

Step1:Given data

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

The Line of Segment Equation

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

$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: find fx

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

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

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

Step3: find fy

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

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

$f_y(1,-5,3)=-10$ $\left(3\right)$

Step4:Find fz:

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

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

$f_z(1,-5,3)=-27$ $\left(4\right)$

Step5:Find (fx,fy,fz)

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

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

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

Step6:Solution

Substituting equation $f_x(1,-5,3)=2$,$f_y(1,-5,3)=-10$,$f_z(1,-5,3)=-27$ and$f\left(x_0,y_0,z_0\right)=-1$i$f_x(1,-5,3)(x-1)+f_y(1,-5,3)(y+5)+f_z(1,-5,3)(z-3)=w-f(1,-5,3)$ get

$\begin{array}{r}2(x-1)-10(y+5)-27(z-3)=w+1\\ 2x-2-10y-50-27z+81=w+1\end{array}$

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

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