# Finding a tangent plane to a level surface

Suppose that i have some function $f:\mathbb{R}^3 \rightarrow \mathbb{R}$, and a point $P$ in space where $f$ is differentiable. Now, i want to find the equation of a plane $L$ such that $L$ is tangent to the level surface where $P$ is located. So, for example: $f(x,y,z) = \frac{1}{x^2 + y + z^2}$, and the point $P=(2,-7,2)$. So, what i did was:

1) Calculate the value of $f$ in $P$. This will be the value of the surface level.

$f(x,y,z) = \frac{1}{1} = 1$.

2) Find the equation for the surface level:

$1 = \frac{1}{x^2+y+z^2} \implies z = \sqrt{1-x^2-y}$

3) Calculate the value of the derivatives:

$\frac{\partial z}{\partial x} = -1$

$\frac{\partial z}{\partial y} = -\frac{1}{4}$

4) Use the equation of a tangent plane:

$z-z_0 = \frac{\partial z}{\partial x}(x-x_0) + \frac{\partial z}{\partial y}(y-y_0)$

so

$z-2 = -(x-2) -\frac{1}{4}(y+7)$.

Is this method correct?

This is correct. You could use the alternate equation

\begin{equation} f_x(x_0,y_0,z_0)(x-x_0) + f_y(x_0,y_0,z_0)(y-y_0) + f_z(x_0,y_0,z_0)(z-z_0)=0 \end{equation}

since the gradient is a normal vector for the tangent plane at $(x_0,y_0,z_0)$ and \begin{eqnarray} \nabla f(x,y,z)&=&\left\langle \frac{-2x}{x^2+y+z^2}, \frac{-1}{x^2+y+z^2}, \frac{-2z}{x^2+y+z^2}\right\rangle \end{eqnarray}

$$\nabla f(2,-7,2)=\left\langle -4,-1,-4\right\rangle$$

but since we can use any other vector which is parallel to this one we may choose

$$N=\left\langle 4,1,4\right\rangle$$

So the equation of the tangent plane is

$$4(x-2)+(y+7)+4(z-2)=0$$

which can be written

$$4x+y+4z-9=0$$