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?
Best Answer
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
$$ 4x+y+4z-9=0$$
which agrees with your result.