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:

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


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

Is this method correct?

| cite | improve this question | | | | |

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.

| cite | improve this answer | | | | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.