# Tangent planes anyone?

• Nov 6th 2008, 12:13 PM
Ash_underpar
Tangent planes anyone?
Find an equation for the tangent plane to the surface x^2+y^2-z=0 at the point (1,1,2)

Any ideas? thanks
• Nov 6th 2008, 02:45 PM
abender
Find an equation for the tangent plane to the surface x^2+y^2-z=0 at the point (1,1,2).

Your solution will be of the form $a(x-x_{0}) + b(y-y_{0}) + c(z-z_{0}) = 0$, where $a,b,c$ are constants.

Note that $x_{0}, y_{0}, \text{and } z_{0}$ are simply the coordinates of $P=(1, 1, 2)$. So, your solution will look something like this:
$a(x-1) + b(y-1) + c(z-2) = 0$.

We want to find the constants a, b, and c. How? We first compute the partial derivatives of our surface equation $x^2+y^2-z=0$.

$\frac{\partial f}{\partial x} = 2x$
$\frac{\partial f}{\partial y} = 2y$
$\frac{\partial f}{\partial z} = -1$

The gradient vector is given by:
$\nabla f(x,y,z) =(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z})$.

So, $\nabla f(x,y,z) = (2x, 2y, -1)$.

Then, plugging in the coordinates of $P=(1,1,2)$, we get:

$\nabla f(1,1,2) = (2, 2, -1).$

So0oo, the constants $a, b, \text{and } c \text{ are } 2, 2, \text{and } -1$, respectively.

Soooooooo, the equation for the tangent plane to the surface $x^2+y^2-z=0 \text{ at the point } (1,1,2)$ is
$2(x-1) + 2(y-1) + -1(z-2) = 0$.

Rearrange this if you so wish.

Enjoy,
-Andy