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

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- Nov 6th 2008, 12:13 PMAsh_underparTangent 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 PMabender
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 $\displaystyle a(x-x_{0}) + b(y-y_{0}) + c(z-z_{0}) = 0 $, where $\displaystyle a,b,c $ are constants.

Note that $\displaystyle x_{0}, y_{0}, \text{and } z_{0} $ are simply the coordinates of $\displaystyle P=(1, 1, 2) $. So, your solution will look something like this:

$\displaystyle 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 $\displaystyle x^2+y^2-z=0 $.

$\displaystyle \frac{\partial f}{\partial x} = 2x $

$\displaystyle \frac{\partial f}{\partial y} = 2y $

$\displaystyle \frac{\partial f}{\partial z} = -1 $

The gradient vector is given by:

$\displaystyle \nabla f(x,y,z) =(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}) $.

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

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

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

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

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

$\displaystyle 2(x-1) + 2(y-1) + -1(z-2) = 0 $.

Rearrange this if you so wish.

Enjoy,

-Andy