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
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