# finding equation of tangent plane

• September 17th 2009, 04:28 PM
superdude
finding equation of tangent plane
$z=x^2+y^2$ at point (3,4,5)
I've taken the partial derivatives and plugged in the value for x and y. I'm not sure what to do next.

So $fx=2x$ and $fy=2y$
• September 17th 2009, 04:33 PM
Calculus26
The normal is

N = -fx i -fy j + k evaulated at (3,4,5)

N = -6 i - 8 j + k

If the normal to a plane is ai+bj+ck then then the eqn is

a(x-x0)+b(y-y0) +z -z0 = 0

-6(x-3) -8(y-4) + z - 5 = 0

Solve for z