$\displaystyle z=x^2+y^2$ 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 $\displaystyle fx=2x$ and $\displaystyle fy=2y$

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- Sep 17th 2009, 03:28 PMsuperdudefinding equation of tangent plane
$\displaystyle z=x^2+y^2$ 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 $\displaystyle fx=2x$ and $\displaystyle fy=2y$ - Sep 17th 2009, 03:33 PMCalculus26
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