Find direction of normal and the equation of the tangent plane at the point.

z=sqrt(4-x^2-y^2)

at (1,1,sqrt(2))

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- Mar 4th 2011, 02:11 AMheatlyPartial Derivitives
Find direction of normal and the equation of the tangent plane at the point.

z=sqrt(4-x^2-y^2)

at (1,1,sqrt(2)) - Mar 4th 2011, 02:25 AMFernandoRevilla
Use the well known result:

$\displaystyle \vec{v}=(f_x(P_0),f_y(P_0),-1)$

is an orthogonal vector to the tangent plane for the surface $\displaystyle S\equiv z=f(x,y)$ at $\displaystyle P_0(x_0,y_0,z_0)\in S$. - Mar 4th 2011, 03:40 AMHallsofIvy
I prefer to think of such a surface as being a level surface for g(x,y,z)= f(x,y)- z= 0 and using the fact that the gradient, $\displaystyle \nabla g= f_x\vec{i}+ f_y\vec{j}- \vec{k}$, is normal to a level surface. Of course, that gives exactly the same thing FernandoRevilla is saying.