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:

is an orthogonal vector to the tangent plane for the surface at . - 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, , is normal to a level surface. Of course, that gives exactly the same thing FernandoRevilla is saying.