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))
Last edited by heatly; March 5th 2011 at 01:54 AM.
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Use the well known result: is an orthogonal vector to the tangent plane for the surface at .
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.
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