how to derive equation of tangent plane to a surface z=f(x,y)
i read it on
http://tutorial.math.lamar.edu/Classes/CalcIII/TangentPlanes.aspx
but he is dividing throughout by c the coefficient of z . that is he is assuming that the coefficient of z is non zero is this true in general?

Question

how to derive equation of tangent plane to a surface z=f(x,y)
i read it on 
http://tutorial.math.lamar.edu/Classes/CalcIII/TangentPlanes.aspx
but he is dividing throughout by c the coefficient of z . that is he is assuming that the coefficient of z is non zero is this true in general?

turingtest · Answer

no, but if the z-coefficient of z is zero you just get$$z_0=a(x-x_0)+b(y-y_0)$$and you can continue the derivation from there. you will just get a plane parallel to the xy-plane

turingtest · Answer

the easier way to get a tangent plane in my opinion is just to take the gradient at that point and construct a plane from that

clearly it is possible to have a gradient with z-component zero, which will cause the z to drop out of the equation for the tangent plane$$\vec n\cdot(\vec P-\vec P_0)$$

turingtest · Answer

$$\vec n\cdot(\vec p-\vec P_0)=0$$I mean...

turingtest · Answer

where$$\vec n=\nabla f(x,y,z)$$

anonymous · Answer

It depends on whom he was speaking to. For high school and undergraduate students, before dividing the equation by c, he should have mentioned the case c = 0 and discuss the case, then divided throughout by c if c is not 0 or 1. For graduate students and beyond, since he was concentrating on the point, the audiences should know the case c = 0 as Turing Test has explained.