2 different parametrizations of a surface gives another normal

Question

I have the group of points $$S =\{(x,y,z)\mid z=4-x^2-y^2, z\geq 0\}$$ Now the first parametrezation is: $f(x,y)=(x,y,4-x^2-y^2)$ such that $0\leq x^2+y^2 \leq 4$ and the normal to this parametrization is: $(2x,2y,1)$ and we can see, at the point $(0,0,4)$ the normal pointing to $z$ axis direction as we want. Now the other param is : $g(r,t)=(r\cos t,r\sin t,4-r^2)$ and $0\leq t \leq 2\pi$ and $0\leq r \leq 2$ now the normal for this param is: $(2r^2 \cos t,2r^2\sin t,r)$ now if we look at the point $(0,0,4)$ which means $r=0$ then we get the normal $(0,0,0)$. Can someone explain me why in 2 different param, we get another normals at the same point? and how the normals relate to each other? it looks like for me as a scalar multiplication , but at this point, i see there is no normal in the second param, its weird , thanks for help!

Answer 1

The normal is given by the cross product $\frac{\partial f}{\partial x} \times \frac{\partial f}{\partial y}$ of two tangent vectors. These tangent vectors depend on the parametrization. Therefore different parametrizations, give different normals.

A normal vector in a point is unique up to a scalar multiple. This multiple can be negative (i.e. the normal points in the other direction), and can be zero, as in the point $r=0$ in your second parametrization. In points where the normal has zero length, the tangent vectors are linearly dependent. Points where the tangent vectors are linearly independent are called regular points. This regularity depends on the parametrization itself, not on the surface.

Added: since $x=r\cos t$ and $y=r\sin t$, the normal vectors for $g$ are $(2r x, 2r y, r)$, so this is the same as the normal vectors for $f$, but up to a multiple $r$.