Proving a Connected Surface is Contained in a Sphere

[bruno321]

Homework Statement

Show that if all normals to a connected surface pass through the origin, the surface is contained in a sphere.

Homework Equations

The Attempt at a Solution

I know a surface is locally the graph of a differentiable function, so in a neighbourhood of a point p, the points satisfy the equation F(x,y,z) = 0. Then a normal vector would be grad(F)(p), and the parameterized normal line would be X(t) = p + t*grad(F)(p).

I know that line passes through the origin, so for some t, X(t)=(0,0,0). But then I am lost. I don't really know how to handle the problem.

I also thought of taking a parameterization of a neighbourhood of p, then a basis for the tangent plane is given by the partial derivatives of the parameterization, and a normal vector is the vector product of those derivatives.

Thanks for any help :)

 

[VKint]

Hint: If v is a vector function on R³, and v(x₀) passes through the origin, then we can write v(x₀) = kx₀ for some scalar k.

 

[Dick]

If the surface is connected then any two points on the surface can connected with a curve c(t). c'(t) is perpendicular to the normal. VKint's point is that c(t) is parallel to the normal. What is the derivative of c(t) dotted with itself?

 

[bruno321]

Well, I managed to do it differently. I didn't realize the exercise preceding this one was going to help me :P

This other exercise said that if S is a connected surface, f: S->R a differentiable function, and the differential of f is always 0, then f is a constant function. This is easily proved using the mean value theorem for one variable and the chain rule.

On the exercise I posted above, then, all I need is to take the norm function squared: f(x)=||x||^2 , which is a differentiable function, and S is a connected surface.

If . is the inner product, then the differential df_p(v) = grad(f)(p).v = 2 p.v = 0 because v is in TpS and p obviously lies on the normal line through p.

Then using the result posted above, f is constant => the norm squared is constant => the norm is constant => the surface is contained on a sphere.