# If all normals to a surface pass through a point => surface contained in a sphere

1. Apr 26, 2009

### bruno321

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. 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 :)

2. Apr 26, 2009

### 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.

3. Apr 26, 2009

### 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?

4. Apr 27, 2009

### 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.