[Differential Geometry of Curves] Prove the set f(p) = 0 is a circle

[cheersdup]

Homework Statement

Consider a function f that can be put in the form f(p) = g(|p|) where g : [0,+∞) -> ℝ is C¹ with g(0) < 0 and g'(t) > 0 for all t ≥ 0

Assume that |∇f(p)| = 1 for all p ≠ 0 and prove that the set f(p) = 0 is a circle.

Homework Equations

Given above

The Attempt at a Solution

I know i can use |∇f(p)| = 1 because some circle parametrization will be (cos(p), sin(p)) but I can't figure out really where to start.

 

[Dick]

Homework Statement

Consider a function f that can be put in the form f(p) = g(|p|) where g : [0,+∞) -> ℝ is C¹ with g(0) < 0 and g'(t) > 0 for all t ≥ 0

Assume that |∇f(p)| = 1 for all p ≠ 0 and prove that the set f(p) = 0 is a circle.

Homework Equations

Given above

The Attempt at a Solution

I know i can use |∇f(p)| = 1 because some circle parametrization will be (cos(p), sin(p)) but I can't figure out really where to start.

I don't see anything too subtle here. If you can show there is exactly one value of c>0 such that g(c)=0, then the solution to f(p)=0 is just |p|=c. Which is a circle, yes?

 

[cheersdup]

I'm sorry, I am terrible at this... how is f(p) = 0 a circle for |p| = c?

 

[Dick]

I'm sorry, I am terrible at this... how is f(p) = 0 a circle for |p| = c?

Are you asking why |p|=c is a circle?

 

[cheersdup]

Essentially, yes. What coordinate system are we working in?

 

[micromass]

Essentially, yes. What coordinate system are we working in?

The normal coordinate system on $\mathbb{R}^2$, I suppose.
Then $|(x,y)|=\sqrt{x^2+y^2}$ is the distance to the origin. So |p|=c says that p has distance c to the origin. So all points with |p|=c are all the points with distance c to the origin, which is a circle, right?

 

[Dick]

Essentially, yes. What coordinate system are we working in?

Any coordinate system. What's the definition of a circle around the origin? |p| is the distance from p to origin, isn't it?