Let be a regular plane curve such that for all . Suppose there is a point such that . Prove that the curvature at that point satisfies .

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- September 4th 2009, 07:58 AMchiph588@Curvature
Let be a regular plane curve such that for all . Suppose there is a point such that . Prove that the curvature at that point satisfies .

- September 4th 2009, 08:16 AMInvisibleMan
Look at it this way, the curve is enclosed inside a circle of radius 1, there's a point in c which is tangent to this circle (at t0), now if it's the tangent circle to this point in the curve then its radius of curvature coinicide with the radius of the circle, otherwise the radius of curvature is smaller than the radius of the unit circle.

- September 4th 2009, 01:39 PMThePerfectHacker
Since is a regular curve it means there exists a function such that is the curve you are working with, and such that where is an interval. Thus, what we are doing is reparametrizing the regular curve into a

__unit-speed__curve. You are told that somewhere on this curve the curve reaches the boundary of the unit disk, so for some we have . The curvature is given by since is a unit-speed curve. We need to show that .

First we make a simple observation that . This is easy to show because since . Now differenciate both sides to get . This tells us that at any point .

Define the function by , that is, is the square of the length of the complex number.

Consider , this is a real-valued function defined on . But since never leaves the disk we have that , but at the problem says that . Therefore, has a local maximum at . This forces, . Now use the multivariable chain rule, , to get and .

We have shown that but we also shown above that , thus, . Thus, where .

Returning back to the condition and substituting the result above we have that . But that means , thus, . - September 4th 2009, 02:28 PMchiph588@
Doesn't have a local maximum at , not a minimum?

- September 4th 2009, 02:53 PMThePerfectHacker
- September 4th 2009, 03:10 PMchiph588@
Not quite, I'm a bit confused as to why the domain of is .

- September 4th 2009, 03:44 PMThePerfectHacker
By definition is the set . We just do not write complex numbers as we write them as . If it helps you can think of as . The only reason why I used the complex number notation was that I started thinking of this problem with complex numbers so I decided to stick to complex numbers.

Here is the theorem that we used.

**Theorem:**Let be a differenciable multivariable function ( ) and let be a differenciable curve in -space. Then is differenciable (in the classical sense, that is, one variable) with derivative .

This should be look like the classical chain rule because is the generalization of 'derivative' in . There are more general chain rules, and if you curious I can show you, but there are just not necessary here.

Now we defined (in your problem) as , that is, the square of the distance. Thus, , so that means, . Therefore, the derivative of would be .