Can Differentiation Under the Integral Sign Solve Regular Curves Family Length?

[Demon117]

Homework Statement

Let [itex]\alpha : I =[a,b]→R^{2}[/itex] be a rgular curve parametrized by arc length [itex]s[/itex]. Let [itex]f:I→R[/itex] be a differentiable function with [itex]f(a)=f(b)=0[/itex]. For small values of [itex]\epsilon[/itex]

[itex]\alpha_{\epsilon}:=\alpha (s) + \epsilon f(s) N(s)[/itex]

defines a one parameter family of regular curves having endpoints [itex]\alpha(a)[/itex] and [itex]\alpha(b)[/itex]. The length of these curves defines a function

[itex]L(\epsilon):=L_{a}^{b}(\alpha_{\epsilon})=\int_{a}^{b} \left|\frac{d}{ds}\alpha_{\epsilon}\right| ds[/itex]

By differentiating under the integral sign, show that

[itex]L'(0)=-\int_{a}^{b}k_{\alpha}f(s)ds[/itex]

where [itex]k_{\alpha}[/itex] is the curvature of the function [itex]\alpha[/itex].

The Attempt at a Solution

My attempt at the solution is somewhat limited to purely differentiating the term inside the integrand. I merely want to check that I am approaching this the right way. What I have so far is the following. Since the term [itex]k_{\alpha}f(s)[/itex] is continuous we can differentiate under the integral sign. We have therefore,

[itex]\frac{d}{ds}\alpha_{\epsilon}=\alpha'(s)+\epsilon [f'(s)N(s)+f(s)N'(s)][/itex]

[itex]=\alpha'(s)+\epsilon [f'(s)N(s)+f(s)(-k_{\alpha}(s)T(s))][/itex]

Where T(s) is the tangent to the curve [itex]\alpha[/itex]. Furthermore, this gives

[itex]\alpha'(s)+\epsilon [f'(s)(\frac{1}{k_{\alpha}(s)}T'(s)+f(s)(-k_{\alpha}(s)\alpha'(s))][/itex]

[itex]=\alpha'(s)+\epsilon [f'(s)(\frac{1}{k_{\alpha}(s)}\alpha''(s)+f(s)(-k_{\alpha}(s)\alpha'(s))][/itex]

Now this is where it gets tricky. I want to show

[itex]L'(0)=-\int_{a}^{b}k_{\alpha}f(s)ds[/itex]

which implies that I look at the derivative of the function at [itex]\epsilon=0[/itex]. But how would one differentiate [itex]\left|\frac{d}{ds}\alpha_{\epsilon}\right|[/itex] with the "modulus" present?
Either I am over complicating this or I have simply forgotten this sort of exercise from previous work. . . Suggestions?

 

[Demon117]

I realize what my problem was now. I took a wrong turn and now realize that I must next differentiate with respect to [itex]\epsilon[/itex] then take the limit as [itex]\epsilon[/itex] approaches zero. Furthermore I could get away with utilizing the definition of the inner product [itex]\left|\frac{d}{ds}\alpha_{\epsilon}(s)\right|=\sqrt{<\frac{d}{ds}\alpha_{\epsilon}(s),\frac{d}{ds}\alpha_{\epsilon}(s)>}[/itex].

Following through with some differentiation on [itex]s[/itex] and then differentiation the inner product with respect to [itex]\epsilon[/itex]; things simplify quite a bit once the limit is taken. The result immediately pops out after some useful inner product applications. Now to write down a rigorous proof of this.