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

[Demon117]

Homework Statement

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

\alpha_{\epsilon}:=\alpha (s) + \epsilon f(s) N(s)

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

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

By differentiating under the integral sign, show that

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

where k_{\alpha} is the curvature of the function \alpha.

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 k_{\alpha}f(s) is continuous we can differentiate under the integral sign. We have therefore,

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

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

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

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

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

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

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

which implies that I look at the derivative of the function at \epsilon=0. But how would one differentiate \left|\frac{d}{ds}\alpha_{\epsilon}\right| 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 \epsilon then take the limit as \epsilon approaches zero. Furthermore I could get away with utilizing the definition of the inner product \left|\frac{d}{ds}\alpha_{\epsilon}(s)\right|=\sqrt{<\frac{d}{ds}\alpha_{\epsilon}(s),\frac{d}{ds}\alpha_{\epsilon}(s)>}.

Following through with some differentiation on s and then differentiation the inner product with respect to \epsilon; 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.