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.