# Prove convergence of formula for length of parameterized curve

In class, our professor was discussing the arc length of a parameterized curve $$x=f(t), y= g(t)$$. In his derivation, he reached the sum - $$\sum_{i=1}^n L_i = \sum_{i=1}^n \sqrt{(\Delta x_i)^2 + (\Delta y_i)^2} = \sum_{i=1}^n \sqrt{(f'(t^*_i)^2 + (g'(t_i^{**})^2} \Delta t_i$$ where $$\Delta x_i = f(t_i) - f(t_{i-1}) = f'(t^*_i) \Delta t_i$$ $$\Delta y_i = g(t_i) - g(t_{i-1}) = g'(t^{**}_i) \Delta t_i$$ He then said that this sum converges to $$\int_a^b \sqrt {[f'(t)]^2 + [g'(t)]^2}$$ when the norm of the partition tends to $$0$$. However, when I asked him the proof, he said that he did not know. Now, as the sum is not a Riemann sum, we can't directly prove convergence. So, how do we go about proving convergence? I don't get this.

• Sep 23, 2023 at 10:05

This has to do with the theorem which says that given (regular) smooth curve $$\Gamma,$$ with a smooth compact parametrization $$\alpha \in C^{1}([a,b],\mathbb{R}^n),$$ with $$\dot{\alpha} \neq \mathcal{O}$$ in $$[a,b]$$ the curve is rectifiable (i.e. it has finite length) and it's given by $$l(\Gamma)=\int\limits_a^b ||\dot{\alpha}(t)||dt=:I.$$

So one proof, illustrated in $$\mathbb{R}^2$$ for technical simplicity is what your professor showed you. For any partition $$\Pi: a=t_0 of $$[a,b].$$ Following the points $$A_0=\alpha(a_1),\ldots,A_{i_0}=\alpha(t_{i_0})$$ on the curve calculating the lenght of piecewise line approximating the lenght of $$\Gamma$$ below is \begin{align*} l_\Pi = \sum_{i=1}^{i_0} || A_i - A_{i-1} || = \sum_{i=1}^{i_0} ||\alpha(t_i)-\alpha(t_{i-1})||&= \sum_{i=1}^{i_0} \sqrt{ [\alpha_1(t_i)-\alpha_1(t_{i-1})]^2 + [\alpha_2(t_i)-\alpha_2(t_{i-1})]^2} \\ &=\sum_{i=1}^{i_0} \sqrt{\dot{\alpha_1}(\xi_i)^2+\dot{\alpha_2}(\eta_i)^2}(t_i - t_{i-1}), \end{align*} where by $$\xi_i,\eta_i \in (t_{i-1},t_i)$$ are obtained by Lagrange mean value theorem. Now this would have be a integral sum if $$\eta_i = \xi_i$$ were the same. It's kind of obvious that this sum if convergent to the same thing, tho, as the integral sum is $$\lim_\limits{\operatorname{diam}(\Pi)\to0}\sum_{i=1}^{i_0} \sqrt{\dot{\alpha_1}(\xi_i)^2+\dot{\alpha_2}(\xi_i)^2}(t_i - t_{i-1})=\lim_{\operatorname{diam}(\Pi)\to 0} \mathfrak{S}_{||\dot{\alpha}(\cdot)||}(\Pi,\{\xi\}_{i=1}^{i_0})=\int_a^b ||\dot{\alpha}(t)||dt,$$ since this is now a integral-Rimann sum of the function $$||\dot{\alpha}(t)||$$ in $$[a,b].$$ More formally the above means that $$\forall \varepsilon > 0 \exists \delta >0$$ s.t. for all partitons $$\Pi$$ of $$[a,b]$$ with $$\operatorname{diam}(\Pi)<\delta$$ we have

$$\left|\mathfrak{S}_{||\dot{\alpha}(\cdot)||}(\Pi,\{\xi\}_{i=1}^{i_0})-\int_a^b ||\dot{\alpha}(t)||dt \right| < \varepsilon.$$

Now for each $$\xi_i$$ of those, denothing with $$\Phi_{\xi_i}(t):=\sqrt{\dot{\alpha}(\xi_i)^2 + \dot{\alpha}(t)^2}$$ is continous function on $$t \in [a,b],$$ in particular in $$t \in [t_{i-1},t_i].$$ From Cantor theorem it's equicontinous in the compact $$[a,b]$$ and so $$\forall \varepsilon > 0 \exists \delta > 0$$ s.t. $$|\Phi_{\xi}(t')-\Phi_{\xi}(t'')| < \varepsilon, \hspace{0.2cm} \forall t',t'' \in [a,b]: \hspace{0.2cm} |t'-t''|<\delta.$$ In particular it gives $$|\Phi_{\xi}(\eta_i)-\Phi_{\xi}(\xi_i)| < \varepsilon,$$ provided that $$\operatorname{diam}(\Pi)<\delta.$$

Now fix $$\varepsilon>0$$ we find $$\delta>0$$ so that $$|\Phi_{\xi}(t')-\Phi_{\xi}(t'')|<\frac{\varepsilon}{2(b-a)}$$ for $$\operatorname{diam}(\Pi)<\delta.$$

Now evaluating \begin{align*} |l_\Pi - \mathfrak{S}_{||\dot{\alpha}(\cdot)||}(\Pi,\xi)|=\left|\sum_{i=1}^{i_0}(\Phi_{\xi_i}(\eta_i) -\Phi_{\xi_i}(\xi_i))(t_i - t_{i-1}) \right| &\leqq \sum_{i=1}^{i_0} |\Phi_{\xi_i}(\eta_i) -\Phi_{\xi_i}(\xi_i)| (t_i-t_{i-1}) \\ &\leqq \frac{\varepsilon}{2(b-a)} (b-a)=\frac{\varepsilon}{2}. \end{align*}

And finally $$|l_{\Pi}-I|=|l_{\Pi}-\mathfrak{S}_{||\dot{\alpha}(\cdot)||}(\Pi,\xi)+\mathfrak{S}_{||\dot{\alpha}(\cdot)||}(\Pi,\xi)-I|<\frac{\varepsilon}{2}+|\mathfrak{S}_{||\dot{\alpha}(\cdot)||}(\Pi,\xi)-I|<\varepsilon,$$ for all partitons $$\Pi$$ with $$\operatorname{diam}(\Pi)< \min\{\delta,\delta'\},$$ where $$\delta'>0$$ is so small that $$|\mathfrak{S}_{||\dot{\alpha}(\cdot)||}(\Pi,\xi)-I|<\frac{\varepsilon}{2}$$ for all $$\Pi$$ with $$\operatorname{diam}(\Pi)<\delta'.$$

This proves $$\lim\limits_{\operatorname{diam}(\Pi)\to 0} l_P = \int_a^b ||\dot{\alpha}(t)||dt.$$