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.


1 Answer 1


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 $\Pi: a=t_0<t_1<\ldots<t_{i_0}=b$ 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 $\lim\limits_{\operatorname{diam}(\Pi)\to 0} l_P = \int_a^b ||\dot{\alpha}(t)||dt.$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .