I have recently looked back on the formula for arc length $$\int_a^b \sqrt{1+f'(x)^2}\ dx .$$ In the case of a circle we have the interval $[-r,r]$ and can rearrange $ r^2 = x^2 + y^2 $ to $y=\sqrt{r^2-x^2}$. Since we consider the principal square root we have the top half of the circle and so can simply multiply by $ 2 $ in our final answer. In this formula we require the function that describes this curve to have a continuous derivative on the closed interval $[a, b]$ we find the arc length in. However, in the case of this "semi-circle" function the derivative $f'(x) = \frac{-x}{\sqrt{r^2 - x^2}}$ has an infinite discontinuity at $x=-r$ and $x=r$, the endpoints of our interval $[-r, r]$ . So why does the integral, when evaluated, give the correct answer ?
-
The arc length of rectifiable curve is a continuous function. More formally if the curve $x=f(t), y=g(t), t\in[a, b] $ with $f, g$ continuous on $[a, b] $ is rectifiable then the arc length of curve from $t=a$ to $t=u$ for $u\in[a, b] $ is a continuous function of $u$. A semicircle is a rectifiable curve and the result applies here. – Paramanand Singh Oct 04 '22 at 07:42
-
1+1 for raising a concern which is not usually mentioned in textbooks. – Paramanand Singh Oct 04 '22 at 07:43
3 Answers
Because you don't need a continuous derivative. You need only a continuously defined tangent, which is allowed to be vertical, on a bounded curve and the integral always converges properly. Since the endpoints here give vertical tangents, they do not really interfere.
- 39,403
-
This is not about geometrical tangent. The arc length formula assumes that $\sqrt{1+(f')^2}$ is Riemann integrable. This does not apply here as $f'$ is unbounded here. – Paramanand Singh Oct 04 '22 at 11:24
-
Riemann integrability covers improper integrals that converge, no? – Oscar Lanzi Oct 04 '22 at 12:10
-
@ParamanandSingh: That is false. In fact if $t\mapsto (t,f(t))$ is absolutely continuous, then is rectifiable, differentiable a.s. and $(1,f'(t))$ is locally integrable (this require a little of Lebesgue integration). – Mittens Oct 05 '22 at 04:33
-
Well in that case it might be better to add the details in the answer for benefit of everyone. – Paramanand Singh Oct 05 '22 at 06:58
Your formula is a particular case of the following definition: the length of a (equivalence class of) $C^1$ parametrized arc of curve is $$\int_{t_0}^{t_1}\sqrt {\left(\frac {\mathrm dx}{\mathrm dt}\right)^2 + \left(\frac {\mathrm dy}{\mathrm dt}\right)^2}\mathrm d t.$$
This makes the length a continuous function of the ends of the arc. Hence your integral, when evaluated on $[-r+\varepsilon,r-\varepsilon]$, ($0<\varepsilon<r$), gives (without singularity) the length of an arc of circle which, as $\varepsilon\to0$, tends to the length of the semi-circle.
- 34,650
-
You have implicitly assumed that arc length of rectifiable curve is a continuous function. More formally if the curve $x=f(t), y=g(t), t\in[a, b] $ with $f, g$ continuous on $[a, b] $ is rectifiable then the arc length of curve from $t=a$ to $t=u$ for $u\in[a, b] $ is a continuous function of $u$. A semicircle is a rectifiable curve and the result applies here. – Paramanand Singh Oct 04 '22 at 07:40
-
I don't think I assumed that. I deduced it from my "more general formula" (for a $C^1$ parametrised curve). – Anne Bauval Oct 04 '22 at 08:05
-
Your first paragraph is the assumption which is same as my comment. – Paramanand Singh Oct 04 '22 at 08:41
-
It was not an assumption but an observation, justified by what followed. But since it does not seem to be clear, I shall permute. – Anne Bauval Oct 04 '22 at 08:44
-
Well the length of the arc tends to the length of semicircle as $\varepsilon\to 0 $ only because of continuity of the arc-length as mentioned in my comment. This is not automatic and needs a proof. I was just trying to write this explicitly via comment. – Paramanand Singh Oct 04 '22 at 08:56
A formal definition of arc length is discussed (along with the integral formula) in this answer.
The key part to understand here is that arc length is a continuous function. To put in symbols let $x=f(t), y=g(t) $ where $f, g$ are continuous on $[a, b] $ define a simple curve $\mathcal{C} $. Further let $\mathcal{C} $ be rectifiable and if $[c, d] \subseteq [a, b] $ then let $L_{\mathcal{C}} ([c, d]) $ denote the arc length of the portion of curve $\mathcal{C} $ corresponding to $t\in [c, d] $.
Then it is a standard result that $L_{\mathcal{C}} ([a, x]) $ and $L_{\mathcal{C}} ([x, b]) $ are continuous functions of $x$ on $[a, b] $. Under certain conditions there is an explicit formula to evaluate the arc length. Thus if $f'(t), g'(t) $ are Riemann integrable on $[c, d] $ (this implies that these derivatives are necessarily bounded) then $$L_{\mathcal{C}} ([c, d]) =\int_c^d\sqrt{(f'(t)) ^2+(g'(t))^2} \, dt\tag{1}$$
Now the curve in question (let us denote it by $\mathcal{C} $) is given by $x=t, y=\sqrt{r^2-t^2},t\in[-r,r]$ and it can be proved that it is rectifiable. Note however that the derivatives $f', g'$ are not bounded on $[-r, r] $ and hence the formula $(1)$ does not apply directly.
But let us observe that the derivatives $f', g'$ are bounded in any interval of the form $[-r+h, r-h] $ where $0<h<r$. We now use continuity of the arc length mentioned earlier. We have for $0<h<r$ \begin{align} L_{\mathcal{C}} ([0,r]) & =\lim_{h\to r^{-}} L_{\mathcal{C}} ([0,h])\notag\\ &=\lim_{h\to r^{-}} \int_0^h\sqrt{(f'(t))^2+(g'(t))^2}\,dt\notag\\ &=\lim_{h\to r^{-}}r \int_0^h\frac{dt}{\sqrt{r^2-t^2}}\,dt\notag\\ &=r\lim_{h\to r^{-}} \int_0^{h/r}\frac{du}{\sqrt{1-u^2}}\text{ (putting }t=ru) \notag\\ &=r\int_0^1\frac{du}{\sqrt{1-u^2}}\notag \end{align} where the last integral exists as an improper Riemann integral. A similar treatment can be provided for $L_{\mathcal{C}} ([-r, 0])$ to reach the same result and adding them we get $$L_{\mathcal{C}} ([-r, r]) =2r\int_0^1\frac{du}{\sqrt{1-u^2}}$$ as the length of a semicircle of radius $r$.
- 87,309