## Calculating Arc-Length of Curves

1. Recall we have some curve usually of the form $y=f(x)$. But sometimes we can write it as $(x(t), y(t))$ where both coordinates are a function of “time” $t$.

Convention: we will set $0\leq t\leq1$. [Caveat: some exercises may have a different range of values for $t$, be careful!]

(This isn’t horribly strange: if we let $x(t)=t$ then we recover the familiar $y=f(x)$ case!)

The problem: how do we calculate the length of a curve $(x(t), y(t))$?

2. Consider the curve

We want to consider its length, and we’re given its coordinates parametrized by “how far along” we are on the curve.

One solution is to make some approximation of the curve as a collection of lines. Why do this? Well, we know how to compute the distance between two points $(x_{0}, y_{0})$ and $(x_{1}, y_{1})$ as:
(1)$\displaystyle \Delta s^{2} = \Delta x^{2}+\Delta y^{2}$
This is just the Pythagorean theorem.

So we make our approximation of our curve, for example:

But look, if we pick some partition of the interval $[0,1]$, then we can rewrite the length of each segment described in Equation (1) as
(2)$\displaystyle \Delta s^{2} = \left[\left(\frac{\Delta x}{\Delta t}\right)^{2}+\left(\frac{\Delta y}{\Delta t}\right)^{2}\right]\Delta t^{2}$
Taking the square-root of both sides
(3)$\displaystyle \Delta s = \sqrt{\left(\frac{\Delta x}{\Delta t}\right)^{2}+\left(\frac{\Delta y}{\Delta t}\right)^{2}}\Delta t$
But look, the length of the approximation is just the sum of all of these guys!

We see that the approximation gets better with more points added. With 9 points, our approximation looks like:

With 17 points it distinguishing the approximation from the curve becomes difficult:

Expressed mathematically, the length is approximately the sum of the hyptoneuse-es of the triangles
(4)$\displaystyle s \approx \sum^{n}_{k=1}\Delta s = \sum^{n}_{k=1}\sqrt{\left(\frac{\Delta x}{\Delta t}\right)^{2}+\left(\frac{\Delta y}{\Delta t}\right)^{2}}\Delta t$
Are we done? No, we only have an approximation!

We have to take the limit $n\to\infty$. Why? Because as we have seen, more points makes the approximation better. So infinitely more points makes our approximation infinitely good! The limit $n\to\infty$ turns this calculation into an integral:
(4′)$\displaystyle s = \int\mathrm{d} s = \int^{1}_{0}\sqrt{\left(\frac{\mathrm{d} x}{\mathrm{d} t}\right)^{2}+\left(\frac{\mathrm{d} y}{\mathrm{d} t}\right)^{2}}\,\mathrm{d} t$
So what’s going on? We take our curve, and divide it into really small segments which Equation (1) describes. When the segments become “infinitesimally” small, we get
(5)$\displaystyle \mathrm{d} s = \sqrt{\left(\frac{\mathrm{d} x}{\mathrm{d} t}\right)^{2}+\left(\frac{\mathrm{d} y}{\mathrm{d} t}\right)^{2}}\,\mathrm{d} t$
We cannot really calculate the left hand side of (5) directly. But we can directly calculate the right hand side!

So we have an effective method to calculate the infinitesimal arc-length of a curve. We just add up all these infinitesimal lengths, and this sum is an integral.

Example 1. Calculate the arc-length of the curve $x(t)=\cos(t)$ and $y(t)=\sin(t)$ for $0\leq t\leq \pi$.

Solution. We calculate out the derivatives first:
(6a)$\displaystyle \frac{\mathrm{d}x(t)}{\mathrm{d}t} = -\sin(t)$ thus $\displaystyle \left(\frac{\mathrm{d}x(t)}{\mathrm{d}t}\right)^{2} = \sin^{2}(t)$
and
(6b)$\displaystyle \frac{\mathrm{d}y(t)}{\mathrm{d}t} = \cos(t)$ and $\displaystyle \left(\frac{\mathrm{d}y(t)}{\mathrm{d}t}\right)^{2} = \cos^{2}(t)$.
Thus we can calculate the differential arc-length as
(7)$\displaystyle \begin{array}{rl}\displaystyle \mathrm{d} s &= \displaystyle \sqrt{\left(\frac{\mathrm{d} x}{\mathrm{d} t}\right)^{2}+\left(\frac{\mathrm{d} y}{\mathrm{d} t}\right)^{2}}\,\mathrm{d} t\\ &=\displaystyle \sqrt{\sin^{2}(t)+\cos^{2}(t)}\,\mathrm{d}t \end{array}$
But we recall the trigonometric identity
$\displaystyle \sin^{2}(t)+\cos^{2}(t)=1$
thus our arc-length differential (7) becomes
(7′)$\displaystyle \mathrm{d} s =\mathrm{d}t$.
Thus we compute the arc-length as
(8)$\displaystyle s =\int^{\pi}_{0}\mathrm{d}t = \pi-0=\pi$.
This concludes our example.

Reflection. This solution seems reasonable, since elementary geometry tells us that the circumference of a circle is $2\pi r$. Since $r=1$, and we consider half the circumference, we should expect the solution be $\pi$. That’s precisely what we get!

Exercise 1. Calculate the arc-length of the curve $y(x)=\ln\bigl(\sec(x)\bigr)$ between $[0,\pi/4]$. [From Math.SE]

[Note this is slightly different, but if we write $x(t)=t$ and $y(t)=\ln\bigl(\sec(t)\bigr)$ we get the same calculation.]

Exercise 2. Calculate the arc-length for $y = \sqrt{4x+1}$ [from Math.SE].

Exercise 3. Calculate the arc-length for the curve described by $y^{3}=x^{2}$ over the interval when $0\leq x\leq1$. [Hint: consider the parametrization $x(t)=t^{2}$, $y(t)=t^{3}$; from Math.SE]

Exercise 4. For $1\leq x\leq 9$ find the arc-length of the curve $y = (1/3)\cdot(x-3)\sqrt{x}$ [from Math.SE].

Exercise 5. For $1\leq x\leq2$, find the arc-length for $\displaystyle y = \frac{x^3}{3} + \frac{1}{4x}$. [From Math.SE]