Arc length in cartesian coordinates

In physics, we often need to calculate the so-called arc length of a curve or generally use the arc length formula. Therefore, we shall calculate it here for the two-dimensional case.

Definition

Let $y(x)$ be a given function that is continuously differentiable at least once. This means, it could also be a physical trajectory. Furthermore, two points $A(x_a,y(x_a))$ and $B(x_b,y(x_b))$ that are part of the function are given. (see sketch)

This implies the following question:

---

How long is the way from $A$ to $B$ along the curve?

---

Formally, one should define precisely, what the "way" is. For this article, "way" shall be the "physical way", i. e. the length of a string whose shape is $y(x)$. This is the arc length ${\mathcal B}_{A,B}$.

The derivation

The line segments $\Delta s_i$ make a polyline of a length, that can be calculated, given the endpoints. This length is an approximation for the actual arc length ${\mathcal B}_{A,B}$, which gets better, when the $\Delta x_i$s (and therefore also the $\Delta y_i$s and $\Delta s_i$s, respectively) get smaller. A specific segment $\Delta s_i$ can be expressed using the Pythagorean theorem:

$$ \begin{align} \Delta s_i=& \sqrt{(\Delta x_i)^2+(\Delta y_i)^2}\\ =&\sqrt{1+\left(\frac{\Delta y_i}{\Delta x_i}\right)^2}\cdot\Delta x_i\\ \end{align} $$

Chunking the interval $[x_a,x_b]$ into $n$ pieces, the arc length ${\mathcal B}_{A,B}$ is approximated by the polyline of length $l_n$:

$$ \begin{align} l_n = &\sum\limits_{i=1}^n{\Delta s_i}\\ = &\sum\limits_{i=1}^n{\sqrt{1+\left(\frac{\Delta y_i}{\Delta x_i}\right)^2}\cdot\Delta x_i}\\ \end{align} $$

In the limit $n\to\infty$, the length $l_n$ becomes the actual arc length.

As $\Delta x_i$, $\Delta y_i$ and $\Delta s_i$ get infinitely small, the sum becomes an integral:

$$ \begin{align} {\mathcal B}_{A,B} =& \lim\limits_{n\to\infty} l_n\\ = & \lim\limits_{n\to\infty} \sum\limits_{i=1}^n{\Delta s_i(n)}\\ = & \int\limits_0^{{\mathcal B}_{A,B}}{ds}\\ = & \int\limits_{x_A}^{x_B}{\sqrt{1+\left(\frac{dy}{dx}\right)^2}dx}\\ \end{align} $$

Thus, the final formula for the arc length is:

$$\boxed{{\mathcal B}_{A,B}= \int\limits_{x_A}^{x_B}{\sqrt{1+\left(y^\prime(x)\right)^2}dx}}$$