Arc Length of a Function

Let $f(x)$ be a differentiable function defined over the interval $[a, b]$ with its derivative $f'(x)$ being continous. Such functions are also called smooth. Our goal is to calculate the arc length of that function from point $(a,f(a))$ to point $(b,f(b))$.

We start by approximating the curve using line segments. To do so, we create a regular partition $X=\{a=x_0, x_1, ... b=x_n\}$. Now for each point $P_i=(x_i, f(x_i))$ we construct line segments from point $P_i$ to $P_{i+1}$.

To find the length $|P_i - P_{i-1}|$ of each line segment, we need to look at the horizontal and vertical change over the distance of each interval $[x_{i-1}, x_i]$. Since we use a regular partition, the horizontal change is $\Delta x = x_i - x_{i-1} = \frac{b-a}{n}$. The vertical change in distance depends on the function value, so $\Delta y_i=f(x_i)-f(x_{i-1})$.

The length of the line segment can be calculated using Pythagorean Theorem:

$$|P_i - P_{i-1}| = \sqrt{(\Delta x)^2 + (\Delta y_i)^2}$$

By Mean Value Theorem we know there is a point $x^*_i\in[x_{i-1}, x_i]$ such that

$$\begin{array}{rl} f'(x_i^*) &= \frac{f(x_i) - f(x_{i-1})}{x_i-x_{i-1}}\\ &= \frac{\Delta y_i}{\Delta x} \end{array}$$

Therefore the length of each line segment can be expressed as

$$\begin{array}{rl} |P_i - P_{i-1}| &= \sqrt{(\Delta x)^2 + (\Delta y_i)^2}\\ &= \sqrt{(\Delta x)^2 + (\Delta x)^2\left(\frac{\Delta y_i}{\Delta x}\right)^2}\\ &= \sqrt{(\Delta x)^2 \left[ 1+\left(\frac{\Delta y_i}{\Delta x}\right)^2\right]}\\ &= \Delta x\sqrt{1+\left(\frac{\Delta y_i}{\Delta x}\right)^2}\\ &= \Delta x\sqrt{1+\left[f'(x_i^*)\right]^2}\\ \end{array}$$

Summing over the lengths of the line segments approximates the arc length of the curve already

$$\begin{array}{rl} L &\approx \sum\limits_{i=1}^n |P_i - P_{i-1}|\\ &= \sum\limits_{i=1}^n \sqrt{1+\left[f'(x_i^*)\right]^2}\Delta x \end{array}$$

The approximation gets better and better the larger $n$ becomes. Taking the limit as $n\to\infty$ we get a Riemann sum that can be translated to a definite integral, which allows us to calculate the exact arc length of the curve

$$\begin{array}{rl} L &= \lim\limits_{n\to\infty}\sum\limits_{i=1}^n \sqrt{1+\left[f'(x_i^*)\right]^2} \Delta x\\ &= \int_a^b \sqrt{1+\left[f'(x)\right]^2} dx \end{array}$$