Q.Find the degree of the angle subtended at the centre of a circle of diameter 50 cm by an arc of length 11 cm.

The Arc Length Formula: Measuring the Unmeasurable

You already know how to find the distance between two points on a straight line — that's just the Pythagorean theorem. But what if the path between them isn't straight? What if it curves like a roller coaster track, a river on a map, or the graph of $y = \sin x$?

That curved distance is called arc length, and the formula that gives it is one of the most elegant applications of calculus.

The Intuition: Straight Lines Approximate Curves

Imagine you're walking along a winding path. If you take a single giant step, you'll cut the corner and miss the true distance. But if you take many tiny steps — each one almost perfectly straight — the sum of those tiny straight steps will be very close to the actual curved distance.

This is the core idea: break a curve into infinitely many infinitesimally small straight pieces, add them up, and let the pieces become infinitely small. That's exactly what an integral does.

For a function $y = f(x)$ from $x = a$ to $x = b$, here's the reasoning:

1. Take a tiny horizontal step $dx$.
2. The corresponding vertical change is $dy = f'(x) \, dx$.
3. The tiny straight piece connecting $(x, f(x))$ to $(x+dx, f(x+dx))$ has length, by Pythagoras:

4. Summing all these tiny lengths from $a$ to $b$ gives the total arc length.

That's the arc length formula for a curve given as $y = f(x)$.

The Precise Statement

Let $f$ be a function whose derivative $f'$ is continuous on the closed interval $[a, b]$. Then the length $L$ of the curve $y = f(x)$ from $x = a$ to $x = b$ is:

The continuity of $f'$ guarantees the curve is "smooth" — no sharp corners or jumps — so the tiny straight pieces genuinely approximate the curve.

What If the Curve Is Given Parametrically?

Sometimes a curve is described by $x = g(t)$, $y = h(t)$ for $t$ from $\alpha$ to $\beta$. The same idea applies: a tiny step in $t$ gives $dx = g'(t) \, dt$ and $dy = h'(t) \, dt$, so the tiny straight piece has length:

Integrating gives:

This is the parametric arc length formula. It's actually more fundamental — the $y = f(x)$ version is just a special case where $x = t$ and $y = f(t)$.

A Quick Example

Find the arc length of $y = \frac{2}{3}x^{3/2}$ from $x = 0$ to $x = 3$.

First, $f'(x) = \frac{2}{3} \cdot \frac{3}{2} x^{1/2} = \sqrt{x}$.

Then:

Let $u = 1 + x$, $du = dx$, limits become $1$ to $4$: …