Q.Find the radius of a circle in which a central angle of $60^\circ$ subtends an arc of length $37.4$ cm. (Use $\pi = \dfrac{22}{7}$.)

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 $4$:

The arc length is $\frac{14}{3}$ units.

Why This Matters

The arc length formula is your first encounter with measuring curved distances analytically. It's the foundation for:

• Finding the length of any smooth curve (not just graphs of functions)
• Computing surface areas of solids of revolution
• Understanding curvature and the geometry of space curves in physics

The formula itself is simple: Pythagoras on an infinitesimal scale, summed up by an integral. That's the entire story.

The arc length formula, derived using integral calculus, extends the NCERT Class 12 Mathematics chapters on Application of Integrals and Differential Equations, and "arc length formula for curves using integration" is a frequently searched topic among JEE Main and JEE Advanced aspirants. Because it combines derivative and integral skills in one problem, this concept regularly appears in "application of integrals important questions" for competitive-exam revision.