Arc Length

The arc length of a curve measures its true length along the curve, not the straight-line distance between its endpoints. The idea is beautifully simple: chop the curve into tiny near-straight pieces, use Pythagoras on each, and integrate. From this one principle flow three formulae — Cartesian, parametric and polar — and a string of elegant results (the catenary's length is $\sinh a$ ; one arch of a cycloid is exactly $8a$ ). This lesson derives the formula rigorously, drills the all-important "simplify under the root" step, and stretches to the elliptic integrals that defeat elementary methods.

1 Spec mapping (AQA 7367)

Arc length is compulsory Pure content for Paper 1 and Paper 2 (each 2 h, 100 marks, 33⅓%, with per-paper weighting AO1 55% / AO2 25% / AO3 20%), in the "Further calculus" strand alongside surface area of revolution, with which it shares the element $\mathrm{d}s$ . The bulk of a question is AO1 — differentiate, substitute into $\int\sqrt{1+(y')^2}\,dx$ , integrate — but the decisive step is almost always AO2: recognising the identity (hyperbolic $1+\sinh^2=\cosh^2$ , or double-angle $1-\cos t = 2\sin^2\tfrac{t}{2}$ ) that collapses the surd into something integrable, and communicating that simplification cleanly. A modelling context (the length of a suspension cable, the distance a wheel-point travels) lifts it into AO3. Arc length feeds directly into the next lesson (surface area $= 2\pi\int y\,\mathrm{d}s$ ) and into polar coordinates, so the $\mathrm{d}s$ idea pays dividends repeatedly.

2 Core theory — deriving $\mathrm{d}s$ from Pythagoras

Take a curve $y=f(x)$ and a small step from $x$ to $x+\delta x$ . The curve rises by $\delta y$ , so the straight chord joining the two points has length, by Pythagoras,

$\delta s \approx \sqrt{(\delta x)^2 + (\delta y)^2} = \sqrt{1 + \left(\frac{\delta y}{\delta x}\right)^{2}}\;\delta x.$

Letting $\delta x \to 0$ turns $\dfrac{\delta y}{\delta x}$ into $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ and $\delta s$ into the arc-length element

$\mathrm{d}s = \sqrt{1 + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^{2}}\;\mathrm{d}x,$

Summing the chords and passing to the limit replaces $\sum\delta s$ by $\int\mathrm{d}s$ , so the total length from $x=a$ to $x=b$ is

$s = \lim_{\delta x\to0}\sum \sqrt{1+\left(\frac{\delta y}{\delta x}\right)^2}\,\delta x = \boxed{\;\int_a^{b} \sqrt{1 + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^{2}}\;\mathrm{d}x\;}.$

This is a genuine limit of a sum (a Riemann sum), not a definition pulled from nowhere: the polygonal path made of tiny chords approximates the smooth curve ever more closely as the chords shrink, and its total length converges to the integral. The approximation $\delta s\approx\sqrt{(\delta x)^2+(\delta y)^2}$ becomes exact in the limit because, for a differentiable curve, each small arc is indistinguishable from its chord to leading order.

The same chord, expressed in a parameter $t$ , gives $\delta s \approx \sqrt{(\delta x)^2+(\delta y)^2}$ , and dividing inside the root by $(\delta t)^2$ yields the parametric form; in polar coordinates $x = r\cos\theta,\ y=r\sin\theta$ a short computation (done in §6) gives the polar form:

$s = \int_{t_1}^{t_2}\sqrt{\left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)^2 + \left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)^2}\;\mathrm{d}t, \qquad s = \int_{\alpha}^{\beta}\sqrt{r^2 + \left(\frac{\mathrm{d}r}{\mathrm{d}\theta}\right)^2}\;\mathrm{d}\theta.$

All three say the same thing: $s = \int \mathrm{d}s$ , where $\mathrm{d}s$ is the hypotenuse of an infinitesimal right triangle. The whole skill is choosing the form that matches how the curve is given, then simplifying the surd before integrating.

Why the surd is unavoidable. Unlike area ( $\int y\,\mathrm{d}x$ ) or volume ( $\pi\int y^2\,\mathrm{d}x$ ), arc length measures the curve's own extent, so the vertical motion $\mathrm{d}y$ must be combined with the horizontal motion $\mathrm{d}x$ — and Pythagoras combines perpendicular displacements through a square root. This is precisely why arc-length integrals are harder than area integrals: the integrand $\sqrt{1+(y')^2}$ is rarely a polynomial, and an exam curve is almost always engineered so that this root collapses to something elementary. Spotting that collapse is the examined skill.

A useful way to read the formula: $\sqrt{1+(y')^2}$ is the stretch factor by which a horizontal step $\mathrm{d}x$ is lengthened into the slanted arc $\mathrm{d}s$ . On a flat part of the curve ( $y'=0$ ) the factor is $1$ (no stretch); on a steep part ( $|y'|$ large) the factor is large, because a small horizontal advance traces a long slanted piece. Integrating this stretch factor across $[a,b]$ accumulates the total length. The same interpretation holds in parametric form, where $\sqrt{(x')^2+(y')^2}$ is the speed of a particle tracing the curve as $t$ advances — and length is "speed integrated over time", exactly as distance $=$ speed $\times$ time generalises to $\int |\mathbf v|\,\mathrm{d}t$ . This is why arc length and kinematics are two views of one idea: the distance a moving point covers along its path is the integral of its speed.

3 Worked examples with M1/A1 mark scheme

Worked Example 1 — a power curve

Find the arc length of $y = x^{3/2}$ from $x=0$ to $x=4$ .

$\frac{\mathrm{d}y}{\mathrm{d}x} = \tfrac32 x^{1/2} \;\implies\; \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 = \tfrac94 x.$

(M1 differentiate and square.) So $s = \displaystyle\int_0^{4}\sqrt{1 + \tfrac94 x}\;\mathrm{d}x$ . Substitute $u = 1+\tfrac94 x$ , $\mathrm{d}u = \tfrac94\,\mathrm{d}x$ ; when $x=0,\ u=1$ and when $x=4,\ u=10$ :

$s = \frac{4}{9}\int_1^{10}\sqrt{u}\;\mathrm{d}u = \frac{4}{9}\left[\frac{2}{3}u^{3/2}\right]_1^{10} = \frac{8}{27}\Big(10^{3/2} - 1\Big).$

(M1 substitution with changed limits; A1 for the bracket.) Hence

$s = \frac{8}{27}\big(10\sqrt{10} - 1\big) \approx 9.073.$

(A1 exact form.)

Worked Example 2 — the catenary

Find the arc length of $y = \cosh x$ from $x=0$ to $x=a$ .

$\frac{\mathrm{d}y}{\mathrm{d}x} = \sinh x \;\implies\; 1 + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 = 1 + \sinh^2 x = \cosh^2 x,$

using the fundamental identity $\cosh^2 x - \sinh^2 x = 1$ . (M1 for the identity — the crux of the question.) Therefore

$s = \int_0^{a}\sqrt{\cosh^2 x}\;\mathrm{d}x = \int_0^{a}\cosh x\;\mathrm{d}x = \big[\sinh x\big]_0^{a} = \sinh a.$

(A1 for $\sinh a$ .) The catenary — the shape of a hanging chain — has this remarkably clean arc length, which is exactly why it recurs throughout applied mathematics.

Worked Example 3 — a straight-line sanity check

Verify the formula on $y = 3x+2$ from $x=0$ to $x=4$ . Here $\dfrac{\mathrm{d}y}{\mathrm{d}x}=3$ , so

$s = \int_0^{4}\sqrt{1+9}\;\mathrm{d}x = \int_0^{4}\sqrt{10}\;\mathrm{d}x = 4\sqrt{10}.$

The endpoints are $(0,2)$ and $(4,14)$ ; the straight-line distance is $\sqrt{4^2 + 12^2} = \sqrt{160} = 4\sqrt{10}$ . The formula agrees — a reassuring check that $\mathrm{d}s$ really is the hypotenuse.

Worked Example 4 — a parametric curve (the cycloid)

Find the arc length of one arch of the cycloid $x = a(t-\sin t),\ y = a(1-\cos t)$ for $0\le t\le 2\pi$ .

$\frac{\mathrm{d}x}{\mathrm{d}t} = a(1-\cos t), \qquad \frac{\mathrm{d}y}{\mathrm{d}t} = a\sin t.$

(M1 differentiate both.) Then

$\left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)^2 + \left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)^2 = a^2(1-\cos t)^2 + a^2\sin^2 t = a^2\big(2 - 2\cos t\big) = 2a^2(1-\cos t).$

Use $1-\cos t = 2\sin^2\tfrac{t}{2}$ : the bracket becomes $4a^2\sin^2\tfrac{t}{2}$ , so $\sqrt{\cdots} = 2a\big|\sin\tfrac{t}{2}\big| = 2a\sin\tfrac{t}{2}$ (since $\sin\tfrac{t}{2}\ge0$ on $[0,2\pi]$ ). (M1 for the half-angle simplification; A1 for the integrand.) Hence

$s = \int_0^{2\pi} 2a\sin\tfrac{t}{2}\;\mathrm{d}t = 2a\Big[-2\cos\tfrac{t}{2}\Big]_0^{2\pi} = 2a\big(2 + 2\big) = 8a.$

(A1 for $8a$ .) One arch has length $8a$ — four times the diameter of the rolling circle, a classic result.

Worked Example 5 — a curve needing a genuine integral

Not every arc length collapses to a one-line antiderivative. Find the length of $y = \tfrac12 x^2$ from $x=0$ to $x=1$ . Here $\dfrac{\mathrm{d}y}{\mathrm{d}x} = x$ , so

$s = \int_0^{1}\sqrt{1 + x^2}\;\mathrm{d}x.$

(M1 set up the integral.) The standard result $\displaystyle\int\sqrt{1+x^2}\,\mathrm{d}x = \tfrac12\Big[x\sqrt{1+x^2} + \operatorname{arsinh}x\Big] + C$ (provable by the substitution $x=\sinh u$ , since $1+\sinh^2 u = \cosh^2 u$ and $\mathrm{d}x = \cosh u\,\mathrm{d}u$ ) gives

$s = \tfrac12\Big[x\sqrt{1+x^2} + \operatorname{arsinh}x\Big]_0^{1} = \tfrac12\Big(\sqrt2 + \operatorname{arsinh}1\Big) = \tfrac12\Big(\sqrt2 + \ln(1+\sqrt2)\Big) \approx 1.148,$

using $\operatorname{arsinh}1 = \ln(1+\sqrt2)$ . (M1 apply the standard integral with limits; A1 exact form.) This is the kind of integral — inverse-hyperbolic, not a perfect square — that examiners reach for when they want to test the harder integration technique rather than a slick simplification.

4 Specimen-style exam question

(Specimen-style — not a real past paper.)

A curve has equation $\displaystyle y = \frac{x^2}{4} - \frac{1}{2}\ln x$ for $1 \le x \le 2$ . (a) Show that $\displaystyle 1 + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 = \left(\frac{x}{2} + \frac{1}{2x}\right)^2$ . (b) Hence find the exact arc length of the curve.

(a) $\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{x}{2} - \dfrac{1}{2x}$ . Squaring, $\left(\dfrac{x}{2}-\dfrac{1}{2x}\right)^2 = \dfrac{x^2}{4} - \dfrac12 + \dfrac{1}{4x^2}$ . Adding $1$ ,

$1 + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 = \frac{x^2}{4} + \frac12 + \frac{1}{4x^2} = \left(\frac{x}{2} + \frac{1}{2x}\right)^2,$

since the last expression expands to exactly that. $\;\square$ (This is the engineered "perfect square" that makes the root vanish — the whole point of part (a).)

(b) Because $\tfrac{x}{2}+\tfrac{1}{2x} > 0$ on $[1,2]$ , the square root is just $\tfrac{x}{2}+\tfrac{1}{2x}$ :

$s = \int_1^{2}\left(\frac{x}{2} + \frac{1}{2x}\right)\mathrm{d}x = \left[\frac{x^2}{4} + \frac12\ln x\right]_1^{2} = \left(1 + \tfrac12\ln 2\right) - \left(\tfrac14 + 0\right) = \frac34 + \frac{1}{2}\ln 2.$

So the arc length is $\dfrac34 + \dfrac12\ln 2$ .

5 Arc length in polar form

Polar curves measure position by a radius $r$ and angle $\theta$ rather than by $(x,y)$ , and they describe many shapes — spirals, cardioids, rose curves — far more naturally than Cartesian equations can. To find their length we exploit the link $x = r\cos\theta,\ y = r\sin\theta$ , treating $\theta$ as the parameter.

For a curve $r = f(\theta)$ , substitute $x = r\cos\theta,\ y = r\sin\theta$ into the parametric formula with parameter $\theta$ . Differentiating (product rule),

$\frac{\mathrm{d}x}{\mathrm{d}\theta} = r'\cos\theta - r\sin\theta, \qquad \frac{\mathrm{d}y}{\mathrm{d}\theta} = r'\sin\theta + r\cos\theta,$

and squaring and adding, the cross-terms cancel and $\sin^2+\cos^2=1$ leaves $\left(\tfrac{\mathrm{d}x}{\mathrm{d}\theta}\right)^2 + \left(\tfrac{\mathrm{d}y}{\mathrm{d}\theta}\right)^2 = r^2 + (r')^2$ . Hence

$\boxed{\;s = \int_{\alpha}^{\beta}\sqrt{r^2 + \left(\frac{\mathrm{d}r}{\mathrm{d}\theta}\right)^2}\;\mathrm{d}\theta\;}.$

Worked Example 6 — the cardioid

Find the total arc length of $r = a(1+\cos\theta)$ . Here $\dfrac{\mathrm{d}r}{\mathrm{d}\theta} = -a\sin\theta$ , so

$r^2 + (r')^2 = a^2(1+\cos\theta)^2 + a^2\sin^2\theta = a^2(2 + 2\cos\theta) = 2a^2(1+\cos\theta).$

Using $1+\cos\theta = 2\cos^2\tfrac{\theta}{2}$ gives $4a^2\cos^2\tfrac{\theta}{2}$ , so $\sqrt{\cdots} = 2a\big|\cos\tfrac{\theta}{2}\big|$ . On $[0,\pi]$ , $\cos\tfrac{\theta}{2}\ge0$ ; the curve is symmetric about the initial line, so

$s = 2\int_0^{\pi} 2a\cos\tfrac{\theta}{2}\;\mathrm{d}\theta = 4a\Big[2\sin\tfrac{\theta}{2}\Big]_0^{\pi} = 4a\cdot 2 = 8a.$

The modulus matters: integrating $2a\cos\tfrac{\theta}{2}$ blindly from $0$ to $2\pi$ would give $0$ (the $\cos\tfrac{\theta}{2}$ goes negative past $\theta=\pi$ ), which is absurd for a length — hence the symmetry split.

Worked Example 7 — the Archimedean spiral

Find the arc length of $r = \theta$ for $0\le\theta\le 2\pi$ . With $\dfrac{\mathrm{d}r}{\mathrm{d}\theta}=1$ ,

$s = \int_0^{2\pi}\sqrt{\theta^2 + 1}\;\mathrm{d}\theta = \left[\frac{\theta}{2}\sqrt{\theta^2+1} + \frac12\ln\!\big(\theta + \sqrt{\theta^2+1}\big)\right]_0^{2\pi},$

the standard integral $\int\sqrt{\theta^2+1}\,\mathrm{d}\theta$ (via $\theta=\sinh u$ or $\operatorname{arsinh}$ ). Evaluating, $s = \pi\sqrt{4\pi^2+1} + \tfrac12\ln\!\big(2\pi + \sqrt{4\pi^2+1}\big) \approx 21.26$ .

5a When the integral is not elementary

Most arc-length integrands $\sqrt{1+(y')^2}$ have no elementary antiderivative — the engineered exam curves are the exception, not the rule. Two famous cases:

The parabola $y=x^2$ gives $\int\sqrt{1+4x^2}\,\mathrm{d}x$ , which is elementary (inverse-hyperbolic, as in Worked Example 5's cousin), but messy.
The sine curve $y=\sin x$ gives $\int\sqrt{1+\cos^2 x}\,\mathrm{d}x$ — an elliptic integral with no elementary form at all (see §11).

When no closed form exists, the length is found numerically, for instance by Simpson's rule applied to $f(x)=\sqrt{1+(y')^2}$ . For $y=\sin x$ over $[0,\pi]$ , a few strips of Simpson's rule give $s\approx 3.820$ . The point for the exam is conceptual: the formula $s=\int\sqrt{1+(y')^2}\,\mathrm{d}x$ is always valid and always sets up correctly; whether it then yields to algebra is a separate question. AQA-set curves are chosen so the algebra works — recognising the perfect square or the hyperbolic identity is the examined skill — but you should understand that arc length is fundamentally an integral that may require numerical evaluation, exactly as the normal-distribution probability integral does in Statistics.

Arc Length

Arc Length

1 Spec mapping (AQA 7367)

2 Core theory — deriving $\mathrm{d}s$ from Pythagoras

3 Worked examples with M1/A1 mark scheme

Worked Example 1 — a power curve

Worked Example 2 — the catenary

Worked Example 3 — a straight-line sanity check

Worked Example 4 — a parametric curve (the cycloid)

Worked Example 5 — a curve needing a genuine integral

4 Specimen-style exam question

5 Arc length in polar form

Worked Example 6 — the cardioid

Worked Example 7 — the Archimedean spiral

5a When the integral is not elementary

6 Synoptic links

More in Mathematics

Arc Length

Arc Length

1 Spec mapping (AQA 7367)

2 Core theory — deriving ds\mathrm{d}sds from Pythagoras

3 Worked examples with M1/A1 mark scheme

Worked Example 1 — a power curve

Worked Example 2 — the catenary

Worked Example 3 — a straight-line sanity check

Worked Example 4 — a parametric curve (the cycloid)

Worked Example 5 — a curve needing a genuine integral

4 Specimen-style exam question

5 Arc length in polar form

Worked Example 6 — the cardioid

Worked Example 7 — the Archimedean spiral

5a When the integral is not elementary

6 Synoptic links

More in Mathematics

2 Core theory — deriving $\mathrm{d}s$ from Pythagoras