Calculus · Applications of the definite integral
How long is a curved path? A ruler only measures straight segments, so we do what surveyors do: replace the curve by many short straight pieces, measure those, and add them up. In the limit, Pythagoras' theorem turns into an integral. This post derives the three arc-length formulas (for \(y = f(x)\), for parametric curves and for polar curves) from that one picture, and works through the classic curves.
Mark points \(M_0, M_1, \dots, M_n\) along the curve and join them by straight segments. The length of this inscribed polygon approximates the length of the curve. As the points get denser, the polygon hugs the curve more and more closely, and the arc length is defined as the limit of the polygon's length (when the longest segment tends to zero).
Each short segment is the hypotenuse of a small right triangle with legs \(\Delta x\) and \(\Delta y\), so its length is \( \sqrt{\Delta x^2 + \Delta y^2} \). In the limit this becomes the arc-length differential:
\[ ds = \sqrt{dx^2 + dy^2}. \]Every formula in this post is this one line, written in different variables.
Derivation. Factor \(dx\) out of the root: \( ds = \sqrt{dx^2 + dy^2} = \sqrt{1 + \big(\frac{dy}{dx}\big)^2}\,dx \). More carefully, by Lagrange's mean value theorem each segment of the polygon satisfies \( \Delta y_k = f'(\xi_k)\Delta x_k \), so its length is exactly \( \sqrt{1 + f'(\xi_k)^2}\,\Delta x_k \). The polygon's length is therefore a Riemann sum of \( \sqrt{1 + f'^2} \), and it converges to the integral.
Reading the formula: the factor \( \sqrt{1 + y'^2} \ge 1 \) is the "stretch" of the curve compared to its shadow on the \(x\)-axis. Where the curve is flat (\(y' = 0\)), arc length and \(dx\) agree; where it is steep, the curve is much longer than its projection. In particular \( s \ge b - a \): the straight segment is the shortest.
Derivation: \( dx = \varphi'(t)\,dt \), \( dy = \psi'(t)\,dt \), so \( ds = \sqrt{\varphi'^2 + \psi'^2}\,dt \). Physically, if \(t\) is time, \( \sqrt{\varphi'^2 + \psi'^2} \) is the speed of a point moving along the curve, and the formula says: distance travelled = integral of speed. Formula 1° is the special case \( x = t \), \( y = f(t) \).
Derivation. A polar curve is a parametric curve with parameter \(\varphi\): \( x = \rho\cos\varphi \), \( y = \rho\sin\varphi \). Then
\[ x' = \rho'\cos\varphi - \rho\sin\varphi, \qquad y' = \rho'\sin\varphi + \rho\cos\varphi, \] \[ x'^2 + y'^2 = \rho'^2(\cos^2\varphi + \sin^2\varphi) + \rho^2(\sin^2\varphi + \cos^2\varphi) = \rho^2 + \rho'^2, \]since the mixed terms cancel. Geometrically: a small step along the curve has a "radial" component \(d\rho\) and a "turning" component \(\rho\,d\varphi\) (an arc of a circle of radius \(\rho\)), at right angles to each other, so \( ds^2 = d\rho^2 + \rho^2\,d\varphi^2 \).
That is why textbook examples use special curves (cycloid, astroid, cardioid, catenary) where \(1 + y'^2\) or \(\rho^2 + \rho'^2\) is a perfect square. In practice, arc length is computed numerically, e.g. with Simpson's rule.
| Curve | Arc-length element \(ds\) | Length |
|---|---|---|
| \( y = f(x) \) | \( \sqrt{1 + y'^2}\,dx \) | \( \int_a^b\sqrt{1 + y'^2}\,dx \) |
| \( x = \varphi(t),\ y = \psi(t) \) | \( \sqrt{\varphi'^2 + \psi'^2}\,dt \) | \( \int_{t_1}^{t_2}\sqrt{\varphi'^2 + \psi'^2}\,dt \) |
| \( \rho = f(\varphi) \) | \( \sqrt{\rho^2 + \rho'^2}\,d\varphi \) | \( \int_{\varphi_1}^{\varphi_2}\sqrt{\rho^2 + \rho'^2}\,d\varphi \) |
| All three come from \( ds^2 = dx^2 + dy^2 \). | ||
| Curve | Length |
|---|---|
| Circle of radius \(r\) | \(2\pi r\) |
| One arch of the cycloid | \(8a\) |
| Astroid | \(6a\) |
| Cardioid \(\rho = a(1 + \cos\varphi)\) | \(8a\) |
| Catenary \(y = \cosh x\), \(0 \le x \le a\) | \(\sinh a\) |
Try each one before opening the solution. More solved problems from this topic are in Applications of Definite Integrals.
\( y' = -\tan x \), \( \sqrt{1 + \tan^2 x} = \frac{1}{\cos x} \). So \( s = \int_0^{\pi/4}\frac{dx}{\cos x} = \Big[\ln\tan\Big(\frac x2 + \frac\pi4\Big)\Big]_0^{\pi/4} = \ln\tan\frac{3\pi}{8} = \ln\big(1 + \sqrt2\big) \approx 0.881 \).
\( y = kx \), \( 0 \le x \le a \): \( s = \int_0^a\sqrt{1 + k^2}\,dx = a\sqrt{1 + k^2} \), which is \( \sqrt{a^2 + (ka)^2} \) by Pythagoras.
\( \rho^2 + \rho'^2 = 4a^2(\cos^2\varphi + \sin^2\varphi) = 4a^2 \); the circle is traced once for \( -\frac\pi2 \le \varphi \le \frac\pi2 \): \( s = 2a\cdot\pi = 2\pi a \).
\( s = a\int_0^{2\pi}\sqrt{\varphi^2 + 1}\,d\varphi = \frac a2\Big[\varphi\sqrt{\varphi^2 + 1} + \ln\big(\varphi + \sqrt{\varphi^2 + 1}\big)\Big]_0^{2\pi} \approx 21.256\,a \).
\( x'^2 + y'^2 = 4t^2 + 4t^4 = 4t^2(1 + t^2) \), \( ds = 2t\sqrt{1 + t^2}\,dt \). \( s = \Big[\frac23(1 + t^2)^{3/2}\Big]_0^{\sqrt3} = \frac23(8 - 1) = \frac{14}{3} \).
Because \( \sqrt{4a^2\sin^2\frac t2} = 2a\big|\sin\frac t2\big| \). Forgetting the absolute value can produce wrong (even zero or negative) lengths when the sign changes on the interval; then the interval must be split.
Arc length is Pythagoras' theorem applied to infinitely many infinitely small triangles: \( ds = \sqrt{dx^2 + dy^2} \). Writing \(dx\) and \(dy\) in terms of \(x\), of a parameter \(t\), or of the polar angle \(\varphi\) gives the three formulas, and the parametric one has a vivid meaning: distance is the integral of speed. The catch is the square root, which makes most arc lengths non-elementary. The curves where it simplifies (cycloid, astroid, cardioid, catenary) have beautiful exact answers, like the cycloid's \(8a\).
Related posts: areas of plane figures (the same "thin pieces" idea), integrating irrational functions, and integrating trigonometric functions.