Arc Length

Formula for y = f(x)

The length of a curve $y = f(x)$ from $x = a$ to $x = b$:

$L = \int_a^b \sqrt{1 + [f'(x)]^2} \, dx = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx$

Formula for x = g(y)

The length of a curve $x = g(y)$ from $y = c$ to $y = d$:

$L = \int_c^d \sqrt{1 + [g'(y)]^2} \, dy = \int_c^d \sqrt{1 + \left(\frac{dx}{dy}\right)^2} \, dy$

Parametric Form

For $x = x(t)$, $y = y(t)$ from $t = \alpha$ to $t = \beta$:

$L = \int_\alpha^\beta \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt$

Differential Form

The arc length differential:

$ds = \sqrt{dx^2 + dy^2} = \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx$

Examples

Example 1: Simple curve

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

$\frac{dy}{dx} = \frac{3}{2}x^{1/2}$

$\left(\frac{dy}{dx}\right)^2 = \frac{9}{4}x$

$L = \int_0^4 \sqrt{1 + \frac{9x}{4}} \, dx$

Let $u = 1 + \frac{9x}{4}$, $du = \frac{9}{4}dx$

$= \frac{4}{9} \int_1^{10} \sqrt{u} \, du = \frac{4}{9} \cdot \frac{2}{3} [u^{3/2}]_1^{10}$

$= \frac{8}{27} [10^{3/2} - 1] = \frac{8}{27} [10\sqrt{10} - 1] \approx 9.07$

Example 2: Curve as x = g(y)

Find the arc length of $x = \frac{y^3}{3} + \frac{1}{4y}$ from $y = 1$ to $y = 2$.

$\frac{dx}{dy} = y^2 - \frac{1}{4y^2}$

$\left(\frac{dx}{dy}\right)^2 = y^4 - \frac{1}{2} + \frac{1}{16y^4}$

$1 + \left(\frac{dx}{dy}\right)^2 = y^4 + \frac{1}{2} + \frac{1}{16y^4} = \left(y^2 + \frac{1}{4y^2}\right)^2$

$L = \int_1^2 \left(y^2 + \frac{1}{4y^2}\right) dy = \left[\frac{y^3}{3} - \frac{1}{4y}\right]_1^2$

$= \left(\frac{8}{3} - \frac{1}{8}\right) - \left(\frac{1}{3} - \frac{1}{4}\right) = \frac{8}{3} - \frac{1}{8} - \frac{1}{3} + \frac{1}{4}$

$= \frac{7}{3} + \frac{1}{8} = \frac{59}{24}$

Example 3: Circle (parametric)

Find the circumference of a circle $x = r \cos(t)$, $y = r \sin(t)$.

$\frac{dx}{dt} = -r \sin(t), \quad \frac{dy}{dt} = r \cos(t)$

$\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 = r^2\sin^2(t) + r^2\cos^2(t) = r^2$

$L = \int_0^{2\pi} \sqrt{r^2} \, dt = \int_0^{2\pi} r \, dt = r \cdot 2\pi = 2\pi r$

Example 4: Parabola

Find the arc length of $y = x^2$ from $x = 0$ to $x = 1$.

$\frac{dy}{dx} = 2x$

$L = \int_0^1 \sqrt{1 + 4x^2} \, dx$

Using trig substitution: $x = \frac{1}{2}\tan(\theta)$

$= \frac{1}{2} \int_0^{\arctan(2)} \sec^3(\theta) \, d\theta$

Using the formula for $\int\sec^3(\theta)$:

$= \frac{1}{4}[\sec(\theta)\tan(\theta) + \ln|\sec(\theta) + \tan(\theta)|]_0^{\arctan(2)}$

At $\theta = \arctan(2)$: $\sec(\theta) = \sqrt{5}$, $\tan(\theta) = 2$

$= \frac{1}{4}[2\sqrt{5} + \ln(\sqrt{5} + 2) - 0] \approx 1.479$

Example 5: Catenary

Find the arc length of $y = \cosh(x)$ from $x = 0$ to $x = 1$.

$\frac{dy}{dx} = \sinh(x)$

$1 + \left(\frac{dy}{dx}\right)^2 = 1 + \sinh^2(x) = \cosh^2(x)$

$L = \int_0^1 \cosh(x) \, dx = [\sinh(x)]_0^1 = \sinh(1) - \sinh(0) = \sinh(1) \approx 1.175$

Example 6: Cycloid

Find one arch of the cycloid: $x = t - \sin(t)$, $y = 1 - \cos(t)$, $0 \leq t \leq 2\pi$.

$\frac{dx}{dt} = 1 - \cos(t), \quad \frac{dy}{dt} = \sin(t)$

$\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 = (1 - \cos t)^2 + \sin^2 t$

$= 1 - 2\cos t + \cos^2 t + \sin^2 t = 2 - 2\cos t = 2(1 - \cos t)$

Using half-angle: $1 - \cos t = 2\sin^2(t/2)$

$\sqrt{2(1 - \cos t)} = 2|\sin(t/2)| = 2\sin(t/2) \text{ for } 0 \leq t \leq 2\pi$

$L = \int_0^{2\pi} 2\sin(t/2) \, dt = [-4\cos(t/2)]_0^{2\pi}$

$= -4(-1) - (-4)(1) = 4 + 4 = 8$