# Polar Equations - Arc Length

The polar equation \(r = f(\theta)\) describes a curve in the Cartesian plane. One may wish to compute the **arclength of this polar curve**, which can be done using the formula below:

If \(r = f(\theta)\) describes a polar curve, the arc of this curve swept out as the angle \(\theta\) varies from \(\theta_1\) to \(\theta_2\) is \[\int_{\theta_1}^{\theta_2}\sqrt{f(\theta)^2 + f'(\theta)^2 } \, d\theta = \int_{\theta_1}^{\theta_2}\sqrt{r^2 + \left( \frac{dr}{d\theta} \right) ^2 } \, d\theta.\]

## Derivation Of Formula

Suppose \(r = f(\theta)\) is a given polar curve, where the angle \(\theta\) varies over, for instance, the interval \([0,2\pi]\). By converting to Cartesian coordinates, one may obtain a parametric form for this curve, namely \[x(\theta) = r \cos(\theta) = f(\theta) \cos(\theta),\] \[y(\theta) = r\sin(\theta) = f(\theta) \sin(\theta).\] Differentiating these equations with the product rule implies \[x'(\theta) = f'(\theta) \cos(\theta) - f(\theta) \sin(\theta),\] \[y'(\theta) = f'(\theta) \sin(\theta) + f(\theta) \cos(\theta).\] In particular, one notes that \[\| (x'(\theta), y'(\theta))\|^2 = x'(\theta)^2 + y'(\theta)^2 \] \[= (f'(\theta) \cos(\theta) - f(\theta) \sin(\theta))^2 + ( f'(\theta) \sin(\theta) + f(\theta) \cos(\theta))^2 = f'(\theta)^2 + f(\theta)^2.\]

Using the formula for parametric arclength, one computes the arclength of this curve to be \[\int_{0}^{2\pi} \sqrt{x'(\theta)^2 + y'(\theta)^2} \, d\theta = \int_{0}^{2\pi} \sqrt{f'(\theta)^2 + f(\theta)^2} \, d\theta = \int_{0}^{2\pi} \sqrt{\left(\frac{dr}{d\theta} \right)^2 + r^2 } \, d\theta,\] as desired.

## Examples And Problems

The line \(y = 1\) has polar equation \(r = \csc(\theta)\). As \(\theta\) varies from \(\pi/4\) to \(3\pi/4\), the curve traces out the line segment from \((-1, 1)\) to \((1, 1)\), which has length \(2\). Prove this using the formula for polar arclength.

Since \[r'(\theta) = \frac{d}{d\theta} \csc(\theta) = -\csc(\theta) \cot(\theta),\] the polar arclength is \[\int_{\pi/4}^{3\pi/4} \sqrt{r(\theta)^2 + r'(\theta)^2} \, d\theta = \int_{\pi/4}^{3\pi/4} \sqrt{\csc^2 (\theta) (1 + \cot^2(\theta))} \, d\theta \] \[ = \int_{\pi/4}^{3\pi/4} \csc^2 (\theta) \, d\theta = -\cot(x) \big\vert_{\pi/4}^{3\pi/4} = 2.\]

The

spiral of Archimedesis the polar curve with equation \(r = \theta\), where \(\theta \in [0,\infty)\). What is the arclength of the portion of this curve traced out as \(\theta\) ranges from \(0\) to \(1\) (in radians)?

We compute \[I = \int_{0}^{1} \sqrt{f(\theta)^2 + f'(\theta)^2} \, d\theta = \int_{0}^{1} \sqrt{\theta^2 + 1} \, d\theta.\] Making the substitution \(\theta = \tan(x)\), we have \(d \theta = \sec^2 (x) \, dx\), hence the integral transforms to \[I = \int_{0}^{\pi/4} \sqrt{\tan^2 (x) + 1} \sec^2 (x) \, dx = \int_{0}^{\pi/4} \sec^3 (x) \, dx,\] where we have simplified using the trigonometric identity \(\tan^2 (x) + 1 = \sec^2 (x)\). Following up with integration by parts, this integral equals \[I = \sec(x) \tan(x) \big\vert_{0}^{\pi/4} - \int_{0}^{\pi/4} \sec (x) \tan^2 (x) \, dx \] \[= \sqrt{2} - \int_{0}^{\pi/4} \sec^2 (x) (\sec^2 (x) - 1) \, dx = \sqrt{2} - I + \int_{0}^{\pi/4} \sec(x) \, dx.\] Since the antiderivative of \(\sec(x)\) is \(\ln|\sec(x) + \tan(x)| + C\), it follows \[ 2I = \sqrt{2} + \ln|\sec(x) + \tan(x)| \big\vert_{0}^{\pi/4} = \sqrt{2} + \ln(\sqrt{2} + 1) \implies I = \frac{1}{2} (\sqrt{2} + \ln(\sqrt{2} + 1)).\]

Consider the polar curve with equation \(r = \dfrac1\theta\). What is the length of the arc traced by this curve as \(\theta\) (measured in radians) varies from \(1\) to \(2\)?

**Cite as:**Polar Equations - Arc Length.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/polar-equations-arc-length/