Polar Equations - Arc Length

Arclengths refer to the lengths of certain curves, sometimes given as the distance between two points. Sometimes arclengths are found in the Cartesian plane with rectangular \((x,y)\) coordinates. But it can also be calculated using polar coordinates. For instance the polar equation \(r = f(\theta)\) describes a curve. The formula for the arclength of this polar curve is given by 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 \[\int_{\theta_1}^{\theta_2}\sqrt{f(\theta)^2 + f'(\theta)^2 } \, d\theta = \int_{\theta_1}^{\theta_2}\sqrt{r^2 + \left( \frac{dr}{d\theta} \right) ^2 } \, d\theta.\]

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: \[\| (x'(\theta), y'(\theta))\|^2 = x'(\theta)^2 + y'(\theta)^2 \] \[= (f'(\theta) \cos(\theta) - f(\theta) \sin(\theta))^2 + ( f'(\theta) \sin(\theta) + f(\theta) \cos(\theta))^2 = f'(\theta)^2 + f(\theta)^2.\]

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.

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.

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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 Archimedes is 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)?

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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 \[ 2I = \sqrt{2} + \ln|\sec(x) + \tan(x)| \big\vert_{0}^{\pi/4} = \sqrt{2} + \ln(\sqrt{2} + 1) \implies I = \frac{1}{2} (\sqrt{2} + \ln(\sqrt{2} + 1)).\]