How do you find the arc length of the cardioid r = 1+cos(theta) from 0 to 2pi?

1 Answer
Oct 25, 2015

$8$

Explanation:

$l = \int \sqrt{{r}^{2} + {\left(\frac{\mathrm{dr}}{d \theta}\right)}^{2}} d \theta$

$r = 1 + \cos \theta \implies \frac{\mathrm{dr}}{d \theta} = - \sin \theta$

$l = 2 {\int}_{0}^{\pi} \sqrt{{\left(1 + \cos \theta\right)}^{2} + {\sin}^{2} \theta} d \theta$

$l = 2 {\int}_{0}^{\pi} \sqrt{1 + 2 \cos \theta + {\cos}^{2} \theta + {\sin}^{2} \theta} d \theta$

$l = 2 {\int}_{0}^{\pi} \sqrt{1 + 2 \cos \theta + 1} d \theta$

$l = 2 {\int}_{0}^{\pi} \sqrt{2 + 2 \cos \theta} d \theta = 2 {\int}_{0}^{\pi} \sqrt{2 \left(1 + \cos \theta\right)} d \theta$

$l = 2 {\int}_{0}^{\pi} \sqrt{4 {\cos}^{2} \left(\frac{\theta}{2}\right)} d \theta = 2 {\int}_{0}^{\pi} 2 \cos \left(\frac{\theta}{2}\right) d \theta$

$l = 8 \sin \left(\frac{\theta}{2}\right) {|}_{0}^{\pi} = 8$