Question

Find the entire length of the cardioid $r=a(1+\cos \theta )$.
Also show that the upper half is bisected by $\theta =\frac{\pi }{3}$

Answer 1

The cardioid is symmetrical about the initial line and for its upper half,$\theta$ increases from $0$ to $\pi$
Also, $\frac{dr}{d\theta }=-a\sin \theta$
$\therefore$Length of the curve$=2{\int }_0^{\pi }\sqrt{[r^2+{\left(\frac{dr}{d\theta }\right)}^2]}d\theta$
$=2{\int }_0^{\pi }\sqrt{{\left[a(1+\cos \theta )\right]}^2+{(-a\sin \theta )}^2}d\theta$
$=a{\int }_0^{\pi }\sqrt{1+{\cos }^2\theta +2\cos \theta +{\sin }^2\theta }d\theta$
We know that ${\sin }^2\theta +{\cos }^2\theta =1$
$\Rightarrow 2a{\int }_0^{\pi }\sqrt{2(1+\cos \theta )}d\theta$
We know that $1+\cos \theta =2{\cos }^2\frac{\theta }{2}$
$=2a{\int }_0^{\pi }\sqrt{2\times 2{\cos }^2\frac{\theta }{2}}d\theta$
$=4a{\int }_0^{\pi }\cos \frac{\theta }{2}d\theta$
$=4a{\left|\frac{\sin \frac{\theta }{2}}{\frac{1}{2}}\right|}_0^{\pi }$
$=8a(\sin \frac{\pi }{2}-\sin 0)$
$=8a$
$\therefore$ Length of upper half of the curve is $4a$
Also length of the arc $AP$ from $0$ to $\frac{\pi }{3}$
$={\int }_0^{\frac{\pi }{3}}\sqrt{2(1+\cos \theta )}d\theta$
$=2a{\int }_0^{\frac{\pi }{3}}\cos \frac{\theta }{2}d\theta$
$=4a{\left|\sin \frac{\theta }{2}\right|}_0^{\frac{\pi }{3}}$
$=4a(\sin \frac{\pi }{6}-0)$
$=4a\times \frac{1}{2}$
$=2a=$half the length of upper half of the cardioid.