Mumbai University > First Year Engineering > sem 2 > Applied Maths 2

Marks : 6

Year : DEC 2015

Given $r=(1+\cos θ)\\ \therefore \dfrac {dr}{d\theta}=(-\sin\theta)$

Perimeter of cardiode = 2 ×perimeter in cardiode

In upper half

Only for upper half θ will vary from 0 to π

$$\therefore Perimeter =2\times \int\limits_0^{\pi}\sqrt{r^2+\Bigg(\dfrac {dr}{d\theta}\Bigg)^2}d\theta $$ $ =2\times \int\limits_0^{\pi}\sqrt{r^2+\sin^2\theta}d\theta\\ =2\times \int\limits_0^{\pi}\sqrt{(1+\cos\theta)^2+\sin^2\theta}\space d\theta\\ =2\times \int\limits_0^{\pi}\sqrt{1+2\cos\theta+\cos^2\theta+\sin^2\theta}\space d\theta \\ =2\times \int\limits_0^{\pi}\sqrt{2+2\cos\theta}\space d\theta \\ =2\times \int\limits_0^{\pi}\sqrt{4\cos^2\dfrac {\theta}2}\space d\theta\\ =4\Bigg[\dfrac {\sin\frac {\theta}2}{\frac 12}\Bigg]_0^{\pi}\\ =8[\sin\dfrac \pi2-\sin\theta]\\ Perimeter=8$