# Draw the graph of r=a(1+sintheta)?

Jun 12, 2016

#### Explanation:

In the function $r = a \left(1 + \sin \theta\right)$, as $\sin \theta$ can take a value between $\left[- 1 , 1\right\}$, $r$ can take values between $\left[0 , 2 a\right]$

Let us select a few points such as $\theta = \left\{0 , \frac{\pi}{3} , \frac{\pi}{2} , \frac{2 \pi}{3} , \pi , \frac{4 \pi}{3} , \frac{3 \pi}{2} , \frac{5 \pi}{3} , 2 \pi\right\}$

Then putting these values of $\theta$ in $r = a \left(1 + \sin \theta\right)$, we get values of $r$ as $\left\{a , \frac{3 a}{2} , 2 a , \frac{3 a}{2} , a , \frac{a}{2} , 0 , \frac{a}{2} , a\right\}$. As $r = 0$ is one of the points, it passes trough origin.

Observe that values move in a cycle and hence the curve is a closed one. Further for each $\theta$ and corresponding $\pi - \theta$ value is same, hence the curve is symmetric around $y$-axis.

Let us also convert them to rectangular coordinates using $x = r \cos \theta$, $y = r \sin \theta$ and ${r}^{2} = {x}^{2} + {y}^{2}$

Hence $r = a \left(1 + \sin \theta\right) \Leftrightarrow {r}^{2} = a r + a r \sin \theta$ or

${x}^{2} + {y}^{2} = a \sqrt{{x}^{2} + {y}^{2}} + a y$

Again note that when $x = 0$, we have ${y}^{2} = 2 a y$ i.e. $y = 0$ or $y = 2 a$ - (relating to $\theta = \frac{\pi}{2}$ and $\theta = \frac{3 \pi}{2}$.

If we assume $a = 4$ the graph appears as below. Such a curve is known as cardioid.

graph{x^2+y^2-4sqrt(x^2+y^2)-4y=0 [-9.71, 10.29, -2, 9]}