91Ó°ÊÓ

$r=\frac{1}{2}+\sin \mathrm{θ},r=1+\sin \mathrm{θ},r=2+\sin \mathrm{θ},r=3+\sin \mathrm{θ}$

Consider the limacon $r=\frac{1}{2}+\sin \mathrm{θ}$

The goal is to figure out how the graphs of the form behave $r=3+b\cos \mathrm{θ}$

Substitute different $\mathrm{θ}$values and find different $r$values.

Consider $\mathrm{θ}=0,\frac{\mathrm{π}}{2},\mathrm{π},\frac{3\mathrm{π}}{2},2\mathrm{π}$

We find the equation's values for various $\mathrm{θ}$values. $r=\frac{1}{2}+\sin \mathrm{θ}$

When $\mathrm{θ}=0$

$r=\frac{1}{2}+\sin 0\left[\text{since}r=\frac{1}{2}+\sin \mathrm{θ}\right]$

$r=\frac{1}{2}+0[$since $\sin 0=0]$

$r=\frac{1}{2}$

Then the coordinate, $(r,\mathrm{θ})=\left(\frac{1}{2},0\right)$

When $\mathrm{θ}=\frac{\mathrm{π}}{2}$

Then the coordinate $(r,\mathrm{θ})=\left(\frac{3}{2},\frac{\mathrm{π}}{2}\right)$

When $\mathrm{θ}=\mathrm{π}$

Then the coordinate $(r,\mathrm{θ})=\left(\frac{1}{2},\mathrm{π}\right)$

When $\mathrm{θ}=\frac{3\mathrm{π}}{2}$

Then the coordinate $(r,\mathrm{θ})=\left(-\frac{1}{2},\frac{3\mathrm{π}}{2}\right)$

When $\mathrm{θ}=2\mathrm{π}$

Then the coordinate $(r,\mathrm{θ})=\left(\frac{1}{2},2\mathrm{π}\right)$

The points are $\left(\frac{1}{2},0\right)\left(\frac{3}{2},\frac{\mathrm{Ï€}}{2}\right)\left(\frac{1}{2},\mathrm{Ï€}\right)\left(-\frac{1}{2},\frac{3\mathrm{Ï€}}{2}\right)\left(\frac{1}{2},2\mathrm{Ï€}\right)$

Similarly, we can find the values of $r=1+\sin \mathrm{θ},r=2+\sin \mathrm{θ}$and $r=3+\sin \mathrm{θ}$

Now draw the different graphs and plot the points on the graph.

Here in the graph as the $a$value increases the Inner loop vanishes. The equation $r=1+\sin \mathrm{θ}$is a cardioid. when $a>1$the graph is a convex. When $a<1$the graph is a dimple. Hence the explanation.