Q 48. - Calculus

Question

Graph the limac¸ons $r = \frac{1}{2}+ \sin \theta , r = 1 + \sin \theta , r = 2 + \sin \theta , r = 3 + \sin \theta$ Make a conjecture about the behavior of graphs of limac¸ons of the form $r = a + \sin \theta$ for various values of a. In particular, try to understand which values of a will give a limac¸on with an inner loop. Which values of a will give a limac¸on with a dimple? Which values of a will give a convex limac¸on?

Step-by-Step Solution

Answer

The graph is convex when $a>1$ The graph is a dimple when $a>1$

Step 1: Given information

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

Step 2: Calculation

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

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

Substitute different $\theta$ values and find different $r$ values.

Consider $\theta =0,\frac{\pi }{2},\pi ,\frac{3\pi }{2},2\pi$

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

When $\theta =0$

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

$r=\frac{1}{2}+0[$ since $\sin 0=0]$

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

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

When $\theta =\frac{\pi }{2}$

$$\begin{gathered} r=\frac{1}{2}+\sin \frac{\pi }{2}\left[sincer=\frac{1}{2}\right. \\ r=\frac{1}{2}+1\left[since\sin \frac{\pi }{2}=1\right] \\ r=\frac{3}{2} \end{gathered}$$

Then the coordinate $(r,\theta )=\left(\frac{3}{2},\frac{\pi }{2}\right)$

When $\theta =\pi$

$$\begin{gathered} r=\frac{1}{2}+\sin \pi \left[sincer=\frac{1}{2}+\sin \theta \right] \\ r=\frac{1}{2}[\text{ since }\sin \pi =0] \end{gathered}$$

Then the coordinate $(r,\theta )=\left(\frac{1}{2},\pi \right)$

When $\theta =\frac{3\pi }{2}$

$$\begin{gathered} r=\frac{1}{2}+\sin \frac{3\pi }{2}\left[\text{ since }r=\frac{1}{2}+\sin \theta \right] \\ r=\frac{1}{2}+(-1)\left[\text{ since }\sin \frac{3\pi }{2}=-1\right] \\ r=-\frac{1}{2} \end{gathered}$$

Then the coordinate $(r,\theta )=\left(-\frac{1}{2},\frac{3\pi }{2}\right)$

When $\theta =2\pi$

$$\begin{gathered} r=\frac{1}{2}+\sin 2\pi \left[\text{ since }r=\frac{1}{2}-\sin \theta \right] \\ r=\frac{1}{2}[\text{ since }\sin 2\pi =0] \end{gathered}$$

Then the coordinate $(r,\theta )=\left(\frac{1}{2},2\pi \right)$

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

Similarly, we can find the values of $r=1+\sin \theta ,r=2+\sin \theta$ and $r=3+\sin \theta$

Step 3: Calculation

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 $r=1+\sin \theta$ is a cardioid. when $a>1$ the graph is a convex. When $a<1$ the graph is a dimple. Hence the explanation.