Chapter H.2, Problem 30E

To determine

To calculate:

The intersection points of the following curve:

  $\begin{array}{l}r=\cos 3\theta \\ r=\sin 3\theta \end{array}$

Answer to Problem 30E

The intersection points of the curves are $\left(\frac{1}{\sqrt{2}},\frac{\pi }{12}\right),\left(\frac{1}{\sqrt{2}},\frac{9\pi }{12}\right),\left(-\frac{1}{\sqrt{2}},\frac{5\pi }{12}\right),\left(0,0\right)$

Explanation of Solution

Given information:

  $\begin{array}{l}r=\cos 3\theta \\ r=\sin 3\theta \end{array}$

Calculation:

The equations of the curves are:

  $\begin{array}{l}r=\cos 3\theta \\ r=\sin 3\theta \end{array}$

So, the curves are plotted as below:

  

Now, need to find the interval in which loop exists:

  $\begin{array}{l}\cos 3\theta =\sin 3\theta \\ \tan 3\theta =1\\ \theta =\frac{\pi }{12},\frac{5\pi }{12},\frac{9\pi }{12},\frac{13\pi }{12},\frac{17\pi }{12},\frac{21\pi }{12}\\ 0\le \theta \le 2\pi \end{array}$

Now, need to find the co-ordinates of $r$ by using the above calculated $\theta$ values:

At $\theta =\frac{\pi }{12}$ : $r=\cos 3\theta =\cos \left(3\times \frac{\pi }{12}\right)=\cos \left(\frac{\pi }{4}\right)=\frac{1}{\sqrt{2}}$ : $\left(r,\theta \right)=\left(\frac{1}{\sqrt{2}},\frac{\pi }{12}\right)$

At $\theta =\frac{5\pi }{12}$ : $r=\cos 3\theta =\cos \left(3\times \frac{5\pi }{12}\right)=\cos \left(\frac{5\pi }{4}\right)=-\frac{1}{\sqrt{2}}$ : $\left(r,\theta \right)=\left(-\frac{1}{\sqrt{2}},\frac{5\pi }{12}\right)$

At $\theta =\frac{9\pi }{12}$ : $r=\cos 3\theta =\cos \left(3\times \frac{9\pi }{12}\right)=\cos \left(\frac{9\pi }{4}\right)=\frac{1}{\sqrt{2}}$ : $\left(r,\theta \right)=\left(\frac{1}{\sqrt{2}},\frac{9\pi }{12}\right)$

At $\theta =\frac{13\pi }{12}$ : $r=\cos 3\theta =\cos \left(3\times \frac{13\pi }{12}\right)=\cos \left(\frac{13\pi }{4}\right)=-\frac{1}{\sqrt{2}}$ : $\left(r,\theta \right)=\left(-\frac{1}{\sqrt{2}},\frac{13\pi }{12}\right)$

At $\theta =\frac{17\pi }{12}$ : $r=\cos 3\theta =\cos \left(3\times \frac{17\pi }{12}\right)=\cos \left(\frac{17\pi }{4}\right)=\frac{1}{\sqrt{2}}$ : $\left(r,\theta \right)=\left(\frac{1}{\sqrt{2}},\frac{17\pi }{12}\right)$

At $\theta =\frac{21\pi }{12}$ : $r=\cos 3\theta =\cos \left(3\times \frac{21\pi }{12}\right)=\cos \left(\frac{21\pi }{4}\right)=-\frac{1}{\sqrt{2}}$ : $\left(r,\theta \right)=\left(\frac{1}{\sqrt{2}},\frac{21\pi }{12}\right)$

So, get that:

In the first quadrant, same point is represented by $\left(\frac{1}{\sqrt{2}},\frac{\pi }{12}\right)$ and $\left(-\frac{1}{\sqrt{2}},\frac{13\pi }{12}\right)$

In the second quadrant, same point is represented by $\left(\frac{1}{\sqrt{2}},\frac{9\pi }{12}\right)$ and $\left(\frac{1}{\sqrt{2}},\frac{21\pi }{12}\right)$

In the third quadrant, same point is represented by $\left(-\frac{1}{\sqrt{2}},\frac{5\pi }{12}\right)$ and $\left(\frac{1}{\sqrt{2}},\frac{17\pi }{12}\right)$

And the origin is also there as another point of intersection: $\left(r,\theta \right)=\left(0,0\right)$

Hence, all the intersection points of the given curves are:

  $\left(\frac{1}{\sqrt{2}},\frac{\pi }{12}\right),\left(\frac{1}{\sqrt{2}},\frac{9\pi }{12}\right),\left(-\frac{1}{\sqrt{2}},\frac{5\pi }{12}\right),\left(0,0\right)$