Two polar curves intersect problem .

Question

Suppose we have two curves given by: $$r=20sin2\theta $$ $$r= 20 cos2\theta$$

Now by solving the equations, we get the solution as $\theta = \frac{\pi}{8}$.

However, on graphing the equations, I can find 8 points of intersection . Where could I have done a mistake?

Answer 1

You have to take into account two issues. One of them is the periodicity of the $\tan$ function. $\tan(2\theta)$ has a periodicity of $\pi/2$, so you get $\pi/8, 5\pi/8, 9\pi/8, 13\pi/8$ as possible solutions in the $[0,2\pi)$ interval. The other issue to consider is that the $r$ in the two equations is not necessarily the same. You also get solution if $r_1=-r_2$ and $\sin(2\theta)=-\cos(2\theta)$. From the definition of $\arctan$ you get $\theta=-\pi/8$, which yields $3\pi/8, 7\pi/8, 11\pi/8, 15\pi/8$ in the above mentioned interval.

Answer 2

You're missing the point that $$ \frac{\pi}{8} $$ is just one of the solutions to this set of polar equations.

Let's find the general solution by setting both values of $r$ as equal to each other:

$$ 20 \sin{2\theta} = 20\cos{2\theta} \implies \sin{2\theta} = \cos{2\theta} $$

Solving this trigonometric equation, we get our general solutions for $\theta$ as:

$$ \theta = \frac{1}{8}(4\pi \ * n \ + \ \pi ) ; \ n \in{Z} $$

Setting $n$ = $0$, you get your first solution: $\frac{\pi}{8}$.

With multiple values of $n$, we get multiple values of $\theta$ and hence you can find $4$ values of $\theta$ in the interval $[0,2\pi]$ and hence you have your multiple points of intersection.