I stumped at one of the exercise in my multivariable calculus textbook.
I try to search online but I can't seem to search
on how answer no 3 and 4 below is derived.
I also plot both of polar coordinates in mathematica but I can't seem to get around on
how to solve them algebraically.
The question and answer as appear in textbook is as below:
The polar equation for the curve $C_{1}$ is $r=\cos 2\theta$,while the curve
$C_{2}$ is described by the polar equation $r=1+\cos\theta$. Find all points
at which $C_{1}$ and $C_{2}$ intersect.
Solution below
1) r=0 (the origin)
2) ($\frac{5-\sqrt{17}}{4}$ , $\pm\arccos \frac{1-\sqrt{17}}{4}$)
(I manage this solve this, just put both equation equal to each other)
I'm at loss what the formula to derive answer for 3 and 4 below.
3) ($-1$ , $\pm 90 ^\circ$)
4) ($\frac{-1}{2}$ , $\pm 60 ^\circ$)
Can anyone point out to me what's the formula used ?
Thanks
Best Answer
The issues come from a lack of uniqueness: $(r,\theta),(-r,\theta+\pi),(r,\theta+2\pi)$ are all the same. Also, all $(0,\theta)$ are the same for any $\theta$. Fortunately these equations are periodic (with period $2\pi$), which simplifies the problem. So you have to do this as well:
$$-\cos(2(\theta+\pi))=\cos(\theta)+1$$
i.e, we need to see where/if one is negative and halfway across the circle, since this will also count as an intersection. $$-\cos(2\theta)=\cos(\theta)+1$$ $$\sin^2(\theta)-\cos^2(\theta)=\cos(\theta)+1$$ $$2\cos^2(\theta)+\cos(\theta)=\cos(\theta)(2\cos(\theta)+1)=0$$ $$\cos(\theta)=0\implies\theta=\pi/2,3\pi/2$$ $$2\cos(\theta)+1=0\implies\theta=2\pi/3, 4\pi/3$$
graphing both should convince you that these, plus the ones you found, give the full set of solutions.