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Find the polar coordinates of the points on $r^2 = a^2 \sin 2\theta$ where the tangent is perpendicular to the initial line ($\theta=0).$

I guess they are assuming $a>0.$

The answer in the back of the book is $\left( a\sqrt{\frac{\sqrt{3}}{{2}}}, \frac{\pi}{6} \right),\ \left( a\sqrt{\frac{\sqrt{3}}{{2}}}, \frac{7\pi}{6} \right),\ \left( 0, \frac{\pi}{2} \right).$

The method they want us to use is to set $\frac{dx}{d\theta}=0$ and see what happens.

I think I can deduce the first two polar coordinates, but I don't how to logically deduce that $\left( 0, \frac{\pi}{2} \right)$ is perpendicular to the initial line because all methods I think of involves dividing by $0.$

Edit: For example, the method suggested by the book says we should do the following:

$x=r\cos\theta = a\cos\theta\sqrt{\sin (2\theta)}.$ Now set $\ \frac{dx}{d\theta} = 0.$ This implies $ \frac{ a\cos(2\theta)\cos\theta }{\sqrt{\sin(2\theta)}} - a\sin\theta\sqrt{\sin(2\theta)} = 0.$

But here we see the problem: this cannot be evaluated at $\theta=\frac{\pi}{2},$ which is one of the answers, because it would be dividing by $0$ which is not allowed.

To elaborate, if you tell me that $\frac{dx}{d\theta} = 0$ when $\theta=\frac{\pi}{2},$ my response is, "no, it is $ \frac{ a\cos\left(2\frac{\pi}{2}\right)\cos\frac{\pi}{2} }{\sqrt{\sin\left(2\frac{\pi}{2}\right)}} - a\sin\frac{\pi}{2}\sqrt{\sin\left(2\frac{\pi}{2}\right)}$ , which is undefined."

I figured out the rough shape of the graph by hand by doing a small table of $(r,\theta)$ pairs, although from doing this it still isn't clear what the derivative(s) is (are) at $r=0.$

Here is an image of the graph when $a=5:$

enter image description here

Note that this is an A Level question where polar coordinates have only just been introduced in this chapter, so overly-advanced solutions are not appropriate.

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  • $\begingroup$ Well, $\frac{d x}{d\theta}=0$ corresponds to $x=\operatorname{const}$ You have to find the equation of the tangent (you can plug into it the coordinates and the expression for the derivative/slope) $\endgroup$
    – Cherry Blossom
    Commented Dec 30, 2023 at 20:37
  • $\begingroup$ @PinkyWay Yes $\frac{dx}{d\theta}$ corresponds to $x=$ const, but I don't get the rest of your comment. Sorry $\endgroup$
    – Adam Rubinson
    Commented Dec 30, 2023 at 20:41
  • $\begingroup$ Have you seen the following formula $y-y_0=f'(x_0)(x-x_0)$? $\endgroup$
    – Cherry Blossom
    Commented Dec 30, 2023 at 21:57
  • $\begingroup$ Yes, but I don't get what you're getting at. I feel like I'm missing something obvious. $\endgroup$
    – Adam Rubinson
    Commented Dec 30, 2023 at 21:57
  • $\begingroup$ Why not just make a change of coordinates " $$\begin{aligned}(x,y)&=(r(\theta)\cos\vartheta,r(\theta)\sin\vartheta)\\&=a\sqrt{\sin(2\vartheta)}(\cos\vartheta,\sin\vartheta)?\\\frac{dy}{dx}&=\frac{dy}{d\vartheta}\cdot\frac{d\vartheta}{dx}\end{aligned}$$ $\endgroup$
    – Cherry Blossom
    Commented Dec 30, 2023 at 22:04

1 Answer 1

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$x=a\sqrt{\sin2\theta}\cos\theta$, $y=a\sqrt{\sin2\theta}\sin\theta$.

$\frac{dx}{dy}=\frac{\frac{dx}{d\theta}}{\frac{dy}{d\theta}}=-\frac{\cos3\theta}{\sin\theta}=0\implies\theta=\frac\pi 6,\frac\pi 2,\frac{7\pi}6.$

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  • $\begingroup$ I did not think of this approach. However, I think $\frac{dx}{d\theta} = \frac{a\cos 2\theta}{\sqrt{\sin 2\theta} } - a\sin\theta\sqrt{\sin 2\theta} $ is not allowed to be evaluated at $\theta=\frac{\pi}{2}.$ $\endgroup$
    – Adam Rubinson
    Commented Dec 30, 2023 at 22:21
  • $\begingroup$ The denominator of $dy/d\theta$ is the same. Maybe we can think of cancellations. @AdamRubinson $\endgroup$
    – Bob Dobbs
    Commented Dec 30, 2023 at 22:26
  • $\begingroup$ It doesn't matter: cancellations don't negate what I said in my previous comment. For example, when $t=5,\ \frac{ \frac{2t}{t-5} }{ \frac{7}{t-5} }$ is not defined. $\endgroup$
    – Adam Rubinson
    Commented Dec 30, 2023 at 22:28
  • $\begingroup$ Removable singularity, it is $10/7.$ @AdamRubinson $\endgroup$
    – Bob Dobbs
    Commented Dec 30, 2023 at 22:32
  • $\begingroup$ I have not learned about removable singularities, and they are certainly well beyond the scope of what is expected to answer this question. For the purposes of A Level maths, $\frac{x}{x}$ is not defined at $x=0.$ So I don't think your method is OK at the level the question is being asked at. Furthermore, your method is not the method suggested in the book, which I outlined in the question. $\endgroup$
    – Adam Rubinson
    Commented Dec 30, 2023 at 22:36

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