# Show that the tangent to $\left(0,\frac{\pi}{2}\right)$ on the polar curve with equation $r^2=a^2\sin 2\theta$ is perpendicular to the initial line

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 $$a=5:$$

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.

• 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) Commented Dec 30, 2023 at 20:37
• @PinkyWay Yes $\frac{dx}{d\theta}$ corresponds to $x=$ const, but I don't get the rest of your comment. Sorry Commented Dec 30, 2023 at 20:41
• Have you seen the following formula $y-y_0=f'(x_0)(x-x_0)$? Commented Dec 30, 2023 at 21:57
• Yes, but I don't get what you're getting at. I feel like I'm missing something obvious. Commented Dec 30, 2023 at 21:57
• 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} Commented Dec 30, 2023 at 22:04

$$x=a\sqrt{\sin2\theta}\cos\theta$$, $$y=a\sqrt{\sin2\theta}\sin\theta$$.
$$\frac{dx}{dy}=\frac{\frac{dx}{d\theta}}{\frac{dy}{d\theta}}=-\frac{\cos3\theta}{\sin\theta}=0\implies\theta=\frac\pi 6,\frac\pi 2,\frac{7\pi}6.$$
• 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 $\theta=\frac{\pi}{2}.$ Commented Dec 30, 2023 at 22:21
• The denominator of $dy/d\theta$ is the same. Maybe we can think of cancellations. @AdamRubinson Commented Dec 30, 2023 at 22:26
• 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. Commented Dec 30, 2023 at 22:28
• Removable singularity, it is $10/7.$ @AdamRubinson Commented Dec 30, 2023 at 22:32
• 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. Commented Dec 30, 2023 at 22:36