<s>[INST] Finding the horizontal and vertical tangents of $r^2=\sin(2\theta)$ Finding the horizontal and vertical tangents of $r^2=\sin(2\theta)$ where we are using polar coordinates.
I was wondering if you guys could show me multiple ways to do this problem. I did it by using implicit differentiation, but whenever I ask questions here I get a bunch of brilliant solutions so I thought I'd ask. Thanks!
 [/INST] $$r^2=\sin(2\theta)$$
$$r^2=2\sin\theta\cos\theta\to r^4=2(r\sin\theta)(r\cos\theta)$$
as $r\cos\theta=x;\;r\sin\theta=y$ we get the cartesian equation
$$(x^2+y^2)^2=2xy\to f(x,y)=(x^2+y^2)^2-2xy=0$$
I use the formula for implicit diferentiation
$$\frac{dy}{dx}=-\frac{\partial f/\partial x}{\partial f/\partial y}$$
So we have
$$\frac{dy}{dx}=\frac{2 y-4 x \left(x^2+y^2\right)}{4 y \left(x^2+y^2\right)-2 x}$$
Horizontal tangents are where derivative is zero.
$$2 y-4 x \left(x^2+y^2\right)=0\to x^2+y^2=\frac{y}{2x}$$
Plug in the equation of the curve
$$\left(\frac{y}{2x}\right)^2=2xy\to y=0;\;y=8x^3$$
The first solution is then $(0,0)$
Plugging $y=8x^3$ in the equation of the curve we get
$$(x^2+64x^6)^2=16x^4\to x^4 \left(64 x^4-3\right) \left(64 x^4+5\right)=0$$
$x=\pm\sqrt[4]{\frac{3}{64}}\approx \pm 0.4653$ and $y\approx \pm 0.806$.
The points where the tangents are vertical are the symmetric wrt line $y=x$ bisector of the first quadrant.
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