All over the net, it is stated that the parametrization of the lemniscate with Cartesian equation $(x^2 + y^2)^2 = 2a^2 (x^2 – y^2)$ is:
$$\varphi: t \mapsto \left(\frac{a\sqrt{2}\cos(t)}{1+\sin^2(t)}, \frac{a\sqrt{2}\cos(t)\sin(t)}{1+\sin^2(t)}\right)$$
I don't understand how to get there…
My method is as follows:
I transform the Cartesian equation to polar coördinates, and I find:
$$r=a\sqrt{2\cos(2\vartheta)}$$
Then replacing $r$ in $(r\cos(\vartheta), r\sin(\vartheta))$ gives:
$$\psi: t \mapsto \left(a\sqrt{2\cos(2t)}\cos(t),a\sqrt{2\cos(2t)}\sin(t) \right)$$
Are these equivalent? (I guess theory says that there must be a certain function which maps $\psi$ into $\varphi$; I can't seem to find which function…)
Best Answer
The "standard" parametrization (the one you find when you look that up) appears to be intended to provide a continuous parametrization around the curve. (This may be connected with the lemniscate being a special case of the class of curves known as "Cassinian ovals" -- but I haven't looked far into that yet.) Using that form for $ \ x(t) \ $ and $ \ y(t) \ $ , we obtain (for $ \ a = 1 \ $)
$$ x^2(t) \ + \ y^2(t) \ = \ 2 \ \frac{\cos^2 t \ (1 + \sin^2 t)}{(1 + \sin^2 t)^2} \ = \ 2 \ \frac{\cos^2 t }{1 + \sin^2 t} \ . $$
EDIT -- Incidentally, this is not an arclength parametrization (which was my early thought). While the integrand doesn't look too bad [ $ \ a \ \sqrt{2} \ \int \frac{d\theta}{\sqrt{\cos 2 \theta}} \ ] $ , the integral function is non-elementary...
The graphs of $ \ x(t) \ , \ y(t) \ , \ $ and $ \ \sqrt{x^2 + y^2} \ $ are shown below.
This covers the curve over the interval $ \ 0 \ \le \ t \ \le \ 2 \pi \ $ as seen here.
The parametrization you found by using polar coordinates is a valid alternative, but because the equation is $ \ r^2 \ = \ 2a^2 \ \cos 2t \ , $ it cannot produce admissible radii for the polar curve on the intervals (in the principal circle) $ \ \frac{\pi}{4} \ < \ t \ < \ \frac{3 \pi}{4} \ $ or $ \ \frac{5\pi}{4} \ < \ t \ < \ \frac{7 \pi}{4} \ . $ By going "a little outside" the principal circle, the two lobes of the lemniscate can be individually covered as shown here (again for $ \ a = 1 \ $ ).
So the parametrization you found is not defined for periodic intervals in the real numbers.