The following differential equation describes the first derivative of an equation that describes an object moving in the X-Y plane along figure of eight with constant speed:
$ 4(x-h)^3 \frac{dx}{dt} = a^2[ 2(x-h)\frac{dx}{dt} - 2(y-k)\frac{dy}{dt} $
The center of the figure of eight is : $(h,k) = (50,70)$ and $(a=10)$ so the equation will be:
$ (x-50)^3 \frac{dx}{dt} = 50[ (x-50)\frac{dx}{dt} - (y-70)\frac{dy}{dt} ]$
suppose that the body first observed at position $(X_S,Y_S) = (30,70)$ and the speed of the body = 150 $m/s$. The position of the body is observed each 10 $seconds$.
Note that the speed is known so :
150 = $ \sqrt{ \frac{dx}{dt}^2 + \frac{dy}{dt}^2 }$
so
$\frac{dx}{dt} = \sqrt{ {150}^2 - \frac{dy}{dt}^2 }$
How to calculate the position $(X,Y)$ of the body each observation by knowing the previous information?