Parameterized ellipse

Question

The general equation of an ellipse with center in the Cartesian axes origin is

$$\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$$

This ellipse can be parameterized by returning it to a circumference in some way: replacing

$$\dfrac{x^2}{a^2} = X$$

$$\dfrac{y^2}{b^2} = Y$$

$$X^2 + Y^2 = 1$$

writing the polar coordinates

$$x = a\rho \cos{\theta}$$

$$y = b\rho \sin{\theta}$$

The vectorial function is

$$\vec{r}(t) = [a\rho \cos{\theta}, b\rho \sin{\theta}]$$

The first derivate

$$\vec{r}\;'(\theta) = [-a\rho \sin{\theta}, b\rho \cos{\theta}]$$

The module

$$||r'(\theta)|| = \sqrt{(-a\rho \sin{\theta})^2 + (b\rho \cos{\theta})^2}$$

Is this parametrization correct? What is the arc length between 0 and 2$\pi$?

Thanks in advance

I don't know what is that $;\rho;$ there in $;x,,y;$ , but drop it and that's the usual parametrization — DonAntonio, Aug 29 '18 at 21:20
What you show is mostly correct, except you don't need the $\rho$ factor. Regarding the arc length, the arc length of the ellipse presents... wait for it... an elliptic integral. In terms of introductory calculus the integral cannot be solved with elementary functions. https://en.wikipedia.org/wiki/Elliptic_integral — Doug M, Aug 29 '18 at 21:40

score 1 · Accepted Answer · answered Aug 29 '18 at 21:28

Your parametrization is valid.

The arc-length integral is not done by finding an antiderivative, but by numerical methods.$$ ||r'(\theta)|| =\sqrt{(-a\rho \sin{\theta})^2 + (b\rho \cos{\theta})^2} $$

Does not have a closed form antiderivative.

That is why we do not have a simple formula for the circumference of an ellipse.

Parameterized ellipse

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