Show that an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2}$ = 1 is a smooth curve

Question

Show that the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2}$ = 1 is a smooth curve by producing an admissable parametrization.

I know the parametrization is $a*cos(t) + i*b*sin(t)$. It looks right but I don't know how to work towards this answer as I suck at parametrization so I don't know how to show

Answer 1

You are on the right track. The parameterization of $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ is:

$$(a\cos(\theta), b\sin(\theta))\text{ for }0\leq 0 < 2\pi$$

Both components of the parameterization are smooth, so it remains to check that the endpoints coincide and are smooth. We have:

$$\lim_{\theta\rightarrow 0^{+}}(a\cos(\theta), b\sin(\theta)) = \lim_{\theta\rightarrow 2\pi^{-}}(a\cos(\theta), b\sin(\theta)) = (a, 0)$$

The directional derivative of this curve is:

$$\frac{d}{d\theta}(a\cos(\theta), b\sin(\theta)) = (-a\sin(\theta), b\cos(\theta))$$

We have:

$$\lim_{\theta\rightarrow 0^{+}}(-a\sin(\theta), b\cos(\theta)) = \lim_{\theta\rightarrow 2\pi^{-}}(-a\sin(\theta), b\cos(\theta)) = (0, b)$$

Thus, $\boxed{\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 \text{ is smooth.}}$