how to calculate the curvature of an ellipse how can I compute the curvative of an ellipse given by $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$
do i need to take $x=a \cos(t)$ and $y=b \sin(t)$?
please show me a way how to solve this? thank you for helping:)
 A: You don't need the unit tangent to get the curvature or parameterization by arc length. It is much simpler to use the following formula:
$$ \kappa = \frac{||v \times v'|| }{||v||^{3}},$$
where
$$ v = (-a\sin(t),b \cos(t)) \qquad \text{and} \qquad v' = (-a\cos(t),-b\sin(t)) $$
and hence
$$ ||v|| = \sqrt{(a^{2}\sin^{2}(t) + b^{2}\cos^{2}(t))} $$
and
\begin{align*}
v \times v'
& = (-a\sin(t), b \cos(t),0) \times (-a \cos(t), -b \sin(t),0) \\
& = (0, 0, a b \sin^{2}(t)+ab\cos^{2}(t))
=(0,0,a b),
\end{align*}
so
$$ ||v \times v'|| = ab.$$
Therefore
$$ \kappa = \frac{||v \times v'|| }{||v||^{3}} = \frac{ab}{\left(\sqrt{a^{2}\sin^{2}(t) + b^{2}\cos^{2}(t)}\right)^{3}} $$
A: We have $\alpha'(t) = \langle -a \sin t, b \cos t \rangle$ and $\alpha''(t) = \langle  -a\cos t , - b \sin t\rangle$, thus $|\alpha'(t)| = \sqrt{a^2\sin^2t + b^2\cos^2t}$. We have that $T(t) = \frac{\alpha'(t)}{|\alpha'(t)|},$ which has length $1$ and is tangent to $\alpha (t).$ $$T(t) = \langle \frac{-a \sin t}{\sqrt{a^2\sin^2t + b^2\cos^2t}}, \frac{b\cos t}{\sqrt{a^2\sin^2t + b^2\cos^2t}}\rangle$$ which  leads to $$\kappa = \frac{|T'(t)|}{|\alpha'(t)| }= \frac{ab}{(\sqrt{a^2\sin^2t + b^2\cos^2t})^3}.$$
