I have solved the following exercise and I would like to have some feedback on my solution, thanks.
Let $a,b>0.$ Find the maximum and minimum speed of the ellipse $\gamma(t)=\left(a\cos(t),b\sin(t)\right)$
My solution:
$\gamma'(t)=\left(-a\sin(t),b\cos(t)\right)$ so the speed is $|\gamma'(t)|=\sqrt{a^2\sin^2(t)+b^2\cos^2(t)}.$
To find maxima and minima it is easier to study the speed squared $F(t):=|\gamma'(t)|^2=a^2\sin^2(t)+b^2\cos^2(t)$ for which $F'(t)=0\Leftrightarrow 2a^2\sin(t)\cos(t)-2b^2\cos(t)\sin(t)=a^2\sin(2t)-b^2\sin(2t)=(a^2-b^2)\sin(2t)=0\Leftrightarrow t=k\pi\text{ or }t=\frac{\pi}{2}+k\pi,\ k\in\mathbb{Z}$ and since $|\gamma'(t=k\pi)|=\sqrt{a^2\sin^2(k\pi)+b^2\cos^2(k\pi)}=\sqrt{b^2}=b$ and $|\gamma'(t=\frac{\pi}{2}+k\pi)|=\sqrt{a^2\sin^2(\frac{\pi}{2}+k\pi)+b^2\cos^2(\frac{\pi}{2}+k\pi)}=\sqrt{a^2}=a$ we have that the speed has maximum value $a$ and minimum value $b$ if $a>b$ (i.e. if the ellipse is horizontal) and it has maximum value $b$ and minimum value $a$ if $a<b$ (i.e. if the ellipse is vertical) and being $\gamma(t=k\pi)=(\pm a,0)$ and $\gamma(t=\frac{\pi}{2}+k\pi)=(0,\pm b)$ we see that the maximum speed is attained where the distance from the center of the ellipse is lowest and the minimum speed is attained where the distance from the center of the ellipse is highest.