Velocity of the Center of a Sliding Ellipse

Probably the easiest way to solve this problem is to fix the ellipse and attach a reference frame (R, d) to a moving point on the perimeter of the fixed ellipse. Since the ellipse is moving counterclockwise, we want our point to move clockwise on the fixed ellipse. And this implies that this moving reference frame attached to the moving point will have an orientation of $\theta = -\dfrac{\pi}{4}$ .

We can choose the x'-axis of the moving frame to point opposite to the time derivative of p , and the normal y'-axis is just a counterclockwise rotation of the x' axis.

Let $s(t)$ be our eccentric angle (a function of time $t$ ), then

$p(t) = ( -a \sin s, - b \cos s)$

the parameter s has ds/dt > 0. Now the velocity of point $p$ is

$p'(t) = \dfrac{dp}{dt}(t) = (-a \cos s, b \sin s ) \dfrac{ds}{dt}$

The rotation matrix of the reference frame attached to the point $p$ has its rotation matrix as follows

$R(\theta) =\begin{bmatrix} \cos \theta && -\sin \theta \\ \sin \theta && \cos \theta \end{bmatrix}$

The first column of the above matrix is the x'-axis, and the second column is the y'-axis.

Comparing the first column of the matrix $R$ with the negative of the vector $p'$ , we deduce that

$\tan \theta = - \dfrac{b}{a} \tan s$

Thus from the given value of $\theta = -\dfrac{\pi}{4}$ , we can compute the eccentric angle $s$ .

Also, by differentiating the above formula w.r.t time, we obtain,

$\sec^2 \theta \dfrac{d\theta}{dt} = - \dfrac{b}{a} \sec^2 s \dfrac{ds}{dt}$

which determines the relation between $\dfrac{d\theta}{dt}$ and $\dfrac{ds}{dt}$ . To determine $\dfrac{ds}{dt}$ we take the magnitude of $\dfrac{dp}{dt}$ and equate it with the given speed of the tangency point. We have

$v = | p'(t)| = \dfrac{ds}{dt} \sqrt{ a^2 \cos^2 s + b^2 \sin^2 s }$

Now, the reference frames relation is

$q = p + R q'$

where $q$ is the coordinate vector of any point in the absolute frame, and $q'$ is the the corresponding coordinate vector with respect to the moving reference frame whose rotation matrix is $R$ .

Differentiating with respect to time,

$\dfrac{dq}{dt} = \dfrac{dp}{dt} + \dfrac{dR}{dt} q' + R \dfrac{dq'}{dt}$

if $q = 0$ (a fixed vector) , then

$0 = \dfrac{dp}{dt} + \dfrac{dR}{dt} q' + R \dfrac{dq'}{dt}$

Hence,

$\dfrac{dq'}{dt} = - R^T \dfrac{dp}{dt} - R^T \dfrac{dR}{dt} q'$

And this completes the solution.