derivative of closed parametric curve

Question

Suppose, a parametric function $\beta:[0,1]\mapsto\mathbb{R}^2$ is a closed curve, that is $\beta(0)=\beta(1)$. For example $\beta(t)=(\sin 2\pi t,\cos 2\pi t)'$.

Then my question: Is the derivative $\beta'(t)$ is also closed curve? For the above example $\beta'(t)=(2\pi\cos 2\pi t,-2\pi \sin 2\pi t)'$.

Could any one please suggest me what is the answer and how to prove it.

Thanks in advance.

Answer 1

This is in general wrong. The curve $[0,2\pi]\ni t\mapsto (\sin (\tfrac{t}{2})\cos (\tfrac{t}{2}),\sin (\tfrac{t}{2}))$ provides a counter-example.

Answer 2

Being closed has nothing to do with it. At any sharp point on the curve, the curve traced by the derivative will be discontinuous. The start/end point is not in any way special, it's just a matter of having a parametric curve with a continuous derivative. You could have a closed curve that is continuously differentiable at start/end point, but has a cusp somewhere else.