Let $v:\Bbb R \to \Bbb R^2$ be $C^{\infty}$. Then $\exists t_0\in (0,1)$ s.t $v(1)-v(0)$ is a scalar multiple of $\frac{dv}{dt}|_{t=t_0}$. True or False: Let $v:\Bbb R \to \Bbb R^2$ be $C^{\infty}$. Then $\exists t_0\in (0,1)$ s.t $v(1)-v(0)$ is a scalar multiple of $\frac{dv}{dt}|_{t=t_0}$.
We know that this is true by Mean value theorem in $\Bbb R \to \Bbb R$ but is it true for $\Bbb R \to \Bbb R^2$
 A: Define $v$ by
$$v(t) = ((2t-1)^2, (2t-1)^3).$$
Then $v(1) - v(0) = (0, 2)$, But it is easy to check that $v'(t)$ is never a scalar multiple of $(0, 2)$. This can be easily seen from the following plot.


The statement can be made true with a bit of correction, which results in a version of Cauchy's mean value theorem.

Let $v : \mathbb{R} \to \mathbb{R}^2$ be $C^{\infty}$ be regular, meaning that $v'(t) \neq 0$ for all $t$. Then there exists $t_0\in(0,1)$ such that $v(1) - v(0)$ is a scalar multiple of $v'(t_0)$.

Proof. Define $w(t) = \det(v(t) - v(0), v(1) - v(0))$. Then $w(0) = w(1) = 0$, and so, there exists $t_0 \in (0, 1)$ such that $w'(t_0) = 0$ by the Rolle's theorem. Since $w'(t_0) = \det(v'(t_0), v(1) - v(0))$, this means that $v'(t_0)$ and $v(1) - v(0)$ are not linearly independent. Since $v'(t_0) \neq 0$, this implies that $v(1) - v(0)$ is a scalar multiple of $v'(t_0)$.
A: HINT: Consider $v(t)=(\cos 2\pi t,\sin 2\pi t)$.
EDIT: If we rule out the case of $v(0)=v(1)$, what does the Cauchy Mean Value Theorem tell us? 
