Another type of derivative, another type of differential equation

Question

Let $\mathbf{v}:(a,b)\to\mathbb{R}^2$ be a continuous function, such that $||\mathbf{v}(t)||=1,\ \forall t\in (a,b)$. Is it possible to find a continuous function $\mathbf{r}:(a,b)\to\mathbb{R}^2$ so that:

$ \mathbf{v}(t_0)=\lim\limits_{t\searrow t_0} \displaystyle \frac{\mathbf{r}(t)-\mathbf{r}(t_0)}{||\mathbf{r}(t)-\mathbf{r}(t_0)||},\forall t_0\in (a,b). $

Answer 1

Let $c={a + b \over 2}$ and define $r(t) = \int_c^t v(\tau) d \tau$. Then you have $\dot{r}(t) = v(t)$ for $t \in (a,b)$.

For $t>t_0$ (and $r(t) \neq r(t_0)$), we have ${ r(t)-r(t_0) \over \| r(t)-r(t_0) \|} = { r(t)-r(t_0) \over t-t_0 } { t-t_0 \over \| r(t)-r(t_0) \|} = { r(t)-r(t_0) \over t-t_0 } {1 \over \| {r(t)-r(t_0) \over t-t_0}\|}$.

Since $\|\dot{r}(t)\|=1$, we have $\lim_{t \downarrow t_0} { r(t)-r(t_0) \over \| r(t)-r(t_0) \| } = \dot{r}(t) \cdot 1 = v(t)$.