I came up with this problem while I was trying to prove the following geometric problem : Let $A, B$ be the distinct points in $\mathbb{R}^2$. If $r(t)$ is the non self-intersecting trajectory of particle in $\mathbb{R}^2$ starts at $A$ and ends at $B$, is there a moment when velocity vector of particle is pointing same direction with the vector $B-A $ $?$, that is, $r'(t)=c(B-A)$ for some $c>0$$?$ This is definitely not true in general for $\mathbb{R}^n$ with $n>2$, but intuitively it must be true for $\mathbb{R}^2$.
First I formalized this problem as follows : Let $r:[a,b] \rightarrow \mathbb{R}^2$ : continuously differentiable unit speed curve in $\mathbb{R}^2$ with $r(a)=(0,0), r(b)=(1,0)$. Then $v(t)=r'(t) \in S^1$. Now letting $r(t)=(r_1(t),r_2(t))$, $v(t)=(r_1'(t),r_2'(t))$ we get differential equation $$ r_1'(t)^2 + r_2'(t)^2 = 1 $$ with boundary conditions $$ r_1(b) = 1\\ r_1(a) = r_2(b) = r_2(a) = 0 $$
Or equivalently we get integral equations $$ \int_a^b r_1'(t) dt = r_1(b) - r_1(a) = 1 \\ \int_a^b r_2'(t) dt = r_2(b) - r_2(a) = 0 \\ r_1'(t)^2 + r_2'(t)^2 = 1 $$
Lastly we can parametrize $v(t)$ by angle function $\theta(t)$ s.t. $v(t)=(r_1'(t),r_2'(t))=(\cos\theta(t),\sin\theta(t))$ and then we get $$ \int_a^b \cos\theta(t) dt = 1 \\ \int_a^b \sin\theta(t) dt = 0 $$
The goal is to prove $\exists t_0 \in [a,b] \ s.t. v(t_0)=(1,0)$.(Under the last formalism it is to prove $\exists t_0 \in [a,b] \ s.t. \theta(t_0) = 2\pi n $ for some $ n\in \mathbb{Z}$.
Now how can I prove this? Also since I have no knowledge about getting information of solution of this type of differential/integral equation, any comment on how/where I can learn about this type of problem would be helpful.
EDIT : I noticed that I need to assume that trajectory is not self-intersecting, that is, $r(t)$ is need to be one-to-one. Otherwise there is an obvious counterexample:
$r(t)$ is not one-to-one" />
You could use the extended mean value theorem, $$ \frac{y(b)-y(a)}{x(b)-x(a)}=\frac{y'(c)}{x'(c)}, $$ which encodes exactly your claim. To avoid the possible division-by-zero in the denominator, use $$ h(t)=\det\pmatrix{1&1&1\\x(a)&x(b)&x(t)\\y(a)&y(b)&y(t)} $$ and use the theorem of Rolle: $h(a)=h(b)=0\implies h'(c)=0$ where $$ h'(c) =\det\pmatrix{1&1&0\\x(a)&x(b)&x'(c)\\y(a)&y(b)&y'(c)} =\det\pmatrix{x(b)-x(a)&x'(c)\\y(b)-y(a)&y'(c)} $$ which also encodes that the tangent at $t=c$ is parallel to $B-A$.