Suppose that $p(t) = (x(t),y(t),z(t))$ is a continuous trajectroy in $R^3$ defined on $t \in [0,t_1]$
Assume that $p(0) = (x_o,y_0,z_0)$, where $(x_o,y_0,z_0)$ is a point in $R^3$.
Let $B$ be a closed ball of radius $r >0$ around the point $(x_0,y_0,z_0)$
Then, the statement that I want to make precise is that:
If the trajectory $p(t)$ leaves the ball $B$, then there exists a $t_0 \in [0,t_1]$ such that
(i) ||$p(t_1) - (x_0,y_0,z_0)$|| $= r$
(ii) for all $0 \le t < t_1$, ||$p(t_1) - (x_0,y_0,z_0)$|| $\le$ r
Roughly speaking, I am trying to say that if the trajectory leaves the ball, then there exists "the first time" that the trajectory intersects the boundary of the ball.
How do I prove this??
It seems intuitively natural, but I am not sure how to prove this precisely.
I think you can invoke the intermediate value theorem (http://en.wikipedia.org/wiki/Intermediate_value_theorem).
Consider the function $f(t)=\Vert (x(t),y(t),z(t))-(x_0,y_0,z_0)\Vert$ which is the distance of the point in the trajectory at the time $t$ from.the center of the ball.
Since the trajectory is continue, $f:\mathbb{R}\to\mathbb{R}$ is a continuous function. In your hypothesis you have that $f(0)=0$ and $f(T)=R>r$ for some $T,R$. For the above mentioned theorem, there exists a $t_0\in(0,T)$ such that $f(t_0)=r$, that is $t_0$ is the moment when the trajectory hits the border of the ball.
(You have to prove that $f$ is continuous, but that follows from the continuity of composition and of the norm)