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This is not to ask for a solution, rather I want comments on the rigor of the step. Thank you for your time!
enter image description here
The graph shows a system of two points at $t$ and $t+dt$, it is a bit exaggerated of course.
The upper point is moving strictly in the $x$ direction and has a constant velocity $u$, while the lower point is always moving towards the upper point with constant velocity $v>u$. I want to find the rate of change of the distance $\mathcal{L}(t)$ which separates them (to find the time of collision $\tau$).

Keeping in mind that $\cos(\pi-\theta)=-\cos(\theta)$ and the cosines' law: $$\begin{align*} \mathcal{L}^2(t+\Delta t)&\approx(u\Delta t)^2+(\mathcal{L}(t)-v\Delta t)^2+2(u\Delta t)(\mathcal{L}(t)-v\Delta t)\cos(\theta(t))\\ &\approx(u\Delta t)^2+\mathcal{L}^2(t)-2v\mathcal{L}(t)\Delta t+(v\Delta t)^2+2u\mathcal{L}(t)\cos(\theta(t))\Delta t-2uv\cos(\theta(t))(\Delta t)^2 \end{align*}$$ $$\begin{align*} \mathcal{L}^2(t+\Delta t)-\mathcal{L}^2(t)&\approx(u\Delta t)^2-2v\mathcal{L}(t)\Delta t+(v\Delta t)^2+2u\mathcal{L}(t)\cos(\theta(t))\Delta t-2uv\cos(\theta(t))(\Delta t)^2\\ \frac{\mathcal{L}^2(t+\Delta t)-\mathcal{L}^2(t)}{\Delta t}&\approx u^2\Delta t-2v\mathcal{L}(t)+v^2\Delta t+2u\mathcal{L}(t)\cos(\theta(t))-2uv\cos(\theta(t))\Delta t \end{align*}$$ Now let $\Delta t\to0$, we have: $$\begin{align*} \frac{d}{dt}\mathcal{L}^2(t)&=-2v\mathcal{L}(t)+2u\mathcal{L}(t)\cos(\theta(t))\\ \Leftrightarrow 2\mathcal{L}(t)\mathcal{L}'(t)&=-2v\mathcal{L}(t)+2u\mathcal{L}(t)\cos(\theta(t))\\ \Leftrightarrow \mathcal{L}'(t)&=-v+u\cos(\theta(t)) \end{align*}$$ My problem is that when $t=\tau$, $\mathcal{L}(\tau)=0$ and so the division is undefined. How can I go around this, or is the simplification valid (why)?

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    $\begingroup$ When $t=\tau$ it is certainly true that $\mathcal{L}(t)=0$, but why do you think that $\mathcal{L}'(t)=0$? Regardless, since you're doing a limiting process, the simplification should be valid regardless... $\endgroup$ Dec 27, 2019 at 21:47
  • $\begingroup$ I have computed the derivative the usual way by considering $\mathcal{L}(t)=\sqrt{(\Delta x(t))^2+(\Delta y(t))^2}$ and found it to be $0$. EDIT: I made a calculation mistake, it is $0/0$ by normal substitution, so nevermind :( $\endgroup$
    – GDGDJKJ
    Dec 28, 2019 at 3:35

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At $\ t=\tau\ $, the derivative of $\ \mathcal{L}\ $ will not be defined, no matter how the points continue to move after that time. A left derivative, $\ \partial_-\mathcal{L}(\tau)=\lim_\limits{\delta\rightarrow0}\frac{0-\mathcal{L}(\tau-\delta)}{\delta}=u\cos(\theta(\tau))-v\ $ does exist, but this is strictly negative (not $0$, as you have supposed), because $\ v>u\ $. Since $\ \mathcal{L}(\tau)=0\ $, and $\ \mathcal{L}(t)\ge0\ $ for any $\ t>\tau\ $ (if, in fact, defined), then if $\ \mathcal{L}\ $ has a right derivative at $\ t=\tau\ $ it must be non-negative, and therefore cannot be equal to the left derivative.

For the purposes of determining the value of $\ \tau\ $, none of this matters, because you only need to use $\ \mathcal{L}'(t)\ $ for $\ t<\tau\ $.

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  • $\begingroup$ So, for solving the problem, it suffices to find $\lim_{x\to\tau^-}\int_0^x\mathcal{L}'(t)$? $\endgroup$
    – GDGDJKJ
    Dec 28, 2019 at 3:48
  • $\begingroup$ Well, you need to find the value of $\ \tau\ $ for which $\ \int_\limits{0}^\tau \mathcal{L}'(t)dt=0\ $, but evaluation of the integral doesn't doesn't depend on the value of its integrand at the end points of the interval of integration (or even require the integrand to have a well-defined value at those points). $\endgroup$ Dec 28, 2019 at 4:03
  • $\begingroup$ You have a point about the endpoints, but I am concerned with the correct way of writing down the equation so that it makes sense. By the way, $\\ \int_\limits{0}^\tau \mathcal{L}'(t)dt=-\mathcal{L}$, it is the accumulation of negative changes to the original distance separating the particles. $\endgroup$
    – GDGDJKJ
    Dec 28, 2019 at 4:25
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    $\begingroup$ Yes, my apologies for the garbled equation. What I meant to write was $\ \mathcal{L}(0)+\int_\limits{0}^\tau \mathcal{L}'(t)dt=0\ $. $\endgroup$ Dec 28, 2019 at 4:37
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Your argument is being applied at a time $t<\tau$. Also, since $\Delta t\to0 $, $ t+\Delta t<\tau$.

Your formula for $\mathcal{L'}(t)$ has therefore been obtained for $t<\tau$ by division by a non-zero quantity.

You are right to be concerned about this formula for $t=\tau$. It does not apply there and this is obvious if you make $u=0$ since then the formula gives the incorrect answer $-v$ instead of $0$.

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