# Division by 0 in a rate of change problem

This is not to ask for a solution, rather I want comments on the rigor of the step. Thank you for your time!

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)?

• 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... Dec 27, 2019 at 21:47
• 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 :( Dec 28, 2019 at 3:35

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\$$.

• So, for solving the problem, it suffices to find $\lim_{x\to\tau^-}\int_0^x\mathcal{L}'(t)$? Dec 28, 2019 at 3:48
• 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). Dec 28, 2019 at 4:03
• 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. Dec 28, 2019 at 4:25
• 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\$. Dec 28, 2019 at 4:37

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 $$0$$.