When a spider threw its web at a fly then one end of the spider's web sticks to the fly and the other one to the table's surface. The fly immediately tries to pull away from the spider's web and flies away with a constant speed while the web's thread began to stretch evenly and indefinitely. The spider scratched its head for a while then managed to assess the speed with which the fly flies away and decided to jump onto this stretching web's thread to crawl all the way to the fly.
Here is the question
What shall be the speed of the spider if it wants to catch up with the fly while crawling along this stretching evenly and indefinitely spider's web thread?
Let $L$ be the starting length of the thread. Let $R(t)$ and $S(t)$ denote the positions of the fly and spider, respectively, at time $t$. Then $R(0) = l$ and $S(0) = 0$. Then let $r$ and $s$ denote the speeds of the fly and spider, respectively.
Since $R'(t) = r$, we have that $R(t) = L + rt$. The differential equation for $S$ is a bit tricker, but it comes out to $$ S'(t) = s + \frac{S(t)}{R(t)} R'(t) \tag{1} $$ or $$ \frac{d}{dt} \left( \frac{S(t)}{R(t)} \right) = \frac{s}{R(t)} \tag{2} $$ You can see either of these formulas directly. For (1), note that there are two contributions to the spider's speed: his own movement $s$, and the movement gained by the string stretching behind him. $\frac{S(t)}{R(t)} R'(t)$ is simply the rate of change in the length of the portion of string behind the spider. For (2), note that if we pretend the string has fixed length $1$, the spider's speed as a percentage of the length of the string is given by his speed divided by the length of the string.
At any rate, (2) gives \begin{align*} \frac{S(t_1)}{R(t_1)} &= \int_0^{t_1} \frac{s}{R(t)} \; dt \\ &= \int_0^{t_1} \frac{s}{L + rt} \; dt \\ &= \frac{s}{r} \ln (L + rt_1) - \frac{s}{r} \ln L \\ &= \frac{s}{r} \ln \left(1 + \frac{rt_1}{L}\right) \\ \end{align*} As long as $s, r, L > 0$, since $\ln$ is unbounded, there will be some time $t_1$ such that $\frac{s}{r} \ln(1 + rt_1 / L) = 1$. Then for this time $t_1$, $S(t_1) = R(t_1)$. Thus the spider will eventually catch the fly no matter its speed.
Specifically, the spider will catch the fly at time $$ t_1 = \frac{L}{r} \left( e^{r/s} - 1 \right). $$