Two koi, Nemo and Dory, live in a pond that’s long and straight. All day long, they swim back and forth, over and over again. Nemo supposedly swims at a speed of exactly 1 mile per hour, while Dory swims faster at 2 miles per hour. That said, you design an experiment to measure their speeds.
You place a sensor in the water that measures the speed of any passing fish and records the event in a log. You collect many such measurements over the course of several days, after which you read out the data and compute the average speed of all the events in the log. If indeed the fish’s speeds were 1 and 2 miles per hour, what average should you expect to get?
Let's assume that the pond has a length of $\ell$ miles. No matter where Nemo and Dory start within the pond or where the sensor is, we know that if the fish's speeds really where 1 and 2 miles per hour, respectively, then after $T = 2 \ell$ hours, then Nemo would pass the sensor exactly twice while Dory would pass the sensor exactly four times. Therefore, the sensor's log would read something like $[ 2, 1, 2, 2, 1, 2 ],$ or any of the other 30 orderings, and thus the average fish speed would be $\frac{ 4 \cdot 2 + 2 \cdot 1 }{4 + 2} = \frac{5}{3}.$
But wait, you say, what if $T$ is not a multiple of $2 \ell$? Let's define the $N_n(T)$ and $N_d(T)$ as the number of times that the sensor picks up Nemo and Demo, respectively. Though obviously counting statistics are defined only on the integers, we see that we should have roughly $$N_n(T) \approx \frac{T}{\ell} \,\, \text{and} \,\, N_d(T) \approx \frac{2T}{\ell}.$$ Therefore, we again arrive at the fact that the average speed recorded by our sensor will be $$\hat{v} = \frac{ 1 \cdot N_n(T) + 2 \cdot N_d(T) }{N_n(T) + N_d(T)} \approx \frac{\frac{T}{\ell} + \frac{4T}{\ell}}{ \frac{T}{\ell} + \frac{2T}{\ell}} = \frac{5}{3}.$$
In general, we see that for any speeds $v_n$ and $v_d,$ that the average speed recorded would be $\hat{v}= \frac{v_n^2 + v_d^2}{v_n + v_d}.$
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