Question: you’re practicing for the forgotten NYC marathon. On the first day you set your treadmill to $9$ minutes per mile. On the second day, you set it to steadily “speed up” over time from $r_0 = 10$ minutes per mile at the beginning down to $r_1 = 8$ minutes per mile at the end (when the run is half over, the treadmill moves at $9$ minutes per mile). Which run took less time?

(FiveThirtyEight)

Solution

What’s interesting here is that the treadmill rate $r$ imposes a different kind of “speed” than is usual. Miles per hour is distance per time, but minutes per mile is time per distance.

Say the run lasts for $T$ minutes total, then the treadmill’s rate a time $t$ after the start of the run is

\[d = r_0 + \left(r_1 - r_0\right)\dfrac{t}{T}.\]

Inharmonious slog

$r$ is the ratio of the incremental time taken to the incremental distance travelled

\[r(t_i) = \dfrac{\Delta t_i}{\Delta x_i}.\]

Inverting, this becomes $\Delta x_i = \Delta t_i / r(t_i)$ so that

\[\begin{align} X_\text{total} &= \sum_i \Delta x_i \\ &= \sum_i \dfrac{\Delta t_i}{r(t_i)} \end{align}\]

or, as the time increments become continuous

\[X_\text{total} = \int\limits_0^{T} \dfrac{dt}{r_0 + \left(r_1 - r_0\right)\dfrac{t}{T}}\]

From $t$ to $r$ and back again

Changing variables to $r = r_0 + \left(r_1-r_0\right)t/T,$ so that $dt = T\, dr/\left(r_1 - r_0\right)$ we find

\[X_\text{total} = \dfrac{T}{r_1 - r_0}\int\limits_{r_0}^{r_1}\dfrac{dr}{r} = \dfrac{1}{r_1 - r_0}T \log \dfrac{r_1}{r_0}\]

which comes to \(X_\text{total} = \dfrac{1}{2\text{ minutes per mile}} T \log \dfrac54 \text{ miles}.\)

Solving for $T,$

\[T = 20\text{ miles}\times 2\text{ minutes per mile} / {\log \frac54} \approx 179.26\text{ minutes}\]

which is just under the three hours it takes when the treadmill is set to a constant rate of $9\text{ minutes per mile}.$

General considerations

Inspecting the steps in the calculation, this undershooting is true whether we accelerate or decelerate the treadmill over time. In fact, the undershoot becomes more substantial as we widen the window around the $9\text{ minutes per mile}.$