Must the speedometer ever read the average speed?

1. 1. Average velocity is a slope

   A tram runs between two stops $600$ m apart in $60$ s. Above is its distance $s$ against time $t$; the secant joining the ends has slope $600/60 = 10$ m/s, its average velocity. Below, that average is a level line on the speedometer's scale.

   Question: Did the tram's speedometer ever read exactly $10$ m/s?

   Yes. Faster than $10$ m/s the whole minute, it would have gone more than $600$ m; slower the whole minute, less. So it was faster at some moments and slower at others, and its speed, changing smoothly, passed $10$ between them. Here it did so twice: on the way up to $15$ m/s and on the way back down.

   The figure: Distance against time for the tram: 600 m in 60 s. The secant from t = 0.0 to t = 60.0 s has slope 10.00 m/s. The speedometer rises from 0 to 15 m/s at t = 30 s and falls back to 0 at t = 60 s. A level line marks the average, 10.00 m/s, from t = 0.0 to t = 60.0 s.

2. 2. A parallel tangent

   The speedometer's reading at time $t$ is $s'(t)$, the slope of the tangent above. So a moment when it reads the average is a moment when the tangent runs parallel to the secant. Drag the moment along either panel.

   Find a moment when the speedometer reads the average, $10$ m/s.

   At $t \approx 12.68$ s the tram is doing $10$ m/s, and its tangent is parallel to the secant. It happens again at $t \approx 47.32$ s, slowing down. The theorem promises at least one such moment between the ends, and says nothing about when.

   $$s'(c) = \frac{s(b) - s(a)}{b - a} \quad \text{for some } c \text{ in } (a, b)$$$s^{\prime }(c)$ $=\frac{s(b)-s(a)}{b-a}$$\text{for some }c\text{ in }(a,b)$

   The figure: Distance against time for the tram: 600 m in 60 s. The secant from t = 0.0 to t = 60.0 s has slope 10.00 m/s. The tangent at t = 12.68 s has slope 10.00 m/s. The speedometer rises from 0 to 15 m/s at t = 30 s and falls back to 0 at t = 60 s. A level line marks the average, 10.00 m/s, from t = 0.0 to t = 60.0 s.

3. 3. A trip with a corner

   A timetable models the same trip crudely: $5$ m/s for $30$ s, then $15$ m/s for $30$ s, switching at once. It covers $600$ m in $60$ s, an average of $10$ m/s again. Using the theorem without checking it, a slip claims the amber moment: $10$ m/s at the switch.

   Question: Did this model's speedometer ever read $10$ m/s?

   No. Its graph has a corner at $t = 30$, where the speed jumps and $s'(30)$ does not exist. The theorem needs $s$ differentiable on $(0, 60)$; that fails, and its promise fails with it. The amber line meets the corner with slope $10$, but a corner has no tangent.

   The figure: Distance against time for the timetable, 5 m/s for 30 s then 15 m/s for 30 s: 600 m in 60 s. The secant from t = 0.0 to t = 60.0 s has slope 10.00 m/s. In amber, a line of slope 10 through the corner at (30, 150): it touches the graph only at the corner and is no tangent. The speedometer reads 5 m/s until t = 30 s and 15 m/s after, jumping at the switch. A level line marks the average, 10.00 m/s, from t = 0.0 to t = 60.0 s. In amber, the reading of 10 m/s at t = 30 s that the slip claims.

4. 4. Check the hypotheses

   The theorem holds on any interval where its hypotheses do: $s$ continuous on $[a, b]$ and differentiable on $(a, b)$. Drag the ends of the secant along the timetable's graph, and watch the average and the speedometer below.

   Move the ends until the theorem applies to the timetable again.

   With the corner left out of $(a, b)$, $s$ is differentiable there and the promise is back: the speed is steady, so the speedometer reads the average at every moment. With the corner inside, the average lies strictly between $5$ and $15$, and is never read.

   $$s \text{ continuous on } [a, b] \text{ and differentiable on } (a, b) \;\Longrightarrow\; s'(c) = \frac{s(b) - s(a)}{b - a} \text{ for some } c \text{ in } (a, b)$$$s\text{ continuous on }[a,b]\text{ and differentiable on }(a,b)$ $⟹s^{\prime }(c)$ $=\frac{s(b)-s(a)}{b-a}\text{ for some }c\text{ in }(a,b)$

   The figure: Distance against time for the timetable, 5 m/s for 30 s then 15 m/s for 30 s: 600 m in 60 s. The secant from t = 30.0 to t = 50.0 s has slope 15.00 m/s. The speedometer reads 5 m/s until t = 30 s and 15 m/s after, jumping at the switch. A level line marks the average, 15.00 m/s, from t = 30.0 to t = 50.0 s.

The film, as said, 1:32

1. 0:00

   A tram runs between two stops $600$ m apart in $60$ s. Above is its distance against time.

2. 0:06

   The secant joining the ends has slope $600/60 = 10$ m/s: the trip's average velocity.

3. 0:11

   Below, the speedometer. It climbs to $15$ m/s and falls back to $0$ at the next stop.

4. 0:16

   It cannot have read above $10$ the whole minute, or below it the whole minute, so somewhere it read exactly $10$.

5. 0:23

   A reading is the slope of a tangent above. Run a moment along the trip.

6. 0:30

   At $t \approx 12.68$ s the speedometer reads $10$ m/s, and the tangent runs parallel to the secant.

7. 0:39

   Again at $t \approx 47.32$ s, slowing down. The theorem promises at least one such moment, and never says when.

8. 0:49

   A cruder timetable for the same trip: $5$ m/s for $30$ s, then $15$ m/s, switching at once. The average is still $10$.

9. 0:57

   Applying the theorem without checking it, a slip claims a reading of $10$ m/s at the switch: the amber moment.

10. 1:03

    But the speed jumps from $5$ to $15$. At the corner $s'(30)$ does not exist, so the theorem never applied.

11. 1:10

    The theorem needs $s$ continuous on $[a, b]$ and differentiable on $(a, b)$. Here the corner is inside.

12. 1:15

    Slide the start to the corner, $a = 30$. Now the speed is a steady $15$ m/s, read at every moment.

13. 1:25

    Wherever a trip is smooth, its speedometer reads the average speed at some moment between the ends: that is the Mean Value Theorem.