Problem 50 worked answer

One Knob, Infinitely Many Histories

Designed for Grade 12 · Uses HS CALC math · About 13–27 minutes

Complete to collectAlexandrite CrystalLevel 2

At a glance

Every parameter value gives $𝑥 𝑎 (0) = 0$ and $𝑥 𝑎 (1) = 1$ .
The velocity is $𝑥' 𝑎 (𝑡) = 1 + 𝑎 (1 - 2 𝑡)$ .
The action function is $𝑆 (𝑎) = 1 + 𝑎 2 3$ .
Since $𝑆' (𝑎) = 2 𝑎 3$ , the unique stationary setting is $𝑎 = 0$ .
That setting gives $𝑥 0 (𝑡) = 𝑡$ , the steady history, and it is the absolute minimum within this family.
The symmetry $𝑆 (𝑎) = 𝑆 (- 𝑎)$ makes early-rush and late-rush histories tie.

Key idea

The parameter $𝑎$ is not the puck's position or time. It is a knob that changes a whole function. Once the action of each function has been computed, the infinitely many-history question becomes an ordinary single-variable stationary-point problem for $𝑆 (𝑎)$ .

Worked solution, part by part

1. Verify the endpoints and find velocity

At the endpoints,

𝑥 𝑎 (0) = 0 + 𝑎 (0) (1) = 0

and

𝑥 𝑎 (1) = 1 + 𝑎 (1) (0) = 1 .

The added term vanishes at both endpoints, so every real value of $𝑎$ gives an allowed history.

Rewrite and differentiate with respect to time:

𝑥 𝑎 (𝑡) = 𝑡 + 𝑎 𝑡 - 𝑎 𝑡 2, 𝑥' 𝑎 (𝑡) = 1 + 𝑎 - 2 𝑎 𝑡 = 1 + 𝑎 (1 - 2 𝑡) .

When $𝑎 > 0$ , the initial velocity $1 + 𝑎$ is larger and the final velocity $1 - 𝑎$ is smaller, so the puck rushes early. When $𝑎 < 0$ , the initial velocity is smaller and the final velocity is larger, so it rushes late.

2. Reduce the family to one action function

Substitute the velocity into the action:

𝑆 (𝑎) = \int 10 [1 + 𝑎 (1 - 2 𝑡)] 2 𝑑 𝑡 = \int 10 [1 + 2 𝑎 (1 - 2 𝑡) + 𝑎 2 (1 - 2 𝑡) 2] 𝑑 𝑡 .

Because $𝑎$ is constant with respect to $𝑡$ , the supplied integrals give

𝑆 (𝑎) = 1 + 2 𝑎 (0) + 𝑎 2 (13) = 1 + 𝑎 2 3 .

The cross term disappears. This cancellation is why the answer depends on $𝑎 2$ , not on the sign of $𝑎$ .

3. Find and interpret stationary action

Differentiate the action function with respect to the parameter:

𝑆' (𝑎) = 2 𝑎 3 .

Therefore $𝑆' (𝑎) = 0$ only when $𝑎 = 0$ .

The corresponding history is

𝑥 0 (𝑡) = 𝑡 .

It moves steadily because $𝑥' 0 (𝑡) = 1$ . Moreover,

𝑆 (𝑎) = 1 + 𝑎 2 3 \geq 1 = 𝑆 (0),

with equality only when $𝑎 = 0$ . Thus this stationary action is also the unique absolute minimum within the supplied family.

4. Explain the opposite-setting tie

Finally,

𝑆 (- 𝑎) = 1 + (- 𝑎) 2 3 = 1 + 𝑎 2 3 = 𝑆 (𝑎) .

The two histories redistribute the same kind of speed variation in opposite time directions. One rushes early and the other rushes late, but squaring and integrating produces the same total action.

Check and completeness

The endpoint computation verifies that every real parameter value is admissible. The derived formula satisfies $𝑆 (𝑎) \geq 1$ for all real $𝑎$ , with equality only at $𝑎 = 0$ , so the classification covers the entire supplied family rather than only the five settings shown in the diagram. The equality $𝑆 (𝑎) = 𝑆 (- 𝑎)$ also confirms the visual time-reversal symmetry.

Teaching note: the doorway to calculus of variations

This worksheet proves stationarity only among histories of the special form $𝑡 + 𝑎 𝑡 (1 - 𝑡)$ . Calculus of variations asks for stationarity under arbitrary small endpoint-fixing changes

𝑥 𝜀 (𝑡) = 𝑥 (𝑡) + 𝜀 𝜂 (𝑡), 𝜂 (0) = 𝜂 (1) = 0 .

The same central move survives: turn the action into a function of the variation size $𝜀$ , then require its derivative at $𝜀 = 0$ to vanish.

Technical fit and rating

MJ HS:CALC.8 · C2 · W2 Standard challenge · Moderate workload

Calculus and continuous change

How the rating works →

Worksheet arc · Synthesis · 3 of 3 on Complete arc

Discover Action

Compress infinitely many endpoint-sharing histories into one action function and find its stationary member.