Trinity de Rale 是一位新晋的装置艺术家。她最新的移动艺术作品由两条固定轨道组成，它们在 $O$ 点处呈 $60^\circ$ 角放置（见图）。一根固定长度为 $L$ cm 的连杆 $PQ$ 放置在轨道上，其两端 $P$ 和 $Q$ 可以在每条轨道上自由滑动。 已知长度 $OP$ 为 $x$ cm，且角 $OPQ$ 为 $\theta$。

1. Construct a perpendicular from $Q$ to the horizontal rail to express vertical and horizontal segments in terms of $L$ and $\theta$.

Let $R$ be the foot of the perpendicular from $Q$ to the horizontal line passing through $OP$.
In the right-angled $\triangle PQR$:
$\sin \theta = \frac{QR}{PQ} \implies QR = L \sin \theta$
$\cos \theta = \frac{PR}{PQ} \implies PR = L \cos \theta$

2. Use right-angled triangle $OQR$ to find $OR$, then sum the segments $OR$ and $PR$ to find $x$.

In the right-angled $\triangle OQR$:
$\tan 60^\circ = \frac{QR}{OR}$
$\sqrt{3} = \frac{L \sin \theta}{OR} \implies OR = \frac{1}{\sqrt{3}} L \sin \theta$
From the diagram, the total length $x$ is the sum of $OR$ and $PR$:
$x = OR + PR$
$x = \frac{1}{\sqrt{3}} L \sin \theta + L \cos \theta$
$x = L\left(\cos\theta + \frac{1}{\sqrt{3}}\sin\theta\right)$

1. Find the initial and final positions of point $P$ to determine the total distance it moves.

At the start ($\theta = 90^\circ$):
$x_{\text{start}} = L\left(\cos 90^\circ + \frac{1}{\sqrt{3}}\sin 90^\circ\right) = \frac{L}{\sqrt{3}}$

At the end ($\theta = 30^\circ$):
$x_{\text{end}} = L\left(\cos 30^\circ + \frac{1}{\sqrt{3}}\sin 30^\circ\right) = L\left(\frac{\sqrt{3}}{2} + \frac{1}{2\sqrt{3}}\right) = \frac{2L}{\sqrt{3}}$

Distance moved by $P = x_{\text{end}} - x_{\text{start}} = \frac{2L}{\sqrt{3}} - \frac{L}{\sqrt{3}} = \frac{L}{\sqrt{3}}$

2. Calculate the constant rate of change of $x$, $\frac{\mathrm{d}x}{\mathrm{d}t}$.

Since the point $P$ moves at a constant rate over 60 seconds:
$\frac{\mathrm{d}x}{\mathrm{d}t} = \frac{\text{Distance}}{\text{Time}} = \frac{\frac{L}{\sqrt{3}}}{60} = \frac{L}{60\sqrt{3}}$

3. Differentiate $x$ with respect to $\theta$.

Given $x = L\left(\cos\theta + \frac{1}{\sqrt{3}}\sin\theta\right)$,
$\frac{\mathrm{d}x}{\mathrm{d}\theta} = L\left(-\sin\theta + \frac{1}{\sqrt{3}}\cos\theta\right)$
(Note: Performing differentiation on trigonometric functions implicitly assumes $\theta$ is measured in radians.)

4. Evaluate $\frac{\mathrm{d}x}{\mathrm{d}\theta}$ at the moment when $\theta = 60^\circ$.

At $\theta = 60^\circ$ (or $\frac{\pi}{3}$ radians):
$\frac{\mathrm{d}x}{\mathrm{d}\theta} = L\left(-\sin 60^\circ + \frac{1}{\sqrt{3}}\cos 60^\circ\right)$
$= L\left(-\frac{\sqrt{3}}{2} + \frac{1}{\sqrt{3}}\left(\frac{1}{2}\right)\right)$
$= L\left(-\frac{3}{2\sqrt{3}} + \frac{1}{2\sqrt{3}}\right) = L\left(-\frac{2}{2\sqrt{3}}\right) = -\frac{L}{\sqrt{3}}$

5. Use the chain rule to find $\frac{\mathrm{d}\theta}{\mathrm{d}t}$ and state the correct units.

Using the chain rule, $\frac{\mathrm{d}x}{\mathrm{d}t} = \frac{\mathrm{d}x}{\mathrm{d}\theta} \times \frac{\mathrm{d}\theta}{\mathrm{d}t}$:
$\frac{L}{60\sqrt{3}} = \left(-\frac{L}{\sqrt{3}}\right) \times \frac{\mathrm{d}\theta}{\mathrm{d}t}$
$\frac{\mathrm{d}\theta}{\mathrm{d}t} = -\frac{1}{60}$
Since we used the standard derivatives of trigonometric functions, the output angle is strictly in radians. Thus, the exact rate of change is $-\frac{1}{60}$ rad s$^{-1}$.

1. Find an expression for the correct position of $P$ at time $t$.

For the correct motion, $P$ moves at a constant speed along the rail.
At $t=0$, $\theta = 90^\circ \implies x_{\text{correct}}(0) = L\left(\cos 90^\circ + \frac{1}{\sqrt{3}}\sin 90^\circ\right) = \frac{L}{\sqrt{3}}$
At $t=60$, $\theta = 30^\circ \implies x_{\text{correct}}(60) = L\left(\cos 30^\circ + \frac{1}{\sqrt{3}}\sin 30^\circ\right) = L\left(\frac{\sqrt{3}}{2} + \frac{1}{2\sqrt{3}}\right) = \frac{2L}{\sqrt{3}}$
The constant speed is $\frac{\frac{2L}{\sqrt{3}} - \frac{L}{\sqrt{3}}}{60} = \frac{L}{60\sqrt{3}}$ cm/s.
Thus, the correct position at time $t$ is:
$x_{\text{correct}}(t) = \frac{L}{\sqrt{3}} + \frac{L}{60\sqrt{3}}t$

2. Find an expression for the incorrect position of $P$ at time $t$.

For the incorrect motion, $\theta$ decreases linearly from $90^\circ$ to $30^\circ$ in 60s.
The constant rate of change of $\theta$ is $\frac{90 - 30}{60} = 1^\circ$ per second.
Thus, at time $t$, $\theta = (90 - t)^\circ$.
Substituting this into the given equation for $x$ gives:
$x_{\text{incorrect}}(t) = L\left(\cos(90 - t)^\circ + \frac{1}{\sqrt{3}}\sin(90 - t)^\circ\right)$
Using complementary angle identities $\cos(90^\circ - \alpha) = \sin \alpha$ and $\sin(90^\circ - \alpha) = \cos \alpha$:
$x_{\text{incorrect}}(t) = L\left(\sin t^\circ + \frac{1}{\sqrt{3}}\cos t^\circ\right)$

3. Formulate the final expression for the distance $D$.

The distance $D$ is the difference between the incorrect position and the correct position.
$D = x_{\text{incorrect}}(t) - x_{\text{correct}}(t)$
$D = L\left(\sin t^\circ + \frac{1}{\sqrt{3}}\cos t^\circ\right) - \left(\frac{L}{\sqrt{3}} + \frac{Lt}{60\sqrt{3}}\right)$
Factoring out $L$ yields:
$D = L\left( \sin t^\circ + \frac{1}{\sqrt{3}}\cos t^\circ - \frac{1}{\sqrt{3}} - \frac{t}{60\sqrt{3}} \right)$

1. Formulate the expression for $D/L$ as a function of time $t$.

From part (c), the incorrect position of $P$ has $\theta(t) = \left(90 - t\right)^\circ$, giving $x_{\text{incorrect}} = L\left( \sin\left(\frac{\pi t}{180}\right) + \frac{1}{\sqrt{3}}\cos\left(\frac{\pi t}{180}\right) \right)$.
The correct position moves at a constant rate from $\frac{L}{\sqrt{3}}$ to $\frac{2L}{\sqrt{3}}$ over 60 seconds, so $x_{\text{correct}} = L\left( \frac{1}{\sqrt{3}} + \frac{t}{60\sqrt{3}} \right)$.

Thus, $\frac{D}{L} = \frac{x_{\text{incorrect}} - x_{\text{correct}}}{L} = \sin\left(\frac{\pi t}{180}\right) + \frac{1}{\sqrt{3}}\cos\left(\frac{\pi t}{180}\right) - \frac{1}{\sqrt{3}} - \frac{t}{60\sqrt{3}}$.

2. Differentiate $\frac{D}{L}$ with respect to $t$ to find the stationary point.

To find the maximum, set $\frac{d}{dt}\left(\frac{D}{L}\right) = 0$:
$\frac{\pi}{180}\cos\left(\frac{\pi t}{180}\right) - \frac{\pi}{180\sqrt{3}}\sin\left(\frac{\pi t}{180}\right) - \frac{1}{60\sqrt{3}} = 0$
$\cos\left(\frac{\pi t}{180}\right) - \frac{1}{\sqrt{3}}\sin\left(\frac{\pi t}{180}\right) = \frac{180}{60\pi\sqrt{3}} = \frac{\sqrt{3}}{\pi}$

3. Solve for $t$ using the R-formula.

Let $\cos\theta - \frac{1}{\sqrt{3}}\sin\theta = R\cos(\theta + \alpha)$.
$R = \sqrt{1^2 + \left(\frac{1}{\sqrt{3}}\right)^2} = \frac{2}{\sqrt{3}}$ and $\tan \alpha = \frac{1}{\sqrt{3}} \implies \alpha = \frac{\pi}{6}$.
$\frac{2}{\sqrt{3}} \cos\left(\frac{\pi t}{180} + \frac{\pi}{6}\right) = \frac{\sqrt{3}}{\pi}$
$\cos\left(\frac{\pi t}{180} + \frac{\pi}{6}\right) = \frac{3}{2\pi}$
$\frac{\pi t}{180} + \frac{\pi}{6} = \arccos\left(\frac{3}{2\pi}\right) \approx 1.07303 \text{ rad}$
$\frac{\pi t}{180} \approx 1.07303 - 0.52360 = 0.54943 \text{ rad} \implies t \approx 31.480 \text{ s}$

4. Evaluate the maximum value of $D/L$.

Substitute $t \approx 31.480$ back into the expression for $\frac{D}{L}$:
$\frac{D}{L} \approx \sin(0.54943) + \frac{1}{\sqrt{3}}\cos(0.54943) - \frac{1}{\sqrt{3}} - \frac{31.480}{60\sqrt{3}}$
$\frac{D}{L} \approx 0.52220 + 0.49238 - 0.57735 - 0.30292 = 0.13431$

(Note: This maximum point can also be found directly by plotting the function $Y_1 = \frac{D}{L}$ on a Graphing Calculator and using the maximum feature.)

The maximum value to 4 decimal places is $0.1343$.