在某栋建筑中，有一个 $L$ 形通道（见下图）。通道 1 的固定宽度为 3 米，与通道 1 垂直的通道 2 的固定宽度为 2 米。点 $A$ 和点 $B$ 分别位于通道 1 和通道 2 的一侧墙壁上，使得线段 $AB$ 接触转角 $O$，且与其中一面墙壁成 $\theta$ 弧度的角（如图所示），其中 $0 \le \theta \le \frac{\pi}{2}$。

1. Express the length of $OB$ in terms of $\theta$.

Let $P$ be the foot of the perpendicular from $O$ to the right wall of Corridor 2.
In $\triangle OPB$, $\angle OPB = \frac{\pi}{2}$ radians and $OP = 2$ m (the width of Corridor 2).
Given that the angle between $OB$ and the vertical right wall is $\theta$, we have $\angle OBP = \theta$.
Using trigonometric ratios:
$\sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{OP}{OB} = \frac{2}{OB}$
$OB = \frac{2}{\sin \theta}$

2. Express the length of $OA$ in terms of $\theta$.

Let $Q$ be the foot of the perpendicular from $O$ to the top wall of Corridor 1.
In $\triangle OQA$, $\angle OQA = \frac{\pi}{2}$ radians and $OQ = 3$ m (the width of Corridor 1).
Draw a vertical line passing through $O$. The angle between $OB$ and the downward vertical is $\theta$. Since $AB$ is a straight line segment, the vertically opposite angle between $OA$ and the upward vertical ($OQ$) is also $\theta$. Thus, $\angle AOQ = \theta$.
Using trigonometric ratios:
$\cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{OQ}{OA} = \frac{3}{OA}$
$OA = \frac{3}{\cos \theta}$

3. Combine the lengths to show the required expression for $AB$ and determine the constants.

Since $A$, $O$, and $B$ are collinear, the total length $AB$ is the sum of $OA$ and $OB$:
$AB = OA + OB$
$AB = \frac{3}{\cos \theta} + \frac{2}{\sin \theta}$
Comparing this with the given expression $AB = \frac{\alpha}{\cos \theta} + \frac{\beta}{\sin \theta}$, we determine that:
$\alpha = 3 \quad \text{and} \quad \beta = 2$

1. Deduce the constants $\alpha$ and $\beta$ for the expression of $AB$ as required from part (a).

From the problem description, $OB$ forms a right triangle with the vertical right wall (horizontal width $2\text{ m}$) with angle $\theta$ at $O$. Thus,
$\sin \theta = \frac{2}{OB} \implies OB = \frac{2}{\sin \theta}$
Similarly, $OA$ makes an angle $\theta$ with the vertical, forming a right triangle with the horizontal top wall (vertical width $3\text{ m}$). Thus,
$\cos \theta = \frac{3}{OA} \implies OA = \frac{3}{\cos \theta}$
Hence, $AB = \frac{3}{\cos \theta} + \frac{2}{\sin \theta} = 3\sec \theta + 2\csc \theta$, which confirms $\alpha = 3$ and $\beta = 2$ from part (a).

2. Find the exact value of $\theta$ at the instant when $OB = 4\text{ m}$.

At the instant when $OB = 4\text{ m}$:
$\frac{2}{\sin \theta} = 4 \implies \sin \theta = \frac{1}{2}$
Given $0 \le \theta \le \frac{\pi}{2}$, we have:
$\theta = \frac{\pi}{6}$

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

$\frac{\mathrm{d}(AB)}{\mathrm{d}\theta} = \frac{\mathrm{d}}{\mathrm{d}\theta} (3\sec \theta + 2\csc \theta)$
$\frac{\mathrm{d}(AB)}{\mathrm{d}\theta} = 3\sec \theta \tan \theta - 2\csc \theta \cot \theta$

4. Evaluate the derivative at $\theta = \frac{\pi}{6}$ and apply the chain rule to find the rate of change of $AB$.

Evaluate the trigonometric ratios at $\theta = \frac{\pi}{6}$:
$\sec\left(\frac{\pi}{6}\right) = \frac{2}{\sqrt{3}}, \quad \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}, \quad \csc\left(\frac{\pi}{6}\right) = 2, \quad \cot\left(\frac{\pi}{6}\right) = \sqrt{3}$
$\frac{\mathrm{d}(AB)}{\mathrm{d}\theta} = 3\left(\frac{2}{\sqrt{3}}\right)\left(\frac{1}{\sqrt{3}}\right) - 2(2)(\sqrt{3}) = 2 - 4\sqrt{3}$
Given that $\theta$ is decreasing at $0.1\text{ rad/s}$, $\frac{\mathrm{d}\theta}{\mathrm{d}t} = -0.1$.
By the chain rule:
$\frac{\mathrm{d}(AB)}{\mathrm{d}t} = \frac{\mathrm{d}(AB)}{\mathrm{d}\theta} \times \frac{\mathrm{d}\theta}{\mathrm{d}t}$
$\frac{\mathrm{d}(AB)}{\mathrm{d}t} = (2 - 4\sqrt{3}) \times (-0.1) = 0.4\sqrt{3} - 0.2$
Thus, the rate of increase of $AB$ is $0.4\sqrt{3} - 0.2 \approx 0.493\text{ m/s}$.

1. Relate the maximum length of the pole $p$ to the length of the line segment $AB$ and set up the expression to minimize.

For the pole to successfully negotiate the corner horizontally, its maximum length $p$ must not exceed the minimum possible length of the line segment $AB$ as it turns.
Using $\alpha = 3$ and $\beta = 2$ derived from part (a), the length is:
$L = \frac{3}{\cos \theta} + \frac{2}{\sin \theta} = 3(\cos \theta)^{-1} + 2(\sin \theta)^{-1}$

2. Differentiate $L$ with respect to $\theta$ and set it to zero to find the stationary point.

$\frac{dL}{d\theta} = 3(-1)(\cos \theta)^{-2}(-\sin \theta) + 2(-1)(\sin \theta)^{-2}(\cos \theta)$
$\frac{dL}{d\theta} = \frac{3\sin \theta}{\cos^2 \theta} - \frac{2\cos \theta}{\sin^2 \theta}$
For a stationary point, we set $\frac{dL}{d\theta} = 0$:
$\frac{3\sin \theta}{\cos^2 \theta} = \frac{2\cos \theta}{\sin^2 \theta}$

3. Solve for $\tan \theta$.

$3\sin^3 \theta = 2\cos^3 \theta$
$\frac{\sin^3 \theta}{\cos^3 \theta} = \frac{2}{3}$
$\tan^3 \theta = \frac{2}{3} \implies \tan \theta = \left(\frac{2}{3}\right)^{\frac{1}{3}}$

4. Use the value of $\tan \theta$ to find exact expressions for $\sin \theta$ and $\cos \theta$.

We can construct a right-angled triangle where the opposite side to $\theta$ is $2^{\frac{1}{3}}$ and the adjacent side is $3^{\frac{1}{3}}$.
The hypotenuse $H$ is:
$H = \sqrt{\left(2^{\frac{1}{3}}\right)^2 + \left(3^{\frac{1}{3}}\right)^2} = \sqrt{2^{\frac{2}{3}} + 3^{\frac{2}{3}}}$
From this triangle:
$\sin \theta = \frac{2^{\frac{1}{3}}}{\sqrt{2^{\frac{2}{3}} + 3^{\frac{2}{3}}}}, \quad \cos \theta = \frac{3^{\frac{1}{3}}}{\sqrt{2^{\frac{2}{3}} + 3^{\frac{2}{3}}}}$

5. Substitute $\sin \theta$ and $\cos \theta$ back into the expression for $L$ and simplify using fractional indices.

$L = \frac{3}{\frac{3^{\frac{1}{3}}}{\sqrt{2^{\frac{2}{3}} + 3^{\frac{2}{3}}}}} + \frac{2}{\frac{2^{\frac{1}{3}}}{\sqrt{2^{\frac{2}{3}} + 3^{\frac{2}{3}}}}}$
$L = 3^{1 - \frac{1}{3}}\sqrt{2^{\frac{2}{3}} + 3^{\frac{2}{3}}} + 2^{1 - \frac{1}{3}}\sqrt{2^{\frac{2}{3}} + 3^{\frac{2}{3}}}$
$L = 3^{\frac{2}{3}}\sqrt{2^{\frac{2}{3}} + 3^{\frac{2}{3}}} + 2^{\frac{2}{3}}\sqrt{2^{\frac{2}{3}} + 3^{\frac{2}{3}}}$

6. Factorise and simplify the expression to the required form to determine $k$.

$L = \left(3^{\frac{2}{3}} + 2^{\frac{2}{3}}\right)\left(2^{\frac{2}{3}} + 3^{\frac{2}{3}}\right)^{\frac{1}{2}}$
$L = \left[ (2)^{\frac{2}{3}} + (3)^{\frac{2}{3}} \right]^{\frac{3}{2}}$
Comparing this to $p = \left[ (2)^k + (3)^k \right]^{\frac{1}{k}}$, we obtain exactly $k = \frac{2}{3}$.

7. Perform the first derivative test to formally verify that this value minimizes $AB$.

Consider the sign of $\frac{dL}{d\theta}$ around the stationary point where $\tan \theta = \left(\frac{2}{3}\right)^{\frac{1}{3}}$.

• For $\theta < \tan^{-1}\left(\left(\frac{2}{3}\right)^{\frac{1}{3}}\right)$, $\tan^3 \theta < \frac{2}{3} \implies 3\sin^3 \theta < 2\cos^3 \theta$, making $\frac{dL}{d\theta} < 0$.
• For $\theta > \tan^{-1}\left(\left(\frac{2}{3}\right)^{\frac{1}{3}}\right)$, $\tan^3 \theta > \frac{2}{3} \implies 3\sin^3 \theta > 2\cos^3 \theta$, making $\frac{dL}{d\theta} > 0$.
  Since $\frac{dL}{d\theta}$ changes from negative to positive, $L$ represents a minimum. Hence, the pole's maximum possible length $p$ is strictly this minimum value of $AB$.