在时刻 $t$（其中 $0 < t < \frac{\pi}{3}$），一个质点沿某路径运动，其参数方程定义为
$x = \cot 3t \quad \text{和} \quad y = 2\csc 3t + 1.$

1. Find the parametric derivatives $\frac{dx}{dt}$ and $\frac{dy}{dt}$.

$x = \cot 3t \implies \frac{dx}{dt} = -3\csc^2 3t$
$y = 2\csc 3t + 1 \implies \frac{dy}{dt} = 2(-3\csc 3t \cot 3t) = -6\csc 3t \cot 3t$

2. Find the gradient of the tangent at point $P$.

$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{-6\csc 3t \cot 3t}{-3\csc^2 3t}$
$= \frac{2\cot 3t}{\csc 3t} = \frac{2\left(\frac{\cos 3t}{\sin 3t}\right)}{\frac{1}{\sin 3t}} = 2 \cos 3t$
At $P$ where $t = p$, the gradient is $m = 2\cos 3p$.

3. Form the equation of the tangent at $P$ and substitute $x = 0$ to find the $y$-intercept.

The equation of the tangent at $P(\cot 3p, 2\csc 3p + 1)$ is:
$y - (2\csc 3p + 1) = 2\cos 3p(x - \cot 3p)$
Since $Q$ lies on the $y$-axis, substitute $x = 0$:
$y = 2\csc 3p + 1 + 2\cos 3p(0 - \cot 3p)$
$y = 2\csc 3p + 1 - 2\cos 3p \cot 3p$

4. Simplify the $y$-coordinate using trigonometric identities to obtain the coordinates of $Q$.

Expressing in terms of sine and cosine:
$y = \frac{2}{\sin 3p} + 1 - 2\cos 3p \left(\frac{\cos 3p}{\sin 3p}\right)$
$y = 1 + \frac{2 - 2\cos^2 3p}{\sin 3p}$
$y = 1 + \frac{2(1 - \cos^2 3p)}{\sin 3p}$
$y = 1 + \frac{2\sin^2 3p}{\sin 3p}$
$y = 1 + 2\sin 3p$
Thus, the coordinates of $Q$ are $(0, 2\sin 3p + 1)$ (shown).

1. Express $s^2$ in terms of $t$ by substituting the parametric equations.

$s^2 = x^2 + (y - 1)^2$
Substituting $x = \cot 3t$ and $y = 2\csc 3t + 1$:
$s^2 = (\cot 3t)^2 + (2\csc 3t + 1 - 1)^2$
$s^2 = \cot^2 3t + 4\csc^2 3t$

2. Differentiate implicitly with respect to $t$.

Differentiating both sides with respect to $t$:
$2s \frac{ds}{dt} = 2\cot 3t \cdot \frac{d}{dt}(\cot 3t) + 4(2\csc 3t) \cdot \frac{d}{dt}(\csc 3t)$
$2s \frac{ds}{dt} = 2\cot 3t (-3\csc^2 3t) + 8\csc 3t (-3\csc 3t \cot 3t)$
$2s \frac{ds}{dt} = -6\cot 3t \csc^2 3t - 24\csc^2 3t \cot 3t$
$2s \frac{ds}{dt} = -30\csc^2 3t \cot 3t$

3. Evaluate $s$ and $\frac{ds}{dt}$ at $t = \frac{\pi}{4}$.

At $t = \frac{\pi}{4}$, $3t = \frac{3\pi}{4}$.
We have $\csc \frac{3\pi}{4} = \sqrt{2}$ and $\cot \frac{3\pi}{4} = -1$.
$s^2 = (-1)^2 + 4(\sqrt{2})^2 = 1 + 8 = 9$
$s = 3 \quad \text{(since } s \ge 0 \text{ as it represents distance)}$
Substitute $s$, $\csc 3t$, and $\cot 3t$ into the differentiated equation:
$2(3) \frac{ds}{dt} = -30(\sqrt{2})^2 (-1)$
$6 \frac{ds}{dt} = -30(2)(-1)$
$6 \frac{ds}{dt} = 60 \implies \frac{ds}{dt} = 10$