Jim 将一根 8 m 长的杆 $XY$ 靠在一面垂直的墙上，如下图所示。点 $O, X$ 和 $Y$ 共面，其中 $O$ 是垂直墙面与平整地面的交点。在时间为 $t$ 秒时，端点 $X$ 和 $Y$ 距离 $O$ 的距离分别为 $x$ 和 $y$ 米。 Jim 提出一个猜想：当杆滑动时，$y$ 减少的速率与 $y$ 成正比。

1. Translate Jim's conjecture into a mathematical equation involving derivatives.

Since $y$ decreases at a rate proportional to $y$, the rate of decrease is $-\frac{\mathrm{d}y}{\mathrm{d}t}$. Therefore,
$-\frac{\mathrm{d}y}{\mathrm{d}t} = k y \implies \frac{\mathrm{d}y}{\mathrm{d}t} = -k y$
where $k$ is a positive constant.

2. Establish the geometric relationship between $x$ and $y$.

By Pythagoras' theorem on the right-angled triangle formed by the wall, ground, and the rod:
$x^2 + y^2 = 8^2$
$x^2 + y^2 = 64$

3. Differentiate implicitly with respect to time $t$ to relate the rates of change.

Differentiating both sides with respect to $t$:
$2x \frac{\mathrm{d}x}{\mathrm{d}t} + 2y \frac{\mathrm{d}y}{\mathrm{d}t} = 0$
$\frac{\mathrm{d}x}{\mathrm{d}t} = -\frac{y}{x} \frac{\mathrm{d}y}{\mathrm{d}t}$

4. Substitute $\frac{\mathrm{d}y}{\mathrm{d}t}$ and eliminate $y$ to show the final expression.

Substitute $\frac{\mathrm{d}y}{\mathrm{d}t} = -k y$ into the equation:
$\frac{\mathrm{d}x}{\mathrm{d}t} = -\frac{y}{x} (-k y) = \frac{k y^2}{x}$
From the Pythagoras' relationship, we know $y^2 = 64 - x^2$. Substituting this in:
$\frac{\mathrm{d}x}{\mathrm{d}t} = \frac{k(64 - x^2)}{x} \quad \text{(Shown)}$

1. Substitute $k=3$ into the given differential equation and separate the variables.

$\frac{\mathrm{d}x}{\mathrm{d}t} = \frac{3(64 - x^2)}{x}$
$\int \frac{x}{64 - x^2} \,\mathrm{d}x = \int 3 \,\mathrm{d}t$

2. Integrate both sides to obtain the general solution.

$-\frac{1}{2}\int \frac{-2x}{64 - x^2} \,\mathrm{d}x = \int 3 \,\mathrm{d}t$
$-\frac{1}{2}\ln(64 - x^2) = 3t + C$

3. Use the initial condition $x = 4$ when $t = 0$ to find the constant of integration $C$.

$-\frac{1}{2}\ln(64 - 4^2) = 3(0) + C$
$C = -\frac{1}{2}\ln 48$
$-\frac{1}{2}\ln(64 - x^2) = 3t - \frac{1}{2}\ln 48$

4. Make $x$ the subject of the equation.

$\ln(64 - x^2) = -6t + \ln 48$
$64 - x^2 = e^{-6t + \ln 48}$
$64 - x^2 = 48e^{-6t}$
$x^2 = 64 - 48e^{-6t}$
$x = \sqrt{64 - 48e^{-6t}}$

5. Relate $x$ and $y$ using Pythagoras' theorem to find $t$ when $y = 3$.

From the right-angled triangle, $x^2 + y^2 = 8^2 = 64$.
When $y = 3$, $x^2 + 3^2 = 64 \implies x^2 = 55$.
Substitute $x^2 = 55$ into the earlier equation:
$55 = 64 - 48e^{-6t}$
$48e^{-6t} = 9 \implies e^{-6t} = \frac{3}{16}$
$-6t = \ln\left(\frac{3}{16}\right)$
$t = -\frac{1}{6}\ln\left(\frac{3}{16}\right) \approx 0.27898$
To 1 decimal place, $t = 0.3$.

1. Concisely derive the expression for $x$ in terms of $t$ (required from part b).

Since the sketch requires the relationship between $x$ and $t$, we concisely re-derive it from the differential equation in (a).
Given $\frac{\mathrm{d}x}{\mathrm{d}t} = \frac{3(64-x^{2})}{x}$, separate the variables:
$\int \frac{x}{64-x^{2}} \, \mathrm{d}x = \int 3 \, \mathrm{d}t$
$-\frac{1}{2} \ln(64-x^{2}) = 3t + C$
Using the initial condition $x=4$ when $t=0$:
$-\frac{1}{2} \ln(64-16) = C \implies C = -\frac{1}{2} \ln 48$
Substituting $C$ back and solving for $x$:
$\ln(64-x^{2}) = -6t + \ln 48 \implies 64-x^{2} = 48e^{-6t}$
Since $x \ge 0$, we have $x = \sqrt{64 - 48e^{-6t}}$.

2. Determine the initial point and asymptotic behaviour of the graph.

When $t = 0$, $x = \sqrt{64 - 48e^{0}} = \sqrt{16} = 4$.
Thus, the $x$-intercept (starting point) is $(0, 4)$.
As $t \to \infty$, $e^{-6t} \to 0$. Therefore, $x \to \sqrt{64} = 8$.
This indicates a horizontal asymptote at $x = 8$.

3. Determine the shape (gradient and concavity) of the curve.

As $t$ increases, $e^{-6t}$ decreases, which makes $64 - 48e^{-6t}$ increase. Hence, $x$ is strictly increasing.
Furthermore, from $\frac{\mathrm{d}x}{\mathrm{d}t} = \frac{3(64-x^{2})}{x}$, as $x$ increases from $4$ towards $8$, the numerator decreases and the denominator increases, meaning the positive gradient $\frac{\mathrm{d}x}{\mathrm{d}t}$ is steadily decreasing towards $0$.
Therefore, the curve is concave down.

1. Determine the theoretical time for the rod to reach the ground based on the model.

For the rod to hit the ground, we require $y = 0$, which corresponds to $x = 8$.
From the equation derived in part (b), $x = \sqrt{64 - 48e^{-6t}}$, or by observing the graph in part (c), $x$ only approaches $8$ asymptotically as $t \to \infty$.

2. Evaluate if this theoretical behaviour aligns with physical reality to draw a conclusion.

In reality, the sliding rod will hit the ground in a finite amount of time. Since the model implies the rod takes an infinite time to reach the ground, it does not accurately represent the physical behaviour of the rod.
Therefore, Jim's conjecture is inappropriate.