Jesse leans a 6 metre rod $XY$ in the position as shown below. At time $t$ seconds, the two ends $X$ and $Y$ are $x$ and $y$ metres from $O$ respectively.

1. Apply Pythagoras' theorem to the right-angled triangle formed by the wall, the ground, and the rod.

In the right-angled triangle $OXY$, the lengths of the two legs are $x$ and $y$, and the hypotenuse is the length of the rod, $6$ m.

By Pythagoras' Theorem:
$x^2 + y^2 = 6^2$
$x^2 + y^2 = 36$

1. Translate the given proportional rate of decrease into a mathematical equation.

Since $y$ decreases at a rate proportional to $y$, the rate of change of $y$ is negative and given by:
$\frac{\mathrm{d}y}{\mathrm{d}t} = -ky$
where $k$ is a constant of proportionality.

2. Differentiate the relationship between $x$ and $y$ with respect to time $t$.

From part (i), we have the relationship $x^2 + y^2 = 36$.
Differentiating implicitly with respect to $t$:
$2x \frac{\mathrm{d}x}{\mathrm{d}t} + 2y \frac{\mathrm{d}y}{\mathrm{d}t} = 0$
$x \frac{\mathrm{d}x}{\mathrm{d}t} = -y \frac{\mathrm{d}y}{\mathrm{d}t}$

3. Substitute $\frac{\mathrm{d}y}{\mathrm{d}t}$ and $y^2$ to obtain the target expression for $\frac{\mathrm{d}x}{\mathrm{d}t}$.

Substitute $\frac{\mathrm{d}y}{\mathrm{d}t} = -ky$ into the connected rates equation:
$x \frac{\mathrm{d}x}{\mathrm{d}t} = -y(-ky)$
$x \frac{\mathrm{d}x}{\mathrm{d}t} = k y^2$
Since $y^2 = 36 - x^2$, we substitute this in to eliminate $y$:
$x \frac{\mathrm{d}x}{\mathrm{d}t} = k(36 - x^2)$
$\frac{\mathrm{d}x}{\mathrm{d}t} = \frac{k(36 - x^2)}{x} \text{ (shown)}$

1. Substitute $k=2$ and separate the variables of the differential equation.

$\frac{\mathrm{d}x}{\mathrm{d}t} = \frac{2(36 - x^2)}{x}$
$\int \frac{x}{36 - x^2} \, \mathrm{d}x = \int 2 \, \mathrm{d}t$

2. Integrate both sides to find the general solution.

$-\frac{1}{2} \ln(36 - x^2) = 2t + C$

3. Apply the initial condition to find the constant of integration.

When $t = 0$, $x = 4$:
$-\frac{1}{2} \ln(36 - 4^2) = 2(0) + C$
$C = -\frac{1}{2} \ln 20$
So, $-\frac{1}{2} \ln(36 - x^2) = 2t - \frac{1}{2} \ln 20$

4. Rearrange to make $x$ the subject of the formula.

$\ln(36 - x^2) = -4t + \ln 20$
$36 - x^2 = e^{-4t + \ln 20}$
$36 - x^2 = 20e^{-4t}$
$x^2 = 36 - 20e^{-4t}$
Since $x$ is a distance ($x > 0$),
$x = \sqrt{36 - 20e^{-4t}}$

5. Use the equation from part (i) to find the value of $x^2$ when $y = 3$, and solve for $t$.

From part (i), $x^2 + y^2 = 36$.
When $y = 3$, $x^2 + 3^2 = 36 \implies x^2 = 27$.
Substitute $x^2 = 27$ into the earlier equation:
$27 = 36 - 20e^{-4t}$
$20e^{-4t} = 9$
$e^{-4t} = 0.45$
$-4t = \ln 0.45$
$t = \frac{\ln 0.45}{-4} \approx 0.1996$
$t = 0.2 \text{ (to 1 d.p.)}$

1. Derive the explicit equation for $x$ in terms of $t$ (solving the required result from part iii).

Given $\frac{\mathrm{d}x}{\mathrm{d}t} = \frac{2(36 - x^2)}{x}$ from parts (ii) and (iii):
$\int \frac{x}{36 - x^2} \,\mathrm{d}x = \int 2 \,\mathrm{d}t$
$-\frac{1}{2}\ln(36 - x^2) = 2t + C$
$36 - x^2 = A e^{-4t}$
Given $x = 4$ when $t = 0$:
$36 - 16 = A \implies A = 20$
$x^2 = 36 - 20e^{-4t}$
Since $x > 0$ as it represents a physical distance,
$x = \sqrt{36 - 20e^{-4t}}$

2. Identify the key features and sketch the graph of $x$ against $t$.

Using $x = \sqrt{36 - 20e^{-4t}}$:

• Initial point: At $t = 0$, $x = \sqrt{16} = 4$.
• Asymptote: As $t \to \infty$, $e^{-4t} \to 0$, so $x \to \sqrt{36} = 6$. Hence, $x = 6$ is a horizontal asymptote.
• Gradient: Since $\frac{\mathrm{d}x}{\mathrm{d}t} > 0$ for $x < 6$, the graph is strictly increasing.

Sketching details:

• Draw the $t$-axis (horizontal) and $x$-axis (vertical).
• Mark the initial point $(0, 4)$.
• Draw a dashed horizontal line at $x = 6$.
• Sketch a smooth concave-down curve starting from $(0, 4)$ that increases and asymptotically approaches the line $x = 6$.

3. Comment on the appropriateness of the model with justification based on the graph.

From the graph, $x$ approaches $6$ asymptotically as $t \to \infty$.
This implies that, under Jesse's model, the rod only fully reaches the ground ($y = 0$, which corresponds to $x = 6$) after an infinite amount of time.
In reality, a slipping rod would completely fall to the ground in a finite amount of time.
Therefore, Jesse's proposed model is not appropriate.