A ladder of length 3.12 m is sliding down a vertical wall such that the foot of the ladder is moving along the floor at a constant rate of 0.2 m/s (see diagram). Find the rate at which the top of the ladder is sliding down the wall when it is 1.2 m above the floor.

1. Establish a mathematical relationship between the horizontal and vertical distances using Pythagoras' theorem.

Let $x$ be the horizontal distance of the foot of the ladder from the wall and $y$ be the vertical height of the top of the ladder.
$x^2 + y^2 = 3.12^2$

2. Calculate the horizontal distance $x$ at the instant when the height $y$ is $1.2$ m.

When $y = 1.2$:
$x^2 + 1.2^2 = 3.12^2$
$x^2 = 9.7344 - 1.44 = 8.2944$
Since $x > 0$,
$x = \sqrt{8.2944} = 2.88 \text{ m}$

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

Differentiating both sides with respect to $t$:
$2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$
$x \frac{dx}{dt} + y \frac{dy}{dt} = 0$

4. Substitute the known values to find the rate at which the ladder is sliding down.

Given that the foot of the ladder moves away at $0.2$ m/s, $\frac{dx}{dt} = 0.2$.
Substituting $x = 2.88$, $y = 1.2$, and $\frac{dx}{dt} = 0.2$:
$(2.88)(0.2) + (1.2)\frac{dy}{dt} = 0$
$0.576 + 1.2\frac{dy}{dt} = 0$
$\frac{dy}{dt} = -\frac{0.576}{1.2} = -0.48$
The negative sign indicates the height is decreasing. Hence, the top of the ladder is sliding down the wall at a rate of $0.48$ m/s.