The figure shows two rods PA and PB which are hinged at P. The rods slide such that P, A and B are on the same vertical plane. When P moves down, A and B will move in opposite directions on the horizontal ground. The horizontal distance between A and B is x metres and P is y metres from the ground.

1. Form an equation using Pythagoras' theorem.

Since P is vertically above the midpoint of $AB$, the horizontal distance from the midpoint of $AB$ to $A$ is $\frac{x}{2}$.

Using Pythagoras' theorem on the right-angled triangle,
$\left(\frac{x}{2}\right)^2+y^2=5^2$
$\left(\frac{x}{2}\right)^2+y^2=25$

2. Rearrange the equation to express $x$ in terms of $y$.

$\left(\frac{x}{2}\right)^2=25-y^2$
$\frac{x}{2}=\sqrt{25-y^2}$
Since $x$ is a distance, take the positive square root.
$x=2\sqrt{25-y^2}$

1. Write the geometric relation between $x$ and $y$.

From the right triangle,
$\left(\frac{x}{2}\right)^2+y^2=25$
so
$\frac{x^2}{4}+y^2=25.$

2. Differentiate implicitly with respect to time.

Differentiate both sides with respect to $t$:
$\frac{1}{2}x\frac{dx}{dt}+2y\frac{dy}{dt}=0.$

3. Substitute $y=3$ and the falling rate of $P$ to find $\dfrac{dx}{dt}$.

Since $P$ is falling at $2\text{ m s}^{-1}$,
$\frac{dy}{dt}=-2.$
When $y=3$,
$\left(\frac{x}{2}\right)^2+3^2=25$
$\left(\frac{x}{2}\right)^2=16$
$x=8.$
Substitute into
$\frac{1}{2}x\frac{dx}{dt}+2y\frac{dy}{dt}=0:$
$\frac{1}{2}(8)\frac{dx}{dt}+2(3)(-2)=0$
$4\frac{dx}{dt}-12=0$
$\frac{dx}{dt}=3.$

1. Write a trigonometric relation connecting $x$ and $\theta$.

From the right triangle with hypotenuse $PA=5$ and adjacent side $\dfrac{x}{2}$,
$\cos\theta=\frac{x/2}{5}=\frac{x}{10}.$

2. Differentiate the relation with respect to time.

Differentiate
$\cos\theta=\frac{x}{10}$
with respect to $t$:
$-\sin\theta\frac{d\theta}{dt}=\frac{1}{10}\frac{dx}{dt}.$

3. Find $\sin\theta$ when $P$ is $3$ metres from the ground.

When $y=3$, from the earlier geometry,
$\left(\frac{x}{2}\right)^2+3^2=25$
$\left(\frac{x}{2}\right)^2=16$
so $\dfrac{x}{2}=4$ and $x=8$.

Thus
$\cos\theta=\frac{8}{10}=\frac45.$
Since $\theta$ is acute,
$\sin\theta=\frac35.$

4. Substitute $\dfrac{dx}{dt}=\dfrac32$ and solve for $\dfrac{d\theta}{dt}$.

Using the answer to part (ii),
$\frac{dx}{dt}=\frac32.$
Substitute into
$-\sin\theta\frac{d\theta}{dt}=\frac{1}{10}\frac{dx}{dt}:$
$-\frac35\frac{d\theta}{dt}=\frac{1}{10}\left(\frac32\right)$
$-\frac35\frac{d\theta}{dt}=\frac{3}{20}$
$\frac{d\theta}{dt}=-\frac{3}{20}\times\frac{5}{3}=-\frac14.$