A luxury clock has an hour hand and a minute hand. The hands are of length $5\text{cm}$ and $10\text{cm}$ respectively and they move smoothly and continuously.

1. Let $\theta$ represent the angle between the hands, where $0\le \theta <2\pi$. Show that

   $$\frac{d\theta }{dt}=\frac{11\pi }{360}{\text{ rad min}}^{-1}$$

2. At $12:20\text{pm}$, find, in ${\text{cm}}^2{\text{ min}}^{-1}$, the rate of change of the area formed by the centre of the clock, the tip of the hour hand, and the tip of the minute hand.

3. Find also the rate of change of the distance between the tip of the hour hand and the tip of the minute hand at this time, in $\text{cm}{\text{ min}}^{-1}$.

In the diagram below, $t$ is measured in minutes, and if you move $t$ you can see the clock hands move between $12:00$ and $13:00.$

If $\alpha$ and $\beta$ are the angles indicated below

then $\theta =\alpha -\beta ,$ so

Now, $\alpha$ will rotate a full $2\pi$ radians in $60$ minutes, so how many radians each minute?

$\beta$ moves more slowly (one hour is $\frac{1}{12}$ of a full turn.)

For (b), the area of a triangle is given by

For (c), use the cosine rule

Find the sum of your answers for (b) and (c), then round to three significant figures, and give your answer below