The rate of change in the area of a circle sector

Question

a circular clock of radius 5 inches .At time t minutes past noon, how fast is the area of the sector of the circle between the hour and minute hand increasing?

I used implicit differentiation on the equation $A=\frac{(r²\theta)}{2}$ with r as a constant and $\frac{d\theta}{dt}=\frac{\pi}{30}$ which yielded $\frac{dA}{dt}=\frac{5\pi}{12}$ which is wrong according to the textbook's guide(solution:$\frac{55\pi}{144}$)

I can't seem to find the stated solution.

Answer 1

$\textbf{The hour hand moves too}$

Just for completeness I'll include a solution along that line of reasoning.

Set at noon, the initial angle is $0^c$, which makes the computation quite easy.

After $t$ minutes the angle covered by the minute hand is $t \cdot \dfrac{2\pi}{60}=\dfrac{t\pi}{30}$

Note that this is the displacement from the 12 mark. This is necessary to be able to compute the required angle.

After $t$ minutes which is $\dfrac{t}{60}$ hours which is $\dfrac{t}{60} \cdot \dfrac1{12}$ of that whole clock face meaning the hour hand is meant to displace by an angle of:

$\dfrac{t}{60} \cdot \dfrac1{12} \cdot 2\pi=\dfrac{t\pi}{360}$

So the displacement of the minute hand from the hour hand given by a time t:

$=\dfrac{t\pi}{30}-\dfrac{t\pi}{360}=\dfrac{11t\pi}{360}$

The area function to be used, as you correctly gave is:

$\dfrac{r^2}2 \times \theta=\dfrac{5^2}{2} \times \dfrac{11t\pi}{360}=\dfrac{55t\pi}{144}$

Hope I haven't spoiled the ending too much.

Answer 2

You also have to account for the movement of the hour hand, which would decrease the rate of change.

In one minute, the minute hand moves forward $1$ unit and the hour hand moves forward $\frac{t}{12}$ units. Hence, $$\frac{dh}{dt}=\frac{1}{12} \implies h(t)= \frac{t}{12} \\ \frac{dm}{dt} = 1 \implies m(t)= t$$

Now, $60$ units correspond to an arc length of $2\pi \times 5$ in, so $1$ unit corresponds to an arc length of $\frac{\pi}{6}$ in.

The arc length of the sector, $$a(t)= \big(m(t)-h(t)\big)\frac{\pi}{6} = \frac{11\pi}{72} \ t$$ and so

$$\theta(t) = \frac{a(t)}{r} = \frac{11\pi}{72\times 5} \ t$$

Therefore, $$\frac{dA}{dt} = \frac{r^2}{2} \frac{d\theta}{dt} = \frac{55\pi}{144} $$