Question

If the minute hand of a clock has length r (in centimetres), find the rate at which it sweeps out area as a function of r.

Short answer

The minutes hand of the clock sweeps across the area is\(\pi {r^2}\;{\rm{c}}{{\rm{m}}^{\rm{2}}}{\rm{/h}}\).

Step-by-step solution

Apply clock rule

The length of minute hand clock is \(r\) cm.

The objective is to find the rate at which minute hand sweeps out area.

The area swept by minute hand is the area of sector in the clock.

\(A = \frac{1}{2}{r^2}\theta \)

Apply chain rule of differentiation

Chain rule of differentiation is\(\frac{{dy}}{{dx}} = \frac{{dy}}{{du}}.\frac{{du}}{{dx}}\).

Differentiate with respect to \(t\).

\(\begin{aligned}\frac{{dA}}{{dt}} &= \frac{d}{{dt}}\left( {\frac{1}{2}{r^2}\theta } \right)\\\frac{{dA}}{{dt}} &= \frac{1}{2}{r^2}\frac{{d\theta }}{{dt}}\end{aligned}\)

Since minute hand rotate \({360^ \circ } = 2\pi \) radian in each hour.

Therefore, \(\frac{{d\theta }}{{dt}} = 2\pi \)rad/h

Substitutes, \(\frac{{d\theta }}{{dt}} = 2\pi \) in \(\frac{{dA}}{{dt}}\).

\(\begin{aligned}\frac{{dA}}{{dt}} &= \frac{1}{2}{r^2}(2\pi )\\\frac{{dA}}{{dt}} &= \pi {r^2}\end{aligned}\)

So, the rate at which the minutes hand of the clock sweeps across the area is\(\pi {r^2}\;{\rm{c}}{{\rm{m}}^{\rm{2}}}{\rm{/h}}\).