Question

The minute hand on a watch is \(8\) mm long and the hour hand is \(4\) mm long. How fast is the distance between the tips of the hands changing at one o’clock?

Short answer

At \(1:00\) the distance between the tips of the hands decreasing at a rate of \(0.005\;{\rm{mm}}/{\rm{s}}\).

Step-by-step solution

Write the formula of Law of Cosines and trigonometric identities

Law of Cosines: \({c^2} = {a^2} + {b^2} - 2ab\cos \theta \)

Construct the diagram according to the given conditions

Step 3 Find at \(1:00\) the distance between the tips of the hands decreasing at a rate of \(0.005\;{\rm{mm/s}}\)

Since the hour hand of a clock takes a round around once every \(12\) hour and in radians \(\frac{{2\pi }}{{12}} = \frac{\pi }{6}\;{\rm{rad}}/{\rm{h}}\).

Also, the minute hand of a clock takes a round around once an hour and in radians \(2\pi \;{\rm{rad}}/{\rm{h}}\). The angle between them is \(\theta \) at the rate is shown below:

\(\begin{aligned}\frac{{d\theta }}{{dt}} &= \frac{\pi }{6} - 2\pi \\ &= - \frac{{11\pi }}{6}\;{\rm{rad}}/{\rm{h}}\end{aligned}\)

Apply the Law of Cosines.

\(\begin{aligned}{l^2} &= {4^2} + {8^2} - 2\left( 4 \right)\left( 8 \right)\cos \theta \\ &= 80 - 64\cos \theta .............\left( 1 \right)\end{aligned}\)

Differentiate the above equation w.r.t \(t\).

\(\begin{aligned}\frac{d}{{dt}}\left( {{l^2}} \right) &= \frac{d}{{dt}}\left( {80 - 64\cos \theta } \right)\\2l\frac{{dl}}{{dt}} &= - 64\left( { - \sin \theta } \right)\frac{{d\theta }}{{dt}}............\left( 2 \right)\end{aligned}\)

At \(1:00\) the angle \(\theta \)between the two hands is one-fifth of the circle then \(\frac{{2\pi }}{{12}} = \frac{\pi }{6}\;{\rm{rad}}\)

At \(1:00\), find \(l\) using equation \(\left( 1 \right)\).

\(\begin{array}{c}l = \sqrt {80 - 64\cos \left( {\frac{\pi }{6}} \right)} \\ = \sqrt {80 - 32\sqrt 3 } \end{array}\)

Substitute the above values in the equation \(\left( 2 \right)\).

\(\begin{aligned}2\left( {\sqrt {80 - 32\sqrt 3 } } \right)\frac{{dl}}{{dt}} &= - 64\left( { - \sin \left( {\frac{\pi }{6}} \right)} \right)\left( { - \frac{{11\pi }}{6}} \right)\\\frac{{dl}}{{dt}} &= \frac{{64\left( {\frac{1}{2}} \right)\left( { - \frac{{11\pi }}{6}} \right)}}{{2\left( {\sqrt {80 - 32\sqrt 3 } } \right)}}\\ &= - \frac{{88\pi }}{{3\sqrt {80 - 32\sqrt 3 } }}\\ \approx - 18.6\;{\rm{mm}}/{\rm{h}}\\ \approx 0.005\;{\rm{mm}}/{\rm{s}}\end{aligned}\)

Thus, At \(1:00\) the distance between the tips of the hands decreases at a rate of \(0.005\;{\rm{mm}}/{\rm{s}}\).