1 problem found
An accurate clock has an hour hand of length \(a\) and a minute hand of length \(b\) (where \(b>a\)), both measured from the pivot at the centre of the clock face. Let \(x\) be the distance between the ends of the hands when the angle between the hands is \(\theta\), where \(0\le\theta < \pi\). Show that the rate of increase of \(x\) is greatest when \(x=(b^2-a^2)^\frac12\). In the case when \(b=2a\) and the clock starts at mid-day (with both hands pointing vertically upwards), show that this occurs for the first time a little less than 11 minutes later.
Solution: The position of the hands are \(\begin{pmatrix} a\sin(-t) \\ a \cos(-t) \end{pmatrix}\) and \(\begin{pmatrix} b\sin(-60t) \\ b \cos(-60t) \end{pmatrix}\), the distance between the hands is \begin{align*} x &= \sqrt{\left ( a \sin t - b \sin 60t\right)^2+\left ( a \cos t - b \cos 60t\right)^2} \\ &= \sqrt{a^2+b^2-2ab\left (\sin t \sin 60t+\cos t \cos 60t \right)} \\ &= \sqrt{a^2+b^2-2ab \cos(59t)} = \sqrt{a^2+b^2-2ab \cos \theta} \\ \\ \frac{\d x}{\d \theta} &= \frac{ab \sin \theta}{ \sqrt{a^2+b^2-2ab \cos \theta}} \\ \frac{\d^2 x}{\d \theta^2} &= \frac{ab \cos \theta\sqrt{a^2+b^2-2ab \cos \theta} - \frac{a^2b^2 \sin^2 \theta}{\sqrt{a^2+b^2-2ab \cos \theta}} }{a^2+b^2-2ab \cos \theta} \\ &= \frac{ab \cos \theta(a^2+b^2-2ab \cos \theta) - a^2b^2 \sin^2 \theta }{(a^2+b^2-2ab \cos \theta)^{3/2}} \\ &= \frac{ab \cos \theta(a^2+b^2-2ab \cos \theta) - a^2b^2(1-\cos^2 \theta)}{(a^2+b^2-2ab \cos \theta)^{3/2}} \\ &= \frac{ab(a^2+b^2) \cos \theta-a^2b^2 \cos \theta- a^2b^2}{(a^2+b^2-2ab \cos \theta)^{3/2}} \\ &= \frac{-ab(a\cos \theta -b)(b \cos \theta - a)}{(a^2+b^2-2ab \cos \theta)^{3/2}} \\ \end{align*} So the rate of increase is largest when \(\cos \theta = \frac{a}{b}\) (since \(\frac{b}{a}\) is impossible. Therefore when \(x = \sqrt{a^2+b^2-2ab \frac{a}{b}} = \sqrt{a^2+b^2-2a^2} = \sqrt{b^2-a^2}\) If \(b = 2a\) then \(\cos \theta = \frac{a}{2a} = \frac12 = \frac{\pi}{3} = 60^\circ\) The relative speed of the hands is \(5.5^\circ\) per minute, so \(\frac{60}{5.5} = \frac{120}{11} \approx 11\) but clearly also less than since \(121 = 11^2\).