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2014 Paper 1 Q4
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 \(121 = 11^2\).
2002 Paper 1 Q3
Show that \((a+b)^2\le 2a^2+2b^2\,\). Find the stationary points on the curve $y=\big(a^2\cos^2\theta +b^2\sin^2\theta\big)^{\frac12} + \big(a^2\sin^2\theta +b^2\cos^2\theta\big)^{\frac12}\,$, where \(a\) and \(b\) are constants. State, with brief reasons, which points are maxima and which are minima. Hence prove that \[ \vert a\vert +\vert b \vert \le \big(a^2\cos^2\theta +b^2\sin^2\theta\big)^{\frac12} + \big(a^2\sin^2\theta +b^2\cos^2\theta\big)^{\frac12} \le \big(2a^2+2b^2\big)^{\frac12} \;. \]
Solution: \begin{align*} && 2a^2+2b^2 &= a^2 + b^2 + (a^2+b^2) \\ &&&\underbrace{\geq}_{AM-GM} a^2+b^2+2\sqrt{a^2b^2} \\ &&&= a^2+b^2 + 2|a||b| \\ &&&\geq a^2+b^2 + 2ab \\ &&&= (a+b)^2 \end{align*} Assume \(a^2 \neq b^2\), otherwise the curve is a constant. \begin{align*} && y & = \big(a^2\cos^2\theta +b^2\sin^2\theta\big)^{\frac12} + \big(a^2\sin^2\theta +b^2\cos^2\theta\big)^{\frac12}\\ && \frac{\d y}{\d \theta} &= \tfrac12 \left (a^2\cos^2\theta +b^2\sin^2\theta \right)^{-\frac12} \cdot (2 \sin \theta \cos \theta (b^2 - a^2)) + \tfrac12 (a^2\sin^2\theta +b^2\cos^2\theta)^{-\frac12} \cdot (2 \sin \theta \cos \theta (a^2 - b^2) \\ &&&= \tfrac12\sin2 \theta (b^2 - a^2) \left ( \left (a^2\cos^2\theta +b^2\sin^2\theta \right)^{-\frac12} - (a^2\sin^2\theta +b^2\cos^2\theta)^{-\frac12}\right) \\ \therefore \frac{\d y}{\d x} = 0 \Rightarrow && \sin 2\theta = 0 & \text{ or } a^2\cos^2\theta +b^2\sin^2\theta = a^2\sin^2\theta +b^2\cos^2\theta \\ \Rightarrow && \theta &= 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi \\ && (a^2-b^2) \cos ^2\theta &= (a^2-b^2) \sin^2 \theta \\ \Rightarrow && \theta &= \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4} \end{align*} WLOG \(b^2 - a^2 > 0\), then the two parts of the derivative look like:
2001 Paper 2 Q5
The curve \(C_1\) passes through the origin in the \(x\)--\(y\) plane and its gradient is given by $$ \frac{\d y}{\d x} =x(1-x^2)\e^{-x^2}. $$ Show that \(C_1\) has a minimum point at the origin and a maximum point at \(\left(1,{\frac12\, \e^{-1}} \right)\). Find the coordinates of the other stationary point. Give a rough sketch of \(C_1\). The curve \(C_2\) passes through the origin and its gradient is given by $$ \frac{\d y}{\d x}= x(1-x^2)\e^{-x^3}. $$ Show that \(C_2\) has a minimum point at the origin and a maximum point at \((1,k)\), where \phantom{} \(k > \frac12 \,\e^{-1}.\) (You need not find \(k\).)
1988 Paper 1 Q2
The function \(\mathrm{f}\) and \(\mathrm{g}\) are related (for all real \(x\)) by \[ \mathrm{g}(x)=\mathrm{f}(x)+\frac{1}{\mathrm{f}(x)}\,. \] Express \(\mathrm{g}'(x)\) and \(\mathrm{g}''(x)\) in terms of \(\mathrm{f}(x)\) and its derivatives. If \(\mathrm{f}(x)=4+\cos2x+2\sin x\), find the stationary points of \(\mathrm{g}\) for \(0\leqslant x\leqslant2\pi,\) and determine which are maxima and which are minima.
Solution: \(g'(x) = f'(x) - \frac{f'(x)}{(f(x))^2} = f'(x) \l 1 - \frac{1}{(f(x))^2} \r\) \(g''(x) = f''(x) - \frac{f''(x)f(x)^2-f'(x)\cdot 2f(x) f'(x)}{(f(x))^4} = f''(x) + \frac{f''(x)f(x)-2(f'(x))^2}{(f(x))^3}\) \begin{align*} f(x) &=4+\cos2x+2\sin x \\ f'(x) &=-2\sin2x+2\cos x \\ f''(x) &= -4\cos2x-2\sin x \end{align*} Therefore, since the stationary points of \(g\), ie points where \(g'(x) = 0\) are where \(f'(x) = 0\) or \(f(x) = \pm 1\) we should look at \begin{align*} && 0 &= f'(x) \\ && 0 &= 2 \cos x - 2 \sin 2x \\ &&&= 2 \cos x - 4 \sin x \cos x \\ &&&= 2\cos x (1 - 2 \sin x) \\ \Rightarrow && x &= \frac{\pi}2, \frac{3\pi}{2}, \frac{\pi}{6}, \frac{5\pi}{6} \end{align*} \begin{align*} && 1 &= f(x) \\ && 1 &= 4 + \cos 2x + 2 \sin x \\ \Rightarrow && \cos 2x = -1,& \sin x = -1 \\ \Rightarrow && x &= \frac{3\pi}{2} \end{align*} which we were already checking. For each of these points we have: \begin{array}{c|c|c|c||c} x & f(x) & f'(x) & f''(x) & g''(x) \\ \hline \frac{\pi}{2} & 5 & 0 & 2 & > 0\\ \frac{3\pi}{2} & 1 & 0 & 6 &> 0\\ \frac{\pi}{6} & 5.5 & 0 & -3 & < 0 \\ \frac{5\pi}{6} & 5.5 & 0 & -3 & < 0\\ \end{array} Therefore \(\frac{\pi}{2}, \frac{3\pi}{2}\) are minimums and \(\frac{\pi}{6}\) and \(\frac{5\pi}{6}\) are maxima.