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1990 Paper 3 Q9
D: 1700.0 B: 1484.7
hyperbolic functions trigonometric chain rule implicit differentiation second derivative proof sech relationship between trig and hyperbolic

The real variables \(\theta\) and \(u\) are related by the equation \(\tan\theta=\sinh u\) and \(0\leqslant\theta<\frac{1}{2}\pi.\) Let \(v=\mathrm{sech}u.\) Prove that

  1. \(v=\cos\theta;\)
  2. \(\dfrac{\mathrm{d}\theta}{\mathrm{d}u}=v;\)
  3. \(\sin2\theta=-2\dfrac{\mathrm{d}v}{\mathrm{d}u}\quad\) and \(\quad\cos2\theta=-\cosh u\dfrac{\mathrm{d}^{2}v}{\mathrm{d}u^{2}};\)
  4. \({\displaystyle \frac{\mathrm{d}u}{\mathrm{d}\theta}\frac{\mathrm{d}^{2}v}{\mathrm{d}\theta^{2}}+\frac{\mathrm{d}v}{\mathrm{d}\theta}\frac{\mathrm{d}^{2}u}{\mathrm{d}\theta^{2}}+\left(\frac{\mathrm{d}u}{\mathrm{d}\theta}\right)^{2}=0.}\)

Solution:

  1. \begin{align*} v &= \mathrm{sech} u \\ &= \frac{1}{\mathrm{cosh } u} \\ &= \frac{1}{\sqrt{1+\mathrm{sinh}^2 u}} \tag{\(u > 0\)} \\ &= \frac{1}{\sqrt{1+\tan^2 \theta}} \\ &= \frac{1}{\sqrt{\mathrm{sec}^2 \theta}} \\ &= \cos \theta \tag{\(0 < \theta < \tfrac{\pi}{2}\)} \end{align*}
  2. \begin{align*} && \tan \theta &= \textrm{sinh} u \\ \underbrace{\Rightarrow}_{\frac{\d}{\d u}} && \sec^2 \theta \cdot \frac{\d \theta}{\d u} &= \cosh u \\ \Rightarrow && \frac{\d \theta}{\d u} &=\cosh u \cdot \cos^2 \theta \\ &&&= \frac{1}{v} \cdot v^2 \\ &&&=v \end{align*}
  3. \begin{align*} \sin 2 \theta &= 2 \sin \theta \cos \theta \\ &= 2 \sin \theta \cdot \frac{\d \theta}{\d u} \\ &= -2 \frac{\d v}{\d \theta} \cdot \frac{\d \theta}{\d u} \tag{\(\cos \theta = v\)} \\ &= -2 \frac{\d v}{\d u} \end{align*} \begin{align*} && \sin 2 \theta &= -2 \frac{\d v}{\d u} \\ \underbrace{\Rightarrow}_{\frac{\d}{\d u}} && 2 \cos 2 \theta \cdot \frac{\d \theta}{\d u} &= -2 \frac{\d^2 v}{\d u^2} \\ \Rightarrow && \cos 2 \theta &= - \frac{\d^2 v}{\d u^2} \frac{1}{v} \\ &&&= -\frac{\d ^2v}{\d u^2} \cosh u \end{align*}
  4. \begin{align*} && \frac{\d u}{\d \theta} &= \frac{1}{v} \\ \Rightarrow && \frac{\d^2 u}{\d \theta^2} &= -\frac{1}{v^2} \frac{\d v}{\d \theta} \\ &&&= \frac{1}{v^2} \sin \theta \\ && \frac{\d v}{\d \theta} &= -\sin \theta \\ \Rightarrow && \frac{\d^2 v}{\d \theta^2} &= -\cos \theta \\ &&&= - v \\ \end{align*} Therefore \begin{align*} \frac{\mathrm{d}u}{\mathrm{d}\theta}\frac{\mathrm{d}^{2}v}{\mathrm{d}\theta^{2}}+\frac{\mathrm{d}v}{\mathrm{d}\theta}\frac{\mathrm{d}^{2}u}{\mathrm{d}\theta^{2}}+\left(\frac{\mathrm{d}u}{\mathrm{d}\theta}\right)^{2} &= \frac{1}{v} \cdot \left (-v\right) + \left ( - \sin \theta \right ) \cdot \left (\frac{1}{v^2} \sin \theta \right) + \frac{1}{v^2} \\ &= -1 + \frac{1-\sin^2 \theta}{v^2} \\ &= -1 + \frac{\cos^2 \theta}{v^2} \\ &= -1 + 1 \\ &= 0 \end{align*}

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