Hyperbolic functions are commonly used to study dynamics and electromagnetic theory, where integrals with a $\sqrt{{{x}^{2}}+1}$ term occurs. The inverse function of the hyperbolic function $\sinh x$ is ${{\sinh }^{-1}}x$. It is given that $y={{\sinh }^{-1}}x=\ln \left( x+\sqrt{{{x}^{2}}+1} \right)$. Show that ${{e}^{y}}\frac{{{\mathrm{d}}^{2}}y}{\mathrm{d}{{x}^{2}}}+{{e}^{y}}{{\left( \frac{\mathrm{d}y}{\mathrm{d}x} \right)}^{2}}={{({{x}^{2}}+1)}^{-\frac{1}{2}}}-{{x}^{2}}{{({{x}^{2}}+1)}^{-\frac{3}{2}}}$. [2] Hence, by further differentiation, find the first two non-zero terms of the Maclaurin’s series for ${{\sinh }^{-1}}x$ in ascending powers of $x$. [4] The diagram shows a right angled tringle $ABC$ with angle $ACB=\frac{\pi }{6}$ radians. $D$ lies on $AB$ produced such angle $BCD=\theta $ radians. Show that $\frac{AB}{AD}=\frac{1}{\sqrt{3}\tan \left( \frac{\pi }{6}+\theta \right)}$ . [1] Given that $\theta$ is sufficiently small for ${{\theta }^{3}}$ and higher powers of $\theta$ to be neglected, show that $\frac{AB}{AD}\approx 1+a\theta +b{{\theta }^{2}}$, where $a$ and $b$ are exact constants to be determined. [4]

2018 NYJC P2 Q4

11 marks

Prelims

Hyperbolic functions are commonly used to study dynamics and electromagnetic theory, where integrals with a $\sqrt{{{x}^{2}}+1}$ term occurs. The inverse function of the hyperbolic function $\sinh x$ is ${{\sinh }^{-1}}x$. It is given that $y={{\sinh }^{-1}}x=\ln \left( x+\sqrt{{{x}^{2}}+1} \right)$.
1. Show that ${{e}^{y}}\frac{{{\mathrm{d}}^{2}}y}{\mathrm{d}{{x}^{2}}}+{{e}^{y}}{{\left( \frac{\mathrm{d}y}{\mathrm{d}x} \right)}^{2}}={{({{x}^{2}}+1)}^{-\frac{1}{2}}}-{{x}^{2}}{{({{x}^{2}}+1)}^{-\frac{3}{2}}}$.[2]
2. Hence, by further differentiation, find the first two non-zero terms of the Maclaurin’s series for ${{\sinh }^{-1}}x$ in ascending powers of $x$.[4]
The diagram shows a right angled tringle $ABC$ with angle $ACB=\frac{\pi }{6}$ radians. $D$ lies on $AB$ produced such angle $BCD=\theta $ radians.
1. Show that $\frac{AB}{AD}=\frac{1}{\sqrt{3}\tan \left( \frac{\pi }{6}+\theta \right)}$ .[1]
2. Given that $\theta$ is sufficiently small for ${{\theta }^{3}}$ and higher powers of $\theta$ to be neglected, show that
  $\frac{AB}{AD}\approx 1+a\theta +b{{\theta }^{2}}$,
  where $a$ and $b$ are exact constants to be determined.[4]

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