Hyperbolic functions – further maths test answers to download as PDF

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1. Answer to Question 1 (4 marks)

   Method: replace each hyperbolic function by its exponential definition, \(\sinh x=\dfrac{e^x-e^{-x}}2\) and \(\cosh x=\dfrac{e^x+e^{-x}}2\), and use \(e^{\ln a}=a\). The identity \(\cosh^2x-\sinh^2x=1\) is the hyperbolic version of \(\cos^2x+\sin^2x=1\). Mark scheme as shown in brackets.

   1. Square \(e^x+e^{-x}\): \(e^{2x}+2+e^{-2x}\), then divide by 4 and multiply by 2. \(2\cosh^2x-1=\dfrac{(e^x+e^{-x})^2}2-1\) [1] \(=\dfrac{e^{2x}+e^{-2x}}2=\cosh2x\) [1]
   2. Use part a) to obtain a quadratic in \(c=\cosh x\); remember that \(\cosh x\ge1\) for every real \(x\). \(2c^2-1=3c-2\) gives \(2c^2-3c+1=0\), \((2c-1)(c-1)=0\) [1]. \(\cosh x\ge1\) so \(\cosh x=1\) and \(x=0\) [1] The root \(c=\tfrac12\) is rejected.

2. Answer to Question 2 (4 marks)

   Method: replace each hyperbolic function by its exponential definition, \(\sinh x=\dfrac{e^x-e^{-x}}2\) and \(\cosh x=\dfrac{e^x+e^{-x}}2\), and use \(e^{\ln a}=a\). The identity \(\cosh^2x-\sinh^2x=1\) is the hyperbolic version of \(\cos^2x+\sin^2x=1\). Multiply by \(e^x\) to obtain a quadratic in \(e^x\). Mark scheme as shown in brackets.

   1. \(\dfrac{2(e^x-e^{-x})+(e^x+e^{-x})}2=2\) [1] gives \(3e^x-e^{-x}=4\) [1]. With \(u=e^x\): \(3u^2-4u-1=0\) [1]. \(u=\dfrac{4+\sqrt{28}}6=\dfrac{2+\sqrt7}3\), so \(x=\ln\dfrac{2+\sqrt7}3\) (negative root rejected) [1]

   Check: with \(e^x=\tfrac{2+\sqrt7}3\approx1.549\), \(2\sinh x+\cosh x\approx0.903+1.097=2\).

3. Answer to Question 3 (4 marks)

   Method: \(\dfrac d{dx}\sinh x=\cosh x\) and \(\dfrac d{dx}\cosh x=\sinh x\) (no minus sign, unlike the trigonometric case), so \(\int\cosh ax\,dx=\dfrac1a\sinh ax+c\) and \(\int\sinh ax\,dx=\dfrac1a\cosh ax+c\). The chain, product and quotient rules work as usual. Mark scheme as shown in brackets.

   1. Take \(u=x\) and \(v=\sinh2x\), so \(v'=2\cosh2x\). \(\dfrac{dy}{dx}=\sinh2x+2x\cosh2x\) [2: 1 product rule, 1 chain rule]
   2. Differentiate \(u^2\) with \(u=\cosh3x\), then use \(2\sinh A\cosh A=\sinh2A\) with \(A=3x\). \(\dfrac{dy}{dx}=2\cosh3x\cdot3\sinh3x=6\sinh3x\cosh3x\) [1] \(=\) \(3\sinh6x\) [1] So \(k=3\).

4. Answer to Question 4 (5 marks)

   Method: \(\dfrac d{dx}\sinh x=\cosh x\) and \(\dfrac d{dx}\cosh x=\sinh x\) (no minus sign, unlike the trigonometric case), so \(\int\cosh ax\,dx=\dfrac1a\sinh ax+c\) and \(\int\sinh ax\,dx=\dfrac1a\cosh ax+c\). The chain, product and quotient rules work as usual. Mark scheme as shown in brackets.

   1. \(\left[\operatorname{arsinh}\dfrac x2\right]_0^2\) [1] \(=\operatorname{arsinh}1=\) \(\ln(1+\sqrt2)\) [1]
   2. Let \(u=x\), \(dv=\sinh x\,dx\): \(\left[x\cosh x\right]_0^1-\int_0^1\cosh x\,dx\) [1] \(=\cosh1-\left[\sinh x\right]_0^1\) [1] \(=\cosh1-\sinh1=\) \(e^{-1}\) [1]

   Answers: a) \(\ln(1+\sqrt2)\approx0.881\); b) the integral equals \(e^{-1}\approx0.368\), since \(\cosh1-\sinh1=e^{-1}\) by the definitions.

5. Answer to Question 5 (3 marks)

   Method: in a model, identify what each variable represents, substitute the given values, and use the properties of \(\cosh\) (minimum 1 at 0, even function) and \(\tanh\) (between \(-1\) and 1, tends to 1) to interpret the answers. Mark scheme as shown in brackets.

   1. Substitute \(x=15\); by symmetry both posts have the same height. \(10\cosh1.5=\) 23.5 m [1] The lowest point of the cable is 10 m high, at \(x=0\).
   2. \(\cosh\) is even, so integrate from 0 to 15 and double. \(\displaystyle\int_{-15}^{15}\cosh\dfrac x{10}dx=2\left[10\sinh\dfrac x{10}\right]_0^{15}\) [1] \(=20\sinh1.5=\) 42.6 m [1] So the cable is about 42.6 m long, compared with a 30 m gap between the posts.

Total : 20 marks