Hyperbolic functions – further maths worksheet with answers to download as PDF

Recall that \(\sinh x=\dfrac{e^x-e^{-x}}2\), \(\cosh x=\dfrac{e^x+e^{-x}}2\) and \(\tanh x=\dfrac{\sinh x}{\cosh x}\). Evaluate each of the following exactly, without a calculator.

1. \(\sinh(\ln3)\)
2. \(\cosh(\ln3)\)
3. \(\tanh(\ln3)\)
4. Verify that \(\cosh^2(\ln3)-\sinh^2(\ln3)=1\).

Use the definitions in terms of exponentials, or Osborn’s rule, for each part.

1. Prove that \(\sinh2x=2\sinh x\cosh x\).
2. Prove that \(\cosh2x=2\cosh^2x-1\).
3. Use Osborn’s rule to convert \(\cos3\theta=4\cos^3\theta-3\cos\theta\) into an identity for \(\cosh3x\).
4. Given that \(\cosh x=\dfrac53\) with \(x>0\), find \(\sinh x\) and \(\tanh x\).

Solve each equation, giving exact answers in terms of natural logarithms.

1. \(\cosh x=2\)
2. \(\sinh x=-\dfrac34\)
3. \(\cosh2x-3\cosh x+2=0\)
4. \(4\cosh x+5\sinh x=3\)

The inverse hyperbolic functions have the logarithmic forms \(\operatorname{arsinh}x=\ln\left(x+\sqrt{x^2+1}\right)\), \(\operatorname{arcosh}x=\ln\left(x+\sqrt{x^2-1}\right)\) for \(x\ge1\), and \(\operatorname{artanh}x=\tfrac12\ln\dfrac{1+x}{1-x}\) for \(|x|<1\).

1. Find \(\operatorname{arsinh}2\) and \(\operatorname{arcosh}3\) in exact form, and to 3 decimal places.
2. Find \(\operatorname{artanh}\tfrac12\) in exact form.
3. Prove the formula for \(\operatorname{arsinh}x\) by letting \(y=\operatorname{arsinh}x\) and solving \(x=\sinh y\) for \(e^y\).

Differentiate each function with respect to \(x\). Use \(\dfrac d{dx}\sinh x=\cosh x\), \(\dfrac d{dx}\cosh x=\sinh x\) and \(\dfrac d{dx}\tanh x=\operatorname{sech}^2x\).

1. \(\sinh3x\)
2. \(x^2\cosh x\)
3. \(\ln(\cosh x)\)
4. \(\tanh2x\) and \(\sinh^2x\)

Find each integral. Use \(\displaystyle\int\dfrac{dx}{\sqrt{x^2+a^2}}=\operatorname{arsinh}\dfrac xa\) and \(\displaystyle\int\dfrac{dx}{\sqrt{x^2-a^2}}=\operatorname{arcosh}\dfrac xa\).

1. \(\displaystyle\int\cosh2x\,dx\)
2. \(\displaystyle\int\sinh^2x\,dx\) (use \(\cosh2x=1+2\sinh^2x\))
3. \(\displaystyle\int_0^1\tanh x\,dx\)
4. \(\displaystyle\int_2^4\dfrac{dx}{\sqrt{x^2-4}}\)

A power line hangs between two pylons 60 m apart and takes the shape of the catenary \(y=20\cosh\dfrac x{20}\), where \(x\) and \(y\) are in metres and the origin is vertically below the lowest point of the cable at ground level directly under its midpoint (so \(-30\le x\le30\)).

1. Find the lowest height of the cable above the ground.
2. Find the height of the cable at each pylon, to 2 decimal places.
3. Find the length of the cable, using the fact that the arc length of a curve is \(\int\sqrt{1+\left(\dfrac{dy}{dx}\right)^2}\,dx\).
4. Find the angle that the cable makes with the horizontal at the pylon.

A garden arch is modelled by \(y=40-5\left(\cosh\dfrac x{10}-1\right)\) for \(y\ge0\), with lengths in metres.

1. Find the height of the arch at its centre.
2. Show that the arch meets the ground where \(\cosh\dfrac x{10}=9\).
3. Hence find the width of the arch at ground level, in exact form and to 1 decimal place.
4. Find the height of the arch at \(x=15\).

A sphere falling from rest through air with resistance proportional to the square of the speed has speed \(v=40\tanh\dfrac{9.8t}{40}\) m s\(^{-1}\) after \(t\) seconds.

1. State the terminal velocity and explain how you can tell from the formula.
2. Find the speed after 3 seconds.
3. Show that the acceleration is \(\dfrac{dv}{dt}=9.8\operatorname{sech}^2\dfrac{9.8t}{40}\) and find its value at \(t=0\).
4. After what time is the speed 99% of the terminal velocity?

The speed \(v\) of a water wave of wavelength \(\lambda\) metres in water of depth \(d\) metres satisfies \(v^2=\dfrac{g\lambda}{2\pi}\tanh\dfrac{2\pi d}\lambda\), with \(g=9.8\) m s\(^{-2}\).

1. Find the speed of a wave of wavelength 20 m in water 5 m deep.
2. Find the speed of the same wave in very deep water, using \(\tanh\to1\).
3. By what percentage is the wave slowed by the shallow water?
4. Explain why waves slow down as they approach a beach.

A metal fin of length 0.5 m cools a computer processor. The temperature excess above the room along the fin is \(T(x)=T_0\dfrac{\cosh m(L-x)}{\cosh mL}\), where \(x\) is the distance from the processor, \(L=0.5\) m, \(m=2\) m\(^{-1}\) and \(T_0=40^\circ\)C is the excess at the base.

1. Verify that \(T(0)=T_0\).
2. Find the temperature excess at the tip \(x=L\).
3. Find the excess halfway along the fin, to 1 decimal place.
4. Show that \(T\) is decreasing along the fin, using the derivative.

In special relativity, velocities in a straight line combine by adding rapidities. A velocity \(v\) is written \(v=c\tanh\phi\), where \(c\) is the speed of light and \(\phi\) is the rapidity. Rapidities add: \(\phi=\phi_1+\phi_2\).

1. Find \(\phi_1\) and \(\phi_2\) for \(v=0.6c\) and \(v=0.8c\), using \(\operatorname{artanh}\).
2. Find \(\phi_1+\phi_2\) and the corresponding speed as a fraction of \(c\).
3. Prove that \(\tanh(a+b)=\dfrac{\tanh a+\tanh b}{1+\tanh a\tanh b}\) and use it to check part (b).
4. Explain why the combined speed is less than \(c\) although \(0.6+0.8>1\).

For the sphere in Question 9, the distance fallen is the integral of the speed.

1. Show that \(\displaystyle\int40\tanh\dfrac{9.8t}{40}\,dt=\dfrac{1600}{9.8}\ln\cosh\dfrac{9.8t}{40}\).
2. Find the distance fallen in the first 10 seconds, to the nearest metre.
3. Compare with the distance \(40\times10=400\) m that would be covered at terminal speed throughout, and explain the difference.

The daily cost, in thousands of pounds, of running a delivery depot is modelled by \(C(x)=3\cosh x-\sinh x\), where \(x\) is a measure of the number of vans on the road.

1. Find \(C'(x)\) and show that the stationary point satisfies \(\tanh x=\tfrac13\).
2. Find the exact value of \(x\) at the stationary point.
3. Find the minimum cost exactly and justify that it is a minimum.
4. Give the minimum cost in pounds.

Let \(y=\operatorname{arcosh}x\) for \(x>1\).

1. Show that \(\cosh y=x\) and differentiate implicitly to show that \(\dfrac{dy}{dx}=\dfrac1{\sinh y}\).
2. Use \(\cosh^2y-\sinh^2y=1\) to show that \(\dfrac{dy}{dx}=\dfrac1{\sqrt{x^2-1}}\).
3. Hence write down \(\displaystyle\int\dfrac{dx}{\sqrt{x^2-4}}\) and evaluate \(\displaystyle\int_2^5\dfrac{dx}{\sqrt{x^2-4}}\) to 4 decimal places.
4. Check the answer to part (c) using the logarithmic form of \(\operatorname{arcosh}\).

Use the substitution \(x=2\sinh u\) to evaluate \(\displaystyle\int_0^2\sqrt{x^2+4}\,dx\).

1. Show that \(dx=2\cosh u\,du\) and \(\sqrt{x^2+4}=2\cosh u\).
2. Find the limits for \(u\).
3. Show that the integral becomes \(\displaystyle\int_0^{\operatorname{arsinh}1}\left(2+2\cosh2u\right)du\).
4. Hence find the exact value of the integral and its value to 3 decimal places.

Consider the equation \(4\sinh x+3\cosh x=6\).

1. Write \(4\sinh x+3\cosh x\) in the form \(R\sinh(x+\alpha)\) and find \(R\) and \(\tanh\alpha\).
2. Solve the equation exactly using \(\alpha=\tfrac12\ln7\).
3. Check by solving with the substitution \(u=e^x\).
4. Show that the equation \(3\sinh x+4\cosh x=2\) has no real solution.

The curve \(C\) has equation \(y=\sinh x-2x\).

1. Find \(\dfrac{dy}{dx}\) and show that the stationary points occur where \(\cosh x=2\).
2. Find the exact coordinates of the stationary points.
3. Use the second derivative to determine their nature.
4. Hence sketch the curve and state how many real roots the equation \(\sinh x=2x\) has.

The curve \(y=\cosh x\) is drawn for \(0\le x\le1\). All lengths are in the same units.

1. Find the exact area under the curve between \(x=0\) and \(x=1\).
2. Show that \(1+\left(\dfrac{dy}{dx}\right)^2=\cosh^2x\) and hence find the exact length of the curve.
3. What do you notice about your answers to (a) and (b)?
4. Find the volume formed when the area in (a) is rotated about the \(x\)-axis, in exact form and to 3 decimal places.

Let \(t=\tanh\dfrac x2\).

1. Show that \(\sinh x=\dfrac{2t}{1-t^2}\) and \(\cosh x=\dfrac{1+t^2}{1-t^2}\).
2. Show that \(\dfrac{dx}{dt}=\dfrac2{1-t^2}\).
3. Hence show that \(\displaystyle\int\operatorname{sech}x\,dx=2\arctan t+C\).
4. Evaluate \(\displaystyle\int_0^\infty\operatorname{sech}x\,dx\) and check numerically that the area under \(\operatorname{sech}x\) for \(0\le x\le6\) is close to your answer.