Hyperbolic functions – further maths worksheet answers to download as PDF

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1. Answer to Question 1

   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 \(\cos^2x+\sin^2x=1\).

   1. \(e^{\ln3}=3\) and \(e^{-\ln3}=\tfrac13\). \(\sinh(\ln3)=\dfrac{3-\frac13}2=\dfrac{8/3}2=\) \(\dfrac43\) So \(\sinh(\ln3)=\tfrac43\).
   2. Use the same two values in the definition of \(\cosh\). \(\cosh(\ln3)=\dfrac{3+\frac13}2=\dfrac{10/3}2=\) \(\dfrac53\) So \(\cosh(\ln3)=\tfrac53\).
   3. Divide the answers to a) and b). \(\tanh(\ln3)=\dfrac{4/3}{5/3}=\) \(\dfrac45\) As expected, \(\tanh\) lies between \(-1\) and 1.
   4. Substitute the answers to a) and b) into the identity. \(\left(\dfrac53\right)^2-\left(\dfrac43\right)^2=\dfrac{25-16}9=\dfrac99=\) 1 ✓ This confirms the fundamental identity for this value of \(x\).

2. Answer to Question 2

   1. \(2\sinh x\cosh x=2\cdot\dfrac{e^x-e^{-x}}2\cdot\dfrac{e^x+e^{-x}}2=\dfrac{e^{2x}-e^{-2x}}2=\) \(\sinh2x\) ✓
   2. \(2\cosh^2x-1=2\cdot\dfrac{e^{2x}+2+e^{-2x}}4-1=\dfrac{e^{2x}+e^{-2x}}2=\) \(\cosh2x\) ✓
   3. Osborn’s rule: replace each trig function by the hyperbolic one and change the sign of any product of two sines. There are none in this identity, so \(\cosh3x=4\cosh^3x-3\cosh x\).
   4. \(\sinh^2x=\cosh^2x-1=\dfrac{25}9-1=\dfrac{16}9\) and \(x>0\) so \(\sinh x=\dfrac43\); \(\tanh x=\dfrac45\).

3. Answer to Question 3

   1. \(e^x+e^{-x}=4\), so \(u^2-4u+1=0\) with \(u=e^x\), \(u=2\pm\sqrt3\). \(x=\pm\ln(2+\sqrt3)\) (since \((2+\sqrt3)(2-\sqrt3)=1\)).
   2. \(e^x-e^{-x}=-\tfrac32\), \(2u^2+3u-2=0\), \((2u-1)(u+2)=0\). Only \(u=\tfrac12\) is positive, so \(x=-\ln2\).
   3. \(2\cosh^2x-1-3\cosh x+2=0\), so \(2\cosh^2x-3\cosh x+1=0\), \((2\cosh x-1)(\cosh x-1)=0\). Since \(\cosh x\ge1\), \(\cosh x=1\) and \(x=0\).
   4. \(\dfrac{4(u+1/u)+5(u-1/u)}2=3\) gives \(9u-\dfrac1u=6\), \(9u^2-6u-1=0\), \(u=\dfrac{6\pm\sqrt{72}}{18}=\dfrac{1\pm\sqrt2}3\). Only the positive root is valid: \(x=\ln\dfrac{1+\sqrt2}3\) (\(\approx-0.2\)).

4. Answer to Question 4

   Method: to evaluate an inverse hyperbolic function exactly, substitute into its logarithmic form. To prove a logarithmic form, write \(x=\sinh y\) in exponentials, multiply by \(e^y\) to get a quadratic in \(e^y\), and reject the root that is not positive.

   1. \(\operatorname{arsinh}2=\ln(2+\sqrt5)=\) 1.444. \(\operatorname{arcosh}3=\ln(3+\sqrt8)=\ln(3+2\sqrt2)=\) 1.763.
   2. \(\tfrac12\ln\dfrac{3/2}{1/2}=\) \(\tfrac12\ln3\) (\(\approx0.549\)).
   3. \(x=\dfrac{e^y-e^{-y}}2\) gives \(e^{2y}-2xe^y-1=0\). With \(w=e^y\): \(w=x\pm\sqrt{x^2+1}\). Since \(w>0\) we take the plus sign, so \(y=\ln\left(x+\sqrt{x^2+1}\right)\) ✓.

5. Answer to Question 5

   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.

   1. The inside function \(3x\) gives a factor 3. Chain rule: \(3\cosh3x\)
   2. Take \(u=x^2\) and \(v=\cosh x\), so \(u'=2x\) and \(v'=\sinh x\). Product rule: \(2x\cosh x+x^2\sinh x\) This can be written \(x(2\cosh x+x\sinh x)\).
   3. Use \(\dfrac d{dx}\ln u=\dfrac{u'}u\) with \(u=\cosh x\). \(\dfrac{\sinh x}{\cosh x}=\) \(\tanh x\) So this is the reverse of \(\int\tanh x\,dx=\ln\cosh x+c\).
   4. Use the chain rule for both; the second uses the identity \(\sinh2x=2\sinh x\cosh x\). \(\dfrac d{dx}\tanh2x=\) \(2\operatorname{sech}^22x\). \(\dfrac d{dx}\sinh^2x=2\sinh x\cosh x=\) \(\sinh2x\).

6. Answer to Question 6

   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.

   1. Integrate \(\cosh\) to \(\sinh\) and divide by 2. \(\tfrac12\sinh2x+C\) Check: differentiating gives \(\cosh2x\).
   2. Rearrange the double angle identity to make \(\sinh^2x\) the subject, then integrate term by term. \(\sinh^2x=\tfrac12(\cosh2x-1)\), so the integral is \(\tfrac14\sinh2x-\tfrac x2+C\)
   3. \(\tanh x=\dfrac{\sinh x}{\cosh x}\) has the numerator equal to the derivative of the denominator. \(\left[\ln\cosh x\right]_0^1=\ln\cosh1=\) 0.434 So the exact value is \(\ln(\cosh1)\).
   4. Use the standard result with \(a=2\), and \(\operatorname{arcosh}1=0\). \(\left[\operatorname{arcosh}\dfrac x2\right]_2^4=\operatorname{arcosh}2-\operatorname{arcosh}1=\ln(2+\sqrt3)-0=\) 1.317 So the exact value is \(\ln(2+\sqrt3)\).

7. Answer to Question 7

   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.

   1. At \(x=0\), \(y=20\cosh0=\) 20 m.
   2. \(y=20\cosh\dfrac{30}{20}=20\cosh1.5=20\times2.3524=\) 47.05 m (so it rises 27.05 m above its lowest point).
   3. Arc length \(s=\int_{-30}^{30}\sqrt{1+\sinh^2\tfrac x{20}}\,dx=\int_{-30}^{30}\cosh\dfrac x{20}\,dx=2\left[20\sinh\dfrac x{20}\right]_0^{30}=40\sinh1.5=\) 85.17 m.
   4. \(\tan\theta=\dfrac{dy}{dx}=\sinh\dfrac{30}{20}=2.129\), so \(\theta=\) 64.8\(^\circ\)

8. Answer to Question 8

   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.

   1. At \(x=0\), \(\cosh0=1\), so \(y=\) 40 m.
   2. \(y=0\) gives \(5\left(\cosh\tfrac x{10}-1\right)=40\), so \(\cosh\tfrac x{10}-1=8\), \(\cosh\tfrac x{10}=9\).
   3. \(\tfrac x{10}=\pm\operatorname{arcosh}9=\pm\ln(9+\sqrt{80})\). So the width is \(2\times10\ln(9+4\sqrt5)=20\ln(9+4\sqrt5)\approx\) 57.7 m.
   4. \(y=40-5(\cosh1.5-1)=40-5(1.3524)=\) 33.2 m.

9. Answer to Question 9

   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.

   1. As \(t\to\infty\), \(\tanh\to1\), so \(v\to\) 40 m s\(^{-1}\).
   2. \(v=40\tanh0.735=40\times0.6263=\) 25.1 m s\(^{-1}\).
   3. \(\dfrac{dv}{dt}=40\cdot\dfrac{9.8}{40}\operatorname{sech}^2\dfrac{9.8t}{40}=9.8\operatorname{sech}^2\dfrac{9.8t}{40}\). At \(t=0\) this is 9.8 m s\(^{-2}\), i.e. \(g\).
   4. \(\tanh\dfrac{9.8t}{40}=0.99\), so \(\dfrac{9.8t}{40}=\operatorname{artanh}0.99=\tfrac12\ln199=2.647\), \(t=\dfrac{2.647\times40}{9.8}=\) 10.8 s.

10. Answer to Question 10

    1. \(\dfrac{2\pi\times5}{20}=1.5708\) and \(\tanh1.5708=0.9172\). \(v^2=\dfrac{9.8\times20}{2\pi}\times0.9172=31.19\times0.9172=28.61\), so \(v=\) 5.35 m s\(^{-1}\).
    2. \(v^2=31.19\), \(v=\) 5.59 m s\(^{-1}\).
    3. \(\dfrac{5.59-5.35}{5.59}=\) 4.2%.
    4. As \(d\) decreases, \(\tanh\dfrac{2\pi d}\lambda\) decreases below 1, so \(v\) decreases: that is why waves bunch up and slow near the shore.

11. Answer to Question 11

    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.

    1. Substitute \(x=0\). \(T(0)=T_0\dfrac{\cosh mL}{\cosh mL}=T_0=\) 40\(^\circ\)C. So the formula gives the correct temperature at the base.
    2. Substitute \(x=L\), so \(m(L-x)=0\) and \(mL=1\). \(T(L)=T_0\dfrac{\cosh0}{\cosh1}=\dfrac{40}{1.543}=\) 25.9\(^\circ\)C. So the tip is about \(25.9^\circ\)C above room temperature.
    3. Substitute \(x=0.25\), so \(m(L-x)=0.5\). \(x=0.25\): \(T=40\dfrac{\cosh0.5}{\cosh1}=40\times\dfrac{1.1276}{1.5431}=\) 29.2\(^\circ\)C. So halfway along, the excess is about \(29.2^\circ\)C.
    4. Differentiate with the chain rule; the inside function \(m(L-x)\) has derivative \(-m\). \(T'(x)=-mT_0\dfrac{\sinh m(L-x)}{\cosh mL}<0\) for \(0\le x\lt L\), so the temperature falls along the fin. Since \(\sinh\) is positive for positive arguments, the derivative is negative.

12. Answer to Question 12

    1. \(\phi_1=\tfrac12\ln\dfrac{1.6}{0.4}=\tfrac12\ln4=\) 0.693; \(\phi_2=\tfrac12\ln\dfrac{1.8}{0.2}=\tfrac12\ln9=\) 1.099.
    2. \(\phi=1.792\) and \(\tanh1.792=\) 0.9459, so \(v=0.946c\).
    3. \(\tanh(a+b)=\dfrac{\sinh a\cosh b+\cosh a\sinh b}{\cosh a\cosh b+\sinh a\sinh b}\); divide top and bottom by \(\cosh a\cosh b\). Then \(\dfrac{0.6+0.8}{1+0.48}=\dfrac{1.4}{1.48}=0.9459\) ✓.
    4. \(\tanh\) is always less than 1, so rapidities of any size give speeds below \(c\). Speeds do not add as ordinary numbers; rapidities do.

13. Answer to Question 13

    1. Let \(k=\dfrac{9.8}{40}\). \(\int40\tanh kt\,dt=\dfrac{40}k\ln\cosh kt=\dfrac{1600}{9.8}\ln\cosh kt\).
    2. \(s(10)=163.27\ln\cosh2.45\). \(\cosh2.45=5.8373\), \(\ln5.8373=1.7640\). So \(s=\) 288 m (to the nearest metre).
    3. At terminal speed the sphere would have gone 400 m, which is more than 288 m, because it takes time to build up to that speed: it is slower during the acceleration phase.

14. Answer to Question 14

    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.

    1. \(C'(x)=3\sinh x-\cosh x=0\) gives \(\tanh x=\dfrac13\).
    2. \(x=\operatorname{artanh}\tfrac13=\tfrac12\ln\dfrac{4/3}{2/3}=\) \(\tfrac12\ln2\).
    3. At this point \(e^x=\sqrt2\), so \(\cosh x=\dfrac{\sqrt2+1/\sqrt2}2=\dfrac3{2\sqrt2}\) and \(\sinh x=\dfrac1{2\sqrt2}\). \(C=\dfrac9{2\sqrt2}-\dfrac1{2\sqrt2}=\dfrac8{2\sqrt2}=\) \(2\sqrt2\). \(C''=3\cosh x-\sinh x=C>0\), so it is a minimum.
    4. \(2\sqrt2\times1000\approx\) £2828.

15. Answer to Question 15

    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.

    1. By definition \(\cosh y=x\). Differentiating, \(\sinh y\dfrac{dy}{dx}=1\), so \(\dfrac{dy}{dx}=\dfrac1{\sinh y}\).
    2. \(\sinh y=\sqrt{\cosh^2y-1}=\sqrt{x^2-1}\) (positive since \(y>0\)). So \(\dfrac{dy}{dx}=\dfrac1{\sqrt{x^2-1}}\).
    3. \(\int\dfrac{dx}{\sqrt{x^2-4}}=\operatorname{arcosh}\dfrac x2+C\). From 2 to 5: \(\operatorname{arcosh}2.5-\operatorname{arcosh}1=\operatorname{arcosh}2.5=\) 1.5668.
    4. \(\ln\left(2.5+\sqrt{6.25-1}\right)=\ln(2.5+2.2913)=\ln4.7913=1.5668\) ✓.

16. Answer to Question 16

    Method: for integrals containing \(\sqrt{x^2+a^2}\), substitute \(x=a\sinh u\): the identity \(1+\sinh^2u=\cosh^2u\) removes the square root. Change the limits and use \(\cosh^2u=\tfrac12(1+\cosh2u)\).

    1. \(\dfrac{dx}{du}=2\cosh u\). \(x^2+4=4\sinh^2u+4=4\cosh^2u\), so \(\sqrt{x^2+4}=2\cosh u\).
    2. \(x=0\) gives \(u=0\); \(x=2\) gives \(\sinh u=1\), \(u=\operatorname{arsinh}1=\ln(1+\sqrt2)\).
    3. The integrand becomes \(2\cosh u\cdot2\cosh u=4\cosh^2u=2(1+\cosh2u)=2+2\cosh2u\).
    4. \(\left[2u+\sinh2u\right]_0^{\operatorname{arsinh}1}=2\operatorname{arsinh}1+2\sinh u\cosh u=2\ln(1+\sqrt2)+2\cdot1\cdot\sqrt2=\) \(2\sqrt2+2\ln(1+\sqrt2)\approx4.591\).

    Answer: \(\displaystyle\int_0^2\sqrt{x^2+4}\,dx=2\sqrt2+2\ln(1+\sqrt2)\approx4.591\).

17. Answer to Question 17

    1. \(R\sinh(x+\alpha)=R\sinh x\cosh\alpha+R\cosh x\sinh\alpha\). So \(R\cosh\alpha=4\), \(R\sinh\alpha=3\). Then \(R^2=16-9=7\), \(R=\sqrt7\), and \(\tanh\alpha=\tfrac34\).
    2. \(\alpha=\operatorname{artanh}\tfrac34=\tfrac12\ln\dfrac{7/4}{1/4}=\tfrac12\ln7\). Then \(\sinh(x+\alpha)=\dfrac6{\sqrt7}\), \(x+\alpha=\ln\left(\dfrac6{\sqrt7}+\sqrt{\dfrac{36}7+1}\right)=\ln\dfrac{6+\sqrt{43}}{\sqrt7}\), so \(x=\ln\dfrac{6+\sqrt{43}}{\sqrt7}-\tfrac12\ln7=\ln\dfrac{6+\sqrt{43}}7\approx0.585\).
    3. \(\dfrac{4(u-1/u)+3(u+1/u)}2=6\) gives \(7u-\dfrac1u=12\), \(7u^2-12u-1=0\), \(u=\dfrac{12+\sqrt{172}}{14}=\dfrac{6+\sqrt{43}}7=1.794\). So \(x=\ln1.794=0.585\) ✓.
    4. Here the coefficient of \(\cosh x\) is larger than that of \(\sinh x\), so write \(4\cosh x+3\sinh x=\sqrt7\cosh(x+\beta)\) with \(\tanh\beta=\tfrac34\). Since \(\cosh(x+\beta)\ge1\), the left side is at least \(\sqrt7\approx2.65\), which is greater than 2. So there is no real solution.

18. Answer to Question 18

    1. \(\dfrac{dy}{dx}=\cosh x-2\), which is 0 when \(\cosh x=2\).
    2. \(x=\pm\ln(2+\sqrt3)\). With \(\cosh x=2\), \(\sinh x=\pm\sqrt3\). The points are \(\left(\ln(2+\sqrt3),\ \sqrt3-2\ln(2+\sqrt3)\right)\approx(1.317,-0.902)\) and \(\left(-\ln(2+\sqrt3),\ 2\ln(2+\sqrt3)-\sqrt3\right)\approx(-1.317,0.902)\).
    3. \(\dfrac{d^2y}{dx^2}=\sinh x\). It is \(+\sqrt3>0\) at the first point (minimum) and \(-\sqrt3<0\) at the second (maximum).
    4. The curve passes through the origin, rises to a local maximum at \(x=-1.317\), falls through a local minimum at \(x=1.317\), then rises. It crosses the \(x\)-axis at 0 and then twice more (at \(x\approx\pm2.18\)), so \(\sinh x=2x\) has 3 real roots.

19. Answer to Question 19

    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.

    1. \(\displaystyle\int_0^1\cosh x\,dx=\sinh1=\) \(\dfrac{e-e^{-1}}2\approx1.175\).
    2. \(\dfrac{dy}{dx}=\sinh x\) and \(1+\sinh^2x=\cosh^2x\). Length \(=\displaystyle\int_0^1\cosh x\,dx=\) \(\sinh1\).
    3. The area and the arc length are equal, a special property of the catenary.
    4. \(V=\pi\displaystyle\int_0^1\cosh^2x\,dx=\dfrac\pi2\int_0^1(1+\cosh2x)dx=\dfrac\pi2\left[x+\dfrac{\sinh2x}2\right]_0^1=\) \(\dfrac\pi2\left(1+\dfrac{\sinh2}2\right)\approx4.418\).

20. Answer to Question 20

    1. \(\sinh x=2\sinh\tfrac x2\cosh\tfrac x2=\dfrac{2\sinh\frac x2\cosh\frac x2}{\cosh^2\frac x2-\sinh^2\frac x2}\) because the denominator equals 1. Dividing top and bottom by \(\cosh^2\tfrac x2\) gives \(\dfrac{2t}{1-t^2}\). Similarly \(\cosh x=\dfrac{\cosh^2\frac x2+\sinh^2\frac x2}{\cosh^2\frac x2-\sinh^2\frac x2}=\dfrac{1+t^2}{1-t^2}\).
    2. \(t=\tanh\tfrac x2\) gives \(\dfrac{dt}{dx}=\tfrac12\operatorname{sech}^2\tfrac x2=\tfrac12(1-t^2)\), so \(\dfrac{dx}{dt}=\dfrac2{1-t^2}\).
    3. \(\operatorname{sech}x=\dfrac{1-t^2}{1+t^2}\), so \(\displaystyle\int\operatorname{sech}x\,dx=\int\dfrac{1-t^2}{1+t^2}\cdot\dfrac2{1-t^2}dt=\int\dfrac2{1+t^2}dt=2\arctan t+C\).
    4. As \(x\to\infty\), \(t\to1\), so the integral from 0 is \(2\arctan1-2\arctan0=\) \(\dfrac\pi2\approx1.5708\). Numerically \(\int_0^6\operatorname{sech}x\,dx=2\arctan(\tanh3)=1.5666\), very close to \(1.5708\).