Applications: Hyperbolic Functions

A hanging cable is a catenary, y = acosh(xa), and the number a is the height of its lowest point above the level the model measures from. The sag is therefore a difference of two heights, not the constant a itself, and the only piece of arithmetic needed is cosh of the half-span divided by a, which is the average of two exponentials.

1. Name the parts of the model. Here a = 100 m. The middle of the span is x = 0, and since cosh 0 = 1 the cable is y = 100 × 1 = 100 m up at its lowest point.
2. The pylons are 100 m apart, so each fixing sits at x = 50 m from the middle, and the number inside the cosh is 50100 = 0.5. Work cosh 0.5 out from the two exponentials it is built from: cosh 0.5 = e^0.5 + e^−0.52 = 1.64872 + 0.606532 = 1.12763.
3. The height of the cable at a fixing is y = 100 × 1.12763 = 112.763 m.
4. (a) The sag is the height at the fixing less the height at the lowest point: 112.763 − 100 = 12.763 m, which is 12.8 m to three significant figures.
5. (b) The fixing is 30 m above the road and the lowest point of the cable hangs 12.763 m below that fixing, so it is 30 − 12.763 = 17.237 m above the road, or 17.2 m to three significant figures. Check: 17.237 + 12.763 = 30 m, which is the height of the fixing, as it must be.

answer(a) 12.8 m; (b) 17.2 m above the road

techniquesinh, cosh and tanh

The sag is again a difference of two heights. The length is the arc length of the curve, and that is where the catenary repays the work: its gradient is a sinh, the identity cosh² u − sinh² u = 1 turns the square root in the arc-length formula into a cosh with no approximation at all, and a cosh integrates back to a sinh.

1. Each post stands at x = 12 m from the middle, so the number inside the cosh is 1220 = 0.6. From the exponentials, cosh 0.6 = e^0.6 + e^−0.62 = 1.82212 + 0.548812 = 1.18547.
2. (a) The chain is 20 × 1.18547 = 23.709 m up at each post and 20 m up at its lowest point, so it sags 23.709 − 20 = 3.709 m, which is 3.71 m to three significant figures.
3. For the length, differentiate the model. The derivative of cosh u is sinh u, so dydx = sinh(x20), and the arc-length integrand is √1 + sinh²(x20). The identity cosh² u − sinh² u = 1 says that root is exactly cosh(x20).
4. Integrating cosh(x20) gives 20sinh(x20), so the length from the middle out to one post is 20sinh 0.6 and the whole chain is twice that. From the exponentials, sinh 0.6 = 1.82212 − 0.548812 = 0.63665.
5. (b) The chain is 2 × 20 × 0.63665 = 25.466 m long, which is 25.5 m to three significant figures. That is 25.466 − 24 = 1.466 m more than the 24 m straight across, so 1.47 m of extra chain. Check: a hanging chain must be longer than the straight line between its ends, and it is.

answer(a) a sag of 3.71 m; (b) 25.5 m of chain, which is 1.47 m more than the 24 m between the posts

techniqueDifferentiating sinh x and cosh x · The Identity cosh²x − sinh²x = 1

At any point the cable pulls along itself, and that pull is the horizontal and vertical pulls combined at right angles. The horizontal pull never changes, so the whole story is in the vertical one, which is a sinh. Combining them would need a square root, but the identity cosh² u − sinh² u = 1 does the root exactly and leaves a cosh.

1. At the lowest point x = 0, and sinh 0 = 0, so there is no vertical pull there at all. (a) The whole pull at the lowest point is the horizontal pull, 12 × 25 = 300 N.
2. At a support x = 20, so the number inside the hyperbolic functions is 2025 = 0.8. From the exponentials, sinh 0.8 = 2.22554 − 0.449332 = 0.88811 and cosh 0.8 = 2.22554 + 0.449332 = 1.33743.
3. The vertical pull at the support is 300 × 0.88811 = 266.4 N, and the horizontal pull there is still 300 N. The two act at right angles, so the pull along the cable is √300² + 266.4².
4. Do that root with the identity rather than with a calculator. Since 300² + (300sinh 0.8)² = 300²(1 + sinh² 0.8) = 300²cosh² 0.8, the pull along the cable is simply 300cosh 0.8.
5. (b) The pull at the support is 300 × 1.33743 = 401.2 N, which is 401 N to three significant figures. Check with Pythagoras' theorem: 300² + 266.4² = 90000 + 70969 = 160969, and √160969 = 401.2 N, the same figure.

answer(a) 300 N; (b) 401 N, made up of a horizontal pull of 300 N and a vertical pull of 266.4 N

techniqueThe Identity cosh²x − sinh²x = 1 · sinh, cosh and tanh

The gradient of a catenary is the tidiest derivative in the subject: differentiating acosh(xa) gives sinh(xa), because the a outside and the 1a from the chain rule cancel. The angle then follows without a second calculation, since the slope triangle has sides 1, sinh u and cosh u, so its sine is tanh u.

1. Differentiate the model. The derivative of cosh u is sinh u, and the chain rule brings out a factor 136, so dydx = 36 × 136sinh(x36) = sinh(x36).
2. At an anchor x = 27, so the number inside the sinh is 2736 = 0.75. From the exponentials, sinh 0.75 = 2.11700 − 0.472372 = 0.82232.
3. (a) The gradient of the cable at the anchor is 0.822 to three significant figures. It is a pure number, a rise of 0.822 m for every meter along.
4. For the angle, draw the slope triangle: it runs 1 across and sinh 0.75 up, so its hypotenuse is √1 + sinh² 0.75, which the identity makes exactly cosh 0.75. Therefore sinθ = sinh 0.75cosh 0.75 = tanh 0.75.
5. (b) tanh 0.75 = 0.822321.29468 = 0.635, so θ = 39.4^° to the nearest tenth of a degree. Check the same angle from the gradient itself: tanθ = 0.82232 also gives θ = 39.4^°.

answer(a) a gradient of 0.822; (b) 39.4^°, since sinθ = tanh 0.75 = 0.635

techniqueDifferentiating sinh x and cosh x · The Identity cosh²x − sinh²x = 1

A resistance that grows with the square of the speed gives a tanh law, and the whole behavior of the boat is the behavior of that one function. Evaluating it is a matter of two exponentials; the settled speed is read off the fact that tanh u climbs toward 1 and never arrives, so the boat has a speed it approaches but never attains.

1. (a) After 8 seconds the number inside the tanh is 88 = 1. Work tanh 1 out from the exponentials: tanh 1 = e¹ − e⁻¹e¹ + e⁻¹ = 2.71828 − 0.367882.71828 + 0.36788 = 2.350403.08616 = 0.76159.
2. The speed is therefore v = 12 × 0.76159 = 9.139, which is 9.14 meters per second to three significant figures.
3. Now look at what tanh u does as u grows. Multiplying the top and the bottom of e^u − e^−ue^u + e^−u by e^−u gives 1 − e^−2u1 + e^−2u. As u grows, e^−2u tends to 0, so tanh u climbs toward 1 and stays below it for every u.
4. (b) The speed is therefore always below 12 × 1 = 12 meters per second, and it closes in on that figure as the seconds pass. The launch settles at 12 meters per second, which it approaches but never reaches.
5. After 16 seconds the number inside the tanh is 2, and tanh 2 = 1 − e⁻⁴1 + e⁻⁴ = 1 − 0.0183161 + 0.018316 = 0.96403, so v = 12 × 0.96403 = 11.568, or 11.6 meters per second. That is 96.4% of the settled speed. Check: 11.568 is below 12, as every speed this model gives must be.

answer(a) 9.14 meters per second; (b) it settles at 12 meters per second, and after 16 seconds it is at 11.6 meters per second, which is 96.4% of that

techniquesinh, cosh and tanh

The formula has two factors: a deep-water part that depends on the wavelength alone, and a tanh that carries everything the depth does. Since tanh of a large number is as good as 1, deep water loses the second factor entirely, and the shelf's effect is read off as the square root of that factor.

1. Work the two pieces out separately. The first is gλ2π = 9.8 × 602π = 5886.28319 = 93.583. The number inside the tanh is 2π × 1260 = 2π5 = 1.25664.
2. Evaluate the tanh from its exponentials, in the form that is easiest here: tanh 1.25664 = 1 − e^−2.513271 + e^−2.51327 = 1 − 0.0810031 + 0.081003 = 0.85013.
3. (a) So c² = 93.583 × 0.85013 = 79.558 and c = √79.558 = 8.9195. The swell travels at 8.92 meters per second.
4. In deep water the depth is large beside the wavelength, so 2π dλ is large and tanh of it differs from 1 by less than the measurements do. The formula then loses the tanh: c² = gλ2π = 93.583, so c = √93.583 = 9.6738 meters per second.
5. (b) In deep water the same swell would run at 9.67 meters per second. Over the shelf it runs at 8.91959.6738 = 0.92203 of that, so the shelf slows it by 7.80%. Check: 0.92203 is the square root of 0.85013, which is exactly what taking a square root of the tanh factor should give.

answer(a) 8.92 meters per second; (b) 9.67 meters per second in deep water, so the shelf slows it by 7.80%

techniquesinh, cosh and tanh

The parabola is not a guess. Expanding each exponential in cosh u = e^u + e^−u2 cancels every odd power and leaves 1 + u²2 + u⁴24 + …, so keeping the first two terms turns the catenary into exactly the draughtsman's parabola. The error is then whatever the terms that were thrown away are worth at the end of the span.

1. See where the parabola comes from. Expanding both exponentials gives cosh u = 1 + u²2 + u⁴24 + …, because every odd power appears once with a plus and once with a minus and cancels.
2. Keep the first two terms only, with u = x40: 40cosh(x40) ≈ 40(1 + x²3200) = 40 + x²80. That is the draughtsman's parabola exactly, so the parabola is the catenary with everything past the squared term thrown away.
3. (a) The catenary's sag at x = 30 is 40(cosh 0.75 − 1). With cosh 0.75 = 2.11700 + 0.472372 = 1.29468, that is 40 × 0.29468 = 11.787 m, which is 11.8 m to three significant figures.
4. The parabola's sag at the same place is 30²80 = 90080 = 11.25 m, since the parabola starts at 40 m as well.
5. (b) The parabola falls short by 11.787 − 11.25 = 0.537 m, about 4.6% of the true sag. Check by putting the next term back: 40 × 0.75⁴24 = 40 × 0.013184 = 0.527 m, and 11.25 + 0.527 = 11.777 m, within a centimeter of the true 11.787 m.

answer(a) 11.8 m; (b) the parabola gives 11.25 m, short by 0.537 m

techniquesinh, cosh and tanh · Differentiating sinh x and cosh x

Reading a height off the model is direct; reading a position back out of it is the inverse problem, and that is what arcosh is for. Written as a logarithm it needs nothing but a square root and a ln. The length along the cable then comes free, because the identity turns it into a difference of two squares.

1. Put the height into the model and make the cosh the subject: 45cosh(x45) = 53, so cosh(x45) = 5345.
2. Turn that round with the inverse hyperbolic cosine, and write it as a logarithm: x45 = arcosh(5345), and arcosh u = ln(u + √u² − 1) for u ≥ 1.
3. Work the root out exactly. (5345)² − 1 = 2809 − 20252025 = 7842025, whose square root is 2845. So u + √u² − 1 = 53 + 2845 = 8145 = 1.8.
4. (a) Therefore x45 = ln 1.8 = 0.587787, and x = 45 × 0.587787 = 26.450 m, which is 26.5 m to three significant figures.
5. (b) The length along the cable from the lowest point is 45sinh(x45), and with e^x/45 = 1.8 that is 45 × 1.8 − 11.82 = 45 × 0.62222 = 28 m. Check with the identity: multiplying cosh² u − sinh² u = 1 by 45² gives 53² − (45sinhx45)² = 45², so the length squared is 2809 − 2025 = 784 and the length is exactly 28 m.

answer(a) 26.5 m; (b) exactly 28 m of cable

techniqueInverse Hyperbolics as Logarithms · Inverse Hyperbolic Functions · The Identity cosh²x − sinh²x = 1

Rapidity is what the inverse hyperbolic tangent is for. Because artanh turns a speed into a number that simply adds, the whole calculation is three lines of logarithms: one for each boost, one for the total. The last step turns that total back into a speed with tanh, and since tanh of any number lies between −1 and 1, the answer is a speed below c however many boosts are applied.

1. Turn the definition round. From vc = tanh w the rapidity is w = artanh(vc), and the logarithm form of that is artanh k = 12ln1 + k1 − k for −1 < k < 1.
2. The first boost has k = 0.6, so w₁ = 12ln1.60.4 = 12ln 4 = ln 2 = 0.6931.
3. The second boost has k = 0.8, so w₂ = 12ln1.80.2 = 12ln 9 = ln 3 = 1.099.
4. (a) Rapidities add, so the total is w = ln 2 + ln 3 = ln 6 = 1.792.
5. (b) Turn the total back into a speed. With e^w = 6, tanh w = 6 − 166 + 16 = 3537 = 0.946, so the particle ends at 3537c, about 0.946c. Check against the usual addition law: 0.6 + 0.81 + 0.6 × 0.8 = 1.41.48 = 3537, the same figure.

answer(a) ln 2 = 0.6931 and ln 3 = 1.099, which add to ln 6 = 1.792; (b) 3537c, which is 0.946c

techniqueInverse Hyperbolics as Logarithms · sinh, cosh and tanh

A circle is parametrized by cosine and sine because cos² + sin² = 1; a hyperbola is parametrized by cosh and sinh for the same reason, because cosh² u − sinh² u = 1. Once the parametrization is in place, moving between a height and a radius is a matter of inverting one hyperbolic function and then using the identity to get the other.

1. Check that the parametrization fits. Putting x = 30cosh u and y = 50sinh u into the left-hand side gives cosh² u − sinh² u, which the identity says is 1 for every u, so every value of u names a point of the outline.
2. (a) At a height of 40 m, 50sinh u = 40, so sinh u = 0.8.
3. Get cosh u from the identity rather than from a calculator: cosh² u = 1 + sinh² u = 1 + 0.64 = 1.64, so cosh u = √1.64 = 1.28062, taking the positive root because cosh u is never negative. The radius there is 30 × 1.28062 = 38.419 m, or 38.4 m to three significant figures.
4. (b) Now the radius is the given: 30cosh u = 45, so cosh u = 1.5. The identity gives sinh² u = 1.5² − 1 = 1.25, so sinh u = √1.25 = 1.11803 for the part of the tower above the waist.
5. The height is 50 × 1.11803 = 55.902 m, so 55.9 m to three significant figures. Check it against the equation of the outline: 45²30² − 55.902²50² = 2.25 − 1.25 = 1, as it should be. In logarithm form the two parameters are arsinh 0.8 = ln(0.8 + 1.28062) = 0.7327 and arcosh 1.5 = ln(1.5 + 1.11803) = 0.9624.

answer(a) 38.4 m; (b) 55.9 m above the waist

techniqueThe Identity cosh²x − sinh²x = 1 · Inverse Hyperbolic Functions · Inverse Hyperbolics as Logarithms