Damnit. On preview the letters didn’t work out right (I used the Windows character map). Anyways, just substitute è for theta and
ö for phi.
------------ Heres my original post:
There’s also another neat thing with the hyperbolic trigs.
For the unit circle, if you draw a line from the origin to the circle, the area of the sector of that circle is given by:
x/(2*Pi) * Pi = x/2 where x = è
For the unit hyperbola, if you draw a line from the origin to a point (cosh(è), sinh(è)), the area between the line and hyperbola/x-axis is also è/2.
To show that the area is è/2:
The slope of the line from the origin to the point (cosh(è),sinh(è)) is m = tanh(è) and the equation is y = xtanh(è) ==> x = y/tanh(è) = ycoth(è). Let x2 be a point on the hyperbola and x1 be a point on the line. Use a horizontal area element for integration, which is:
dA = (x2 - x1)dy = (sqrt(1+y^2) - y*coth(è))dy
Integrate and then,
A = Integral[ (sqrt(1 + y^2) - y*coth(è))dy, {0, sinhè}]
Read: A is the integral of (sqrt(1 + y^2) - y*coth(è))dy from 0 to sinhè (where the hyperbola and line meet). Note: it’s sinhè and not coshè because a horizontal strip is being used.
First, split up the integral:
A = INT[ sqrt(1+y^2)dy ] - cothèINT[ ydy ] (both on the same interval)
= INT[ sqrt(1 + y^2)dy ] - cothè(y^2/2)
= … - cothè*(sinh^2(è) - 0)/2
= INT[ sqrt(1 + y^2)dy ] - 1/2cothèsinh^2(è)
Now, substitute y = sinhö <==> dy = coshö*dö.
y = 0 corresponds to ö = 0 and y = sinhè corresponds to ö = è.
Now the area is:
A = Integral[ (coshöcoshödö), {0, è}] - 1/2cothèsinh^2(è)
= INT[ cosh^2ö*dö ] - …
but cosh2ö = 2cosh^2ö - 1 (another interesting property of the hyperbolic trigs – sharing similar identities, although not for all identities, like when you have an implied or explicit product of two sines, which is linked to cosx - isin(x) because (isin(x))^2 = -sin^2(x).
anyways…
A = INT[1/2*(1 + cosh2ö)] - …
= 1/2[ö + 1/2sinh2ö] on [0,è] - 1/2coshè/sinhè*sinh^2(è)
= è/2 + 1/4sinh2è - 1/2coshèsinhè
= è/2 + 1/4(2coshèsinhè) - 1/2coshèsinhè
= è/2
And yes, I know I didn’t type all that out too well (especially on writing out my intervals, and skipping it altogether several times), but it’s still understandable and the result is interesting.
A summary of the hyperbolic trigs:
- The area between the graph and a line from the origin are related to that of a circle.
- The hyperbolic trigs have identities very similar to those of circular trigs (you can find one from the other using Osborne’s Rule).
- cosh(x) is the even part of e^x; sinh(x) is the odd part of e^x.
- You can express the inverse hyperbolics in terms of logs using a bit of algebra.
- The antiderivatives of several common functions involve hyperbolic trigs (look up an antiderivative table).
… and there are more I’m sure.
Also, the curve y = cosh(x) is called a caternary, and is the shape of a hanging heavy wire – it minimizes the potential energy of the wire. At least, that’s what our prof said. There are also some interesting relationships if you take the solid of revolution about the x-axis of cosh(x).