I was thinking again about the bent rail problem from a couple of days ago, and got to wondering what the height would be if it was an inverted catenary arc instead of a circular arc. I was led to this idea by one of my students who also had suggested that it might NOT be circular....and I had no way to assure him it would be... I have no reason to assume it is a catenary either, but it was a curve I felt I needed to know more about so I used it as a self-teaching exercise.
At that point I realized I didn't know much about the catenary curve in general. (what I mostly knew is about its history and can be found here). So I set out to see if I could learn a little more and figure out how to solve the question of how high would the inverted catenary reach. Below is what I have attempted to do... the answer is close enough to the circular amount of bowing that I wonder if it can be right, so if you are up on this stuff, please comment.
I knew that the catenary curve y= a cosh(x/a) has a minimum at (0,a). For students who haven't seen "cosh" that's just a hyperbolic cosine function (much like the circular cosine function in many ways, and often pronounced to rhyme with "gosh"... see an explanation of hyperbolic functions here). It can be replaced with an alternate form that may be more palatable to your eyes:
Of course it is easy to see that the vertex is at (0,a). What I had never known, and find particularly wonderful about the curve, is that for any catenary, the area under the curve from x=c to x=d divided by the constant a of the equation is the length of the curve.....A good calculus student can show this is true pretty easily with the exponential form since the derivative of cosh(x) is sinh(x).... and for hyperbolas we have cosh^2 - sinh^2 = 1 instead of the addition. That makes the integral for arc length the same as the integral for area between the curve and the x-axis.
creation of the word scientist. I just found that there is also a Whewell equation for the catenary which gives the slope of the tangent of the catenary as a function of the distance from the origin, s. Tan(T)= s/a. That means in the catenary version of our inverted rail, the angle between the curve and the horizontal secant is just over .03 radians.... I think I did that correctly.