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Homework Help: How to find the equation for an equation from a hyperbolic graph

  1. Apr 24, 2010 #1
    I need to do a project where i hang a chain from two points and find the hyperbolic function for it. The problem is, its a complete do it yourself project and i we never learned hyperbolic functions in class. That means i need to learn how to find hyperbolic fuctions from a set of data that i collect myself without even knowing the first thing about hyperbolic functions.

    My question is, where can i learn all the basics about hypebolic functions enough so to learn how to fit an equation?
  2. jcsd
  3. Apr 24, 2010 #2
  4. Apr 24, 2010 #3
    How exactly do these hyperbolic equations involving e relate to those that are in the form:

    [tex]\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1[/tex]
  5. Apr 24, 2010 #4
    wait a second, this shouldnt be in calculus and beyond, im only in precalc, can someone delete or lock this. or just move it to the correct place?
  6. Apr 24, 2010 #5
    If you let x = cosh t and y = sinh t, you get the equation for a hyperbola [tex]y^2-x^2=1[/tex].

    Assuming that the two points are at the same potential in an uniform gravitational field, the hanging chain's shape can be shown to be of the form y = a cosh bx, so, you just have to gather enough data points to solve for a and b in this equation. You could also derive an explicit formula for a and b given the length of the chain and the distance between the two points, but this requires calculus.
  7. Apr 24, 2010 #6
    Okay, so how exactly could i form an equation for hyperbolic cosine using data that i collect?

    Secondly, i dont understand how the a value over 2 in the front of the equation and the a value under the x and -x can be the same.
    In this equation dont you need those two variables to be different in order to stretch the graph how you want? Why are the values the same?
    Last edited: Apr 24, 2010
  8. Apr 24, 2010 #7
    Well, you have to start with a general equation and fit it to your data. For example, y = a cosh bx. Take any two points from your data, and you have two equations:
    [tex]y_1 = a cosh bx_1[/tex]
    [tex]y_2 = a cosh bx_2[/tex]
    These can easily be combined into one equation which can be solved numerically. I can't think of any algebraic way to solve it.

    If you experiment, you'll see that the shape of the chain always remains the same, only the scale and height changes.

    If x is half the distance between the points, there are infinitely many solutions (y,a), corresponding to various lengths of the chain, to the equation [tex]y=a cosh\frac x a[/tex], since this is one equation in two variables, y and a.
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