Hyperbolic Cosine curve fitting

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Homework Help Overview

The original poster attempts to fit a hyperbolic cosine curve to a hyperbola defined by a vertex at (0,0) and a point at (4,7). The problem involves using the equation for a hyperbola and determining the constant that modifies its shape.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the algebraic attempts to solve for the constant 'a' by substituting known values into the hyperbolic cosine equation. Some suggest graphing the function to find the value of 'a' and question the nature of the equation, noting it describes a catenary rather than a hyperbola.

Discussion Status

The discussion includes various approaches to finding the value of 'a', with some participants suggesting graphing as a method to aid in the solution. There is an acknowledgment that the equation may not have a closed-form solution, and participants are exploring the implications of this observation.

Contextual Notes

Participants note that the solution value of 'a' is not an integer, and there is a discussion about the characteristics of the equation being used, specifically its classification as a catenary rather than a hyperbola.

Denyven
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Homework Statement


I need to fit a curve using cosh to a hyperbola with a vertex of (0,0) and a point at (4,7).

The scanned worksheet can be found here
http://img519.imageshack.us/i/scan0001gu.jpg/"
http://img192.imageshack.us/i/scan0002uz.jpg/"

Homework Equations


y=a cosh (\frac{x}{a})-a=\frac{a}{2}(e^{\frac{x}{a}}+e^{\frac{-x}{a}})-a "is the formula for a hyperbola at a vertex of 0,0. a is a constant that modifies the shape" that is what the assignment said exactaly

The Attempt at a Solution


I plugged in 4 for x and 7 for y and attempted to solve algebraically, I just got stuck. then I plugged the equation I attempted to solve into wolfram alpha and got the response no integer solution.

http://www.wolframalpha.com/input/?i=7%3D(a/2)(e^(4/a)%2Be^(-4/a))%2Ba+solve+for+a
 
Last edited by a moderator:
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The solution value of a isn't an integer.

Try graphing the function

<br /> y=x cosh (\frac{4}{x})-x-7<br />

and use that to help you find the right value of a.
 
There is probably no closed-form solution, and besides, this formula doesn't describe a hyperbola, but a catenary.
 
hgfalling said:
The solution value of a isn't an integer.

Try graphing the function

<br /> y=x cosh (\frac{4}{x})-x-7<br />

and use that to help you find the right value of a.

Thanks so much for your help! I graphed that, and since you solved to equation for 0 then turned 0 into y I looked for the x intercept of the graph and the value fit, thank you so much!
But just for the sake of knowing, how can you guys tell if an equation has no closed form solution?
 
Last edited:

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