Solving a System of Inverse Hyperbolic Trigonometry Equations ... with Natural Logs and the Quadratic EquationDate: 03/28/2010 at 01:50:57 From: Ben Subject: solve 2 simultaneous eqns w/ 2 unknowns involving cosh How would one solve the following set of simultaneous equations? 7 = a cosh(-x/a) - a 10 = a cosh((5 - x)/a) - a There are 2 equations and 2 unknowns, so I figure that it is possible, but I run into trouble when trying to tackle the simultaneous equations with my usual methods. cosh(x) = (e^x + e^-x) / 2; but when I attempt to solve the simultaneous equations, my usual go-to methods fail -- namely, multiplying one equation by a scalar, then adding together; and solving for one variable and substituting. It all boils down to how one cannot pull an x out of the cosh function, with the e^x and e^-x. I'm stuck. Date: 03/28/2010 at 20:29:09 From: Doctor Vogler Subject: Re: solve 2 simultaneous eqns w/ 2 unknowns involving cosh Hi Ben, Thanks for writing to Dr Math. That's an interesting question. In the end, I don't think you'll be able to solve this algebraically. But your observation about what it "boils down to" does suggest a way forward. Again, I don't think this will lead to an algebraic solution, but it should make calculating a numerical solution much faster. To compute y = arccosh(z), or to solve the equation cosh(y) = z for the variable y, you proceed as follows: cosh(y) = z 2 cosh(y) = 2z e^y + e^(-y) = 2z (e^y)^2 + 1 = 2z e^y (e^y)^2 - 2z(e^y) + 1 = 0 This is a quadratic equation in e^y, the solution of which, according to the quadratic formula, is e^y = z +/- sqrt(z^2 - 1). Notice that e^-y = 1/(e^y), and 1 / [z + sqrt(z^2 - 1)] = z - sqrt(z^2 - 1), So the two solutions are ... ln [z + sqrt(z^2 - 1)] ... and its negative, ln [z - sqrt(z^2 - 1)]. To sum up, we have ... y = arccosh(z) = ln [z + sqrt(z^2 - 1)], ... which is the positive inverse cosh (much like square root, this usually refers to the *positive* square root); and the other value, the cosh of which is z, is -y. If you have any questions about this or need more help, please write back and show me what you have been able to do, and I will try to offer further suggestions. - Doctor Vogler, The Math Forum http://mathforum.org/dr.math/ |
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