I posted this question on the physics stack exchange, but I have got no answers in a few days. I was hoping that the mathematicians would be more helpful! I am trying to solve exercise 7 from chapter 2 of Goldstein, Poole, and Safko's Classical Mechanics, 3rd edition. I am getting the exact opposite of what it is asking and I don't know why. Can someone please help? The problem statement is:
In example 2 of section 2.2, we considered the problem of the minimum surface of revolution. Examine the symmetric case $ x_1=x_2 $, $ y_2=-y_1>0 $, and express the condition for the parameter $ a $ as a transcendental equation in terms of the dimensionless quantities $ k=x_2/a $ and $ \alpha=y_2/x_2 $. Show that for $ \alpha $ greater than a certain value $ \alpha_0 $ two values of $ k $ are possible, for $\alpha=\alpha_0 $ only one value of $ k $ is possible, while if $ \alpha<\alpha_0 $ no real value of $ k $ (or $ a $) can be found, so that no catenary solution exists in this region. Find the value $ \alpha_0 $, numerically if necessary.
Using the given relations and parameters $ x_1=x_2,y_1=-y_2,\alpha=y_2/x_2,k=x_2/a $, we have $$ y_1=-y_2=-a\cosh^{-1}(\frac{x_1}{a})-b $$ $$ y_2=a\cosh^{-1}(\frac{x_1}{a})+b $$ $$ x_1=a\cosh(\frac{-y_2-b}{a}) $$ $$ x_2=a\cosh(\frac{y_2-b}{a}) $$ It follows that $ b=0 $. Then using the given parameter relationships, we find that $$ k=\cosh(\alpha k) $$ $$ \alpha=\frac{\cosh^{-1}(k)}{k}. $$ We can plot $ \alpha $ as a function of $ k $ (see below), and noting that in the limit as $ k\rightarrow\infty $, $ \alpha\rightarrow0 $, we see that for $ \alpha $ less than the plot maximum, there are 2 values of $ k $ for each $ \alpha $. When $ \alpha $ equals the plot maximum, there is only one $ k $, and anything greater than this value, there are no real $ k $ (or $ a $). Solving numerically in Mathematica, we find the plot maximum to be about $ 0.6627 $, which is our $ \alpha_0 $.
So this is the opposite of what was expected! Can someone point out my errors, please! Thank you!
You are right! Just get a Manipulate plot of the function
$$f(k)=k-\cosh(\alpha k)$$
in the manipulated variable $\alpha$. Then observe how many zeros are there.You will observe that a symmetric range is obtained, i.e. $-\alpha_0<\alpha<\alpha_0$ in which $\alpha_0$ is the value you have found and is such a range there are twho solutions in the $k$ variable. So to say, the book is wrong.