Hyperbolic Trigonometry: The Hanging Chain

Section 8.4 Hyperbolic Trigonometry: The Hanging Chain

We commented in Section 5.10 that Galileo believed erroneously that a chain hanging from two pegs falls naturally into the shape of a parabola. In Problem 5.10.3 we showed that Galileo was wrong but we have not yet addressed the question: What is the shape of a hanging chain?

Recall that the hanging chain must satisfy the differential equation

\begin{equation*} \dfdxn{y}{x}{2} = \frac{w}{H}\sqrt{1+\left(\dfdx{y}{x}\right)^2} \end{equation*}

where \(w\) is the weight density of the chain and \(H\) is the (constant) magnitude of the horizontal tension. In Problem 8.4.1 we will see that although the natural exponential function \(y=e^x\) is not the solution itself, it is the key to the solution.

Problem 8.4.1. The Shape of a Hanging Chain.

The Calculus in this problem is pretty straightforward but the Algebra gets a bit messy. To avoid some of this mess we will solve the special case where \(\frac{w}{H}=1\) in parts (a), (b), and (c). In part (d) we’ll use our solution of this special case to solve the original problem.

So assume that \(\frac{w}{H}=1\text{.}\) For this special case we will call our variables \(X\) and \(Y\text{,}\) rather than \(x\) and \(y\text{.}\) The reasons for this will become clear in part (d).

(a)

Show that

\begin{equation*} 1+\left(\frac{e^X-e^{-X}}{2}\right)^2=\left(\frac{e^X+e^{-X}}{2}\right)^2\text{.} \end{equation*}

Hint.

It might help to make the substitution \(e^X=a\) just to make things “easier on the eyes.” That makes \(e^{-X}=\frac1a.\)

(b)

Show that

\begin{equation*} \dfdx{\left(\frac12\left(e^X+e^{-X}\right)\right)}{X} =\frac12 \left(e^X-e^{-X}\right) \end{equation*}

and that

\begin{equation*} \dfdx{\left(\frac12\left(e^X-e^{-X}\right)\right)}{X} =\frac12 \left(e^X+e^{-X}\right). \end{equation*}

(c)

Show that the curve \(Y=\frac12 \left( e^X+e^{-X}\right) \) satisfies the differential equation

\begin{equation*} \dfdxn{Y}{X}{2}=\sqrt{1+\left(\dfdx{Y}{X} \right)^2 }. \end{equation*}

(d)

Now let’s tackle the original problem. Show that

\begin{equation} y=\frac{H}{w} \left(\frac{e^{\frac{wx}{H}}+e^{-\frac{wx}{H}}}{2}\right)\tag{8.14} \end{equation}

satisfies the equation

\begin{equation*} \dfdxn{y}{x}{2}=\frac{w}{H} \sqrt{1+\left(\dfdx{y}{x}\right)^2} \end{equation*}

where \(w\) is the weight density of the chain, and \(H\) is the constant (magnitude of the) horizontal tension.

Hint.

This is essentially the same computation you did for the special case, but the Algebra is messier. Consider making it “easier on the eyes” with the substitutions

\begin{equation*} X=\frac{wx}{H} \end{equation*}

and

\begin{equation*} Y(X)=\frac12\left(e^X+e^{-X}\right). \end{equation*}

We already know from part 8.4.1.c that \(\dfdxn{Y}{X}{2}=\sqrt{1+\left(\dfdx{Y}{X}\right)^2}\text{.}\) Notice that \(y=\frac{H}{w}Y(X)\) and use this to show that \(\dfdx{y}{x}=\dfdx{Y}{X}\text{ and that } \dfdxn{y}{x}{2}=\frac{w}{H}\dfdxn{Y}{X}{2}.\)

(e)

Assume that \(w=1 \text{,}\) and \(H=1\) and graph the curve given in equation (8.14). Does it look like a hanging chain? What happens to the graph if we use \(w=1\) and \(H=2\text{,}\) or \(w=2\) and \(H=1\text{?}\) Does this make sense physically? Why or why not?

The expressions \(\frac12\left(e^x+e^{-x}\right)\) and \(\frac12\left(e^x-e^{-x}\right)\) arise frequently in many scientific and engineering problems and have been named the hyberbolic cosine and hyberbolic sine and are denoted \(\cosh(x)\) and \(\sinh(x)\text{,}\) respectively.These names break with the longstanding mathematical tradition of giving important functions three letter names. Why do you suppose we would do that?

Drill 8.4.2.

Recall that the unit circle, \(x^2+y^2=1\text{,}\) is parameterized by

\begin{equation*} P(t)= \ParamEqThree {\cos(t)} {\sin(t)} { 0\leq t\lt 2\pi} \end{equation*}

since

\begin{equation*} x^2+y^2= \cos^2(t)+\sin^2(t)=1. \end{equation*}

Analogously, show that the unit hyperbola, \(x^2-y^2=1\) is parameterized by

\begin{equation*} P(t)= \begin{Bmatrix} \cosh(t)\\ \sinh(t)\\ {-\infty\lt t\lt \infty} \end{Bmatrix}. \end{equation*}

That is, show that \(x^2-y^2=\cosh^2(t)-\sinh^2(t)=1\text{.}\)

The obvious similarity between the formulas

\begin{align*} \cos^2(x)+\sin^2(x)=1\amp{}\amp{} \text{and} \amp{}\amp{}\cosh^2(x)-\sinh^2(x)=1 \end{align*}

is not a coincidence. It also explains why they are called Hyperbolic Trigonometric Functions.

The many scientific and engineering problems that involve the hyperbolic functions are slightly outside the scope of this text. So we will only examine some of their more elementary properties in the next problem.

Definition 8.4.3. The Hyperbolic Trigonometric Functions.

In addition to \(\cosh(x)\) and \(\sinh(x)\) we make the following definitions which are clearly modeled on the definitions of the trigonometric functions:

\begin{align*} \tanh(x) \amp{}=\frac{\sinh(x)}{\cosh{(x)}}, \amp{}\coth(x)\amp{} =\frac{\cosh(x)}{\sinh{(x)}},\\ \sech(x) \amp{}=\frac{1}{\cosh(x)},\amp{}\csch(x) \amp{}=\frac{1}{\sinh(x)}. \end{align*}

Problem 8.4.4.

(a)

Show that

\(\displaystyle \dfdx{(\cosh(x))}{x} = \sinh(x)\)
\(\displaystyle \dfdx{(\sinh(x))}{x} = \cosh(x)\)
\(\displaystyle \dfdx{(\tanh(x))}{x} = \sech^2(x)\)
\(\displaystyle \dfdx{(\sech(x))}{x} = -\sech(x)\tanh(x)\)
\(\displaystyle \dfdx{(\coth(x))}{x} = -\csch^2(x)\)
\(\displaystyle \dfdx{(\csch(x))}{x} = -\csch(x)\coth(x)\)

(b)

Show that: \(\sinh(-x)=-\sinh(x)\)

(c)

Show that: \(\cosh(-x)=\cosh(x)\)

(d)

Show that: \(\sinh(x\pm y)=\sinh(x)\cosh(y)\pm \cosh(x)\sinh(y).\)

(e)

Show that: \(\cosh(x\pm y)=\cosh(x)\cosh(y) \pm \sinh(x)\sinh(y).\)

(f)

Compare each of the identities above with the corresponding trigonometric identity.

Drill 8.4.5.

Show that \(y=\cosh(x)\) and \(y=\sinh(x)\) both satisfy the differential equation:

\begin{equation*} \dfdxn{y}{x}{2} = y \end{equation*}

and compare this with Problem 6.2.9.

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