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### Hyperbolic Functions

Package of problems

## Solution

### Setting up the problem: Tension in a rope

This diagram shows a rope hanging. We have drawn axes so that the rope hangs in the $(x,y)$-plane with the $y$-axis vertically upwards and the lowest point of the rope on the $y$-axis.

Here, $T_0$ is the tension at the lowest point of the rope, while $T$ is the tension at a point $P$ which is at a distance of $s$ along the rope from the lowest point. At $P$, the rope makes an angle of $\psi$ with the horizontal.

The rope has weight $w$ per unit length.

Prove that $\tan\psi=\Bigl(\dfrac{w}{T_0}\Bigr)s$.

Resolving vertically and horizontally on the portion of rope between the lowest point and $P$, we find: \begin{align*} \mathcal{R}(\uparrow)&& T\sin\psi - ws &= 0\\ \mathcal{R}(\rightarrow)&& T\cos\psi - T_0 &= 0 \end{align*}

(The weight of the rope is its length $s$ times the weight per unit length $w$.) Rearranging to get $T\sin\psi=ws$ and $T\cos\psi=T_0$, we can divide to get $\tan\psi=\Bigl(\dfrac{w}{T_0}\Bigr)s$ as required.

From now on, we’ll make our life easier by assuming that $w/T_0=1$, so that the equation becomes $\tan\psi=s$.

If the rope has the equation $y=f(x)$, prove that $f'(x)=s$.

The gradient of the curve at $P$ is $f'(x)=\tan\psi$, so this follows immediately from $\tan\psi=s$.

### An iterative approach

We’ll start by assuming a very simple (wrong) formula for the rope length $s$, use that to work out a first approximation for $f(x)$, and improve it step-by-step.

We will also assume that the rope has a $y$-intercept of $1$.

#### Step 1

Assume that $s=x$, that is, assume the weight is evenly distributed horizontally. (This is the case for ropes holding up bridges.)

What is $f_1(x)$ in this case, where $f'_1(x)=s$?

Using $s=x$, we can integrate $f'_1(x)=x$ to get $f_1(x)=\frac{1}{2}x^2+c$. Using $f_1(0)=1$, we get $f_1(x)=1+\frac{1}{2}x^2$.

#### Step 2

Now assume that the weight is distributed uniformly along the shape given by the answer to Step 1, that is to say, $s$ is now the arc length of $y=f_1(x)$.

What is $f_2(x)$ approximately in this case, where $f'_2(x)=s$? Work out an answer as a polynomial, valid for small values of $x$, including one term more than in $f_1(x)$.

The relevant version of the formula for arc length in this context is $\int\sqrt{1+(f'_1(x))^2}\,dx=\int\sqrt{1+x^2}\,dx.$

We would like a polynomial approximation to this, so we expand the square root using the binomial theorem: \begin{align*} \sqrt{1+x^2}&=(1+x^2)^{\frac{1}{2}}\\ &=1+\frac{1}{2}x^2+\frac{(\frac{1}{2})(-\frac{1}{2})}{2!}(x^2)^2+\cdots\\ &=1+\frac{1}{2}x^2-\frac{1}{8}x^4+\cdots \end{align*}

So integrating gives us a formula for $s$: $s=x+\frac{1}{6}x^3-\frac{1}{40}x^5+\cdots$ Finally, we can now solve the equation $f'_2(x)=s$, where $f_2(0)=1$, to get $f_2(x)=1+\frac{1}{2}x^2+\frac{1}{24}x^4-\frac{1}{240}x^6+\cdots$ We only want one term more than $f_1(x)$, though, so we get $f_2(x)=1+\frac{1}{2}x^2+\frac{1}{24}x^4.$

#### Step 3

Now assume that the weight is distributed uniformly along the shape given by the answer to Step 2, that is to say, $s$ is now the arc length of $y=f_2(x)$.

What is $f_3(x)$ approximately in this case, where $f'_3(x)=s$? Work out an answer as a polynomial, valid for small values of $x$, including one term more than in $f_2(x)$.

We follow the same procedure as in step 2.

We start by finding $s$, working up to the term in $x^4$: \begin{align*} s &= \int\sqrt{1+(f'_2(x))^2}\,dx\\ &= \int\sqrt{1+(x+\tfrac{1}{6}x^3)^2}\,dx\\ &= \int 1+\frac{1}{2}(x+\tfrac{1}{6}x^3)^2+ \frac{(\frac{1}{2})(-\frac{1}{2})}{2!}(x+\tfrac{1}{6}x^3)^4+\cdots\\ &= \int 1+\frac{1}{2}(x^2+\tfrac{2}{6}x^4+\tfrac{1}{36}x^6)- \frac{1}{8}(x^4+\cdots)+\cdots\\ &= \int 1+\frac{1}{2}x^2+\frac{1}{24}x^4+\cdots\\ &= x+\frac{1}{6}x^3+\frac{1}{120}x^5+\cdots \end{align*}

and since $f'_3(x)=s$, it follows that $f_3(x)=1+\frac{1}{2}x^2+\frac{1}{24}x^4+\frac{1}{720}x^6= 1+\frac{x^2}{2!}+\frac{x^4}{4!}+\frac{x^6}{6!},$ giving one term more than for $f_2(x)$.

#### In the limit

1. Can you predict $f_4(x)$? What about $f_5(x)$? And how will the sequence of polynomials $f_1(x)$, $f_2(x)$, $f_3(x)$, … continue further?

Following the pattern, it would seem likely that \begin{align*} f_4(x)&=1+\frac{x^2}{2!}+\frac{x^4}{4!}+\frac{x^6}{6!}+ \frac{x^8}{8!}\\ f_5(x)&=1+\frac{x^2}{2!}+\frac{x^4}{4!}+\frac{x^6}{6!}+ \frac{x^8}{8!}+\frac{x^{10}}{10!}\\ \dots\\ f_n(x)&=1+\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots+\frac{x^{2n}}{(2n)!}\\ \end{align*}
2. What is the power series, $f(x)$, that this sequence is tending to? What do we call this function?

It appears to be $f(x)=1+\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots+\frac{x^{2n}}{(2n)!}+ \cdots$

This function is $\cosh x$, the hyperbolic cosine. It is defined either by this power series or by the identity $\cosh x=\frac{e^x+e^{-x}}{2}.$

Interestingly, this power series is valid for all values of $x$, not only values less than $1$.

3. Can you prove that this limit function $f(x)$ satisfies the equation $f'(x)=s$, where $s$ is the arc length of $y=f(x)$?

The limit function, $y=f(x)=\cosh x$, has arc length from $0$ to $x$ given by \begin{align*} s&=\int_0^x \sqrt{1+(f'(x))^2}\,dx\\ &=\int_0^x \sqrt{1+(\sinh x)^2}\,dx\\ &=\int_0^x \cosh x\,dx\\ &=[\sinh x]_0^x\\ &=\sinh x \end{align*}

And $f'(x)=\sinh x$, so $f'(x)=s$.

We have been a bit sloppy with notation here, using $x$ as both one of the limits and also as the variable in the integral. If we were to be more careful, we would have changed the variable in the integral to a different letter such as $t$.

4. In the above procedure, what would Step $n$ have been? Can you prove that Step $n$ would have given the function $f_n(x)$ which you predicted?

Step $n$ would read:

“Now assume that the weight is distributed uniformly along the shape given by the answer to Step $n-1$, that is to say, $s$ is now the arc length of $y=f_{n-1}(x)$.

What is $f_n(x)$ approximately in this case, where $f'_n(x)=s$? Work out an answer as a polynomial, valid for small values of $x$, including one term more than in $f_{n-1}(x)$."

Since the construction is by iteration, the proof is by induction.

We assume that the previous step (Step $n-1$) gave us the expected function, that is, we assume that $f_{n-1}(x)=1+\dfrac{x^2}{2!}+\cdots+\dfrac{x^{2(n-1)}}{(2(n-1))!}.$ Then to find $f_n(x)$, we begin by working out the arc length of $y=f_{n-1}(x)$ from $0$ to $x$: $s=\int_0^x \sqrt{1+(f'_{n-1}(x))^2}\,dx.$ But this integral is going to be very hard to calculate directly: we will need to differentiate $f_{n-1}(x)$, square it, and then use the binomial theorem to expand the square root.

However, we can use what we know about $f_{n-1}(x)$ to sidestep many of these difficulties: $f_{n-1}(x)$ is the expansion of $\cosh x$ up to the term in $x^{2n-2}$.

Therefore, $f'_{n-1}(x)$ is the expansion of $\sinh x$ up to the term in $x^{2n-3}$. (We can work this out directly by differentiating $f_{n-1}(x)$ if we wish.) In fact, it even agrees up to the term in $x^{2n-2}$, which is zero.

Therefore $1+(f'_{n-1}(x))^2$ will agree with the expansion of $1+\sinh^2 x=\cosh^2 x$ up to at least the term in $x^{2n-2}$.

When we then use the binomial theorem to expand $\sqrt{1+(f'_{n-1}(x))^2}$ as a power series in $x$, it will have to agree with $\cosh x$ up to at least the term in $x^{2n-2}$, since the expression inside the square root agrees up to at least this term.

Thus when we integrate this to get $f'_n(x)=s$, it will agree with $\int\cosh x\,dx=\sinh x$ up to at least the term in $x^{2n-1}$. (And the constant term in the integral is zero, as the integral is from $0$ to $x$, which agrees with the zero constant term in the expansion of $\sinh x$.)

Then we integrate this final expression to get $f_n(x)$, so this will agree with $\int\sinh x\,dx=\cosh x$ up to at least the term in $x^{2n}$, and the constant term is also correct as $f_n(0)=1$.

Since we let $f_n(x)$ be the polynomial with just one term more than $f_{n-1}(x)$, we find that $f_n(x)$ is the polynomial approximation to $\cosh x$ with terms up to $x^{2n}$, that is, $f_n(x)=1+\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots+\frac{x^{2n}}{(2n)!}$ as we wanted.

1. Why do you think we assumed that the curve has a $y$-intercept of $1$?

The limiting function is $\cosh x$, and $\cosh 0=1$. Had we chosen a different $y$-intercept, say $0$, the limiting function would have been $\cosh x-1$, which would have been a little bit less pretty.

### A direct approach

Can you solve the equation $f'(x)=s$ directly?

Using the formula for arc length, this differential equation becomes $f'(x)=\int_0^x \sqrt{1+(f'(t))^2}\,dt.$ (Here we’re being good and not using $x$ as the variable in the integration, to avoid confusing ourselves.) This equation may seem very hard at first, but we might notice that $f(x)$ itself never appears, just its derivative.

So let’s write $p(x)=f'(x)$, to get $p(x)=\int_0^x \sqrt{1+(p(t))^2}\,dt,$ which looks marginally simpler as there’s no differentiation involved.

We still have this pesky integral, but we can get rid of that by differentiating both sides with respect to $x$, and making use of the fundamental theorem of calculus: $p'(x)=\sqrt{1+(p(x))^2}.$ No integral left!

We could try squaring both sides to get rid of the square root, but then we’d have $(p'(x))^2$, a squared derivative, which is not nice.

If we write $p'(x)$ as $\frac{dp}{dx}$ things might look more familiar (and we’ll write $p$ in place of $p(x)$, too): $\frac{dp}{dx}=\sqrt{1+p^2}.$ This is a separable differential equation! So separating the variables, we get $\int \frac{1}{\sqrt{1+p^2}}\,dp=\int 1\,dx.$ The right hand side is easy. For the left hand side, we use the substitution $p=\sinh u$, so $\frac{dp}{du}=\cosh u$, giving $\int \frac{1}{\sqrt{1+\sinh^2 u}}\cosh u\,du=x,$ which simplifies to $\int 1\,du=x,$ so $u=x+c$, giving $\arsinh p=x+c$ or $p=\sinh(x+c)$.

Now we can also work out from our original differential equation that $p(0)=0$, so $c=0$ and $p=\sinh x$.

But $p=f'(x)$, so $f'(x)=\sinh x$, giving $f(x)=\cosh x+C$.

Finally, using $f(0)=1$ gives $C=0$, so $f(x)=\cosh x$.

More generally, can you solve the equation $f'(x)=\dfrac{s}{a}$?

The argument is very similar to the above; one just has to take care with the extra $a$ term. The solution is $f(x)=a\cosh(\frac{x}{a})$.