Written by Colin+ in further pure 2, integration.

This problem came via the lovely @realityminus3 and caused me no end of problems - although I got there in the end. I thought it'd be useful to look at not just the answer, but the mistakes I made on the way. Maths is usually presented as 'here's what you do, ta-da!', without any sense of the struggles that even decent mathematicians have on the way.

So, here's the problem:

A curve is defined by $y = \frac 12 \ln \left( \tanh (x) \right)$. Points $A$ and $B$, with $x$-coordinates $\ln(2)$ and $\ln(4)$, respectively, lie on the curve. Find the arc length $\vec {AB}$ in the form $p \ln(q)$, where $p$ and $q$ are rational.

Helpfully, the question asks you to show that $\dydx = \frac {1}{\sinh(2x)}$, which I did for the sake of form:

$\dydx = \frac 12 \frac 1{\tanh(x)} \sech^2(x) = \frac {\cosh(x)}{2\sinh(x) \sech^2(x)} = \frac{1}{\sinh(2x)}$.

I then wrote down the arclength formula: $L = \int_a^b \sqrt{1 + \left(\dydx\right)^2} \, \d x$. That's in the formula book, but I know it because it's related to Pythagoras: I think of the infinitesimal distance along the curve is $\sqrt{\d x^2 +\d y^2}$, and the rest is tidying it up into proper maths ($\d x$s don't hang around on their own, except in an integral. If you 'divide' the things inside the square root by $\d x^2$ and multiply outside by $\d x$ - something I'm sure will horrify @realityminus3 - you've got something equivalent.)

In honesty, that's where the problem started: although I *wrote down* the formula correctly, I then forgot to square the $\sinh$ and got into all sorts of pickles trying to sort it out. That's the second time in a week forgetting to square something has cost me half an hour.

**Lesson 1:** Make sure you've squared all the things that should be squared.

If I still kept an idiot list, that would now go on it.

Once I sorted that out, the integral became a lot nicer: the thing under the square root is $1 + \frac 1{\sinh^2(2x)} = \frac{\sinh^2(2x) + 1}{\sinh^2(2x)} \\= \frac {\cosh^2(2x)}{\sinh^2(2x)} = \coth^2(2x)$

That means the integral is now $\int_{\ln(2)}^{\ln(4)} \coth(2x) \d x$. A spot of function-derivative (or substituting $u = \sinh(2x)$) gives $\left[\frac 12 \ln \left| \sinh(2x) \right| \right]_{\ln(2)}^{\ln(4)}$.

This is where the complacency set in. Dear readers: I'm sorry to disappoint you, I took my eye off of the ball. I worked out $\sinh(2\ln(2))$, correctly, to be $\frac{15}{8}$. I then worked out $\sinh(2\ln(4))$ to be $\frac{63}{16}$. Our survey (or rather Wolfram Alpha) said, heehaw. I apologise whole-heartedly for my lack of professionalism: $2\ln(4)$ is not the same thing as $\ln(8)$. No. It's $\ln(16)$, which means $\sinh(2\ln(4)) = \frac{255}{32}$.

**Lesson 2:** Don't get cocky.

That one was already on the idiot list.

To finish it off: $\left[\frac 12 \ln \left| \sinh(2x) \right| \right]_{\ln(2)}^{\ln(4)} = \frac{1}{2}\left[ \ln \left( \frac {255}{32}\right) - \ln \left( \frac{15}{8} \right) \right] \\ = \frac{1}{2} \ln \frac{17}{4}$.

The most important insight I can give, though, is that after I got my answer, it smelt funny. I wasn't certain I'd done it right, so I plopped it into Wolfram Alpha to check I was right - and I wasn't. So I went back and found my mistake.

**Lesson 3:** Check your work.

Obviously, you don't have WA available in your exam; in an exam, wouldn't have been doing my $\sinh$es in my head.

## Caro_lann

Nice description of process “@icecolbeveridge: [FCM] “A little biter of a question”: http://t.co/TN0BZbcJqS”

## RealityMinus3

@icecolbeveridge Hahaha! I thought that sounded like me. 🙂 On my phone it renders as so: even funnier! Apt 😉 http://t.co/vIiTNS61G7

## RealityMinus3

@icecolbeveridge But yes, you HAVE horrified me with your treatment of dx O.o

## MathbloggingAll

“A little biter of a question” http://t.co/9fGgLk8geu

## srcav

Love this from @icecolbeveridge – a little jealous I’m not teaching fp2 this year! http://t.co/0r4cM3FEol