# How do you find the arc length of the curve y=e^(-x)+1/4e^x from [0,1]?

Jun 19, 2015

The arc length turns out to be identical to simply integrating the original function. It is:
$\frac{e}{4} - \frac{1}{e} + \frac{3}{4} \approx 1.06169$

How you do it is written below:

The arc length formula is derived from a "dynamic" distance formula with an independently increasing x value and a y value that varies with a single-valued function:

$D \left(x\right) = \sqrt{{\left(\Delta x\right)}^{2} + {\left(\Delta y\right)}^{2}}$

But you know that $\frac{\Delta y}{\Delta x}$ on really small intervals is equivalent to $\frac{\mathrm{dy}}{\mathrm{dx}}$. Thus:

$D \left(x\right) = \sqrt{{\left(\Delta x\right)}^{2} + {\left(\Delta y\right)}^{2} / {\left(\Delta x\right)}^{2} \cdot {\left(\Delta x\right)}^{2}}$

$= {\sum}_{x} \sqrt{1 + {\left(\frac{\Delta y}{\Delta x}\right)}^{2}} \Delta x$

$= {\int}_{a}^{b} \sqrt{1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}} \mathrm{dx}$

And now we have the formula!

$y = {e}^{- x} + \frac{1}{4} {e}^{x}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = - {e}^{- x} + \frac{1}{4} {e}^{x}$

${\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2} = \left(- {e}^{- x} + \frac{1}{4} {e}^{x}\right) \left(- {e}^{- x} + \frac{1}{4} {e}^{x}\right)$

$= {\left(- {e}^{- x}\right)}^{2} - \frac{1}{4} \left({e}^{x} {e}^{- x}\right) - \frac{1}{4} \left({e}^{x} {e}^{- x}\right) + \frac{1}{16} {\left({e}^{x}\right)}^{2}$

$= {e}^{- 2 x} - \frac{1}{2} \left(1\right) + \frac{1}{16} {e}^{2 x}$

$= \frac{1}{16} {e}^{2 x} + {e}^{- 2 x} - \frac{1}{2}$

Now let's simplify it in context.

$D \left(x\right) = s = {\int}_{a}^{b} \sqrt{1 + \frac{1}{16} {e}^{2 x} + {e}^{- 2 x} - \frac{1}{2}} \mathrm{dx}$

$= {\int}_{0}^{1} \sqrt{\frac{1}{16} {e}^{2 x} + {e}^{- 2 x} + \frac{1}{2}} \mathrm{dx}$

My best guess on where to go next is getting some sort of perfect square. I'm thinking it might be a quartic, though. Let's just try things.

Getting a common denominator:
$= {\int}_{0}^{1} \sqrt{\frac{{e}^{2 x} + 16 {e}^{- 2 x} + 8}{16}} \mathrm{dx}$

Factoring out $\sqrt{\frac{1}{16}}$:
$= \frac{1}{4} {\int}_{0}^{1} \sqrt{{e}^{2 x} + 16 {e}^{- 2 x} + 8} \mathrm{dx}$

Factoring out $\sqrt{{e}^{2 x}}$:
= 1/4int_0^1 sqrt(e^(2x)(1 + 16e^(-4x) + 8e^(-2x))dx

$= \frac{1}{4} {\int}_{0}^{1} {e}^{x} \sqrt{1 + 16 {e}^{- 4 x} + 8 {e}^{- 2 x}} \mathrm{dx}$

Factoring out $\sqrt{16}$:
$= {\int}_{0}^{1} {e}^{x} \sqrt{{e}^{- 4 x} + \frac{1}{2} {e}^{- 2 x} + \frac{1}{16}} \mathrm{dx}$

...This looks promising. What if we just let $u = {e}^{x}$?

$\implies {\int}_{a}^{b} u \sqrt{{u}^{- 4} + \frac{1}{2} {u}^{- 2} + \frac{1}{16}} \mathrm{dx}$

Now let's attempt to factor this. ... Nice, it worked!

$= {\left({u}^{- 2} + \frac{1}{4}\right)}^{2}$

Sweet, so now we can put the real letters back in.

$= {\int}_{0}^{1} {e}^{x} \left({e}^{- 2 x} + \frac{1}{4}\right) \mathrm{dx}$

This looks much better. Now let's just multiply it out and solve it. Notice though that Wolfram Alpha keeps saying that $\sqrt{{e}^{2 x}} \ne | {e}^{x} |$, but we all know that's not right; ${e}^{x}$ is always, always positive! Anyways...

$= {\int}_{0}^{1} {e}^{- x} + \frac{1}{4} {e}^{x} \mathrm{dx}$

...wait a minute. Are you serious? That's what we started with! Wow.

$= \left[- {e}^{- x} + \frac{1}{4} {e}^{x}\right] {|}_{0}^{1}$

$= \left[- {e}^{- 1} + \frac{1}{4} {e}^{1}\right] - \left[- {e}^{- 0} + \frac{1}{4} {e}^{0}\right]$

$= \left[- \frac{1}{e} + \frac{1}{4} e\right] - \left[- 1 + \frac{1}{4}\right]$

$= - \frac{1}{e} + \frac{1}{4} e + 1 - \frac{1}{4}$

$= \frac{e}{4} - \frac{1}{e} + \frac{3}{4} \approx 1.06169$

tl;dr: You can see that here.