# How do you find the arc length of the curve y = 4 ln((x/4)^(2) - 1) from [7,8]?

Jul 17, 2015

The final answer can be seen here.

The general formula for the arc length is as follows:

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

$s = D \left(x\right) = {\sum}_{a}^{b} \sqrt{{\left(\Delta x\right)}^{2} + {\left(\Delta y\right)}^{2} / {\left(\Delta x\right)}^{2} \cdot {\left(\Delta x\right)}^{2}}$

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

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

Thus, take the derivative and simplify.

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

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

$= \frac{x}{2 {\left(\frac{x}{4}\right)}^{2} - 2}$

$= \frac{x}{{x}^{2} / 8 - 2}$

$= \frac{8 x}{{x}^{2} - 16}$

Now, plug it in and square it:
$s = {\int}_{7}^{8} \sqrt{1 + {\left(\frac{8 x}{{x}^{2} - 16}\right)}^{2}} \mathrm{dx}$

$= {\int}_{7}^{8} \sqrt{1 + \frac{64 {x}^{2}}{{x}^{2} - 16} ^ 2} \mathrm{dx}$

$= {\int}_{7}^{8} \sqrt{\frac{{\left({x}^{2} - 16\right)}^{2} + 64 {x}^{2}}{{x}^{2} - 16} ^ 2} \mathrm{dx}$

If we multiply this out, we should find that something like $- 32 {x}^{2}$ adds with the $64 {x}^{2}$ for a nice and sneaky shift into a perfect square.

${\left({x}^{2} - 16\right)}^{2} = {x}^{4} - 32 {x}^{2} + 256$

${\left({x}^{2} - 16\right)}^{2} + 64 {x}^{2} = {x}^{4} + 32 {x}^{2} + 256$

$= {\left({x}^{2} + 16\right)}^{2}$

Much better. Now we can get rid of that ugly square root.

$= {\int}_{7}^{8} \sqrt{\frac{{\left({x}^{2} + 16\right)}^{2}}{{x}^{2} - 16} ^ 2} \mathrm{dx}$

$= {\int}_{7}^{8} \frac{{x}^{2} + 16}{{x}^{2} - 16} \mathrm{dx}$

Then some manipulation to make this evaluation easier...ish.

$= {\int}_{7}^{8} \frac{{x}^{2} + 16 - 16 + 16}{{x}^{2} - 16} \mathrm{dx}$

$= {\int}_{7}^{8} \frac{{x}^{2} - 16 + 32}{{x}^{2} - 16} \mathrm{dx}$

$= {\int}_{7}^{8} \mathrm{dx} + {\int}_{7}^{8} \frac{32}{\left(x + 4\right) \left(x - 4\right)} \mathrm{dx}$

Looks like we probably have to do Partial Fraction Decomposition on this, unfortunately. Oh well.

$\int \frac{32}{\left(x + 4\right) \left(x - 4\right)} = \frac{A}{x + 4} + \frac{B}{x - 4}$

$= \frac{A \left(x - 4\right) + B \left(x + 4\right)}{\left(x + 4\right) \left(x - 4\right)}$

$= \frac{A x - 4 A + B x + 4 B}{\left(x + 4\right) \left(x - 4\right)}$

$= \frac{\left(A + B\right) x + \left(- 4 A + 4 B\right)}{\left(x + 4\right) \left(x - 4\right)}$

Thus, equating it back to the original equation:

$A + B = 0$
$A = - B$

$- 4 A + 4 B = 32$
$- A + B = 8$
$2 B = 8$
$B = 4 \to A = - 4$

Not too bad, actually. Now we have, overall:

$= {\int}_{7}^{8} \mathrm{dx} + \left(- {\int}_{7}^{8} \frac{4}{x + 4} \mathrm{dx} + \int \frac{4}{x - 4} \mathrm{dx}\right)$

$= {\int}_{7}^{8} \mathrm{dx} - {\int}_{7}^{8} \frac{4}{x + 4} \mathrm{dx} + \int \frac{4}{x - 4} \mathrm{dx}$

$= \left[x\right] {|}_{7}^{8}$ $- \left[4 \ln | x + 4 |\right] {|}_{7}^{8}$ $+ \left[4 \ln | x - 4 |\right] {|}_{7}^{8}$

$= \left(8 - 7\right) - \left(4 \ln 12 - 4 \ln 11\right) + \left(4 \ln 4 - 4 \ln 3\right)$

$\approx 1.8027 \text{u}$

$\textcolor{b l u e}{1 - 4 \left(\ln 12 - \ln 11 - \ln 4 + \ln 3\right)}$