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

Jun 11, 2015

Basically, you use the arc length formula:
$s = {\int}_{a}^{b} \sqrt{1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}} \mathrm{dx}$

And you have to simplify down to a perfect square and then take the square root. The simplification is the hard part. Afterwards it's very simple (keep reading).

You can find the derivation for the arc length at the bottom if you don't remember it or don't have it derived.

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

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

Multiply stuff out and add everything together:
$= {\int}_{1}^{e} \sqrt{{x}^{2} / 4 + \frac{1}{2} + \frac{1}{4 {x}^{2}}} \mathrm{dx}$

Small trick with multiplication:
$= \frac{1}{2} {\int}_{1}^{e} 2 \sqrt{{x}^{2} / 4 + \frac{1}{2} + \frac{1}{4 {x}^{2}}} \mathrm{dx}$

Another small trick with square roots:
$= \frac{1}{2} {\int}_{1}^{e} \cancel{\sqrt{4}} \sqrt{{x}^{2} / \cancel{4} + {\cancel{\frac{1}{2}}}^{2} + \frac{1}{\cancel{4} {x}^{2}}} \mathrm{dx}$

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

Making it look nicer:
$= \frac{1}{2} {\int}_{1}^{e} \sqrt{\frac{{x}^{4} + 2 {x}^{2} + 1}{{x}^{2}}} \mathrm{dx}$

Since we are in a positive domain, all this works out to be:

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

Alright, this looks good to go. Thankfully, we have a perfect square now, AND we know that $x > 0$, always, in this domain.

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

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

$= \frac{1}{2} \left[{x}^{2} / 2 + \ln | x |\right] {|}_{1}^{e}$

$= \frac{1}{2} \left[\left({\left(e\right)}^{2} / 2 + {\cancel{\ln | e |}}^{1}\right) - \left({\left(1\right)}^{2} / 2 + {\cancel{\ln | 1 |}}^{0}\right)\right]$

$= \frac{1}{2} \left({e}^{2} / 2 + \frac{1}{2}\right)$

$= \frac{1}{4} \left({e}^{2} + 1\right) \approx 2.09726$

Derivation
The arc length can be derived, if you forget the (rather simple) formula. Since it really is just an accumulation of really short straight lines, you can use the distance formula and modify it in such a way that it works for a specific variable using the definition of the derivative.

$D = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$

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

We can let $x$ vary as it may, because we're assuming that the arc length is of a function that more than likely acts according to the "vertical line test", which means a continuously increasing $x$ (never decreasing AND increasing in the same graph) but a varying $y$. What we need to do is modify how $\Delta y$ is written to allow for any such variation in $y$ that matches how our function behaves. Recall the Leibnitz notation of the derivative:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\Delta y}{\Delta x}$, but varying over infinitely small intervals

Thus:
${\left(\Delta y\right)}^{2} = {\left(\frac{\Delta y}{\Delta x}\right)}^{2} \cdot {\left(\Delta x\right)}^{2}$

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

Since ${\left(\Delta x\right)}^{2}$ is always increasing, it will always be positive on a graph where $x$ increases from left to right, so there's no need to put it as $| \Delta x |$ when pulling it out of the square root. Now let's call the arc length $s$, like in $s = r \theta$, and add a $\sum$ symbol to accumulate the tiny, short straight lines into an arc length:

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

Finally, this can be converted to an integral by using $\Delta x$ as $\mathrm{dx}$, etc., and replacing the sum with an integral symbol:

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