# How do you find the arc length of the curve sqrt(4-x^2) from [-2,2]?

Jul 28, 2015

The answer turned out to be $2 \pi$ units.

The arc length is essentially the usage of the distance formula with a small, independent change in x and a small change in y, where y changes according to $f \left(x\right)$. Thus:

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

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

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

Make it a sum, and you've got the arc length over an interval $\left[a , b\right]$.

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

Turn it into an integral expression to get:
$s = {\int}_{a}^{b} \sqrt{1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}} \mathrm{dx}$

So, we could take the derivative first to get:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 \sqrt{4 - {x}^{2}}} \cdot - 2 x = \frac{- x}{\sqrt{4 - {x}^{2}}}$

Then, let's square it and plug it in.

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

$\to s = {\int}_{a}^{b} \sqrt{1 + {x}^{2} / \left(4 - {x}^{2}\right)} \mathrm{dx}$

Cross-multiply:
$= {\int}_{a}^{b} \sqrt{\frac{4 - {x}^{2} + {x}^{2}}{4 - {x}^{2}}} \mathrm{dx}$

$= {\int}_{a}^{b} \sqrt{\frac{4}{4 - {x}^{2}}} \mathrm{dx}$

$= 2 {\int}_{a}^{b} \frac{1}{\sqrt{4 - {x}^{2}}} \mathrm{dx}$

And we've got ourselves a u-substitution, if you know that $\frac{d}{\mathrm{dx}} \left[\arcsin \left(x\right)\right] = \frac{1}{\sqrt{1 - {x}^{2}}}$.

$= 2 {\int}_{a}^{b} \frac{1}{\sqrt{4}} \cdot \frac{1}{\sqrt{1 - {x}^{2} / 4}} \mathrm{dx}$

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

Let $u = \frac{x}{2}$ and you get:

$\mathrm{dx} = 2 \mathrm{du}$

$= 2 {\int}_{a}^{b} \frac{1}{\sqrt{1 - {u}^{2}}} \mathrm{du}$

$= \left[2 \arcsin \left(\frac{x}{2}\right)\right] {|}_{- 2}^{2}$

$= 2 \arcsin \left(\frac{2}{2}\right) - 2 \arcsin \left(\frac{- 2}{2}\right)$

$= 2 \arcsin \left(1\right) - 2 \arcsin \left(- 1\right)$

Let:
$y = \arcsin \left(1\right) \to \sin y = 1 \to y = \frac{\pi}{2}$

$y = \arcsin \left(- 1\right) \to \sin y = - 1 \to y = \frac{3 \pi}{2}$

$\to 2 \left(\frac{\pi}{2}\right) - 2 \left(\frac{3 \pi}{2}\right) \to 2 \left(\frac{\pi}{2}\right) - 2 \left(\frac{- \pi}{2}\right)$
(so that our answer is positive)

$= \pi + \pi = \textcolor{b l u e}{2 \pi \text{ u}}$