# How do you find the length of the curve defined by f(x) = x^2 on the x-interval (0, 3)?

Oct 30, 2015

The formula for the arc length for a curve is:

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

which is a modified version of the distance formula implementing the limit definition of the derivative (zoom in very close, and the graph looks linear), so $\Delta y$ and $\Delta x$ are very small. This way, you accumulate very short straight lines along the curve to fit the curve.

So, we need to find the derivative, and then square it:

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

Plugging it in:

$s = {\int}_{0}^{3} \sqrt{1 + 4 {x}^{2}} \mathrm{dx}$

Assuming you are in your first year of Calculus, this is not something you might have been taught how to do. So maybe you should evaluate this on your calculator or on Wolfram Alpha numerically.

But, if you have heard of trigonometric substitution, that is what I would do.

$= 2 {\int}_{0}^{3} \sqrt{{x}^{2} + \frac{1}{4}} \mathrm{dx}$

With this, let:
$x = a \tan \theta$ where $a = \sqrt{\frac{1}{4}} = \frac{1}{2}$
$\mathrm{dx} = \frac{1}{2} {\sec}^{2} \theta d \theta$
$\sqrt{{x}^{2} + \frac{1}{4}} = \sqrt{\frac{1}{4} {\tan}^{2} \theta + \frac{1}{4}} = \frac{1}{2} \sec \theta$

So you now have the substituted integral (ignoring the bounds for now):

$\implies 2 \int \frac{1}{2} \sec \theta \cdot \frac{1}{2} {\sec}^{2} \theta d \theta$

$= \frac{1}{2} \int {\sec}^{3} \theta d \theta$

To do this difficult integral, we have to use integration by parts...

int udv = ?

Let:
$u = \sec \theta$
$\mathrm{dv} = {\sec}^{2} \theta d \theta$
$\mathrm{du} = \sec \theta \tan \theta d \theta$
$v = \tan \theta$

$= \sec \theta \tan \theta - \int \sec \theta {\tan}^{2} \theta d \theta$

$= \sec \theta \tan \theta - \int \sec \theta \left({\sec}^{2} \theta - 1\right) \mathrm{dx}$

$= \sec \theta \tan \theta - \int {\sec}^{3} \theta d \theta + \int \sec \theta d \theta$

$2 \int {\sec}^{3} \theta d \theta = \sec \theta \tan \theta + \int \sec \theta d \theta$

$\int {\sec}^{3} \theta \mathrm{dx} = \frac{1}{2} \left(\sec \theta \tan \theta + \ln | \sec \theta + \tan \theta |\right)$

Thus:

$= \frac{1}{2} \int {\sec}^{3} \theta d \theta$

$= \frac{1}{4} \left(\sec \theta \tan \theta + \ln | \sec \theta + \tan \theta |\right)$

Using the substitutions from before:

$\sec \theta = \sqrt{4 {x}^{2} + 1}$
$\tan \theta = 2 x$

$\implies | \left[\frac{1}{4} \left(2 \sqrt{4 {x}^{2} + 1} x + \ln \left(\sqrt{4 {x}^{2} + 1} + 2 x\right)\right)\right] {|}_{0}^{3}$

$= \frac{1}{4} \left(2 \sqrt{4 {\left(3\right)}^{2} + 1} \left(3\right) + \ln \left(\sqrt{4 {\left(3\right)}^{2} + 1} + 2 \left(3\right)\right)\right) - {\cancel{\frac{1}{4} \left(2 \sqrt{4 {\left(0\right)}^{2} + 1} \left(0\right) + \ln \left(\sqrt{4 {\left(0\right)}^{2} + 1} + 2 \left(0\right)\right)\right)}}^{0}$

$= \textcolor{b l u e}{\frac{1}{4} \left(6 \sqrt{37} + \ln \left(\sqrt{37} + 6\right)\right)}$

$\approx \textcolor{b l u e}{{\text{9.7471 u}}^{2}}$