How do you find the length of the curve for y=x^2 for (0, 3)?

Nov 3, 2016

Arc Length $= \frac{1}{4} {\sinh}^{-} 1 6 + \frac{3}{2} \sqrt{37}$

Explanation:

The Arc Length $l$ is given by the integration formula
$l = {\int}_{a}^{b} \sqrt{1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}} \mathrm{dx}$

With $y = {x}^{2} \implies \frac{\mathrm{dy}}{\mathrm{dx}} = 2 x$. And so:

$l = {\int}_{0}^{3} \sqrt{1 + {\left(2 x\right)}^{2}} \mathrm{dx}$
$\therefore l = {\int}_{0}^{3} \sqrt{1 + 4 {x}^{2}} \mathrm{dx}$

I will quote the result, but if you want to see how to perform the integration, please use this link

$l = {\left[{\sinh}^{-} 1 \frac{2 x}{4} + \frac{x \sqrt{4 {x}^{2} + 1}}{2}\right]}_{0}^{3}$
$\therefore l = \left({\sinh}^{-} 1 \frac{6}{4} + \frac{3 \sqrt{36 + 1}}{2}\right) - \left({\sinh}^{-} 1 \frac{0}{4} + 0\right)$
$\therefore l = \left({\sinh}^{-} 1 \frac{6}{4} + \frac{3 \sqrt{37}}{2}\right) - \left(0 + 0\right)$
$\therefore l = \frac{1}{4} {\sinh}^{-} 1 6 + \frac{3}{2} \sqrt{37}$