What is the arc length of f(x)=x^2/sqrt(7-x^2) on x in [0,1]?

Jun 1, 2018

The arc length of a function $f$ on the interval $\left[a , b\right]$ is given by:

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

Here, $f \left(x\right) = {x}^{2} / \sqrt{7 - {x}^{2}}$ and I leave it to you to determine that $f ' \left(x\right) = \frac{x \left(14 - {x}^{2}\right)}{7 - {x}^{2}} ^ \left(\frac{3}{2}\right)$.

Then, the arc length desired is:

$s = {\int}_{0}^{1} \sqrt{1 + \frac{{x}^{2} {\left(14 - {x}^{2}\right)}^{2}}{7 - {x}^{2}} ^ 3} \mathrm{dx}$

This has no closed form. Use a calculator to find:

$s \approx 1.10458$

Jun 1, 2018

$\approx 1.10458$

Explanation:

We have
$f \left(x\right) = {x}^{2} / \sqrt{7 - {x}^{2}}$
so
$f ' \left(x\right) = \frac{2 x \cdot \sqrt{7 - {x}^{2}} - {x}^{2} \cdot \frac{1}{2} \cdot {\left(7 - {x}^{2}\right)}^{- \frac{1}{2}} \left(- 2 x\right)}{7 - {x}^{2}}$
Multiplying numerator and denominator by
$\sqrt{7 - {x}^{2}}$
we get
$f ' \left(x\right) = \frac{2 x \left(7 - {x}^{2}\right) + {x}^{3}}{\left(7 - {x}^{2}\right) \cdot \sqrt{7 - {x}^{2}}}$
so
f'(x)=(14x-x^3)/((sqrt(7-x^2)(7-x^2))
Using the Formula
$s = {\int}_{0}^{1} \sqrt{1 + f ' {\left(x\right)}^{2}} \mathrm{dx}$
we get by a numerical method $\approx 1.10458$