What is the arclength of f(x)=(x-2)/(x^2+3) on x in [-1,0]?

Jun 9, 2017

$s \approx 1.013$ using technology.

In cartesian,

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

So, for $f \left(x\right) = \frac{x - 2}{{x}^{2} + 3}$:

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

The derivative of this function is:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(x - 2\right) \cdot \frac{d}{\mathrm{dx}} \left[\frac{1}{{x}^{2} + 3}\right] + \frac{1}{{x}^{2} + 3} \cdot {\cancel{\frac{d}{\mathrm{dx}} \left[x - 2\right]}}^{1}$

$= - \frac{2 x \left(x - 2\right)}{{x}^{2} + 3} ^ 2 + \frac{1}{{x}^{2} + 3}$

$= \frac{3 - 2 x \left(x - 2\right) + {x}^{2}}{{x}^{2} + 3} ^ 2$

$= \frac{- {x}^{2} + 4 x + 3}{{x}^{2} + 3} ^ 2$

So, the square of it is:

${\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2} = {\left(\frac{- {x}^{2} + 4 x + 3}{{x}^{2} + 3} ^ 2\right)}^{2}$

Plugging it in:

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

... Nope, not doable with elementary functions.

So, we will just use Wolfram Alpha to obtain:

$\textcolor{b l u e}{s \approx 1.013}$