# What is the arclength of (1/t^2,1/t) on t in [1,2]?

Mar 24, 2018

$\approx 0.905$

#### Explanation:

For the parametric equation

$x - \frac{1}{t} ^ 2 , \quad y = \frac{1}{t}$

we have

$\mathrm{dx} = - \frac{2}{t} ^ 3 \mathrm{dt} , \quad \mathrm{dy} = - \frac{1}{t} ^ 2 \mathrm{dt}$

Thus, the infinitesimal arc length is

$\mathrm{ds} = \sqrt{{\mathrm{dx}}^{2} + {\mathrm{dy}}^{2}} = \sqrt{\frac{4}{t} ^ 6 + \frac{1}{t} ^ 4} \mathrm{dt} = \sqrt{4 + {t}^{2}} \frac{\mathrm{dt}}{t} ^ 3$

Thus, the required arc length is

${\int}_{1}^{2} \sqrt{4 + {t}^{2}} \frac{\mathrm{dt}}{t} ^ 3$

To evaluate this, lets try the substitution $4 + {t}^{2} = {u}^{2}$. Then $2 t \mathrm{dt} = 2 u \mathrm{du}$. We also have $u = \sqrt{5}$ and $u = \sqrt{8}$ at $t = 1$ and $t = 2$, respectively.

Thus the arc length is

${\int}_{\sqrt{5}}^{\sqrt{8}} u \frac{u \mathrm{du}}{{u}^{2} - 4} ^ 2$

We carry out the partial fractions expansion

${u}^{2} / {\left({u}^{2} - 4\right)}^{2} = \frac{A}{u - 2} + \frac{B}{u + 2} + \frac{C}{u - 2} ^ 2 + \frac{D}{u + 2} ^ 2$

${u}^{2} = A \left(u - 2\right) {\left(u + 2\right)}^{2} + B {\left(u - 2\right)}^{2} \left(u + 2\right) + C {\left(u + 2\right)}^{2} + D {\left(u - 2\right)}^{2}$

Substituting $u = 2$ and $u = - 2$, respectively, gives ${2}^{2} = 16 C$ and ${2}^{2} = 16 D$, leading to

$C = D = \frac{1}{4}$

${u}^{2} = A \left(u - 2\right) {\left(u + 2\right)}^{2} + B {\left(u - 2\right)}^{2} \left(u + 2\right) + \frac{1}{4} \left({\left(u + 2\right)}^{2} + {\left(u - 2\right)}^{2}\right)$
$q \quad = A \left(u - 2\right) {\left(u + 2\right)}^{2} + B {\left(u - 2\right)}^{2} \left(u + 2\right) + \frac{1}{2} \left({u}^{2} + 4\right)$

and so

$\frac{1}{2} \left({u}^{2} - 4\right) = A \left(u - 2\right) {\left(u + 2\right)}^{2} + B {\left(u - 2\right)}^{2} \left(u + 2\right) \implies$
$\frac{1}{2} = A \left(u + 2\right) + B \left(u - 2\right)$

Substituting $u = \pm 2$ in the above leads to

$A = \frac{1}{8} , \quad B = - \frac{1}{8}$

So,

${u}^{2} / {\left({u}^{2} - 4\right)}^{2} = \frac{1}{8} \frac{1}{u - 2} - \frac{1}{8} \frac{1}{u + 2} + \frac{1}{4} \frac{1}{u - 2} ^ 2 + \frac{1}{4} \frac{1}{u + 2} ^ 2$

and the arc length is

${\int}_{\sqrt{5}}^{\sqrt{8}} \left(\frac{1}{8} \frac{1}{u - 2} - \frac{1}{8} \frac{1}{u + 2} - \frac{1}{4} \frac{1}{u - 2} ^ 2 - \frac{1}{4} \frac{1}{u + 2} ^ 2\right) \mathrm{du}$
$q \quad = {\left(\frac{1}{8} \ln | \frac{u - 2}{u + 2} | - \frac{1}{4} \frac{1}{u - 2} - \frac{1}{4} \frac{1}{u + 2}\right)}_{\sqrt{5}}^{\sqrt{8}}$

$q \quad = {\left(\frac{1}{8} \ln | \frac{u - 2}{u + 2} | - \frac{1}{2} \frac{u}{{u}^{2} - 4}\right)}_{\sqrt{5}}^{\sqrt{8}} \approx 0.905$