# What is the arclength of f(t) = (t^3-1,t^2-1) on t in [2,3]?

May 9, 2018

$\frac{1}{27} \left[\sqrt{\left[{85}^{3}\right]} - \sqrt{\left[{40}^{3}\right]}\right]$

#### Explanation:

Given, $f \left[t\right]$ = [t^3-1, t^2-1]  i.e, $\left[x = {t}^{3} - 1 , y = {t}^{2} - 1\right]$.....$\left[1\right]$
Differentiating w.r.t. $t$,

$\frac{\mathrm{dx}}{\mathrm{dt}} = 3 {t}^{2} , \frac{\mathrm{dy}}{\mathrm{dt}} = 2 t$.....$\left[2\right]$. The arc length of a curve in parametric form is given by, $L$=${\int}_{2}^{3} \sqrt{{\left[\frac{\mathrm{dx}}{\mathrm{dt}}\right]}^{2} + {\left[\frac{\mathrm{dy}}{\mathrm{dt}}\right]}^{2}} \mathrm{dt}$, substituting $\frac{\mathrm{dx}}{\mathrm{dt}} \mathmr{and} \frac{\mathrm{dy}}{\mathrm{dt}}$ from ....$\left[2\right]$ into the length formula $L$ will give,

$L$=${\int}_{2}^{3} \sqrt{9 {t}^{4} + 4 {t}^{2} \mathrm{dt}}$ = ${\int}_{2}^{3} \sqrt{{t}^{2} \left[9 {t}^{2} + 4\right]}$=${\int}_{2}^{3} t \sqrt{9 {t}^{2} + 4} \mathrm{dt}$.

We can now make a substitution by letting $\left[u = 9 {t}^{2} + 4\right]$........$\left[3\right]$. Differentiating .....$\left[3\right]$ w.r.t $x$, $\frac{\mathrm{du}}{\mathrm{dt}} = 18 t$, .i.e, $\frac{\mathrm{du}}{18 t}$=$\mathrm{dt}$.

So we now have $L$=$\int t \frac{\sqrt{u}}{18 t} \mathrm{dt}$ = $\frac{1}{18} \int \sqrt{u} \mathrm{dt}$ [ since the terms in $t$will cancel],

And after integrating w.r.t. $t$ we are left with $\frac{1}{18} \left[\frac{2}{3}\right] \sqrt{{u}^{3}}$ = 1/27sqrt[u^3 we now have change the bounds of integration, From ......$\left[3\right]$, $u = 9 {t}^{2} + 4$, so when $t = 2 , u = 40$, when $t = 3 , u = 85$.

So length $L$=$\frac{1}{27} \left[\sqrt{{85}^{3}} - \sqrt{{40}^{3}}\right]$ units.