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

Jan 17, 2017

Arc Length $= 1.1404$ (4dp)

#### Explanation:

The Arc Length for a Parametric Curve is given by

$L = {\int}_{\alpha}^{\beta} \sqrt{{\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)}^{2} + {\left(\frac{\mathrm{dy}}{\mathrm{dt}}\right)}^{2}} \setminus \mathrm{dt}$

So in this problem we have

$x = {t}^{3} - {t}^{2} + t \implies \frac{\mathrm{dx}}{\mathrm{dt}} = 3 {t}^{2} - 2 t + 1$
$y = t - {t}^{2} \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \implies \frac{\mathrm{dy}}{\mathrm{dt}} = 1 - 2 t$

So the Arc Length is;

$L = {\int}_{0}^{1} \sqrt{{\left(3 {t}^{2} - 2 t + 1\right)}^{2} + {\left(1 - 2 t\right)}^{2}} \setminus \mathrm{dt}$
$\setminus \setminus \setminus = {\int}_{0}^{1} \sqrt{\left(9 {t}^{4} - 12 {t}^{3} + 10 {t}^{2} - 4 t + 1\right) + \left(4 {t}^{2} - 4 t + 1\right)} \setminus \mathrm{dt}$
$\setminus \setminus \setminus = {\int}_{0}^{1} \sqrt{9 {t}^{4} - 12 {t}^{3} + 14 {t}^{2} - 8 t + 2} \setminus \mathrm{dt}$

Which we can evaluate using Numerical Techniques to get:

$L = 1.14038999 \ldots$