# How do you find the distance travelled from t=0 to t=3 by a particle whose motion is given by the parametric equations x=5t^2, y=t^3?

May 14, 2018

$s = \frac{181 \sqrt{181} - 1000}{27} \approx 53.15$

#### Explanation:

We have parametric equations:

$\left\{\begin{matrix}x = 5 {t}^{2} \\ y = {t}^{3}\end{matrix}\right.$

defining the motion of a particle from $t = 0$ to $t = 3$, so the total distance travelled is the arclength, which we calculate for parametric equations using:

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

$\setminus \setminus = {\int}_{0}^{3} \setminus \sqrt{{\left(10 t\right)}^{2} + {\left(3 {t}^{2}\right)}^{2}} \setminus \mathrm{dt}$

$\setminus \setminus = {\int}_{0}^{3} \setminus \sqrt{{t}^{2} \left(100 + 9 {t}^{2}\right)} \setminus \mathrm{dt}$

$\setminus \setminus = {\int}_{0}^{3} \setminus t \sqrt{100 + 9 {t}^{2}} \setminus \mathrm{dt}$

In order to evaluate this integral, we can perform a substitution, Let

$u = 9 {t}^{2} + 100 \implies \frac{\mathrm{du}}{\mathrm{dt}} = 18 t$

And we change the limits of integration:

$t = \left\{\begin{matrix}0 \\ 3\end{matrix}\right. \implies u = \left\{\begin{matrix}100 \\ 181\end{matrix}\right.$

So then:

$s = {\int}_{100}^{181} \setminus \left(\frac{1}{18}\right) \setminus \sqrt{u} \setminus \mathrm{dt}$

$\setminus \setminus = \frac{1}{27} \setminus \left({181}^{\frac{3}{2}} - {100}^{\frac{3}{2}}\right)$

$\setminus \setminus = \frac{181 \sqrt{181} - 1000}{27}$

$\setminus \setminus \approx 53.15$