# How do you find the distance travelled from t=0 to t=1 by a particle whose motion is given by x=4(1-t)^(3/2), y=2t^(3/2)?

Sep 14, 2017

You can integrate the speed of travel to get a distance of $\frac{14}{3}$.

#### Explanation:

The speed is the length of the velocity vector. It is equal to $\sqrt{{\left(x ' \left(t\right)\right)}^{2} + {\left(y ' \left(t\right)\right)}^{2}}$.

We differentiate to get

$x ' \left(t\right) = 6 {\left(1 - t\right)}^{\frac{1}{2}} \cdot \left(- 1\right)$ and $y ' \left(t\right) = 3 {t}^{\frac{1}{2}}$.

Therefore, ${\left(x ' \left(t\right)\right)}^{2} + {\left(y ' \left(t\right)\right)}^{2} = 36 \left(1 - t\right) + 9 t = 36 - 27 t$ and the speed is $\sqrt{36 - 27 t}$.

The distance traveled is therefore ${\int}_{0}^{1} \sqrt{36 - 27 t} \setminus \mathrm{dt}$.

This integral can be done by the substitution $u = 36 - 27 t , \mathrm{du} = - 27 \mathrm{dt}$. We then change the limits of integration appropriately and then switch the order of the limits of integration like this (note that $36 - 27 \cdot 0 = 36$ and $36 - 27 \cdot 1 = 9$):

$- \frac{1}{27} {\int}_{36}^{9} {u}^{\frac{1}{2}} \mathrm{du} = \frac{1}{27} {\int}_{9}^{36} {u}^{\frac{1}{2}} \mathrm{du}$.

Now use the Fundamental Theorem of Calculus to get

$\frac{2}{81} {u}^{\frac{3}{2}} {|}_{9}^{36} = \frac{2}{81} \left({36}^{\frac{3}{2}} - {9}^{\frac{3}{2}}\right) = \frac{2}{81} \left(216 - 27\right) = \frac{378}{81} = \frac{14}{3.}$