Question

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)`?

Answer 1

You can integrate the speed of travel to get a distance of `14/3`.

Explanation:

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

We differentiate to get

`x'(t)=6(1-t)^{1/2}*(-1)` and `y'(t)=3t^{1/2}`.

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

The distance traveled is therefore `int_{0}^{1}sqrt{36-27t}\ dt`.

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

`-1/27 int_{36}^{9}u^{1/2}du=1/27 int_{9}^{36}u^{1/2}du`.

Now use the Fundamental Theorem of Calculus to get

`2/81 u^{3/2}|_{9}^{36}=2/81(36^{3/2}-9^{3/2})=2/81(216-27)=378/81=14/3.`