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)?

1 Answer
Sep 14, 2017

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.