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.#