Question

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`?

Answer 1

`s = (181sqrt(181)-1000)/27 ~~ 53.15`

Explanation:

We have parametric equations:

`{ (x=5t^2), (y=t^3) :}`

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 \ sqrt( (dx/dt)^2+(dy/dt)^2 ) \ dt`

`\ \ = int_0^3 \ sqrt( (10t)^2+(3t^2)^2 ) \ dt`

`\ \ = int_0^3 \ sqrt( t^2(100+9t^2 )) \ dt`

`\ \ = int_0^3 \ tsqrt( 100+9t^2 ) \ dt`

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

`u = 9t^2 +100 => (du)/dt = 18t`

And we change the limits of integration:

`t={ (0), (3) :} => u={ (100), (181) :}`

So then:

`s = int_100^181 \ (1/18) \ sqrt( u ) \ dt`

`\ \ = 1/27 \ (181^(3/2) - 100^(3/2))`

`\ \ = (181sqrt(181)-1000)/27`

`\ \ ~~ 53.15`