We now use the formula to find the distance travelled by the car.
In this case, we have:
`x(t) = 20+0.2t^3`, so `(dx)/(dt)=0.6t^2`
`y(t) = 20t − 2t^2` giving `(dy)/(dt)=20-4t`
Our lower and upper limits for this example are `t = 0` to `t = 8`.
Substituting these into the distance formula gives:
`text[length]=r` `=int_0^8 sqrt[(0.6t^2)^2+(20-4t)^2]`
Using a computer algebra system (see the answer in Wolfram|Alpha) gives us the length `144.7\ "m"`.
Our answer is reasonable and is consistent with our earlier estimate.