Question

For `f(t)= (te^(3t),t^2-t)` what is the distance between `f(2)` and `f(5)`?

Answer 1

`f(t) = (te^(3t),t^2-1)`

`t=2 => f(2) = (2e^6, 4-2) = (2e^6, 2)`
`t=5 => f(5) = (5e^15, 25-5) = (5e^15, 20)`

So then By Pythagoras, the distance, `d`, between these points is given by:

`d^2 = (5e^15-2e^6)^2 + (20-2)^2`

I'll leave you to work out `d` using a calculator