Question

For `f(t)= (1/(2t-3), te^t )` what is the distance between `f(0)` and `f(1)`?

Answer 1

`sqrt(4/9 + e^2)`

Explanation:

we have `f(t)= (1/(2t-3), te^t)`,so:

`f(0) = (-1/3,0)`
`f(1)=(-1,e)`

So then by Pythagoras, the distance, `d`, between `f(0)` and `f(1)` is given by:

`d^2 = (-1-(-1/3))^2 + (e-0)^2`
`\ \ \ = (-2/3)^2 + (e)^2`
`\ \ \ = 4/9 + e^2`

And so:

`d = sqrt(4/9 + e^2)`