Question

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

Answer 1

`2/3sqrt(10)`

Explanation:

If `f(t)=(sqrt(t)/(t+1),t^2-t)`
then
`color(white)("XXX")f(0)=(sqrt(0)/(0+1),0^2-0) =(0,0)`

`color(white)("XXX")f(2)=(sqrt(2)/(2+1),2^2-2)=(sqrt(2)/3,2)`

The distance between `f(0)` and `f(2)` can be calculated using the Pythagorean Theorem
`color(white)("XXX")d=sqrt((sqrt(2)/3-0)^2+(2-0)^2)`

`color(white)("XXX")=sqrt(4/9+4)`

`color(white)("XXX")=sqrt(40/9)`

`color(white)("XXX")=2/3sqrt(10)`