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

I got #sqrt((2 - e + 1/e)^2 + 4)#, or #~~2.03#.

This was given in parametric coordinates, that is, when we say something like, #x = f(t)# and #y = g(t)#.

By implication, it's saying that:

#x = t - e^(t-1)#
#y = t + 1#

and it's asking what the distance between two #\mathbf("("x,y")")# coordinates is, if one of them has #t = 0# and the other has #t = 2#. We can determine values for #x# and #y#, and treat it like any other distance calculation for #x# and #y#.

The distance formula is just the Pythagorean theorem applied to coordinates:

#\mathbf(D = sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2))#

You're basically finding the distance between two points, where the horizontal and vertical distances are how you mapped your #(x,y)# coordinates.

We assigned #x_1# and #y_1# to #t = 0#, and #x_2# and #y_2# to #t = 2#. Now, let's examine #t = 0#:

#color(green)(x_1) = (0) - e^((0)-1) = -e^(-1) = color(green)(-1/e)#
#color(green)(y_1) = (0) + 1 = color(green)(1)#

And for #t = 2#:

#color(green)(x_2) = (2) - e^((2)-1) = color(green)(2-e)#
#color(green)(y_2) = (2) + 1 = color(green)(3)#

So really, in #(x,y)# coordinates, you're finding the distance from #(-1/e,1)# to #(2-e,3)#.

Once we got that translated to Cartesian coordinates, it's back to regular algebra. :) Therefore, the distance from #f(t=0)# to #f(t=2)# is:

#color(blue)(D_(f(0))^(f(2))) = sqrt(((2-e) - (-1/e))^2 + (3 - 1)^2)#

#= sqrt((2 - e + 1/e)^2 + 4)#

# ~~ color(blue)(2.03)#