What is the arclength of #f(t) = (t-e^(t),t-2/t+3)# on #t in [1,2]#?

1 Answer
Jan 17, 2018

#~4.298#

Explanation:

If we have a really small change in #t#, we will travel according to the following equation:
#ds^2 = dx^2 + dy^2 = [((dx)/(dt))^2 +((dy)/(dt))^2]dt^2 #
i.e.
#ds = dt sqrt(((dx)/(dt))^2 +((dy)/(dt))^2) #

The total length is therefore
#int_(t_0)^(t_f) ds = int_(t_0)^(t_f) sqrt(((dx)/(dt))^2 +((dy)/(dt))^2) dt#

Calculating the values,
#(dx)/(dt) = 1 - e^t#
#(dy)/(dt) = 1 + 2/t^2 #

Therefore, plugging in the numbers, we have the equation
#int_(1)^(2) sqrt((( 1 - e^t )^2 +(1 + 2/t^2)^2) dt #

We can't solve this exactly (or at least not in any easy way if it is technically possible), so we plug this into a calculator such as WolframAlpha, getting the value around #4.298#.