How do you find the length of the curve #x=e^t+e^-t#, #y=5-2t#, where #0<=t<=3# ?

1 Answer
Aug 20, 2014

The answer is #e^3-e^(-3)#.

Recall that the arclength for parametric curves is:

#L=int_a^b sqrt(((dx)/(dt))^2+((dy)/(dt))^2)dt#

So,

#(dx)/(dt)=e^t-e^(-t)#
#(dy)/(dt)=-2#

Now substituting:

#L=int_0^3 sqrt((e^t-e^(-t))^2+(-2)^2)dt#
#=int_0^3 sqrt(e^(2t)-2+e^(-2t)+4)dt# expand
#=int_0^3 sqrt(e^(2t)+2+e^(-2t))dt# simplify
#=int_0^3 sqrt((e^t+e^(-t))^2)dt# factor
#=int_0^3 (e^t+e^(-t))dt# simplify
#=e^t-e^(-t)|_0^3# integrate
#=e^3-e^(-3)-(e^0-e^0)# evaluate
#=e^3-e^(-3)#

Note that there aren't many questions that can be solved algebraically. Please note the pattern of this problem because most algebraic solutions have this form.