What is the arclength of #(t-3t^2,t^2-t)# on #t in [1,2]#?
1 Answer
May 19, 2018
Explanation:
#f(t)=(t-3t^2,t^2-t)#
#f'(t)=(1-6t,2t-1)#
Arclength is given by:
#L=int_1^2sqrt((1-6t)^2+(2t-1)^2)dt#
Expand the squares and combine terms:
#L=int_1^2sqrt(40t^2-16t+2)dt#
Complete the square in the square root:
#L=sqrt(2/5)int_1^2sqrt((10t-2)^2+1)dt#
Apply the substitution
#L=1/(5sqrt10)intsec^3thetad theta#
This is a known integral:
#L=1/(10sqrt10)[secthetatantheta+ln|sectheta+tantheta|]#
Reverse the substitution:
#L=1/(10sqrt10)[(10t-2)sqrt((10t-2)^2+1)+ln|(10t-2)+sqrt((10t-2)^2+1)|]_1^2#
Hence
#L=1/(10sqrt10){90sqrt13-8sqrt65+ln((18+5sqrt13)/(8+sqrt65))}#
Simplify:
#L=1/10(9sqrt130-4sqrt26)+1/(10sqrt10)ln((18+5sqrt13)/(8+sqrt65))#