What is the arclength of (t-3t^2,t^2-t)(t−3t2,t2−t) on t in [1,2]t∈[1,2]?
1 Answer
May 19, 2018
Explanation:
f(t)=(t-3t^2,t^2-t)f(t)=(t−3t2,t2−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))
