What is the arc length of f(t)=(t^2-4t,5-t) f(t)=(t2−4t,5−t) over t in [3,4] t∈[3,4]?
1 Answer
Mar 16, 2018
Explanation:
f(t)=(t^2−4t,5−t)f(t)=(t2−4t,5−t)
f'(t)=(2t−4,−1)
Arc length is given by:
L=int_3^4sqrt((2t-4)^2+1)dt
Apply the substitution
L=1/2intsec^3thetad theta
This is a known integral:
L=1/4[secthetatantheta+ln|sectheta+tantheta|]_3^4
Reverse the substitution:
L=1/4[(2t-4)sqrt((2t-4)^2+1)+ln|(2t-4)+sqrt((2t-4)^2+1)|]_3^4
Insert the limits of integration:
L=1/2(2sqrt17-sqrt5)+1/4ln((4+sqrt17)/(2+sqrt5))