What is the arclength of f(t) = (-(t+3)^2,3t-4)f(t)=(−(t+3)2,3t−4) on t in [0,1]t∈[0,1]?
1 Answer
Jun 15, 2018
Explanation:
f(t)=(−(t+3)2,3t−4)f(t)=(−(t+3)2,3t−4)
f'(t)=(−2(t+3),3)
Arclength is given by:
L=int_0^1sqrt(4(t+3)^2+9)dt
Apply the substitution
L=9/2intsec^3thetad theta
This is a known integral. If you do not have it memorized apply integration by parts or look it up in a table of integrals:
L=9/4[secthetatantheta+ln|sectheta+tantheta|]
Reverse the substitution:
L=[1/2(t+3)sqrt(4(t+3)^2+9)+9/4ln|2(t+3)+sqrt(4(t+3)^2+9)|]_0^1
Insert the limits of integration:
L=2sqrt73-9/2sqrt5+9/4ln((8+sqrt73)/(6+3sqrt5))