What is the arclength of #f(t) = (-(t+3)^2,3t-4)# on #t in [0,1]#?
1 Answer
Jun 15, 2018
Explanation:
#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))#