What is the arclength of #(t^2-t,t^2-1)# on #t in [-1,1]#?
1 Answer
Explanation:
#f(t)=(t^2-t,t^2-1)#
#f'(t)=(2t-1,2t)#
Arclength is given by:
#L=int_-1^1sqrt((2t-1)^2+4t^2)dt#
Expand the square:
#L=int_-1^1sqrt(8t^2-4t+1)dt#
Complete the square:
#L=1/sqrt2int_-1^1sqrt((4t-1)^2+1)dt#
Apply the substitution
#L=1/(4sqrt2)intsec^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=1/(8sqrt2)[secthetatantheta+ln|sectheta+tantheta|]#
Reverse the substitution:
#L=1/(8sqrt2)[(4t-1)sqrt((4t-1)^2+1)+ln|4t-1+sqrt((4t-1)^2+1)|]_-1^1#
Insert the limits of integration:
#L=1/(8sqrt2)(3sqrt10+5sqrt26+ln((3+sqrt10)/(-5+sqrt26)))#
Hence:
#L=1/8(3sqrt5+5sqrt13)+1/(8sqrt2)ln((sqrt10+3)/(sqrt26-5))#
