What is the arclength of #(t-3t^2,t^2-t)# on #t in [1,2]#?

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Answer 1 · 2018-05-19T13:40:21

#L=1/10(9sqrt130-4sqrt26)+1/(10sqrt10)ln((18+5sqrt13)/(8+sqrt65))# units.

#f(t)=(t-3t^2,t^2-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 #10t-2=tantheta#:

#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))#