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:

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

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