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

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Answer 1 · 2018-06-16T08:34:32

$L=1/8(3sqrt5+5sqrt13)+1/(8sqrt2)ln((sqrt10+3)/(sqrt26-5))$ units.

$f(t)=(t^2-t,t^2-1)$

$f'(t)=(2t-1,2t)$

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 $4t-1=tantheta$ :

$L=1/(4sqrt2)intsec^3thetad theta$

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:

$L=1/8(3sqrt5+5sqrt13)+1/(8sqrt2)ln((sqrt10+3)/(sqrt26-5))$

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