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

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Answer 1 · 2018-03-08T03:55:55

The arclength is $1/2(2sqrt17-sqrt5)+1/4ln((4+sqrt17)/(2+sqrt5))$ units.

$f(t)=(t-3,t^2)$

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

Arclength is given by:

$L=int_1^2sqrt(1+4t^2)dt$

Apply the substitution $2t=tantheta$ :

$L=1/2intsec^3thetad theta$

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:

$L=1/4[secthetatantheta+ln|sectheta+tantheta|]$

Reverse the substitution:

$L=1/4[2tsqrt(1+4t^2)+ln|2t+sqrt(1+4t^2)|]_1^2$

Insert the limits of integration:

$L=1/2(2sqrt17-sqrt5)+1/4ln((4+sqrt17)/(2+sqrt5))$

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