Given, #f[t]# = #[t^3-1, t^2-1]
# i.e, #[x=t^3-1, y=t^2-1]#.....#[1]#
Differentiating w.r.t. #t#,
#dx/dt=3t^2, dy/dt=2t#.....#[2]#. The arc length of a curve in parametric form is given by, #L#=#int_2^3sqrt[[dx/dt]^2+[dy/dt]^2]dt#, substituting #dx/dt and dy/dt# from ....#[2]# into the length formula #L# will give,
#L#=#int_2^3sqrt[9t^4+4t^2dt]# = #int_2^3sqrt[t^2[9t^2+4]]#=#int_2^3tsqrt[9t^2+4]dt#.
We can now make a substitution by letting #[u=9t^2+4]#........#[3]#. Differentiating .....#[3]# w.r.t #x#, #du/dt= 18t#, .i.e, #[du]/[18t]#=#dt#.
So we now have #L#=#intt sqrt[u]/[18t]dt# = #1/18intsqrt[u]dt# [ since the terms in #t#will cancel],
And after integrating w.r.t. #t# we are left with #1/18 [2/3]sqrt[u^3]# = #1/27sqrt[u^3# we now have change the bounds of integration, From ......#[3]#, #u=9t^2+4#, so when #t=2, u=40#, when #t=3, u=85#.
So length #L#=#1/27[sqrt[85^3]-sqrt[40^3]]# units.
