Question

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

Answer 1

`1/27[sqrt[[85^3]]-sqrt[[40^3]]]`

Explanation:

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.