# How do you find the length of the curve x=1+3t^2, y=4+2t^3, where 0<=t<=1 ?

Oct 10, 2014

By taking the derivative with respect to $t$,

$\left\{\begin{matrix}x ' \left(t\right) = 6 t \\ y ' \left(t\right) = 6 {t}^{2}\end{matrix}\right.$

Let us now find the length $L$ of the curve.

$L = {\int}_{0}^{1} \sqrt{{\left[x ' \left(t\right)\right]}^{2} + {\left[y ' \left(t\right)\right]}^{2}} \mathrm{dt}$

$= {\int}_{0}^{1} \sqrt{{6}^{2} {t}^{2} + {6}^{2} {t}^{4}} \mathrm{dt}$

by pulling $6 t$ out of the square-root,

$= {\int}_{0}^{1} 6 t \sqrt{1 + {t}^{2}} \mathrm{dt}$

by rewriting a bit further,

$= 3 {\int}_{0}^{1} 2 t {\left(1 + {t}^{2}\right)}^{\frac{1}{2}} \mathrm{dt}$

by General Power Rule,

$= 3 {\left[\frac{2}{3} {\left(1 + {t}^{2}\right)}^{\frac{3}{2}}\right]}_{0}^{1} = 2 \left({2}^{\frac{3}{2}} - 1\right)$

I hope that this was helpful.