Answer
$70\sqrt 5$
Work Step by Step
$x=6t^2$, $y=2t^3$, $1\leq t \leq4$
The arc length of a parametric curve is calculated using the formula:
$L=\int_{a}^{b}\sqrt ((\frac{dx}{dt})^2+(\frac{dy}{dt})^2)dt$
$\frac{dx}{dt}=12t$
$\frac{dy}{dt}=6t^2$
$L=\int_{1}^{4}\sqrt ((12t)^2+(6t^2)^2)dt$
$L=\int_{1}^{4}\sqrt (144t^2+36t^4)dt$
$L=\int_{1}^{4}\sqrt (36t^2(4+t^2))dt$
$L=\int_{1}^{4}6t\sqrt (4+t^2)dt$
Use u-sub:
$let$ $u = 4+t^2$
$du=2tdt$
$\frac{1}{2}du=tdt$
$u(1)=5$
$u(4)=20$
$L=\frac{1}{2}\int_{5}^{20}6\sqrt u du$
$L=3[\frac{2}{3}u^{\frac{3}{2}}]_{5}^{20}$
$L=2[u^{\frac{3}{2}}]_{5}^{20}$
$L=2[20^{\frac{3}{2}}-5^{\frac{3}{2}}]$
$L=2[(4\times5)^{\frac{3}{2}}-5\sqrt 5]$
$L=2[8(5\sqrt 5)-5\sqrt 5]$
$L=2[40\sqrt 5-5\sqrt 5]$
$L=70\sqrt 5$
