Question

How do you find the length of the curve x=3t−t3x=3t-t^3, y=3t2y=3t^2, where 0≤t≤3√0<=t<=sqrt(3) ?

Answer 1

It's simple, you just have to do this:

$L=\int\limits_{a}^{b}\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}~dt$

$x=3t-t^3$

$\frac{dx}{dt}=3-3t^2$

$y=3t^2$

$\frac{dy}{dt}=6t$

replacing

$L=\int\limits_{0}^{\sqrt{3}}\sqrt{\left(3-3t^2\right)^2+\left(6t\right)^2}~dt$

$L=\int\limits_{0}^{\sqrt{3}}\sqrt{9-18t^2+9t^4+36t^2}~dt$

$L=\int\limits_{0}^{\sqrt{3}}\sqrt{9+9t^4+18t^2}~dt$

$L=\int\limits_{0}^{\sqrt{3}}\sqrt{(3t^2+3)^2}~dt$

$L=\int\limits_{0}^{\sqrt{3}}(3t^2+3)~dt$

$L=t^3+3t|_0^{\sqrt{3}}$

$\boxed{\boxed{L=3\sqrt{3}+3\sqrt{3}=6\sqrt{3}}}$