Answer :
It's simple, you just have to do this:
[tex]L=\int\limits_{a}^{b}\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}~dt[/tex]
[tex]x=3t-t^3[/tex]
[tex]\frac{dx}{dt}=3-3t^2[/tex]
[tex]y=3t^2[/tex]
[tex]\frac{dy}{dt}=6t[/tex]
replacing
[tex]L=\int\limits_{0}^{\sqrt{3}}\sqrt{\left(3-3t^2\right)^2+\left(6t\right)^2}~dt[/tex]
[tex]L=\int\limits_{0}^{\sqrt{3}}\sqrt{9-18t^2+9t^4+36t^2}~dt[/tex]
[tex]L=\int\limits_{0}^{\sqrt{3}}\sqrt{9+9t^4+18t^2}~dt[/tex]
[tex]L=\int\limits_{0}^{\sqrt{3}}\sqrt{(3t^2+3)^2}~dt[/tex]
[tex]L=\int\limits_{0}^{\sqrt{3}}(3t^2+3)~dt[/tex]
[tex]L=t^3+3t|_0^{\sqrt{3}}[/tex]
[tex]\boxed{\boxed{L=3\sqrt{3}+3\sqrt{3}=6\sqrt{3}}}[/tex]
[tex]L=\int\limits_{a}^{b}\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}~dt[/tex]
[tex]x=3t-t^3[/tex]
[tex]\frac{dx}{dt}=3-3t^2[/tex]
[tex]y=3t^2[/tex]
[tex]\frac{dy}{dt}=6t[/tex]
replacing
[tex]L=\int\limits_{0}^{\sqrt{3}}\sqrt{\left(3-3t^2\right)^2+\left(6t\right)^2}~dt[/tex]
[tex]L=\int\limits_{0}^{\sqrt{3}}\sqrt{9-18t^2+9t^4+36t^2}~dt[/tex]
[tex]L=\int\limits_{0}^{\sqrt{3}}\sqrt{9+9t^4+18t^2}~dt[/tex]
[tex]L=\int\limits_{0}^{\sqrt{3}}\sqrt{(3t^2+3)^2}~dt[/tex]
[tex]L=\int\limits_{0}^{\sqrt{3}}(3t^2+3)~dt[/tex]
[tex]L=t^3+3t|_0^{\sqrt{3}}[/tex]
[tex]\boxed{\boxed{L=3\sqrt{3}+3\sqrt{3}=6\sqrt{3}}}[/tex]
