Find the arclength parametrization of the curve $$r(t) = (t^2, t^3 ), t>=0$$
I find the integral of |r'(t)| = ${1\over9} \left( {(4+9t^2)^3\over 3} -{4^3\over 3}\right)$
The integral looks horrible and I don't know what to do next. Can anyone help me?
If you have $x(t)$ and $y(t)$, the arc length is given by $$L=\int \sqrt{x'(t)^2+y'(t)^2}dt$$ So, in your case, $x(t)=t^2$, $y(t)=t^3$, $x'(t)=2t$, $y'(t)=3t^2$ and then $$L=\int \sqrt{9 t^4+4 t^2}dt=\int t\sqrt{9 t^2+4 }dt$$ in which you could notice that $t$ looks very much as the derivative of $(9t^2+4)$ and, so, a $u$ substitution could make the problem easy to solve.
