find the exact arc length of the parametric curve:
x=t^2cost, y=t^2sint, 0<=t<=2
how do i go about solving the above question?
$\displaystyle \displaystyle L = \int_{\alpha}^{\beta} \sqrt{\bigg(\frac{dx}{dt}\bigg)^2 + \bigg(\frac{dy}{dt}\bigg)^2} \, dt $
for $\displaystyle \displaystyle\alpha \leq t \leq \beta $ (so in your case $\displaystyle \alpha = 0 \, , \beta = 2 $
So, find $\displaystyle \frac{dx}{dt} $ and $\displaystyle \frac{dy}{dx} $, then plug them into that equation, simplify the integrand as much as possible, and then integrate it with the limits.
Also see this picture
This is easier to calculate the integral $\displaystyle \int\limits_0^2 {t\sqrt {{t^2} + 4} dt}$, using this substitution $\displaystyle \sqrt {{t^2} + 4} = u$.
$\displaystyle \int\limits_0^2 {t\sqrt {{t^2} + 4} dt} = \left\{ \begin{gathered}\sqrt {{t^2} + 4} = u, \hfill \\tdt = udu, \hfill \\0 \leqslant t \leqslant 2, \hfill \\2 \leqslant u \leqslant 2\sqrt 2 \hfill \\ \end{gathered} \right\} = \int\limits_2^{2\sqrt 2 } {{u^2}du} = \left. {\frac{{{u^3}}}{3}} \right|_2^{2\sqrt 2 } =$
$\displaystyle = \frac{1}{3}\left( {8\sqrt 8 - 8} \right) = \frac{8}{3}\left( {2\sqrt 2 - 1} \right) \approx 4.8758{\text{ }}\left( {{\text{linear units}}} \right).$