# Finding arc lengths

• Aug 6th 2009, 01:35 AM
Jonny-123
Finding arc lengths
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?
• Aug 6th 2009, 01:43 AM
Mush
$\displaystyle \displaystyle L = \int_{\alpha}^{\beta} \sqrt{\bigg(\frac{dx}{dt}\bigg)^2 + \bigg(\frac{dy}{dt}\bigg)^2} \, dt$

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.
• Aug 6th 2009, 02:01 AM
Calculus26
See attachment
• Aug 6th 2009, 07:28 AM
DeMath
Quote:

Originally Posted by Jonny-123
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?

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 $\displaystyle \sqrt {{t^2} + 4} = u$.
$\displaystyle \int\limits_0^2 {t\sqrt {{t^2} + 4} dt} = \left\{ \begin{gathered}\sqrt {{t^2} + 4} = u, \hfill \\tdt = udu, \hfill \\0 \leqslant t \leqslant 2, \hfill \\2 \leqslant u \leqslant 2\sqrt 2 \hfill \\ \end{gathered} \right\} = \int\limits_2^{2\sqrt 2 } {{u^2}du} = \left. {\frac{{{u^3}}}{3}} \right|_2^{2\sqrt 2 } =$
$\displaystyle = \frac{1}{3}\left( {8\sqrt 8 - 8} \right) = \frac{8}{3}\left( {2\sqrt 2 - 1} \right) \approx 4.8758{\text{ }}\left( {{\text{linear units}}} \right).$