1. ## Derivative

If $\displaystyle r(t) = |\bold{r}(t)| = \sqrt{(t^2-t)^2 + t^2(2t-t^2)}$ then does $\displaystyle \frac{dr}{dt} = \frac{1}{2} \left[ (t^2-t)^{2} + t^{2}(2t-t^{2}) \right]^{-\frac{1}{2}} \cdot \left [ 2(2t-1) + 2t(2-2t) \right ]$?

2. $\displaystyle \left( {t^2 - t} \right)^2 + t^2 \left( {2t - t^2 } \right) = t^2 \left( {t - 1} \right)^2 + t^2 \left( {2t - t^2 } \right) = t^2 \Big[ {\left( {t - 1} \right)^2 + 2t - t^2 } \Big].$

So $\displaystyle \left( {t^2 - t} \right)^2 + t^2 \left( {2t - t^2 } \right) = t^2 .$ Hence, considering $\displaystyle t>0$ your derivative is equal to 1.

3. To find $\displaystyle \frac{d \theta}{dt}$ I used the formula $\displaystyle \frac{d \theta}{dt} = \frac{x(dy/dt) - y(dx/dt)}{x^2+y^2}$.

I got a messy expression.

Basically I want to find the velocity and acceleration in terms of $\displaystyle \bold{u}_r$ and $\displaystyle \bold{u}_\theta$, where $\displaystyle \bold{r}(t) = (t^2-t, t \sqrt{2t-t^2})$