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Thread: Derivative

  1. #1
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    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 ] $?
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  2. #2
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    Krizalid's Avatar
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    $\displaystyle \left( {t^2 - t} \right)^2 + t^2 \left( {2t - t^2 } \right) = t^2 \left( {t - 1} \right)^2 + t^2 \left( {2t - t^2 } \right) = t^2 \Big[ {\left( {t - 1} \right)^2 + 2t - t^2 } \Big].$

    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.
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  3. #3
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    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}) $
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