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Math Help - Derivative

  1. #1
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    Derivative

    If  r(t) = |\bold{r}(t)| = \sqrt{(t^2-t)^2 + t^2(2t-t^2)} then does  \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
    Math Engineering Student
    Krizalid's Avatar
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    \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 \left( {t^2  - t} \right)^2  + t^2 \left( {2t - t^2 } \right) = t^2 . Hence, considering t>0 your derivative is equal to 1.
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  3. #3
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    To find  \frac{d \theta}{dt} I used the formula  \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  \bold{u}_r and  \bold{u}_\theta , where  \bold{r}(t) = (t^2-t, t \sqrt{2t-t^2})
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