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Thread: Polar coords and differentiation

  1. #1
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    Polar coords and differentiation

    Hey guys, got a polar coordinates question Im stuck on.

    A particle is moving with constant velocity $\displaystyle \vec{v} = u\hat{i}$ along the line $\displaystyle y = 2$. Describe $\displaystyle \vec{v}$ in polar coordinates.

    I worked out that $\displaystyle r_t = \sqrt{(vt)^2 + 4}$ and differentiating that shouldnt be too much of a problem but Im stuck with theta

    Ive got $\displaystyle \theta_t = \arctan\frac{y}{x} = \arctan\frac{2}{vt}$
    but Im guessing I cant just differentiate this using $\displaystyle \frac{d}{dx} arctan(x) = \frac{1}{x^2 + 1}$, since theres $\displaystyle \frac{1}{t}$ in there instead of $\displaystyle t$.

    I worked through it using $\displaystyle \frac{d\theta}{dt} = \frac{1}{\frac{dt}{d\theta}}$ and ended up with

    $\displaystyle \displaystyle{\frac{d\theta}{dt}}=\frac{1}{\frac{d }{d\theta}(\frac{2}{\vec{v}}\frac{1}{tan\theta})}= \frac{1}{-\frac{2}{\vec{v}}\frac{sec^2\theta}{tan^2\theta}}$

    $\displaystyle \frac{d\theta}{dt}=-\frac{\vec{v}tan^2\theta}{2sec^2\theta}=-\frac{\vec{v}}{2}\tan^2(\theta)cos^2(\theta)$

    but I'm not sure how to get this in terms of t or if I even worked it out right. Also Im hoping I can replace my $\displaystyle \vec{v}$s in the formulas with $\displaystyle u$, I think I only need the magnitude for polar coords?

    any help would be appreciated greatly!
    Last edited by Snoopey; Oct 26th 2008 at 01:55 PM. Reason: added a bunch of latex code
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