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Math Help - Polar differential problem with my workings...

  1. #1
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    Polar differential problem with my workings...

    Hi All,

    I'm having problems with this question and need someone to look at my workings and give a gentle push in the right direction. If anyone can help I'd be most grateful.

    Calculate \frac{dy}{dx}for the curve whose equation in polar coordinates is r=1+sin^2\theta

    So my understanding is that polar coords are given as r,\theta so not quite sure what to do with this. Would I rearrange to find \theta such that \theta = sin^{-1}(\pm\sqrt{r-1}) and keep r as it is????

    Or is it using r and  \theta with  x = rcos \theta & y=rsin \theta

    Then find  \frac{dx}{d\theta} and hence \frac{d\theta}{dx} and  \frac{dy}{d\theta} and finally get the result \frac{dy}{dx} =-cot\theta

    Best regards, Felix
    Last edited by FelixHelix; November 7th 2011 at 01:39 PM.
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  2. #2
    MHF Contributor alexmahone's Avatar
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    Re: Polar differential problem with my workings...

    Quote Originally Posted by FelixHelix View Post
    Hi All,

    I'm having problems with this question and need someone to look at my workings and give a gentle push in the right direction. If anyone can help I'd be most grateful.

    Calculate \frac{dy}{dx}for the curve whose equation in polar coordinates is r=1+sin^2\theta

    So my understanding is that polar coords are given as r,\theta so not quite sure what to do with this. Would I rearrange to find \theta such that \theta = sin^{-1}(\pm\sqrt{r-1}) and keep r as it is????

    Or is it using r and  \theta with  x = rcos \theta & y=rsin \theta

    Then find  \frac{dx}{d\theta} and hence \frac{d\theta}{dx} and  \frac{dy}{d\theta} and finally get the result \frac{dy}{dx} -cot\theta

    Best regards, Felix
    \frac{dx}{d\theta}=-r\sin \theta+\cos \theta\frac{dr}{d\theta}

    \frac{dy}{d\theta}=r\cos \theta+\sin \theta\frac{dr}{d\theta}

    Note that \frac{dr}{d\theta}\neq 0 in this case.
    Last edited by alexmahone; November 7th 2011 at 01:51 PM.
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  3. #3
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    Re: Polar differential problem with my workings...

    I'm not sure what you have done here. Can you expalin? Also I'm guessing that that is a typo on your second line and should read \frac{dy}{d\theta}...
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  4. #4
    MHF Contributor alexmahone's Avatar
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    Re: Polar differential problem with my workings...

    Quote Originally Posted by FelixHelix View Post
    I'm not sure what you have done here. Can you expalin? Also I'm guessing that that is a typo on your second line and should read \frac{dy}{d\theta}...
    Sorry about the typo.

    I have merely used the product rule to differentiate x=r\cos \theta and y=r\sin \theta.
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  5. #5
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    Re: Polar differential problem with my workings...

    Quote Originally Posted by FelixHelix View Post
    Hi All,

    I'm having problems with this question and need someone to look at my workings and give a gentle push in the right direction. If anyone can help I'd be most grateful.

    Calculate \frac{dy}{dx}for the curve whose equation in polar coordinates is r=1+sin^2\theta

    So my understanding is that polar coords are given as r,\theta so not quite sure what to do with this. Would I rearrange to find \theta such that \theta = sin^{-1}(\pm\sqrt{r-1}) and keep r as it is????

    Or is it using r and  \theta with  x = rcos \theta & y=rsin \theta

    Then find  \frac{dx}{d\theta} and hence \frac{d\theta}{dx} and  \frac{dy}{d\theta} and finally get the result \frac{dy}{dx} =-cot\theta

    Best regards, Felix
    You should know that \displaystyle x = r\cos{\theta}, y = r\sin{\theta}, x^2 + y^2 = r^2.

    So

    \displaystyle \begin{align*} r &= 1 + \sin^2{\theta} \\ \sqrt{x^2 + y^2} &= 1 + \left(\frac{y}{r}\right)^2 \\ \sqrt{x^2 + y^2} &= 1 + \frac{y^2}{r^2} \\ \sqrt{x^2 + y^2} &= 1 + \frac{y^2}{x^2 + y^2} \\ \sqrt{x^2 + y^2} &= \frac{x^2 + 2y^2}{x^2 + y^2} \\ x^2 + y^2 &= \left(\frac{x^2 + 2y^2}{x^2 + y^2}\right)^2 \\ x^2 + y^2 &= \frac{\left(x^2 + 2y^2\right)^2}{\left(x^2 + y^2\right)^2} \\ \left(x^2 + y^2\right)^3 &= \left(x^2 + 2y^2\right)^2 \\ \frac{d}{dx}\left[\left(x^2 + y^2\right)^3\right] &= \frac{d}{dx}\left[\left(x^2 + 2y^2\right)^2\right] \\ 3\left(x^2 + y^2\right)^2\left(2x + 2y\,\frac{dy}{dx}\right) &= 2\left(x^2 + 2y^2\right)\left(2x + 4y\,\frac{dy}{dx}\right) \\ 6x\left(x^2 + y^2\right)^2 + 6y\left(x^2 + y^2\right)^2\frac{dy}{dx} &= 4x\left(x^2 + 2y^2\right) + 8y\left(x^2 + 2y^2\right)\frac{dy}{dx} \\ 6y\left(x^2 + y^2\right)^2\frac{dy}{dx} - 8y\left(x^2 + 2y^2\right)\frac{dy}{dx} &= 4x\left(x^2 + 2y^2\right) - 6x\left(x^2 + y^2\right)^2\end{align*}

    \begin{align*} 2y\left[3\left(x^2 + y^2\right)^2 - 4\left(x^2 + 2y^2\right)\right]\frac{dy}{dx} &= 2x\left[2\left(x^2 + 2y^2\right) - 3\left(x^2 + y^2\right)^2 \right] \\ \frac{dy}{dx} &= \frac{2x\left[2\left(x^2 + 2y^2\right) - 3\left(x^2 + y^2\right)^2\right]}{2y\left[3\left(x^2 + y^2\right)^2 - 4\left(x^2 + 2y^2\right)\right]} \\ \frac{dy}{dx} &= \frac{x\left[2\left(x^2 + 2y^2\right) - 3\left(x^2 + y^2\right)^2\right]}{y\left[3\left(x^2 + y^2\right)^2 - 4\left(x^2 + 2y^2\right)\right]} \end{align*}
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