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Thread: Polar to Cartesian Equations

  1. #1
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    Polar to Cartesian Equations

    Instructions:
    Replace the polar equations in Exercises 23-48 by equivalent Cartesian equations. Then describe or identify the graph.

    40. $\displaystyle \cos^2(\theta) = \sin^2(\theta)$

    I think I have it solved, but I'm unsure whether I am correct or not.. All problems in this section have an $\displaystyle r$ to convert to Cartesian (ie $\displaystyle r\cos(\theta) = x$). However, this problem does not.

    So I multiplied the entire problem by $\displaystyle \frac {r^2}{r^2}$ which gave me the following:

    $\displaystyle \frac {r^2\cos^2(\theta)}{r^2} = \frac {r^2\sin^2(theta)}{r^2}$
    $\displaystyle \sqrt{\frac {r^2\cos^2(\theta)}{r^2}} = \sqrt{\frac {r^2\sin^2(theta)}{r^2}}$
    $\displaystyle \pm \frac {r\cos(\theta)}{r} = \pm \frac {r\sin(\theta)}{r}$
    $\displaystyle \pm \frac {x}{r} = \pm \frac {y}{r}$
    $\displaystyle \pm x = \pm y$

    Is this correct? Can I drop the $\displaystyle \pm$ sign? And if so, why?

    Thanks!
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  2. #2
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    Quote Originally Posted by Calix View Post
    Instructions:
    Replace the polar equations in Exercises 23-48 by equivalent Cartesian equations. Then describe or identify the graph.

    40. $\displaystyle \cos^2(\theta) = \sin^2(\theta)$

    I think I have it solved, but I'm unsure whether I am correct or not.. All problems in this section have an $\displaystyle r$ to convert to Cartesian (ie $\displaystyle r\cos(\theta) = x$). However, this problem does not.

    So I multiplied the entire problem by $\displaystyle \frac {r^2}{r^2}$ which gave me the following:

    $\displaystyle \frac {r^2\cos^2(\theta)}{r^2} = \frac {r^2\sin^2(theta)}{r^2}$
    $\displaystyle \sqrt{\frac {r^2\cos^2(\theta)}{r^2}} = \sqrt{\frac {r^2\sin^2(theta)}{r^2}}$
    $\displaystyle \pm \frac {r\cos(\theta)}{r} = \pm \frac {r\sin(\theta)}{r}$
    $\displaystyle \pm \frac {x}{r} = \pm \frac {y}{r}$
    $\displaystyle \pm x = \pm y$

    Is this correct? Can I drop the $\displaystyle \pm$ sign? And if so, why?

    Thanks!
    $\displaystyle x = r \cos \theta \Rightarrow \cos \theta = \frac{x}{r}$.

    $\displaystyle y = r \sin \theta \Rightarrow \sin \theta = \frac{y}{r}$.

    Therefore $\displaystyle \cos^2 \theta = \sin^2 \theta \Rightarrow \frac{x^2}{r^2} = \frac{y^2}{r^2} \Rightarrow y^2 = x^2 \Rightarrow y = \pm x, ~ r \neq 0$.

    This answer makes perfect sense since the gradient of these two lines is equal to $\displaystyle \pm 1$ and so the angle they make with the x-axis is $\displaystyle \frac{\pi}{4}, ~ \frac{3 \pi}{4}, ~ \frac{5\pi}{4}$ and $\displaystyle \frac{7\pi}{4}$.
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  3. #3
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    Hah, I knew I missed something pretty obvious. Going on only a few hours of sleep..

    Thanks a lot!
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