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Math Help - Conversion of polar equations into cartesian equation

  1. #1
    Member roshanhero's Avatar
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    Conversion of polar equations into cartesian equation

    Find the cartesian equations of these polar equations-
    1.r=a(theta)/(theta-1)
    2.r(1-e^theta)=a
    3.r(theta)cos(theta)=acos2(theta)
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  2. #2
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    Hello, roshanhero!

    These are particularly ugly problems!
    . . Very little simpliying is possible.

    We need: . \begin{array}{ccc}r^2 &=&x^2+y^2 \\<br />
r\cos\theta &=& x \\ r\sin\theta &=& y \\ \theta &=&\arctan(\frac{y}{x}) \end{array}


    Find the cartesian equations of these polar equations:

    1)\;r\:=\:\frac{a\theta}{\theta-1}

    We have: . \sqrt{x^2+y^2} \;-\;\frac{a\arctan\left(\frac{y}{x}\right)}{\arctan\  left(\frac{y}{x}\right) - 1}



    2)\;r\left(1-e^{\theta}\right) \:=\:a

    We have: . \sqrt{x^2+y^2}\left(1 - e^{\arctan(\frac{y}{x})}\right) \:=\:a



    3)\; r\!\cdot\!\theta\!\cdot\!\cos\theta \:=\:a\cos2\theta

    Multiply by r^2\!:\;\;r^3\!\cdot\!\theta\!\cdot\!\cos\theta \;=\;ar^2\!\cdot\!\cos2\theta

    . . . . . . r^2\!\cdot\!\theta\!\cdot\!r\cos\theta \;=\;ar^2(2\cos^2\!\theta - 1)

    . . . . . . r^2\!\cdot\!\theta\!\cdot\!r\cos\theta \;= \;2ar^2\cos^2\!\theta - ar^2

    . . . . . . r^2\!\cdot\!\theta\!\cdot\!r\cos\theta \;=\;2a(r\cos\theta)^2 - ar^2

    . . (x^2+y^2)\!\cdot\!\arctan(\frac{y}{x})\!\cdot\!x \;=\;2ax^2 - a(x^2+y^2)

    . . x(x^2+y^2)\arctan\left(\frac{y}{x}\right) \;=<br />
\;a(x^2-y^2)

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  3. #3
    Member roshanhero's Avatar
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    Can't we have the form that contains only x and y terms?
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  4. #4
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    Quote Originally Posted by roshanhero View Post
    Can't we have the form that contains only x and y terms?

    Isn't that what I found?

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