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Math Help - CONVERTING b/w POLAR equations & RECTANGULAR equations

  1. #1
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    CONVERTING b/w POLAR equations & RECTANGULAR equations

    I 'd like to see the steps it takes to solve each of these problems. Thanks!
    Polar to Rectangular:

    1.) r2 = sin2θ

    2.) r = 2secθ ( i got x=2, not sure if it's right)

    3.) r = 6/(2cosθ - 3sinθ)

    Rectangular to Polar:

    1.) 2xy=1

    2.) y2- 8y - 16 = 0

    3.) x2 + y2 - 2ay = 0

    4.) x2 = y3 ( my final answer is r= cot2θcscθ )
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  2. #2
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    Re: CONVERTING b/w POLAR equations & RECTANGULAR equations

    Hello, Crysland!

    Polar to Rectangular:

    (1)\; r^2\:=\:\sin2\theta

    We have: . . . . . . . r^2 \:=\:2\sin\theta\cos\theta

    Multiply by r^2\!:\qquad\;\; r^4 \:=\:2r^2\sin\theta\cos\theta

    . . . . . . . . . . . . . (r^2)^2 \:=\:2(r\cos\theta)(r\sin\theta)

    Substitute: . (x^2+y^2)^2 \:=\:2xy




    (2)\; r \:=\:2\sec\theta
    ( I got x=2.) . Right!

    \text{We have: }\:r \:=\:\dfrac{2}{\cos\theta} \quad\Rightarrow\quad r\cos\theta \:=\:2 \quad\Rightarrow\quad x \:=\:2




    (3)\; r \:=\:\dfrac{6}{2\cos\theta - 3\sin\theta}

    \begin{array}{ccc}\text{We have:} & r(2\cos\theta - 3\sin\theta) \:=\:6 \\ \\ & 2r\cos\theta - 3r\sin\theta \:=\:6 \\ \\ & 2x - 3y \:=\:6 \end{array}




    Rectangular to Polar:

    (1)\; 2xy\:=\:1

    2(r\cos\theta)(r\sin\theta) \:=\:1 \quad\Rightarrow\quad 2r^2 \:=\:\frac{1}{\sin\theta\cos\theta} \quad\Rightarrow\quad r^2 \:=\:\frac{1}{2\sin\theta\cos\theta}

    . . . . . . . . . . . . . . . . . \Rightarrow\quad r^2 \:=\:\frac{1}{\sin2\theta} \quad\Rightarrow\quad r^2 \:=\:\csc2\theta




    (2)\; y^2- 8y - 16 \:=\: 0

    (r\sin\theta)^2 - 8(r\sin\theta) - 16 \:=\:0 \quad\Rightarrow\quad r^2\sin^2\theta - 8r\sin\theta - 16 \:=\:0

    Quadratic Formula:

    . . r \;=\;\dfrac{8\sin\theta \pm \sqrt{64\sin^2\theta + 64\sin^2\theta}}{2\sin^2\theta} \;=\;\dfrac{8\sin\theta \pm\sqrt{128\sin^2\theta}}{2\sin^2\theta}

    . . r \;=\;\dfrac{8\sin\theta \pm 8\sqrt{2}\sin\theta}{2\sin^2\theta} \;=\;\dfrac{8\sin\theta(1 \pm\sqrt{2})}{2\sin^2\theta} \;=\;\dfrac{4(1\pm\sqrt{2})}{\sin\theta}

    . . r \;=\;4(1\pm\sqrt{2})\csc\theta




    (3)\; x^2 + y^2 - 2ay \:=\: 0

    We have: . r^2 - 2ar\sin\theta \:=\:0 \quad\Rightarrow\quad r(r - 2a\sin\theta) \:=\:0

    Then: . r-2a\sin\theta \:=\:0 \quad\Rightarrow\quad r \;=\;2a\sin\theta

    (We can disregard r = 0.)




    (4)\;x^2 \:=\:  y^3
    (My final answer is: . r\:=\:\cot^2\theta\csc\theta . Yes!

    We have: . (r\cos\theta)^2 \;=\;(r\sin\theta)^3 \quad\Rightarrow\quad r^2\cos^2\theta \;=\;r^3\sin^3\theta

    . . . . . . . . . \cos^2\theta \;=\;r\sin^3\theta \quad\Rightarrow\quad \dfrac{\cos^2\theta}{\sin^3\theta} \;=\;r

    . . . . . . . . . . r \;=\;\frac{\cos^2\theta}{\sin^2\theta}\cdot\frac{1  }{\sin\theta} \quad\Rightarrow\quad r \;=\;\cot^2\theta\csc\theta

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