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Curve C is given in Polar Coordinates by the equation r=2+3sin(theta)

  1. Apr 28, 2013 #1
    1. The problem statement, all variables and given/known data

    Curve C is given in Polar Coordinates by the equation r=2+3sinθ.
    Consider the usual Cartesian plane and take O as the pole and the positive x-axis as the polar axis.

    Find points on the curve C where the tangent lines are horizontal or vertical and sketch the curve C.

    2. Relevant equations

    [itex]x^{2}[/itex]+[itex]y^{2}[/itex]=[itex]r^{2}[/itex]
    x=rcosθ
    y=rsinθ
    tanθ=[itex]\frac{y}{x}[/itex]


    3. The attempt at a solution

    PART 1
    For some reason I feel like the addition of 2 is throwing me off

    r=2+3[itex]\frac{y}{r}[/itex]
    [itex]r^{2}[/itex]=2+3y
    [itex]x^{2}[/itex]+[itex]y^{2}[/itex]=2+3y
    [itex]x^{2}[/itex]+[itex]y^{2}[/itex]-3y=2
    [itex]x^{2}[/itex]+[itex]y^{2}[/itex]-3y+([itex]\frac{-3}{2})^{2}[/itex]=2+([itex]\frac{-3}{2})^{2}[/itex]
    [itex]x^{2}[/itex]+(y-[itex]\frac{3}{2}[/itex][itex])^{2}[/itex]=[itex]\frac{17}{4}[/itex]??

    I don't know where to go from the last line above for the center, maybe ([itex]\frac{3}{2}[/itex],0)??....

    PART 2
    I know also that I am supposed to take
    [itex]\frac{∂r}{∂θ}[/itex] which is 3cosθ

    when I take
    [itex]\frac{∂x}{∂θ}[/itex] do I take the derivative of x=2+3([itex]\frac{y}{r}[/itex])([itex]\frac{x}{r}[/itex])?? And similarly for [itex]\frac{∂y}{∂θ}[/itex].

    Lastly I know I have to take [itex]\frac{∂y}{∂x}[/itex] which I hope I can easily do after I sort out the issue above.

    Thank you
     
  2. jcsd
  3. Apr 28, 2013 #2

    Office_Shredder

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    [itex] r = 2+3\frac{y}{r}[/itex] should become [itex] r^2 = 2r+3y [/itex]

    You don't need to convert your curve to cartesian coordinates to sketch it though.... you can plot them directly by finding the location of a bunch of points and drawing a curve through them

    For [itex] \frac{\partial x}{\partial \theta}[/itex] You should use [itex] x = r\cos(\theta) [/itex] and do the product rule
     
  4. Apr 28, 2013 #3
    What do I end up doing with the 2r now?

    All of my examples from class always end up looking like
    [itex]r^{2}[/itex]=(some coefficient)(a variable)
    we never have a term with r remaining
     
  5. Apr 28, 2013 #4
    I guess I could divide by 2 and get r by itself

    r=([itex]\frac{x}{2})^{2}[/itex]+([itex]\frac{y}{2})^{2}[/itex]-[itex]\frac{3y}{2}[/itex]

    but I don't know what I would do with that.
     
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