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Homework Help: Equation of tangent line (rec. form) to a polar curve

  1. Jun 4, 2010 #1
    Equation of tangent line (rec. form) to a polar curve!!

    1. The problem statement, all variables and given/known data

    Quesiton:

    Find the rectangular form of the equation of the tangent line to the polar curve r=cos^3(theta) at the point corresponding to theta=pi/4


    2. Relevant equations



    3. The attempt at a solution

    How to do that?

    I mean finding it in RECTANGULAR FORM !


    i know that
    [tex]\frac{dy}{dx}=\frac{\frac{dr}{d\theta} sin(\theta)+rcos(\theta)}{\frac{dr}{d\theta} cos(\theta) - r sin(\theta)}[/tex]

    I will calculate dy/dx at theta=pi/4 , and this is easy ..

    The problem here is that, how can I find the equation of the tangent line in RECATNGULAR FORM ??

    The equation of the tangent line is :

    y-y1=m(x-x1)

    m = the slope , and this one will be calculated by using the formula ..

    but what about x1 and y1?
     
  2. jcsd
  3. Jun 4, 2010 #2

    tiny-tim

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    Hi System! :smile:

    (have a pi: π and a theta: θ and try using the X2 and X2 tags just above the Reply box :wink:)
    (x1,y1) will be (r,θ) at θ = π/4. :wink:
     
  4. Jun 4, 2010 #3
    Re: Equation of tangent line (rec. form) to a polar curve!!

    I do not think so.
    so y=pi/4 ?! :S
     
  5. Jun 4, 2010 #4

    tiny-tim

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    (what happened to that π i gave you? :confused:)

    No, you have to convert (r,θ) to (x,y) at θ = π/4.
     
  6. Jun 4, 2010 #5
    Re: Equation of tangent line (rec. form) to a polar curve!!

    ohhh i see now
    x = r cosθ = cos^4 θ
    y = r sinθ = cos^3θ sinθ

    I will evaluate them at θ=pi/4 to get x1 & y1
    and I have m from the formula of dy/dx in polar

    I will substitute x1,y1 & m in the line equation and I will be finish, right?
     
  7. Jun 5, 2010 #6

    tiny-tim

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    (just got up :zzz: …)

    Right! :biggrin:
     
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