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Cardiod parametric equation problem

  1. Sep 28, 2013 #1
    The problem statement, all variables and given/known data.
    Let ##C_a## a cardioid given in polar coordinates by ##r_a= a+cos(\theta)## with a being a parameter, and ##\theta## ##\in [0,2\Pi]##

    a)Prove that, for a>1, ##C_a## is a smooth curve.
    b)Calculate the tangent line to the curve ##C_a## in the cartesian coordinates point ##(x,y)=(0,a)##

    My problem is I don't understand how this curve is, I mean, I googled and saw how it looks like, but I don't know the parameterization.

    By the way, I accidently pressed enter, so I opened an empty thread and I couldn't check the spell in the title, it says "cardiod"", when it should be "cardioid"", does anyone know how can I correct the title?
     
    Last edited: Sep 28, 2013
  2. jcsd
  3. Sep 28, 2013 #2

    LCKurtz

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    I wouldn't worry to much about the misspelling in the title but you can ask a moderator to do it for you if it bothers you. Not sure what is bothering you about the parameterization. It is just a polar equation in the usual form ##r = f(\theta)##. Surely you have studied polar coordinates and their graphs??
     
  4. Sep 28, 2013 #3

    Zondrina

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    For part (a), does the tangent vector vanish at all or is it continuous?
     
  5. Sep 29, 2013 #4
    I got confused, the parameterization in polar coordinates is ##ψ(t)=(f(\theta)cos(\theta),f(\theta)sin(\theta))##
     
  6. Sep 30, 2013 #5
    Sorry, I don't understant the question. The tangent vector doesn't vanish if the curve is smooth, I'm not sure if that what you've asked.
     
  7. Sep 30, 2013 #6

    Zondrina

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    Exactly, if the tangent vector doesn't vanish anywhere, then it is continuous. That is the exact condition you need for the curve to be smooth.
     
  8. Sep 30, 2013 #7

    LCKurtz

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    Use either ##\theta## or ##t## for the parameter, not both. But yes, for a curve given as ##r = f(\theta)##, that is the x-y parameterization$$
    \Psi(\theta) = \langle f(\theta)\cos(\theta), f(\theta)\sin(\theta)\rangle$$You can get the slope by$$
    \frac{dy}{dx}= \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}$$
     
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