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Obtain the differential equation of the family of plane curves

  1. Jun 15, 2014 #1
    1. The problem statement, all variables and given/known data

    Obtain the differential equation of the family of plane curves described:
    Circles tangent to the x-axis.

    2. Relevant equations
    [itex](x-h)^2 + (y-k)^2 = r^2[/itex]

    3. The attempt at a solution
    I tried to answer this question using the same way I did on a problem very similar to this (Circles with fixed radius r and tangent to the x-axis), but now I'm getting a different answer.
    The answer provided by the book for the problem above is [itex][1+(y')^2]^3 = [yy''+1+(y')^2]^2[/itex]. I have no idea how it's done.

    I want to ask the difference between the ways of how to solve these two problems:
    (1) circles tangent to the x-axis.
    (2) circles with fixed radius tangent to the axis.

    I can solve question (2) because of the hint that [itex]h=r[/itex], but doing the same with question (1) doesn't seem to work and it's making me crazy already. Please give me some clue on how to solve this one. Thanks a bunch!
     
    Last edited: Jun 15, 2014
  2. jcsd
  3. Jun 15, 2014 #2

    LCKurtz

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    I haven't worked it all the way out for 1, but I will get you started. For circles tangent to the x axis, you don't know the point of tangency, call it ##(a,0)## or the radius ##r##. The general equation for that family of circles would be ##(x-a)^2 + (y-r)^2 = r^2##. You can differentiate that twice, implicitly with respect to ##x##:$$
    2(x-a) + 2(y-r)y' = 0$$ $$
    2 +2y' + 2(y-r)y'' = 0$$Now your problem becomes using your equations to get rid of the ##r## and ##a##, or maybe easier, get rid of ##(x-a)## and ##y-r## by expressing them in terms of ##y## and its derivatives.
     
  4. Jun 15, 2014 #3
    Thanks! I'll be working on it now. :)

    UPDATE: I followed your instructions and had a lot of substitutions. Finally, I got it right! Thank you so much!
     
    Last edited: Jun 15, 2014
  5. Aug 7, 2016 #4
    is the second derivative of the equation's right?
     
  6. Aug 7, 2016 #5

    LCKurtz

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    No. Good catch. Apparently the OP got the idea anyway though.
     
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