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Implicit Graphing

  1. Nov 6, 2009 #1
    Graphing Implicit Function

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

    Graph (x^2+Y^2)^2=4xy
    A)Find Y'
    B)Find all points that have vertical tangent lines.
    C)Find all points that have horizontal tangent lines.

    3. The attempt at a solution

    I feel like I'm going nuts on this problem, y' is (-x^3-xy^2+y)/(y^3+x^2y-x) right?
    I used a graphing program and it looks like an 8 reflected over y=x, but I just can't understand why I'm not getting the right answers for the HTL and the VTL.
     
    Last edited: Nov 6, 2009
  2. jcsd
  3. Nov 6, 2009 #2

    LCKurtz

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    Your derivative is correct. Try factoring an x out of the first two terms of the numerator and a y out of the first two terms of the denominator and replace the x2 + y2 with square root of 4xy from the original equation and see if that helps.
     
  4. Nov 7, 2009 #3
    alright ill give it a try
     
  5. Nov 7, 2009 #4
    Im still getting stumped, if I do that, I can turn the top half into y=4x^3, and the bottom half into x=4y^3, but obviously those wont give me an answers, Ive looked at the graph and there are HTL at (+-.66,+-1.16) and VTL at (+-1.16,+-.66)
     
  6. Nov 7, 2009 #5

    HallsofIvy

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    Vertical tangent lines are where the denominator of the derivative, [itex]y^3+x^2y-x[/itex] is 0 and horizontal tangent lines are where the numerator of the derivative, [itex]-x^3-xy^2+y[/itex] is 0. For each of those, you still have [itex](x^2+y^2)^2=4xy[/itex] so in each case you have two equations to solve for x and y.

    For vertical tangent lines solve the system
    [itex]y^3+x^2y-x= 0[/itex]
    [itex](x^2+y^2)^2=4xy[/itex].

    For horizontal tangent lines solve the system
    [itex]-x^3-xy^2+y= 0[/itex]
    [itex](x^2+y^2)^2=4xy[/itex].
     
    Last edited: Nov 7, 2009
  7. Nov 7, 2009 #6

    Mentallic

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    You're on the right track. For the horizontal asymptote given as [itex]y=4x^3[/itex] you need to solve is simultaneously with the curve [itex](x^2+y^2)^2=4xy[/itex] as hallsofivy has said.
    Same thing goes for the vertical asymptote.
     
  8. Nov 7, 2009 #7
    ahh ill give it another try, thanks guys
     
  9. Nov 7, 2009 #8

    HallsofIvy

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    A little additional point: The first equation for vertical tangent lines can be written as [itex]y^3+ x^2y- x= y(y^2+ x^2)- y= 0[/itex] so [itex]x^2+ y^2= y/y= 1[/itex]
     
  10. Nov 7, 2009 #9
    I had to talk a little break from it but I just came back to it and got it right, thanks pf!
     
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