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Homework Help: Find the coordinates of all points

  1. Dec 12, 2005 #1
    without using a graphing calculator
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    How would you find the coordinates of all points(x,y) of the tangent lines of a relation such as (x^2)*y - y^3 = 8 (btw this graph looks like 3 parabolas oppening, one to the right, left and one downwards) where the tangent line is horzontal ?

    Would using implicit differentiation work getting y'=(-2xy)/((x^2) -3y^2) then equating it to zero(slope) ?

    Consider x^3 + y^3 - 9xy = 0 how can you isolate the y on one side in the form y = x ?
     
  2. jcsd
  3. Dec 13, 2005 #2

    benorin

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    Homework Helper

    Consider the curve given implicitly by:

    [tex]x^{2}y-y^{3}=8[/tex]

    and the derivative

    [tex]y^{\prime}=\frac{-2xy}{x^{2}-3y^{2}}[/tex]

    horizontal tangent lines occur when [itex]y^{\prime}=0[/itex], so set [itex]-2xy^2=0 \Rightarrow x=0\mbox{ or }y=0[/itex], and we know that [itex]y\neq 0 [/itex] at all points on the curve, but [itex]x=0[/itex] is a point on the curve, namely (0,-2), where we do indeed have a horizontal tangent line to the curve.
     
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