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Implicit function

  1. Oct 14, 2007 #1
    1. The problem statement, all variables and given/known data
    a. Find dy/dx given that x[tex]^{2}[/tex]=y[tex]^{2}[/tex]-4x=7y=15
    b. under what conditions on x and/or y is the tangent line to this curve horizontal? vertical?

    2. The attempt at a solution

    I did solve the first question by simply using implicit fuction.


    above is the answer that i ended up with

    but im kinda stuck with the second question
    if the tangent line is horizonat i guess the slope has to be zero
    so i just set [tex]\frac{4-2x}{9y}[/tex]=0 and
    i ended up with x=2
    but i don't know what to do with "vertical" part and
    i'm not sure if i got the "horizonal" part either.
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Oct 14, 2007 #2
    for a that's right, just when you wrote out the question you forgot some +'s and put = instead.

    yes horizontal tangent is when the derivative =0. so x=2 is current for that.
    vertical tangent occurs when the the derivative = +-infinity.

    and if you have a fraction when does it = infinity?
  4. Oct 14, 2007 #3
    If the question is supposed to be x²+y²-4x+7y = 15 then I believe that your answer for y' is incorrect.

    Note that your original equation is that of a circle which has been shifted. To write it in standard form, (x-a)²+(y-b)²=c² you can complete the square (I'm getting to a nice geometrical result, just hang on :D)

    (x²-4x+4) -4 + (y²+7y + 49/4) - 49/4 = 15
    (x-2)² + (y+7/2)² = 15 +4 + 49/4
    (x-2)² + (y+7/2)² = [sqrt(125)/2]²

    So your original equation is a circle of radius [sqrt(125)/2], centered at (2,-7/2). From this you should gather that there are 2 points that have a horizontal tangent, and two points that have a vertical one.
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