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Tangent line equation

  1. Jun 28, 2011 #1
    1. The problem statement, all variables and given/known data

    Find the equation for the tangent line at (2,2)

    f(x)=xy+y^3=12

    2. Relevant equations



    3. The attempt at a solution

    I really feel like I know what I'm doing here, but the key disagrees.

    My equation for the tangent slope comes out to be:

    (-y)/(x+3y^2)

    when solved for my (2,2), I get a slope of -1/7

    Yielding an entire equation of y= -1/7x + 12/7
     
  2. jcsd
  3. Jun 28, 2011 #2

    Dick

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    Your slope looks ok. But (2,2) isn't on your tangent line. How can that be?
     
  4. Jun 28, 2011 #3
    If my slope is correct, but my tangent line equation is wrong, I can only assume I did something wrong in y - y = M(x-x)?
     
  5. Jun 28, 2011 #4

    Dick

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    Sure. What did you use for the fixed y and x? Remember this line is supposed to go through (2,2).
     
  6. Jun 28, 2011 #5
    I used y - 2 = M (x - 2), y = 2 and x = 2.

    What else could I have used? I think I'm getting confused..
     
  7. Jun 28, 2011 #6

    Dick

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    (y-2)=(-1/7)*(x-2) isn't the same line as y= -1/7x + 12/7, which you said was the answer you got. That's the only thing that's confusing for me.
     
  8. Jun 28, 2011 #7
    Ahh, I think I see now, a sign error?

    y - 2 = -1/7x + 2/7
    y = -1/7x + 16/7
     
  9. Jun 28, 2011 #8

    Dick

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    Yes. Sign error. Looks ok now. If you'd post all of your work to begin with it would be easier to diagnose these things.
     
  10. Jun 28, 2011 #9
    Sorry about that, but thanks for the help!
     
  11. Jun 28, 2011 #10

    Dick

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    No problem. Just suggesting a way to get faster service!
     
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