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Help finding equation of the tangent line to the graph

  1. Jun 9, 2008 #1
    1. The problem statement, all variables and given/known data
    Find the equation of the tangent line to the graph of x^2 - xy + y^2 = 19 (where y=y(x)) at (3,-2).


    2. Relevant equations



    3. The attempt at a solution

    So, this is what I did:
    d/dx(x^2-xy+y^2) = (19)d/dx

    2x*dx/dx-dy/dx+dy/dx+2y(dy/dx)=0

    2x+2y(dy/dx)=0

    And I am stuck, the answer to this question is:
    y+2=8/7(x-3)

    and I don't know how to get the slope.
     
  2. jcsd
  3. Jun 9, 2008 #2

    Gib Z

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    Well you have your values of x and y, (3,-2). Sub that into the 2x+2y(dy/dx)=0 you got to solve for dy/dx, then you have the gradient of the tangent. Now use the gradient-point form of a straight line to find the equation of the tangent.
     
  4. Jun 9, 2008 #3

    HallsofIvy

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    d(x^2)/dx= 2x and d(y^2)/dx= 2y(dy/dx) but d(-xy)/dx is NOT -dy/dx+ dy/dx!

     
  5. Jun 9, 2008 #4
    I am pretty sure I got the x^2 and the y^2 part right.

    I am just having trouble taking the deriv of the bolded part:
    x^2-xy+y^2=19

    Do I apply the product rule for the d/dx(-xy)?
     
  6. Jun 9, 2008 #5

    HallsofIvy

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    Yes! It's a product, isn't it?
     
  7. Jun 9, 2008 #6
    Yeah, but the thing is, I tried applying the product rule, and my results were different, I answer did not match with the answer key, so somewhere within the product rule, I screwed up somewhere, so how do you take d/dx(-xy)?
     
  8. Jun 10, 2008 #7

    Redbelly98

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    The product rule:
    d/dx(u v)
    = (du/dx) v + u (dv/dx)

    In this case, uv = xy, so ... ?
     
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