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Find dy/dx and then find slope of tangent(s).

  1. Jan 25, 2012 #1
    1. The problem statement, all variables and given/known data
    a) If 2y2-xy-x2=0 , find dy/dx.
    b) What is/are the slope(s) of the tangent(s)at x=1?

    2. Relevant equations

    3. The attempt at a solution
    Using implicit differentiation:

    a) (4y)dy/dx-(x)dy/dx-y(1)-2x=0

    (4y)dy/dx-(x)dy/dx=2x+y

    (4y-x)dy/dx=2x+y

    dy/dx=(2x+y)/(4y-x)

    Assuming that I found dy/dx correctly (please let me know if I did not) I am having some trouble seeing where to go from here to complete part b). If I had x and y coordinates I could work through it but with only an x value it is proving difficult for me.

    My latest attempt was to try and solve for y using the original equation.

    b) 2y2-xy-x2=0

    At x=1, 2y2-(1)y-(1)2=0

    Leaving me with:
    2y2-y=1

    And that's where I ended up, stuck and admittedly confused.
    Any help or nudges in the right direction would be greatly appreciated.

    Thanks,
    -Trevor
     
  2. jcsd
  3. Jan 25, 2012 #2

    vela

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    Hint: You have a quadratic equation, 2y2-y-1=0.
     
  4. Jan 25, 2012 #3
    Ah yes I do!
    So from 2y2-y-1=0
    I get (2y+1)(y-1)=0

    I can use this to find my y coordinates and then use the x and y values with my previously found dy/dx to get my slope at both points, correct?
    I calculated a slope of 1 at (1,1) and a slope of -1/2 at (1,-1/2).


    Thanks a lot!
     
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