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Equation to the tangent

  1. Jun 16, 2015 #1
    1. The problem statement, all variables and given/known data
    The cissoid of Diocles is given by the relation y2(2-x) = x3. Find the equation to the tangent line to the curve at the point (1,1).

    2. Relevant equations


    3. The attempt at a solution

    Solution


    d/dx [ y2(2-x) ] = d/dx [ x3 ]
    2y dy/dx (2-x) + y2(-1) = 3x2
    Therefore, dy/dx = 3x2+y2 / 2y(2-x)

    m = dy/dx | (1,1) = 3+1/2(1) = 2 so y=2x+c and y(1) = 1, therefore 1=2+c ==> c=-1

    Equation is y=2x-1

    My attempt
    y2(2-x) = x3
    2y2-xy2-x2=0
    d/dx [2y2-xy2-x3 = d/dx [0]
    4y*dy/dx-y2-2xy*dy/dx-3x2 = 0
    dy/dx [ 2xy - 4y ]-y2-3x2
    dy/dx = -y2-3x2 / 2xy-4y

    My solution gives m=1, therefore y=mx+c, 1=1*1+c, c=0 y=1...

    The problem I'm having is that I don't understand why I can't expand the brackets in the original relation, y2(2-x) to be 2y2-xy2 or subtract x3 from both sides to make the equation 2y2-xy2-x2=0, which in my mind would make the problem easier to solve.
     
  2. jcsd
  3. Jun 16, 2015 #2

    SammyS

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    You need to use adequate parentheses.
    Your method also gives dy/dx = (-1 - 3(1) ) / ( 2(1)(1) - 4(1)) = (-4)/(-2) = 2
     
  4. Jun 17, 2015 #3
    @ (1,1)

    dy/dx = (-y2-3x2)/( 2xy-4y)

    dy/dx = (-12-3(-1)2) / (2(1)(1)-4(1))
    = (1-3) / (2-4)
    = -2 / -2
    = 0


    I see what I've done wrong. -1^2 = -1 ... but doesn't -1^2 = -1*-1 = 1 ???
     
  5. Jun 17, 2015 #4

    SammyS

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    That's not (-y)2 , it's -(y2) .
     
  6. Jun 17, 2015 #5
    dy/dx = - (y2-3x2) / ( 2xy-4y)

    like that?
     
  7. Jun 17, 2015 #6

    Mark44

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    No. -12 means - (12). If you want the square of -1, you need to write (-1)2
     
  8. Jun 17, 2015 #7
    ohhh ok. Is that the same for say, -x2... i.e. -x2 means -(x2)?
     
  9. Jun 17, 2015 #8

    Mark44

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    Yes
     
  10. Jun 17, 2015 #9

    Ray Vickson

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    You can do this; what makes you think otherwise?
     
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