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Parametric differentiation

  1. Apr 20, 2010 #1
    1. The problem statement, all variables and given/known data
    If x(t) = cos2t ; y(t) = t - tant

    Show that:
    (dy/dx)^2 = 1- x / 4(1+x)^3


    3. The attempt at a solution

    dy/dx = dy/dt * dt/dx = (dy/dt) / (dx/dt)

    dy/dt = 1- sect^2 ; dx/dt = -2sin2t

    So (dy/dx)^2 = (1 - sect^2)^2 / (4 sin2t^2)

    From here i'm not too sure on how to get to the answer.
     
  2. jcsd
  3. Apr 20, 2010 #2

    Mark44

    Staff: Mentor

    The statement you are supposed to prove has (dy/dx)^2 as a function of x. In your version, you have (dy/dx)^2 as a function of t. Solve the equation x(t)= cos(2t) for t and replace in your equation for (dy/dx)^2.

    I have a question about what you need to prove, which you have written as (dy/dx)^2 = 1- x / 4(1+x)^3.

    If this is
    [tex](dy/dx)^2 = 1 - \frac{x}{4(1 + x)^3}[/tex]
    then you have written it correctly.

    OTOH, if you meant
    [tex](dy/dx)^2 = \frac{1 - x}{4(1 + x)^3}[/tex]
    then you should have put parentheses around 1 - x to show that it's the numerator.
     
  4. Apr 20, 2010 #3
    Sorry was the latter.

    So I replace t with (arccosx)/2 ?
     
  5. Apr 20, 2010 #4

    Mark44

    Staff: Mentor

    Yes.
     
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