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Homework Help: Parametric equations

  1. Apr 1, 2005 #1
    find [tex]\frac{d^2y}{dx^2}[/tex] as a function of [tex]t[/tex], for the given the parametric equations:

    x = 2-4*cos(t)
    y= 4+cos(t)^2

    [tex]\frac{d^2y}{dx^2}[/tex] = _______

    dy/dt = [tex]-2*cos(t)*sin(t)[/tex]
    second derv. [tex]2*sin(t)^2-2*cos(t)^2[/tex]

    dx/dt = [tex]4*sin(t)[/tex]
    second derv. [tex]4*cos(t)[/tex]

    [tex]\frac{d^2y}{dx^2}[/tex] = [tex]\frac{2*sin(t)^2-2*cos(t)^2}{4*cos(t)}[/tex]

    what did i do wrong?
     
  2. jcsd
  3. Apr 1, 2005 #2
    [tex]\frac{d^2y}{dx^2} \neq \frac{\frac{d^2y}{dt^2}}{\frac{d^2x}{dt^2}}[/tex]

    Here is how I would do it.

    [tex]x = 2 - 4\cos(t)[/tex]
    [tex]4\cos(t) = 2-x[/tex]
    [tex]t = \arccos\left(\frac{2-x}{4}\right)[/tex]

    Now we have [tex]t[/tex] explicitly.

    Substituite this back in for [tex]t[/tex] and we have

    [tex]y = 4 + \cos\left(\arccos\left(\frac{2-x}{4}\right)\right)^2[/tex]
    [tex]\sqrt{y} = 2 + \cos\left(\arccos\left(\frac{2-x}{4}\right)\right)[/tex]
    [tex]\sqrt{y} = 2 + \left(\frac{2-x}{4}\right)[/tex]
    [tex]\sqrt{y} = \frac{3}{2} - \frac{x}{4}[/tex]
    [tex]y = \frac{9}{4} - \frac{x^2}{16}[/tex]

    Now differentiate.

    [tex]\frac{dy}{dx} = -\frac{x}{8}[/tex]
    [tex]\frac{d^2y}{dx^2} = -\frac{1}{8}[/tex]
     
  4. Apr 1, 2005 #3
    Also in general, for y(t) and x(t),

    [tex] \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} [/tex]

    [tex] \frac{d^2y}{dx^2} = \frac{\frac{d^2y}{dt^2}}{(\frac{dx}{dt})^2} [/tex]

    Simplifying the RHS of the second equation you get:

    [tex] \frac{d^2y}{dx^2} = \frac{\frac{d^2y}{dt^2}}{(\frac{dx}{dt})^2} = \frac{d^2y}{dt^2} \frac{dt^2}{dx^2} [/tex] and by cancelling you get the desired result [tex] \frac{d^2y}{dx^2} [/tex]
     
  5. Apr 1, 2005 #4
    awesome thanks
     
  6. Apr 2, 2005 #5

    xanthym

    User Avatar
    Science Advisor

    whozum:
    Highlighted equation above for (d2y/dx2) is NOT correct. For parametric equations {y=y(t) & x=x(t)}, the second derivative is given by:

    [tex] :(2): \ \ \ \ \ \color{blue} \frac{d^2y}{dx^2} \ = \ \left( \frac{d}{dt} \left( \frac{dy}{dx} \right) \right) / \left( \frac{dx}{dt} \right) [/tex]

    where (dy/dx) is function of "t" obtained from the correct Equation #1 above.


    ~~
     
    Last edited: Apr 2, 2005
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