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Tangent Lines of Parametric Equations

  1. Oct 31, 2007 #1
    My book really doesn't go into a lot of depth but I was wondering if this is correct

    If we are asked to find the tangent line of a specific value of t for a given parametric equation then we can find the equation of the tangent line in either rectangular or parametric functions.

    Rectangular Mode
    We need dy/dx and the point at the specific t value, say [itex](x_o,y_o)[/itex] is our point.
    The tangent line is:

    [tex]y-y_o = \frac{dy}{dx}(x-x_o)[/tex]

    Parametric Mode
    We need dy/dt, dx/dt, and the point at the specific t value, once again say [itex](x_o,y_o)[/itex].

    [tex]x(t) = \frac{dx}{dt}t+x_o[/tex]

    [tex]y(t) = \frac{dy}{dt}t+y_o[/tex]

    Is that correct?

    Also could someone explain how we derive dy/dx and d^2y/dx^2?
  2. jcsd
  3. Oct 31, 2007 #2


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    From the slope of the function to which the tangent line is tangent at (x_0, y_0).
  4. Nov 1, 2007 #3
    Ok, thanks^^

    I should be more specific:

    My text gives this definition for dy/dx

    "The Chain Rule states that the derivative dy/dx for the parametric curve is the ratio of dy/dt to dx/dt." But I cannot see where they got that from. They then go on to say "d^2y/dx^2 can be derived in the same manner". Could you explain it a little more, I'm a bit confused? >.>
  5. Nov 2, 2007 #4


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    Let D be the partial derivative operator.

    If y(t) = f(x(t)), or with some abuse of notation y(t) = y(x(t)), how do you write Dy/Dt?
  6. Nov 3, 2007 #5


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    The chain rule. dy/dx= dy/dt dt/dx= (dy/dt)/(dx/dt). Saying "d^2y/dx^2 can be derived n the same manner" is little misleading- it's much more complicated. Notice tha they don't give that formula!

    d^2y/dx^2= d(dy/dx)/dx= (1/(dx/dt)) d((dy/dt)/(dx/dt))/dt and you have to use the quotient rule for the last part.
  7. Nov 4, 2007 #6
    Ah seeing y(t) = f(x(t)) makes it clear to me

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

    Which we rearrange to get:

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