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Partial Derivative of a Parametric Equation

  1. Apr 25, 2006 #1
    I'm getting confused over a few points on the derivative of a parametric equation.
    Say we the world line of a particle are represented by coordinates [tex] x^i [/tex]. We then parametrize this world line by the parameter t. [tex] x^i = f^i(t) [/tex].

    Now here is where I get confused. The partial derivative [tex] \frac {\partial x^i}{\partial t} [/tex] should be zero since x is an independent coordinate and has no explicit time dependence. However, if I take the partial of the RHS above, clearly this is nonzero.

    Moreover we define [tex] \dot x^i [/tex] as [tex] \dot x^i = \frac{d x^i}{d t} [/tex]. It seems the RHS will be the same if we take a partial or total derivative, yet the LHS will be zero if we take [tex] \frac {\partial x^i}{\partial t} [/tex] but [tex] \dot x^i [/tex] if we take [tex] \dot x^i = \frac{d x^i}{d t} [/tex]. Is my question clear? Thanks.
    Last edited: Apr 25, 2006
  2. jcsd
  3. Apr 25, 2006 #2
    Except t isn't a coordinate, it's a parameter. x does have an explicit time dependance, you've given it one when you equated x to f(t). x is not an independant coordinate in this circumstance, t is. That's like saying y=f(x), so [tex]\frac{dy}{dx}[/tex] must be zero because y is an independant coordinate.

    The total and partial derivatives with respect to t are the same here because [itex]f:\mathbb{R}\rightarrow\mathbb{R}[/itex] is a one dimensional function.
    Last edited: Apr 25, 2006
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