1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Partial Derivative of a Parametric Equation
  1. Parametric equations (Replies: 4)