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Difference between vector and parametric differentiation

  1. Mar 9, 2015 #1
    This might seem like a naive question to ask, but a full explanation of why these two concepts are different would be welcome. I am confused because parametric equations are ##y = 8t^2## and ##x = 5t##, but at the same time, these two equations can describe the ##x## and ##y## components of a vector. Parametric differentiation is defined as ##\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{\frac{\mathrm{d} y}{\mathrm{d} t}}{\frac{\mathrm{d} x}{\mathrm{d} t}}##, while vector differentiation is defined as ##\frac{\mathrm{d} \vec{r}}{\mathrm{d} t} = \left \langle \frac{\mathrm{d} x}{\mathrm{d} t}, \frac{\mathrm{d} y}{\mathrm{d} t} \right \rangle##
     
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  3. Mar 9, 2015 #2

    Svein

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    Not exactly. As t varies, the coordinates describe a curve in the xy -plane.
    If we assume that t stands for time, [itex]\frac{d\vec{r}}{dt} [/itex] represents the velocity at a give time (velocity has a magnitude and a direction). This quantity is fundamental in classical differential geometry.
     
  4. Mar 9, 2015 #3

    LCKurtz

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    If you eliminate the parameter ##t## and get ##y## in terms of ##x## then ##\frac{dy}{dx}## represents the slope of the curve. That is what you are calculating when you do$$
    \frac{dy}{dx} =\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$That is not the same thing as ##\frac{dr}{dt}## which is the velocity the point moves along the curve as a function of ##t##. Different things, different formulas.
     
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