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Taylor expansion

  1. Dec 20, 2015 #1
  2. jcsd
  3. Dec 20, 2015 #2

    blue_leaf77

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    ##F(t) = f(x(t),y(t))##, so despite written to be a function of ##t##, the functional form of ##F(t)## is not written explicitly as a function of ##t##.
     
  4. Dec 20, 2015 #3
    Why is it not a function of t? I am still new to this so do you have something i can read on about this?
    Besides thanks for your response!
     
  5. Dec 20, 2015 #4

    blue_leaf77

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    It's not an explicit function of ##t##. When you want to do chain rule, you have to pay attention on which variables are written explicitly, despite whether or not these variables are functions of yet another variable. For example take ##F(t) = xy^2## where ##x = \sqrt{t}## and ##y=t-2##. If you want to calculate ##dF/dt##, you can either first express ##x## and ##y## in terms of ##t## and then differentiate w.r.t. ##t## or let ##F## be expressed in ##x## and ##y## then use the chain rule
    $$
    \frac{dF}{dt} = \frac{\partial F}{\partial x} \frac{dx}{dt} + \frac{\partial F}{\partial y} \frac{dy}{dt} .
    $$
    Both answers should be identical.
    I guess this problem should belong to multivariate calculus.
     
  6. Dec 20, 2015 #5
    Thanks a lot! I now understand.
     
  7. Dec 20, 2015 #6

    Svein

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    I do not agree with the formula shown in the OP (dimensional analysis again). The picture states that [itex]\frac{d}{dt}x(t) = \Delta x [/itex] which is obviously wrong. If x is distance and t is time, it tries to assert that velocity equals a (short) distance. The correct statement is [itex]\frac{d}{dt}x(t) \cdot \Delta t= \Delta x [/itex].
    I can agree with the second line, but the third line is pure nonsense.
     
  8. Dec 20, 2015 #7

    blue_leaf77

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    In the linked pdf file, the author for some reason sets ##\Delta t = 1##.
     
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