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Meaning of delta?

  1. Feb 12, 2010 #1
    I've come across a very ambiguous statement in my notes on implicit functions (part of the partial derivatives part of the course). I'll write out the preceding explanation but the problematic line is marked by *

    "Sometimes we can define a function z=z(x, y) only in implicit form, i.e. through an equation F(x, y, z) = 0.
    It is not always possible to solve this equation for z and obtain the function z=f(x,y).

    In order to calculate the derivatives of a function defined implicitly we note that from the above equation it follows that:
    * [tex]\delta[/tex]F=0 [tex]\Rightarrow[/tex] [tex]\delta[/tex]F = Fx[tex]\delta[/tex]x + Fy[tex]\delta[/tex]y + Fz[tex]\delta[/tex]z = 0.
    Or by taking differentials,
    Fxdx + Fydy + Fzdz = 0"

    My main problem is understanding how [tex]\delta[/tex]x can stand on its own (above used as a factor). Is it just the same as ∆x, i.e. a change in x and not a derivative?

    Also, how the [tex]\delta[/tex] expressions change to d expressions in the second line is unclear to me...
  2. jcsd
  3. Feb 12, 2010 #2
    There is some ambiguity in the meaning of delta. Ask a physicist and he/she will tell you that [tex]\delta[/tex] means a small change whereas [tex]\Delta[/tex] corresponds to a large change.
  4. Feb 13, 2010 #3
    i believe that your usage of ∂ is in the form of a partial derivative
    and that you are using partial differentials in order to gain a total differential of the function F(x,y,z)
    Last edited: Feb 13, 2010
  5. Feb 13, 2010 #4
    The problem is that the Fx term implies ∂F/∂x and so on, whereas these individual partial derivative are then multiplied by the [tex]\delta[/tex]x and so on terms - this is what confuses me.

    So is it a partial derivative multiplied by a small change?
  6. Feb 13, 2010 #5


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    Yes, [itex]\delta x[/itex] here just means a small change in x. Taking the limit as [itex]\delta x[/itex] becomes "infinitesmal" gives you dx.
  7. Feb 13, 2010 #6
    Ok, so what does dx stand for then? It is again multiplied by Fx, giving ∂F/∂x · dx. Since it isn't a double derivative of F with respect to x (especially since ∂ =/= d), then what is it's meaning here?
  8. Feb 13, 2010 #7


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    dx is a "differential".
  9. Feb 13, 2010 #8
    So is it similar to "an infinitesimal change in x" but now it's an "infinitesimally small difference (ie, variation or difference) in x"?
  10. Feb 13, 2010 #9
    well when you are taking differentials, you add limits to your small change in the value as they tend to zero
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