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What is the difference between dy/dx, Δy/Δx, δy/δx and ∂y/∂x?

  1. Jan 30, 2013 #1
    Just in case the symbols do not appear correctly they are d, upper case delta, lower case delta, and lower case del.

    Also, what is the difference between dy, Δy, δy and ∂y when they are on their own?

    I think δy (lower case delta) is a infinitesimally small change in y, where as Δy (uppercase delta) is just y2 - y1 where the difference can be much larger.

    Is this correct?

    Also how does this differ from dy and ∂y?

    Note:
    I understand calculus and rates of change, I just do not know the difference between these different symbols and forms of differentials.
     
  2. jcsd
  3. Jan 30, 2013 #2

    Simon Bridge

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    I'll give it a go: in order -
    dy/dx : is the gradient of the tangent at a point on the curve y=f(x)
    Δy/Δx : is the gradient of a line through two points on the curve y=f(x)
    δy/δx is the gradient of the line between two ponts on the curve y=f(x) which are close together
    ∂y/∂x is the gradient of the tangent through a point on the surface y=f(x,z,...) in the direction of the x axis.

    The lower case delta just indicates a small change - not an infinitesimally small change.
    It's a short-hand notation whose meaning depends on the context.

    You will get a better understanding of the others when you see the more general forms - like the gradient operator, and the relationship to line and surface integrals.
     
  4. Jan 30, 2013 #3
    To add to Simon's post, δy/δx can also be a functional derivative. This is often used in calculus of variations / field theory.
     
  5. Jan 30, 2013 #4

    Simon Bridge

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  6. Feb 4, 2013 #5
    Thanks guys, just one more question, what do they mean on there on, for example I see equations like
    ∂r =∂xcosx +∂ysiny
    PD = ΔV (I understand this one (v1-v2))

    Whats the difference?

    Thanks
     
  7. Feb 4, 2013 #6

    Simon Bridge

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    If it was ##\Delta r = \Delta x \cos x + \Delta y \sin y## would that be clear?

    The ##\partial## is just the ##d## where you have functions of more than one variable, and the ##d## is just a ##\Delta## in the infinitesimal limit.

    In your example, the equation is a relationship between partials and it needs to be operated upon to get something you can use.
    i.e. divide through by ##\partial t## or (or ##\partial x##) or put an integral sign in front and see what happens.

    A classic example comes from finding the area under a function (y=f(x) say, between limits a and b)... you are used to the shortcut. In fact, what you do is divide the total area into loads of small areas and add them up. You write this: ##A=\int dA## (the integral sign refers to the sum over very small bits.)

    In Cartesian coordinates, you'd notice that ##dA=dx.dy## and this lets you rewrite the last integral as $$A=\int_a^b\int_0^{f(x)}dy.dx=\int_a^b f(x).dx$$ ... which is the form you are probably used to.
     
  8. Feb 4, 2013 #7
    Just a note, you may also see things like $$ \partial_x F(x,y,z) $$ which means the partial derivative of the function with respect to [itex]x[/itex]. Newer texts often use this notation.

    I've never something like
    Are you sure that's what the book/paper said? It was always my impression that for partial derivatives the variable of differentiation had to be explicit.
     
  9. Feb 4, 2013 #8

    Simon Bridge

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    That bothered me too. I would have expected to see dr, dx and dy in that context - just assumed sloppy notation.
     
  10. Feb 5, 2013 #9
    Thanks, and yeah I wrote down the example wrong, it was actually:

    ∂r = ∂xcosθ + ∂ysinθ

    It was on a page explaining the derivation of the gradient function, where the gradient = ∂z/∂r

    I still do not understand exactly what's happening on this page but now I at least understand the notation, so if I read it a few more times I should be good :)

    Thanks!
     
  11. Feb 5, 2013 #10
    Oh wait I read it wrong again, it's actually:

    ∂r = ∂xcosθ = ∂ysinθ

    damn dyslexia
     
  12. Feb 5, 2013 #11
    This makes a lot more sense now
     
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