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Operating with dx and dy?

  1. Dec 7, 2007 #1
    1. Why can dx and dy be moved around in DE (a bit but not exactly like fractions)?


    2. What is the justification that allows one to operate with dx and dy after the integral signs as if they were differentials?
     
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  3. Dec 8, 2007 #2

    EnumaElish

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    1. they represent "small changes."
    2. they are differentials. For example, G(x) = [itex]\int[/itex]g(t)F'(t)dt means G(x) = [itex]\int[/itex]g(t)dF(t) where dF(t)/dt = F'(t).
     
  4. Dec 8, 2007 #3

    Gib Z

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    Because their quotient is the limit of a fraction.
     
  5. Dec 8, 2007 #4

    WWGD

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    Overall, in a rigurous way, the dx's are differential forms. There is a whole theory
    of differential form that explains them in detail.
    Still, In some cases this movement of the dy's and dx's is more heuristic and
    loose than rigorous. In the case I think you are describing, it is the case of
    separation of variables.
    Wikipedia has a nice page on them:


    http://en.wikipedia.org/wiki/Separation_of_variables
     
  6. Dec 8, 2007 #5
    that doesn't explain to me why you can break them up. in fact it implies you cannot because the limit does not break up across division/multiplication.
     
  7. Dec 9, 2007 #6

    HallsofIvy

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    That's true. However, any calculus book I have seen defines the differentials, dx and dy, separately from dy/dx. That is, given a derivative dy/dx, dx is defined simply as a "symbol" and then dy is defined by dy= (dy/dx) dx.

    I'm a bit puzzled at your assertion "the limit does not break up across division/multiplication." It is certainly true that
    [tex]\lim_{x\rightarrow a}f(x)g(x)= (\lim_{x\rightarrow a}f(x))(\lim_{x\rightarrow a}g(x))[/tex]
    and that
    [tex]\lim_{x\rightarrow a} \frac {f(x)} {g(x)} = /frac{\lim_{x\rightarrow a}f(x)}{\lim_{x\rightarrow a}g(x)}[/tex]
    as long as the denominator is not 0. Of course that is true for all derivative calculations but the point is still that we can go back "before" the limit, use the fraction property and then take the limit again. That is, we can always treat a derivative as if it were a fraction and the "dy", "dx" notation is defined to make use of that.
     
    Last edited by a moderator: Dec 15, 2007
  8. Dec 9, 2007 #7
    hmm i guess i forgot that limit property. but what do you mean there in bold?
     
    Last edited by a moderator: Dec 15, 2007
  9. Dec 15, 2007 #8

    HallsofIvy

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    For example to prove the chain rule, dy/dx= dy/du du/dx, when u is a differentiable function of x, you would set up the "difference quotient" defining dy/dx:
    [tex]\frac{dy}{dx}= \lim_{h\rightarrow 0}\frac{y(x+h)- y(x)}[/tex]
    write that as
    [tex]\lim_{h\rightarrow 0}\frac{y(x+h)- y(x)}{u(x+h)- u(x)}\frac{u(x+h)- u(x)}{h}[/tex]
    where I have used exactly the "property of fractions", that we can cancel the two "u(x+h)- u(x)" terms, that we "want" to use in saying dy/dx= (dy/du)(du/dx), that we can cancel the two "du" terms.
    But since u is differentiable it is also continuous: as h goes to 0, u(x+h)-u(x)= u(x)- u(x)= 0. Renaming the "u(x+h)- u(x)" term in the denominator "k", we have y(x)= y(u(x)) so y(x+h)= y(u(x+h))= y(u(x)+ u(x+h)- u(x))= y(u+ k). We can separate that limit into two limits:
    [tex]\left[\lim_{k\rightarrow 0}\frac{y(u+k)- y(u)}{k}\right]\left[\lim_{h\rightarrow 0}\frac{u(x+h)- u(x)}{h}\right][/tex]
    Taking each of those limits give (dy/du)(du/dx).

    We use the "fraction property" on the difference quotient and then take the limit. A derivative is not a fraction but can always be treated like one. Using the "differential" notation, defining dy and dx as separate (symbolic) things with the property that dy divided by dx is the derivative, makes that "formal".
     
  10. Dec 16, 2007 #9

    arildno

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    However, a potent reminder that derivatives in general cannot be regarded naively as fractions is given by the classical relationship:
    [tex]\frac{\partial{x}}{\partial{y}}\frac{\partial{y}}{\partial{z}}\frac{\partial{z}}{\partial{x}}=-1[/tex]
    :smile:
     
  11. Dec 16, 2007 #10
    Because they are numbers. [itex]dx[/itex] and [itex]dy[/itex] are tangent space coordinates.
     
  12. Dec 16, 2007 #11
    Arildno : why is this true ?
     
  13. Dec 17, 2007 #12

    arildno

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    It is a consequence of the implicit function theorem:

    Let G(x,y,z) be a function, and consider the equation:

    [tex]G(x,y,z)=0 (*)[/tex]

    The implicit function theorem now states that in a vicinity (i.e some open neighbourhood U) of a particular solution of (*) [itex](x_{0},y_{0},z_{0})[/itex], we may solve for ONE of the variables in terms of the other two (under fairly mild restrictions), for example:
    [tex]x=X(y,z), x_{0}=X(y_{0},z_{0})[/tex], or for that matter y=Y(x,z), z=Z(x,y), where X, Y, Z are functions.

    Now, within U, the following expressions are identities:
    [tex]G(X(y,z),y,z)=0,G(x,Y(x,z),z)=0,G(x,y,Z(x,y)=0[/tex]

    Because they are identities, we may differentiate them, and gain, for example:
    [tex]\frac{\partial{G}}{\partial{x}}\frac{\partial{X}}{\partial{y}}+\frac{\partial{G}}{\partial{y}}=0\to\frac{\partial{X}}{\partial{y}}=-\frac{\frac{\partial{G}}{\partial{y}}}{\frac{\partial{G}}{\partial{x}}}[/tex]

    [tex]\frac{\partial{G}}{\partial{y}}\frac{\partial{Y}}{\partial{z}}+\frac{\partial{G}}{\partial{z}}=0\to\frac{\partial{Y}}{\partial{z}}=-\frac{\frac{\partial{G}}{\partial{z}}}{\frac{\partial{G}}{\partial{y}}}[/tex]
    [tex]\frac{\partial{G}}{\partial{z}}\frac{\partial{Z}}{\partial{x}}+\frac{\partial{G}}{\partial{x}}=0\to\frac{\partial{Z}}{\partial{x}}=-\frac{\frac{\partial{G}}{\partial{x}}}{\frac{\partial{G}}{\partial{z}}}[/tex]

    Thus, we gain:
    [tex]\frac{\partial{X}}{\partial{y}}\frac{\partial{Y}}{\partial{z}}\frac{\partial{Z}}{\partial{x}}=(-\frac{\frac{\partial{G}}{\partial{y}}}{\frac{\partial{G}}{\partial{x}}})(-\frac{\frac{\partial{G}}{\partial{z}}}{\frac{\partial{G}}{\partial{y}}})(-\frac{\frac{\partial{G}}{\partial{x}}}{\frac{\partial{G}}{\partial{z}}}=-1[/tex]
    as stated.


    As an example, let G(x,y,z)=ax+by+cz, so that the equation ax+by+cz=0 describes a plane.

    We thereby have:
    [tex]X(y,z)=-\frac{b}{a}y-\frac{c}{a}z,Y(x,z)=-\frac{a}{b}x-\frac{c}{b}z,Z(x,y)=-\frac{a}{c}x-\frac{b}{c}y[/tex]
    And we have:
    [tex]\frac{\partial{X}}{\partial{y}}=-\frac{b}{a},\frac{\partial{Y}}{\partial{z}}=-\frac{c}{b},\frac{\partial{Z}}{\partial{x}}=-\frac{a}{c}[/tex]
    and you see that the product of these three quantities equals -1.
     
    Last edited: Dec 17, 2007
  14. Dec 17, 2007 #13

    WWGD

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    Nice explanations. But there are restrictions to the manipulation:

    We cannot, e.g. conclude from y=x^2 and dy/dx=2x, that dx/dy=1/2x.
     
  15. Dec 17, 2007 #14

    WWGD

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    Aren't they cotangent space coordinates, dual to del/delx and del/dely ?.
     
  16. Dec 17, 2007 #15

    arildno

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    Except at x=0, we can:
    Let
    [tex]G(x,y)=y-x^{2}[/tex]
    Then, we have at G=0:
    [tex]\frac{\partial{G}}{\partial{y}}+\frac{\partial{G}}{\partial{x}}\frac{dx}{dy}=0\to{1}-2x\frac{dx}{dy}=0\to\frac{dx}{dy}=\frac{1}{2x}[/tex]

    This is meaningful at all points except for x=0
     
  17. Dec 17, 2007 #16
    Strangely enough, the small change terms like dx and dy can be treated as fractions in some cases, but not all.
     
  18. Dec 17, 2007 #17

    HallsofIvy

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    Can you give an example?
     
  19. Dec 17, 2007 #18
    No, [itex]dx[/itex] and [itex]dy[/itex] are linear coordinates on the tangent space. More precisely, [itex]dx,dy[/itex] is the linearised version of the original [itex]x,y[/itex]-coordinate system.

    Example:
    Take polar coordinates [itex]r,\theta[/itex] in the plane. The coordinate system picks out basis vectors [itex]\mathbf{e}_r[/itex] and [itex]\mathbf{e}_\theta[/itex] at every point. So any tangent vector [itex]\mathbf{v}[/itex] decomposes into

    [tex]\mathbf{v} = dr\,\mathbf{e}_r + d\theta\,\mathbf{e}_\theta[/itex]
     
  20. Dec 17, 2007 #19

    WWGD

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    Yes, sorry for my ignorance, I realized this --dy/dx=f =>dx/dy=1/f --is _always_true in a neighborhood of every x for which f'(x)=0, by the inverse function theorem, and as someone above said, it is true everywhere for f(x)=x^2, except at a point (x=0),
    and it may at times be globally true , e.g, for f(x)=x .
    I hope I will be more carefully after I sleep, now that the finals are over.

    I guess the truth of the statement also agrees with another perspective:
    dy/dx near a point is the slope of the tangent line at x to f(x), so that at
    each point we approximate f(x) by dy/dx=lim(deltax->0) delta(y)/delta(x).


    Sorry, I am still trying to learn latex. Hopefully for my next post.
     
  21. Dec 17, 2007 #20

    WWGD

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    So, Order of Things, please let me see if I understand this well:

    The linearization gives you coordinates along the tangent plane (since you

    have dx, dy ; two coordinates) of the curve (embedded in IR^2), at each tangent

    space. If this is the case, then I assume your curve (of which you are giving the

    coordinates) is differentiable?. e.g, for y=x^2 , you work with tangent spaces

    at each point and dy=2xdx, so your linearized coordinates in this tangent space

    (xo,yo) would be (yo,2xo), i.e, the linearized version gives the line y=2x

    as local coordinates?
     
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