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Problem with differential/integral definitions

  1. Jun 15, 2014 #1
    If work ##W = \Delta E = \int_{s} \vec{F} \cdot d\vec{s}##, so work can't be ##\frac{dW}{d\vec{s}} = \vec{F}## like is here: http://en.wikipedia.org/wiki/Work_(physics)#Path_dependence

    bcb9806b128c9984c673ed9d244f8ffa.png

    cause this implies that ##W = \int \vec{F} \cdot d\vec{s} = E##, but the work is the variation of the energy, and not the energy.

    Again, the impulse ##\vec{J} = \Delta \vec{p}##, if I says that ##\frac{d\vec{J}}{dt} = \vec{F}##, this implies that the impulse ##\vec{J} = \int \vec{F} dt = \vec{p}##, but the impulse is the variation of the momentum and not the momentum.

    So, how can I administer these operations of a coherent way?
     
  2. jcsd
  3. Jun 15, 2014 #2

    Simon Bridge

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    In that last integral, you left off the limits of the integration.

    The E in that last one is th energy change between the limits of the integration.
    You are comparing apples and oranges.

    You should take care that the equations are being used in the correct context.
     
  4. Jun 16, 2014 #3
    But my second example is irrefutable! If ##\frac{d\vec{J}}{dt} = \vec{F}##, how I will know that ##\vec{J} = \int \vec{F} dt## or ##\vec{J} = \int_t \vec{F} dt## ? "By context"... ok, but exist situations where the context is unknown, for example, ##\frac{\partial S}{\partial t} = L##, I don't know if the action is the variation of the lagrangian or is the primitive of the lagrangian, and the equation don't says this too.
     
  5. Jun 17, 2014 #4

    Simon Bridge

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    What is it I'm supposed to be refuting?

    There will be no physical situation where the context is unknown - someone set up the experiment, someone made some observations and measurements.

    In your example (lagrangian/action relation) the equation will not be presented by itself - there is always some body of text surrounding it describing the context. By itself it is a meaningless collection of symbols. Even if you know that S is the action and L the lagrangian, all it does is limit the ways in which either may be found from the other.

    [You will have noticed from reading these forums that it is common for people to post a bunch of equations as their "working" - and the common follow-up response is to ask for the reasoning behind them. This is why.]

    It may be that the context is unclear to the reader of, say, a report or a paper - but that is a communication problem, not a physics problem.
     
  6. Jun 18, 2014 #5
    I'm not convinced of your idea and I think that we not be speaking in the same language. What I wants to mean is: the equation, the mathematical equation, works correctly without need of physical interpretations, cause the operations are defined so that to no exit 2 differing interpretations, ie, can't exist double interpretation in math.

    So, if the derivate of A results B, then the integral (indefinite) of B results A. Happens that exist situations where the derivative of A results B, but the definite integral B results A.

    Answer me you, given this differential equation: ##\frac{d}{dt} Y(t) = X(t)##, the solution for Y(t) is: ##Y(t) = \int X(t) dt## or is: ##Y(t) = \Delta X^{(-1)}(t)## ?
     
  7. Jun 18, 2014 #6

    Dale

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    You are missing the constant of integration. Put it in and all is well.
     
  8. Jun 18, 2014 #7
    I'm not missing the consant. By definition: ##\int f(x) dx = F(x) + C##, the integral already represents the antiderivative + the constant. Anyway, with all respect, this is not the principal question!
     
  9. Jun 18, 2014 #8
    Now arise me a ideia: if the indefinite integral of something is different of the definite integral, is like if exist 2 distinct operations... Well, but of fact exist 2 distinct derivative, the derivative and the exterior derivative. Would the exterior derivative the inverse of the definite integral and the derivative the inverse of the integral?
     
  10. Jun 18, 2014 #9
    You clearly are missing the constant of integration and it is an important point. We have two kinds of integrals, definite and indefinite. The definite integral is a value so
    ##\int_a^b \vec{F}\cdot\vec{ds}=E(b)-E(a)=\Delta E##
    is the change in the energy between the initial point on the path and the end point on the path. This is work.

    Then you have the indefinite integral version. In this case the answer is a function and not a value. Usually it is only written to show the mathematical relationship between force and energy, but you could interpret it as
    ##\int\vec{F}\cdot\vec{ds}=E(s)+C##
    which is the energy at a point along the path. The ##C## is important if you are interested in an absolute value of the energy. The important differences are a) this is a class of functions that determines the energy as a function of position and ##C## must be chosen to correspond to the initial conditions b) the indefinite integral is really used as a mathematical tool to help evaluate definite integrals.
     
  11. Jun 18, 2014 #10
    This is nonsensical. The derivative (wrt the integrating variable) of the definite integral is always zero. Review the FTC and recognize that the indefinite integral is really just the antiderivative. That is,

    ##\int f(x)dx## is the class of functions that are, by definition, antiderivatives of ##f(x)##.

    ##\int_a^x f(t)dt## is a function of ##x## that is (often) defined using Riemann sums and and turns out to differentiate to ##f(x)##.
     
  12. Jun 18, 2014 #11

    UltrafastPED

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    The notation may be the issue; blame Leibniz!

    Derivative: dy/dx = infinitesimal ratio of change in y wrt x; the slope of the tangent line

    Definite integral: ∫ Y(x) dx; the infinitesimal sums of slices of width dx for Y(x), representing the area under the curve, Y(x), between two specified points, (a,b). Except I left the end points off of the integral sign .. the elongated ∫ is the letter s, for "summa".

    Anti-derivative: Y(x) = dy/dx ... easy to find Y(x) given y, but not obvious going in the other direction! Note that you get the same Y(x) for y(x) + C for any constant C.

    Fundamental theorem of calculus: for the interval (a,b) we get ∫ Y(x) dx = y(a) - y(b).

    Notation: the indefinite integral, ∫ Y(x) dx = y(x) + C, is used to represent the anti-derivative; this is because it reminds you of the connection proven by the fundamental theorem of calculus.


    Using this notation Leibniz (and his followers) could start with Y(x) = dy/dx, then write
    Y(x) dx = dy, and then slap an integral sign on both sides:
    ∫Y(x) dx = ∫dy, noting that if the integration range on the LHS is (a,b), then the RHS is (y(a),y(b)).

    A very slick mnemonic device!

    Isaac Newton had his own notation; the only part we use today is the dot notation for derivatives wrt time, and that mostly in physics. Leonard Euler introduced the y' notation for derivatives, which is convenient in the expression of differential equations.
     
  13. Jun 18, 2014 #12
    "The wrong solution for the right problem is light years better than the right solution for the wrong problem." - Russell Ackoff

    The doubt I that did I open this topic none answered!!! I continue no understandig how I can pass of this step ##\frac{\partial S}{\partial t} = L##; ##\frac{d W}{d\vec{s}}=\vec{F}##; for this step: ##S = \int_{t_0}^{t_1} L dt##; ##W = \int_{\vec{s}_0}^{\vec{s}_1} \vec{F} \cdot d\vec{s}##. Because when I have ##\frac{dy}{dx}=y'##, the inverse equation is: ##y= \int y' dx## and not: ##y=\int_{x_0}^{x_1} y' dx##.
     
  14. Jun 18, 2014 #13

    Dale

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    "You can lead a horse to water, but you cannot make it drink" English Proverb

    You have received good and consistent answers from multiple sources. If the answers are answering the wrong question then you are not asking clearly. None of us are paid to answer your questions nor to read your mind.

    Try to have a more grateful attitude for the free advice you have been generously given and take a little more responsibility for any miscommunication. This thread is closed.

    No, it is not. The inverse equation is ##y= \int y' dx + C##
     
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