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Why are we integrating?

  1. May 21, 2012 #1
    I would like to take a physics example:- Suppose force is acting and is a fucntion of distance x from origin. So F = kx

    Generally when we have to find the force on a say, a stick then we say - lets suppose a small element dx and then small force dF on this element is kdx

    I don't understand this part - how is dF is equal to small force. For me dF means a small change in force.
    If you could explain me this rigorously it would be great help
    Thank You!
  2. jcsd
  3. May 21, 2012 #2
    Basically for the same reason the Derivative of position with respect to time is velocity, and that's basically due to the definition of average velocity.
  4. May 22, 2012 #3
    In order for change to occur, it must be relative to something else. dF is the statement of an 'infinitesimal quantity', or a differential. If one considers the change of force relative to distance from the origin (position), then that would be represented by [itex]\frac{dF}{dx}[/itex]. Specifically, this is how force is instantaneously changing with respect to position at a particular point x. It is computed by evaluating the derivative at x.

    Before I get to the crux of your question, regarding integration, let me make some comments about the physics. 'The force on a stick' can have numerous interpretations. One may refer to some force distribution over the surface of the stick, or a single force applied instantaneously at a single point. Let's first consider force applied at a single point. Note that this is an idealization: force is not applied instantaneously nor is it ever localized precisely at a single infinitesimal point. Masses that are treated like this are sometimes referred to as point-masses. This idealization is very useful for conceptual purposes and computational ones. By computing with the center of mass of an object, one can get fairly realistic results in certain cases.

    Now let's consider the second: some force distribution on the stick. Let's assume that the force varies as you wrote: F = kx. In this scenario, this is the force on the stick. The force on the stick increases continuously and linearly in the direction x. Thus at x = 1, F = k; x = 2, F = 2k, and so forth. Remember as well that the force must have some direction in order for it to make full sense. So in this case, the integral of F across the length of the stick would provide the sum of all of the force (summed in the 'continuous manner' an integral does) on the stick. In this case, it would be kx^2/2 evaluated appropriately. This is why it can be conceptually thought of as summing up all of the dFs.

    Now to the crux of your question, which we delved into briefly above. dF can be thought of as an infinitesimal force since it is a differential quantity which is carefully defined via limits. The integral as described can be considered as the limiting case of a Riemann Sum where [itex]\Delta[/itex]F → dF in the limit. It is not the change of force for the reasons described in the first paragraph.

    Let's walk through an approximate sum of the mechanical potential energy a spring has if displaced and see how it yields the integral as a limiting case. The spring will be linear in the sense that the restoring force at a point x will be proportional to that distance x from the equilibrium point: F(x) = -kx. Assume the spring has been displaced a distance y from its equilibrium point. To compute the potential energy, it is irrelevant to us how the spring got into its present configuration, we only care to compute the energy it takes to get to that position; potential energy depends on state not on path. Thus we can compute the energy necessary to directly move the spring from its equilibrium point until point y and that will provide the quantity we want. We can do this by summing the energy necessary to move the spring up to each point preceding y until it reaches y. Of course, the difficulty is that the force changes at every point! One strategy would be to partition the interval from 0 to y into equal lengths of Δx and to assume the force constant along these partitions. So imagine you split the interval 0 to y into 3 pieces, then Δx = y/3. Then the total energy could be estimated as: -F(y/3)Δx + -F(2y/3)Δx + -F(y)Δx = (1/3 + 2/3 + 1)kyΔx. Notice how it's intuitive that if y is large, like 10 feet, this is likely to be a very bad approximation, but if it is is small, like 1 inch, it may not be so bad. This intuition captures the fact that a linear approximation made at a point of a function is nearly always better the closer one is to the point, e.g., the tangent line to a parabola will only be close to it within a short distance. Then it's also intuitive that if you were to make this partition smaller, and smaller, the approximation would improve at each point and hence the approximate sum would. Then, it's plausible, possibly intuitive, that if you were to make Δx infinitesimal, the sum would no longer be approximate but exact. This is the argument behind deriving a Riemann Integral from a Riemann Sum. The notation for when the limit of Δx goes to zero is dx. Thus the idea of summing up the 'dxs' or the 'dFs' is analogous to summing up each displacement or force along some interval. In the example above, one summed up all of the energy differentials, which were: dE = F(x)dx. This strategy represents an important idea in calculus and applicable elsewhere: in order to handle something that is changing, break it up into smaller pieces and assume it is not changing there, since that can be handled. Conquer the dynamic whole by dividing into static parts.
  5. May 22, 2012 #4
    Thank You!
  6. May 26, 2012 #5
    No, we don't say dF here!. The force is finite, kx, and the displacement is infinitesimal. In other problems, the force is infinitesimal, and the displacement is finite, for instance lifting water out of an odd shaped pool is a common problem from calculus, when doing work problems there. So instead of

    [tex]\int F\ dx,[/tex]

    we might write

    [tex]\int dF\ \Delta x.[/tex]

    You can always conceptually write [itex]W=\int dW[/itex], and by that I mean, the total work is the integral of all the infinitesimal works. In some cases the infinitesimal work [itex]dW[/itex] is due to a finite force acting over a small displacement, which is usually the only type you'll see in physics, while other times it may be better to look at it as taking a small bit of mass through a finite distance, in which case the force is infinitesimal.

    You might write

    [tex]dW(y)=dF(y)\ \Delta y(y)=dm(y)\ g\ \Delta y(y)=\rho\ dV(y)\ g\ \Delta y(y)
    =\rho\ A(y)\ dy\ g\ \Delta y(y).[/tex]

    So the cross sectional area, thus the infinitesimal volume, mass, and force all depend on [itex]y[/itex], and so might the displacement that that section of water has to travel through, but which is finite number. Thus the infinitesimal amount of work should depend on [itex]y[/itex]
    Last edited: May 26, 2012
  7. May 26, 2012 #6
    In other words, rereading your question, sometimes dF might be a small change in force, but it could also be a small force exerted by something, for instance an infinitesimal amount of water has an infinitesimal amount of gravitational force exerted on.

    What I was worried about is, usually in calculations you don't consider an infinitesimal amount of force exerted over an infinitesimal distance.
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