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Simple doubt in application of calculus

  1. Aug 8, 2013 #1
    I have done quite a few introductory physics problems where calculus is involved .When refreshing the concept of work done by variable force,I find myself unsure about an application of calculus.The underlying concept which is puzzling me is at the very root of calculus .The work done by the force is given by dW =f(x)dx ,where we take f(x) to be constant over dx . An example is the spring force f(x)=kx.

    How and why do we take f(x) to remain same over dx i.e function f remains same over the interval from x to x+dx ? Function f has a definite value for x i.e f(x) but how does it have same value as x changes to x+dx .May be the infinitesimal change dx has something to do with it which i am unable to understand.

    I would deeply appreciate if someone could clarify this doubt in a simple language .
  2. jcsd
  3. Aug 8, 2013 #2


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    It's the same reason we use rectangular strips of varying lengths to determine the area under a curve. The widths of the individual strips are a constant value, dx, while the lengths of the strips vary as the value of the height of the curve from the x-axis, which is f(x). The area of each rectangle is f(x)dx, and by adding the areas of all of these strips, we can approximate the total area under the curve from x = a to x = b. The magical part of this process is in the integration: as the widths of the strips, dx, gets smaller and smaller, we use more and more strips to approximate the area. If the integral converges, our crude approximations to the area under the curve will approach a limiting value.

    For the spring problem, we are arbitrarily saying that the value of the force f(x) is roughly constant at a certain value of the deflection of the spring, x, in other words, f(x+dx) is roughly the same value of f(x). Remember, it's just an approximation, like finding the area under a curve. As dx gets smaller, f(x+dx) tends to approach f(x).
  4. Aug 8, 2013 #3

    Stephen Tashi

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    The modern point of view is not that the function f(x) necessarily remains constant over any finite interval (and the modern point of view doesn't discuss "infinitesimal" intervals). In introductory calculus an integral is approximated as a sequence of finite Riemann sums and these sums use finite intervals. Taking the limit of this sequence of sums as the lengths of the intervals used approaches zero is defined as the correct process for getting the work done by the force. The motivation for the definition is the belief that the Riemann sums with finite intervals are a the appropriate way to approximate the work.

    If you are studying a book that phrases things in terms of "dW" and "dx" then you have the usual problem of trying to understand reasoning that is stated terms of "differentials" or "infinitesimals" when no precise definition of these things has been given. See the thread: https://www.physicsforums.com/showthread.php?t=701423&highlight=calculus which, though doesn't provide any consensus about the matter, at least will make it clear that reasoning with "dW" and "dx" isn't presented in the same precise language as, say, an epsilon-delta proof about a limit is presented.
  5. Aug 8, 2013 #4
    Thanks SteamKing ad Stephan

    but my doubt still remains uncleared.

    What i inferred is f(x) is assumed constant over Δx such that dW≈f(x)Δx is an approximate value.And when we integrate ,with Δx→0,then this approximate value turns into an exact value .

  6. Aug 9, 2013 #5

    Stephen Tashi

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    Do you mean that the sums approach a limit as the number of rectangles approaches infinity and that this limit is regarded as the exact work that is done? Yes, that is correct. It is not correct to say that the area of a single rectangle in the sum approaches some "exact" amount of work. The area of each rectangle approaches zero as the number of rectangles approaches infinity.
  7. Aug 9, 2013 #6
    Thanks Stephen :)

    Interesting .

    Do you mean dW→0 and W→some definite value ?
    Last edited: Aug 9, 2013
  8. Aug 10, 2013 #7
    I am having difficulty understanding how the area of each rectangle approaches zero whereas sum of the areas approach some non zero value .

    Could somebody reflect upon this ?
  9. Aug 10, 2013 #8


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    ^Just addition. Many small things added up need not be small. Imagine you have some large rock, if you break it into many small pieces and add up the mass it will be the same as the large rock.
  10. Aug 10, 2013 #9
    Thanks lurflurf :)

    In the case of work done by spring dW→0 and W→some definite value . Right?
  11. Aug 15, 2013 #10
    That is correct, as you may know,

    [itex]\int F(x)dx = \int kx dx[/itex]

    So then solve as needed
  12. Aug 16, 2013 #11
    Thanks saminator910

    When we say Δx→0,Δx approaches 0.It is very close to 0,but never equal to 0. Right?
  13. Aug 16, 2013 #12
    I think of it like this, [itex]\Delta x[/itex] becomes an infinitesimal, infinitely close to zero, then when you have split [itex][a,b][/itex] into an infinite amount of rectangles and add them up, your essentially "cancelling out" the infinity. I use "cancelling out" very loosely in this context.
  14. Aug 16, 2013 #13
    Is it correct to say Δx never equals 0 ? Very close to zero but not zero.
  15. Aug 16, 2013 #14


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    This is a useful intuition, I think. Technically it is not correct for at least two reasons, because if Δx were very small but not 0 we would still not get the limiting value, and to make it infinitely small needs a new number system (the hyperreals) and it isn't all too clear what such infinitesimal magnitudes represent. Can they even be called magnitudes?

    But yeah, when you calculate limits, this is a useful intuition.
  16. Aug 16, 2013 #15
    It is like calculating the distance from velocity and time. You are driving with a car and accelerating. At any instant, the distance traveled per unit time is v, but v is changing. If you drive at 10 m/s, you might roughly say that during the next second you will travel 10 meters, but that is not exact, because your speed changes during that second. But if you would say that you will travel 1 centimeter during the next millisecond, you would expect it to be much more accurate, because the speed would change much less during a millisecond. For more accuracy you would need to consider smaller and smaller time intervals. And if the intervals are so small that every moment has its own speed then the distance calculation will be exact.

    For example, you travel 10 m/s and your acceleration is roughly 3 m/s^2. The acceleration is changing too, but not too fast.
    So during the first second your traveled distanced is roughly somewhere between 10 and 13 meters, because your speed changes from 10 to 13 m/s. So the error might be approximately 3 m.
    Dividing the 1 second into milliseconds, during the first millisecond the speed is somewhere between 10 and 10.003 m/s, during the second millisecond it is somewhere between 10.003 and 10.006 m/s and so on. So at every instant the speed might be approximately 0.003 m/s wrong so during the whole second the error in distance might be 0.003 m. So you see that dividing the second into 1000 intervals decreases the error roughly 1000 times.
  17. Aug 17, 2013 #16
    Thanks Verty

    And it is Δx which tends to 0 not dx . dx is a notation which is used along with ∫ .Right?

    I am trying to understand calculus as it applies to solving physics problems.
  18. Aug 17, 2013 #17


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    Correct. It is referring to the secant/tangent idea, the limit of the secants is the tangent, I assume you have seen that before.

    If you've not seen that, this video explains it, eg at 11:00.
    Last edited: Aug 17, 2013
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