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Problem with delta x

  1. Jun 15, 2014 #1
    The question

    In my work [itex]\mu[/itex] is the mass per unit length, therefore I believe I can say [tex]m=\mu\Delta x[/tex]because Michael Fowler from the University of Virginia does the same at http://galileo.phys.virginia.edu/classes/152.mf1i.spring02/AnalyzingWaves.htm (the 2nd line bellow the graph)

    I start from[tex]\int^{b}_{0}\sqrt{1+\cos^{2}\left( x\right) }dx=\dfrac{\mu B}{N}[/tex] and since it is nonsensical to say [tex]\int^{b}_{0}\sqrt{1+\cos^{2}\left( x\right) }\left( dx\right)^{2}=\dfrac {mB}{N}[/tex] for a few reasons, I converted the definite integral into a Riemann sum and got [tex]\lim_{n\rightarrow\infty }\sum^{n}_{i=1}\left[\sqrt{1+\cos^{2}\left( \dfrac {b_i}{n}\right) }\right]\left( \dfrac{b}{n}\right)^{2}=\dfrac{mB}{N}[/tex]
    Is this correct? Since I am dealing with [itex]\Delta x[/itex] and not [itex]dx[/itex] (which deals with infinitely small) I can bring [itex]\Delta x[/itex] to the other side from [tex]\dfrac {m B}{N\Delta x}[/tex] right? (Also [itex]\dfrac {b}{n}=\Delta x[/itex] in the Riemann sum)
     
    Last edited: Jun 15, 2014
  2. jcsd
  3. Jun 15, 2014 #2
    The Riemann sum version of your original equation is $$\lim_{n\rightarrow\infty }\sum^{n}_{i=1}\left[\sqrt{1+\cos^{2}\left( \dfrac {b_i}{n}\right) }\left( \dfrac{b}{n}\right)\right]=\dfrac{\mu B}{N}$$. If we multiply both sides by ##\Delta x=\dfrac {b}{n}##, we'd have $$\dfrac {b}{n}\lim_{n\rightarrow\infty }\sum^{n}_{i=1}\left[\sqrt{1+\cos^{2}\left( \dfrac {b_i}{n}\right) }\left( \dfrac{b}{n}\right)\right]=\dfrac{m B}{N}$$ yes? Now we can't just push that ##\dfrac {b}{n}## inside a limit in ##n##, just like we couldn't pull the ##\dfrac {b}{n}## that's originally inside the limit out.
     
  4. Jun 15, 2014 #3
    Thanks alot!
     
  5. Jun 15, 2014 #4
    But what does b/n mean by itself, does n still go to infinity for b/n if it is outside the limit?
     
  6. Jun 15, 2014 #5

    AlephZero

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    You can copy a bad example if you like, but it would be more logical to write ##\Delta m=\mu\Delta x##.

    Maybe Mr Fowler though it was too hard for his students to recognize using ##F=ma## unless the mass was literally called ##m## amd the force was literally called ##F##. That's just spoon-feeding IMO.
     
  7. Jun 15, 2014 #6
    Yes. Writing ##dm = \mu dx## seems more appropriate, where ##dm## is the mass of a small piece of wire or string or whatever we have.

    If the question is from a textbook/homework it could be useful if you provided the full question. What are ##N## and ##B## for example? Constants?

    When you write
    \begin{equation*}
    \int_0^b \sqrt{1 + \cos^2(x)} \, dx = \frac{\mu B}{N}
    \end{equation*}
    the LHS depends on ##b## only, so you are claiming that ##\mu B N^{-1}## depends only on ##b##? This seems a little strange to me if ##\mu## is mass per unit length. I would expect ##\mu = \mu(x)##.

    In your previous threat on this, I asked if you had started with something like
    \begin{align*}
    dm &= udx \\
    \mu &= \mu(x),
    \end{align*}
    and you said I was right. It would seem you have taken a step back from this now.

    I kind of get the feeling that the issues you have stem from an error in the setup at an early stage. If you provide the full question, starting from the beginning, it might be easier to help you.
     
    Last edited: Jun 15, 2014
  8. Jun 15, 2014 #7

    SammyS

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    @ Quesadilla:
    Thanks for mentioning the other thread, and the need to post the full question.


    @ Rochefort:
    It's not a good idea to start a new thread on the same issue as another thread you have started. In fact it may be viewed as a violation of the rules of this Forum.
     
  9. Jun 16, 2014 #8
    I wasn't aware I was violating the rules, I won't repeat the same mistake.

    Also I hope you understand that I don't want to post the whole question as I want to solve it before I share it. But I will follow that [itex]\Delta m=\mu \Delta x[/itex]

    So let me rewrite the problem one more time but this time I will introduce the variables [itex]A[/itex] and [itex]C[/itex] where [tex]\int^{b}_{0}\sqrt{1+\cos^{2}\left( x\right) }dx=\dfrac {\mu BC}{2Am}[/tex] (which should't really effect my question.) Now when I covert the definite integral into a Riemann sum I get [tex]\lim_{n\rightarrow \infty }\sum^{n}_{i=1}\left[\sqrt{1+\cos^{2}\left(\dfrac {b_{i}}{n}\right) }\left(\dfrac{b}{n}\right)\right] =\dfrac{\mu BC}{2Am}[/tex] where [itex]\dfrac {b}{n}=\Delta x[/itex]. So when I substitute [itex]\dfrac {\Delta m}{\Delta x}[/itex] for [itex]\mu[/itex]and bring [itex]\Delta x[/itex] to the other side, I am left with [tex]\dfrac {b}{n}\lim _{n\rightarrow \infty }\sum^{n}_{i=1}\left[\sqrt {1+\cos^{2}\left( \dfrac{b_{i}}{n}\right) }\left(\dfrac{b}{n}\right)\right]=\dfrac {\Delta mBC}{2Am}[/tex] Assuming all the maths is correct I would like someone to answer my last two questions:

    1) Since the [itex]\dfrac {b}{n}[/itex](furthest to the left) is outside the [itex]\lim_{n\rightarrow \infty }[/itex] is it still effected by it? If not what does the [itex]n[/itex] stand for?
    2) How can I get rid of the [itex]m[/itex]s on the right side of the equation?
     
  10. Jun 17, 2014 #9
    Sorry. I didn't mean to abandon this thread.

    1) No. The ##\frac{b}{n}## outside the limit is not affected by the limit. Without any additional context with which to interpret your problem, I have no way of telling you what the ##n## means in ##\frac{b}{n}## other than the standard "number of equal partitions of the interval ##[0,b]##" Riemann sum business, but I doubt that's what you're looking for.

    2) ##\Delta m## and ##m## are two different, apparently independent variables. It's unclear to me why you would expect (or desire) to be be able to "get rid" of them.

    While I respect your desire to work through as much of the problem as possible on your own, I think you are overestimating our ability to help you with so little context regarding the nature of your problem and the meaning of the variables.
     
  11. Jun 17, 2014 #10
    Yes. We would like to help you, but it becomes too much of a guessing game. Since I am here again, I will still try to provide a little more input.

    Firstly, I want to note that
    \begin{equation*}
    f(b) := \int_0^b \sqrt{1 + \cos^2 (x)} \, dx
    \end{equation*}
    depends on ##b## and only ##b##. We also have that ## 0 \leq f(b) \leq \sqrt{2}b##.

    Secondly, you introduced an ##m## in the denominator. Is this ##m## related to ##\Delta m##? In particular, is ##\Delta m ## the mass of a small fraction of an object and ##m## the full mass of the same object? Also, if you introduce variables ##A,B## we need to know what they depend on, or at least if they depend on ##x## or ##m \,(\Delta m)##.

    Lastly, and most importantly, what are you trying to do? Compute ##m##, the mass of the entire object, that is ## m = \int dm = \sum_i \Delta m_i##?
     
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