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Help with basic calculus

  1. Sep 28, 2009 #1
    I have been going over some lecture notes and have some questions about some of the mathematics shown in these notes.

    They start off with the following equation:

    [tex]\delta F_x = \dot{m} \frac{dV_x}{ds} \delta s[/tex]

    and then impose the limit as [tex]\delta s \rightarrow 0[/tex], and gets:

    [tex]\frac{dF_x}{ds} = \dot{m}\frac{dV_x}{ds}[/tex]

    I guess I am kind of confused as to how [tex]\delta s \rightarrow 0[/tex] would form the differential seen on the left hand side of the equation.

    Then the next line of the notes goes from the previous equation to:

    [tex]dF_x = \dot{m}dV_x[/tex]

    Is this because both sides have been integrated with respect to s? Or have I missed something else here too?

    Sorry if these seem like simple questions, but I guess I am not as confident with limits and the difference between [tex]\Delta[/tex], [tex]\delta[/tex] and [tex]d[/tex] as I thought I was.

    Any help would be appreciated.

    Thanks,

    Ryan
     
  2. jcsd
  3. Sep 29, 2009 #2

    arildno

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    Well, from your first equation, you may re-write this as:
    [tex]\frac{\delta{F}_{x}}{\delta{s}}=\dot{m}\frac{dV_{x}}{ds}[/tex]
    When now taking the limit as [itex]\delta\to{0}[/itex], your second equation should appear, agreed?
     
  4. Sep 29, 2009 #3
    I had wondered if it was just as simple as that or whether or not I was missing something, thanks for that.

    But I am still stuck with the second part of my question. If i integrate both sides with respect to s, I get:

    [tex]\int \frac{dF_x}{ds} ds = \dot{m} \int \frac{dV_x}{ds} ds[/tex]

    Which can then be simplified to:

    [tex]\int dF_x = \dot{m} \int dV_x[/tex]

    But this will not give the answer I got from the course notes.

    The only way I can think of to get that answer is to multiply through by [tex]ds[/tex], but I didn't think that you were "allowed" to do this. Please correct me if I am wrong.

    Thanks,

    Ryan
     
  5. Sep 29, 2009 #4

    arildno

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    Indeed.
    And at your level, this is the ONLY rigorous manner inb which you should think of equality between differentials.
    The two integrals will have the same value.

    It is possible to make an entirely different, equally rigorous maths in which differentials can be PROPERLY defined, but this is not basic level courses (and what you learn at basic level is NOT invalid, or of little use, but you've got to start somewhere, leaving other stuff out!)


    Again, fine thinking!

    But, the clever thing about the chain rule of differentiation, CONNECTED with the integral above will yield the same results AS IF you could have multiplied with ds!

    Thus, somewhat paradoxically, a formally meaningless operation (multiplying with ds) necessarily yields good results.

    (The basic reason why it is meaningless is that "ds" is not a real number, and hence can't be multiplied with. The other way I mentioned basically creates another "number system" within, indeed, multiplication of differentials is a meaningful operation.)

    Hope this helped!
     
  6. Sep 29, 2009 #5
    Yeah thanks that does help, but would it be possible to provide an example on how the chain rule would be used in this context? I think this may help me understand it a bit better.

    Thanks,

    Ryan
     
  7. Sep 29, 2009 #6

    arildno

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    Okay.

    Let us have a separable differential equation:
    [tex]\frac{dy}{dx}=\frac{1}{g(y(x))}[/tex]
    where g(y) is some function of y.

    We may solve this as follws:
    [tex]g(y(x))\frac{dy}{dx}=1[/tex],
    or, by integration:
    [tex]\int_{x_{0}}^{X}g(y(x))\frac{dy}{dx}dx=\int_{x_{0}}^{X}1dx=X-x_{0}[/tex]
    where I use X as an arbitrary x-value, x_0 as some initial value.
    Supposing there is a function G(y), so that [tex]\frac{dG}{dy}=g(y)[/itex].

    Then, by the technique of substitution (i.e, the inverse chain rule, if you like!),
    we may rewrite the left-hand side as:
    [tex]G(y(X))-g(y(x_{0})=\int_{x_{0}}^{X}\frac{dG}{dy}\frac{dy}{dx}dx[/tex],

    yielding the solution in (implicit form for Y=y(X), y_0=y(x_0)):
    [tex]G(Y)-G(y_{0})=X-x_{0}[/tex]

    Now, all this is very cumbersome!

    Instead, we cut over the niceties as follows:
    [tex]\frac{dy}{dx}=\frac{1}{g(y)}[/tex]
    we multiply with g(y)dx, yielding:
    [tex]g(y)dy=dx[/tex]
    and we then integrate the left-hand side from y_0 to Y, and the righthand-side from x_0 to X, giving us the same answer.
     
  8. Sep 29, 2009 #7
    Right, now I understand! Thank you very much for all your help arildno!
     
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