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What whould intigration of this result yield?

  1. May 2, 2009 #1
    I got a function -

    [tex]\frac {v - v_c}{cr} = \frac {dv_c}{dt}[/tex]

    Suppose I integrate this so as to get the original function...apparently the original function seems vc...so I should get this function after integrating; since we have dt in the denominator, it means the original function vc was differentiated WRT t...so I gotta integrate it WRT t to get vc.

    Am I right till now?

    But there's no t towards the LHS...is there anyway to integrate this...I mean I'm confused with this part...what should be done?
  2. jcsd
  3. May 3, 2009 #2


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    You just need to "separate" the variables by writing the derivative in terms of differentials- that is, separate [itex]dv_c/dt[/itex] into [itex]dv_c[/itex] and dt:
    [tex](v- v_c)dt= cr dc_v[/tex]
    [tex]dt= \frac{cr}{v- v_c}dv_c= -\frac{cr}{v_c- v}dv_c[/tex]
    and integrate both sides, the left with respect to t and the right with respect to [itex]v_c[/itex], of course.
  4. May 3, 2009 #3
    My question is more conceptual actually.

    What do you mean by [tex]\int dt[/tex]?...this is indefinite integral, its primary job is to anti-derivative...so how can a differentiated function like 'dt' exist?...it should be dx/dt or dy/dt etc...and the whole of dx/dt or dy/dt should be integrated to give the original result + c...so what do we mean by [tex]\int dt[/tex] ?
  5. May 3, 2009 #4


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    "dt" is not a "differentiated function"- it is not a function at all, it is a "differential". Most people are introduced to "differentials" in first semester Calculus. I would be surprised if you were not.

    As far as [itex]\int dt[/itex] is concerned, that is, of course, t+ constant, the anti-derivative of 1. [itex]\int dt= \int f(t) dt[/itex] with f(t)= 1.
  6. May 4, 2009 #5
    For starters I was taught advanced calculus without being told what is calculus...so :biggrin: the same people explaining that concept is a far off thing...they don't know themselves!

    Yes, I'm aware of the fact that dt is a value rather than a function (that's what I asked)...so that's the question, how do you integrate a value?

    I mean how do you solely integrate dx or dt etc...following the definition of an antidrivative?...or in that case how do get the undifferentiated function of a 'dx'.

    Suppose y = x

    Then dy/dx = 1...so if we integrate this differentiated function -

    [tex]\int frac{dy}{dx} = \int 1[/tex]

    y = x

    This is by following the definition of integration, i.e if we integrate dy/dx...we should get y...its apparent that y is the original function...and then we integrate the RHS by rules of integration to get the original function as a function of x.
    What do you mean by making dy = dx and then integrating?...integrating dy or dx will be like integrating an un differentiated function (since its a value actually)...so what's the logic behind doing this?
    Last edited: May 4, 2009
  7. May 4, 2009 #6


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    Another way of thinking about this is to simply integrate both sides with respect to x. For example, consider the differential equation,

    [tex]\frac{dy}{dx} = f\left(x\right)\cdot g\left(y\right)[/tex]

    Now divide through by g(y) (assuming that g(y) has no roots of course),

    [tex]\frac{1}{g\left(y\right)}\frac{dy}{dx} = f\left(x\right)[/tex]

    Now we integrate both sides of the equation with respect to x,

    [tex]\int \frac{1}{g\left(y\right)}\frac{dy}{dx}\;dx = \int f\left(x\right)\; dx[/tex]

    Which yields,

    [tex]\int \frac{1}{g\left(y\right)}\;dy = \int f\left(x\right)\; dx[/tex]

    Does that help?
  8. May 4, 2009 #7
    How did dx just pop by?...what utility does dx have here and what does the above expression mean (like we're here we have integrated..............so as to get the original function...etc...)
  9. May 4, 2009 #8
    The differential of a function is defined as follows (with respect to x.)

    dx is a variable representing the “change in x”, also called “delta x”.

    Then, given a function f(x), we define the differential of f , notated df, as:

    df = f ’ (x) * dx

    So, notice that given the function g(x)=x, we get g ‘ (x) = 1 and so

    dg= g ‘ (x) * dx = 1 * dx = dx
    [The “1” is hidden by our standard abbreviation for 1 * A = A.]

    So, the differential of the function x is dx.

    When we integrate such an expression, we are asking for the anti-derivative. What are the function(s) f such that they have the differential f ‘ (x) * dx?

    So, when you “integrate” dx you get x + C.

    Note: the antiderivative is usually written in terms of the variable and so, given dy we are assuming the variable is y. Thus, when you integrate dy you get y + C.

    But, if you integrate 2xdx you get x^2 + C.

    So, given that dy = 2xdx, we can ask for the antiderivatives and we get

    y + D = x^2 + B, or simply y = x^2 + C
  10. May 4, 2009 #9


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    Or do you mean something else?
  11. May 4, 2009 #10
    Humm...that solves the problem...thanks everyone!
  12. May 5, 2009 #11
    Wait...why do we integrate it WRT t?...why not dvc?
  13. May 6, 2009 #12
    Should have not allowed this thread to end.
  14. May 7, 2009 #13


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    As Halls said, we integrate the left wrt t and the right wrt vc.
  15. May 8, 2009 #14
    Yeah that's the question.

    The RHS is a function of vc...so basically it means t was differentiated WRT vc...so to get the original function, we should integrate it WRT vc...why will it be WRT t?

    Explicitly, integrating WRT t will be like this -

    Suppose we have a function [tex]/frac {dy}{dx} = f'(x)[/tex]
    [tex]dy = f'(x) dx[/tex]

    Now of course we gotta integrate this WRT x; from the original question integrating WRT t will be like integrating WRT y in this explicit example.
  16. May 8, 2009 #15


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    Take a look at my earlier post, we actually integrate both side with respect to t. Saying 'we integrate the left wrt ... and the right wrt ...' is just skipping a few steps.
  17. May 8, 2009 #16


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    It has been said many times, but integrating the LHS with respect to time and the RHS with respect to dvc means:

    \int dt=\int -\frac{cr}{v_c- v}dv_c
  18. May 9, 2009 #17

    Sorry, I overlooked that.

    I'm extremely confused...I'll think about this later. :cry:
  19. May 9, 2009 #18
    Ok...since the function is differentiated WRT one term, to get the original function, we should integrate it with one term.

    Else there will be no difference if a function f(x) is differentiated WRT x and if a function f(y) is differentiated WRT y.
  20. May 9, 2009 #19


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    What do you mean by "integrate it with one term"?

    Exactly right. There is no difference.
  21. May 10, 2009 #20


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    As I said in me earlier post, we do integrate the expression with respect to the 'original' variable of integration. See my earlier post:
    Notice that the differential equation was form by taking the derivative of some function with respect to x. And as you correctly say, to obtain the original function back (up to some constant), we must integrate with respect to the same variable.

    I think that we're going round in circles here. Can I ask that you take a look at my posthttps://www.physicsforums.com/showpost.php?p=2184386&postcount=6" and identify exactly which step(s) you have a problem with?
    Last edited by a moderator: Apr 24, 2017
  22. May 10, 2009 #21
    I seriously don't know. :confused:

    All I think is that if a function has been differentiated WRT a variable...it should be integrated with the same variable to get the original function.

    Ok...wrong example...sorry about that, but this is totally........:confused:

    How can dy/dx = dx/dy?

    Considering a straight line -
    Suppose, if y changes by 5...x changes by 2.

    So dy/dx will mean 5/2 and dx/dy will be 2/5...how can they be the same?

    Moreover if dy/dx exists, dx/dy does not (right) (considering the same function).

    Suppose we have a function y = f(x).
    Of course we're gonna differentiate this WRT x...how can we differentiate WRT y?

    After differentiating we get dy = f'(x)dx.

    To get the original function we should integrate it WRT the the same variable it was differentiated with, following what I think -

    So according to me integrating with both the terms will be wrong.

    The standard technique will be correct if what I think -

    is wrong; i.e "if a function has been differentiated WRT a variable...it should be integrated with both the variable to get the original function." is correct.
  23. May 10, 2009 #22


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    Okay, you meant "with respect to the same variable", not "term". Now, I get you.

    As Hurkyl said, If dy/dx= f(x)g(y) then we can write
    [tex]\frac{1}{g(y)}\frac{dy}{dx}= f(x)[/tex]
    and integrate both sides with respect to x:
    [tex]\int \frac{1}{g(y)}\frac{dy}{dx} dx= \int f(x)dx[/tex]
    If G(y) is an anti-derivative of [itex]1/g(y)[/itex], then, by the chain rule,
    [tex]\frac{dG}{dx}= \frac{dG}{dy}\frac{dy}{dx}= \frac{1}{g(y)}\frac{dy}{dx}[/tex]
    so that
    [tex]\int \frac{1}{g(y)} \frac{dy}{dx} dx= G(y)+ C= \int \frac{1}{g(y)} dy[/tex]

    What I gave earlier, using differentials, was "shorthand" for that.

    Only if y= f(x) is a function that is its own inverse. But no one has said that in general.

    If y= f(x), then dy/dx= f'(x). But if f is an invertible function (one-to-one and onto) then x= f-1(y) and dx/dy= f-1' (y). It is shown in Calculus I that, in this case
    [tex]\frac{dy}{dx}= \frac{1}{\frac{dx}{dy}}[/tex]

    I still do not know where you got "dy/dx= dx/dy".
    I did say, earlier, that there is no difference between df(x)/dx and df(y)/dy- if f(x)= x2 then df/dx=2x, f(y)= y2, df/dy= 2y- but that is NOT the same as "dy/dx= dx/dy"! If y= x2, then dy/dx= 2x, x= y1/2, dx/dy= (1/2)y-1/2= 1/2y1/2= 1/(2x)= 1/(dy/dx).

    "with both the variable"?
    Last edited by a moderator: May 10, 2009
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