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Integration related to physics doubt

  1. Sep 3, 2012 #1
    1. The problem statement, all variables and given/known data
    I am a high school student and I am very fond of calculus. I know we get the distance by calculating the area under the v-t graph. But this is my take on another way to get an equation
    The original differential equation is
    dx/dt=v(t)

    multiplying both sides by dt,
    dx=v(t)dt

    Integrating on both sides,
    ∫dx=∫v(t)dt

    Hence we get
    x+c=F(t)+C
    where F(t) consists of variables only.
    but

    x=∫v(t)dt =F(t)+C

    so i am getting x+c=x but since c is a constant and not necessarily zero x+c might not be equal to x.

    Help!!!!!!!!!!!!!!! Spending sleepless nights on this problem
    2. Relevant equations
    x=∫v(t)dt

    dx/dt=v(t)

    3. The attempt at a solution
    Tried my best. Couldn't reach anywhere.
     
  2. jcsd
  3. Sep 3, 2012 #2
    That's correct.


    Here's where you went wrong. How did x+c become just x? ∫dx=x+c as you indicated in the first quote, so x=∫v(t)dt is not true.
    Also, your "c" and "C" both represent the same initial distance in this case, so you can subtract C from both sides to get x=F(t).
     
  4. Sep 3, 2012 #3
    First off be careful using F(t) as an arbitrary function as F(t) generally denotes Force as a function of time. Your issue here is that x is being used for two different purposes on the LHS and RHS. On the RHS x denotes the distance traveled due to time (x(t)) and c is the initial starting point. Sum them up and you get the total displacement. However, on the LHS x refers to the total displacement where F(t) is the integral of dx/dt without the constant and c is the constant. The easiest way to see that the equations actually are the same is just evaluate the interior of the integral on the RHS.
    i.e dx= v(t) dt = dx
     
  5. Sep 3, 2012 #4

    Ray Vickson

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    Constants of integration are essentially arbitrary (and may have arbitrary names). Would it still trouble you if I wrote x=∫v(t)dt = F(t)+K? Now we would just have K = C-c.

    To eliminate such sources of ambiguity (in cases where it matters at all), go with *definite* integration, so if the object starts from x(0) = a the position at time t is
    [tex] x(t) = a + \int_0^t v(s) \, ds.[/tex] Notice that I have avoided writing [itex]\int_0^t v(t) \, dt,[/itex] which some people will do. This would be a good thing to avoid---it makes the same symbol 't' perform double duty in the same problem---once as a limit of integration, and once as a dummy integration variable. The point is that [itex]\int_0^t v(s) \, ds = \int_0^t v(w) \, dw = \int_0^t v(\text{anything})\, d\text{anything}.[/itex]

    RGV
     
  6. Sep 4, 2012 #5
    Mr. Ray Vickson : You said that C-c=k but Mr. Nessdude14 said that c's are equal giving k as zero. Who is correct???

    and I appreciate the definite integration idea but I also wanna know how I could do it through indefinite integration

    Mr. Nessdude14: If c=C then in this case F(t) consists only of variables dependent on t right??? does that mean the x does not involve a constant?
    in this case is x the change in distance or total distance???

    Mr. dot.hack: thanks for the pointer about F(t) . also does x here mean total displacement or change in displacement???

    Anyway, thanks for all the help you people have put in.
     
  7. Sep 4, 2012 #6

    HallsofIvy

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    Mathematically, if two people are doing the same integral, one gets f(x)+ c and the other gets f(x)+ C, "c" and "C" might well be different, might well be the same.

    However, in applications, there has to be some addition condition that determines the "constant of integration". Initially, you said "I know we get the distance by calculating the area under the v-t graph". Exactly what distance are you talking about?
    If v(t) is the speed of a moving object, [itex]\int f(t)dt[/itex] is NOT the distance a an obect moves. It is, rather, the x coordinate of the position of the object- and that depends on both the speed and the initial position. You would have to know the initial position in order to determine the constant. If you were to do the problem as a definite integral, [itex]\int_{t_0}^{t_1} v(t)dt[/itex], there is no constant of integration and you woud have calculated the distance travelled between [itex]t_0[/itex] and [itex]t_1[/itex].
     
  8. Sep 4, 2012 #7
    Halls of Ivy: Is the c always the initial position and the rest of the integral always the change in x-coordinate of the position ???

    Thanks for the help!!!
     
  9. Sep 4, 2012 #8
    Yes; whenever you integrate velocity with respect to time, the "C" constant represents initial position. The only reason that I was able to say your "C" and "c" values were equal is because within the constraints of the problem, they represent the same initial position.
     
  10. Sep 4, 2012 #9

    Ray Vickson

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    If you write x = F(t) (instead of F(t)+C), does that make a difference? NO, it does not! F(t) = ∫ v(t) dt is defined as any antiderivative of v; that is, F(t) is any function such that dF(t)/dt = v(t). That means that F(t) itself can contain an arbitrary constant. So, if v(t) = a*t (a = constant), F(t) could be (1/2)*a*t^2, or it could be 762 + (1/2)*a*t^2, or it could be -2012 + (1/2)*a*t^2. or be (1/2)*a*t^2 + C.

    Usually in integration tables or in answers delivered by computer algebra systems, for example, a specific formula is written for F(t) and the constant of integration is omitted. The user is supposed to understand that a constant of integration is missing and must by determined by some other problem information, or be left unevaluated.

    Anyway, if Person A writes x+c = F+C and Person B writes x = F+c, there is NO contradiction: just because both persons use the sam letter C does not mean they are the same constant. We really ought to write ca+ x = F + Ca and x = F + Cb, to distinguish between the expressions by A and B.

    RGV
     
  11. Sep 4, 2012 #10
    I finally understood!!!

    Thanks a lot for those all who have contributed to this thread and helped clear my doubt!!!
     
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