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Integrations and derivatives in physics

  1. Sep 19, 2004 #1
    Can someone tell me (or make a list) of what you can integrate and find the derivatve of in physics to get something else. For example if you integrate velocity you will find acceleration and if you take the derivative of velocity you will get displacement. They are not in my book.

    ~Thanks
     
  2. jcsd
  3. Sep 19, 2004 #2
    1. Distance is the total length that an object in motion covers. Displacement is a vector quantity that indicates the change in position that an object moves in a particular direction. Average speed is the distance covered per unit time. Average velocity is the displacement divided by the time interval.



    2. The slope of a position-time graph gives the velocity. The slope of a velocity-time graph gives the acceleration.



    3. The area under the curve of a velocity-time graph is the displacement. The area under the curve of an acceleration-time graph is the velocity.
     
  4. Sep 19, 2004 #3
    What about the ones like: the resulting velocity = the integral of the acceleration vector plus the initial velocity vector (or whatever it is) for motion in 2D.
     
    Last edited: Sep 19, 2004
  5. Sep 19, 2004 #4
    a = acceleration
    v = velocity
    s = displacement

    [itex]
    a = \frac{dv}{dt}
    [/itex]

    [itex]
    v = \frac{ds}{dt} = \int{a}{\;dt} \n
    [/itex]

    [itex]
    s = \int{v}{\;dt}
    [/itex]
     
  6. Sep 19, 2004 #5
    But with those 3 formulas how would you know the resulting velocity = the integral of the acceleration vector plus the initial velocity vector?

    Like how do you know when you integrate the acceleration vector you have to add the initial velocities?
    Example:
    a= 4i +2j = (4.2)
    when you integrate it you get
    v= (4t +vi, 2t + vj)
    where vi and vj are the initial velocities in the x and j directions.
     
    Last edited: Sep 19, 2004
  7. Sep 19, 2004 #6
    Are you talking about 2D motion (vectors)?

    As in, when velocity of a particle moving in the cartesian coordinates (x,y) is given by:

    [itex]
    v = \frac{dx}{dt}i + \frac{dy}{dt}j = \dot{x}i + \dot{y}j
    [/itex]
     
  8. Sep 19, 2004 #7
    yeah i'm talking about 2d motion with vectors
     
  9. Sep 19, 2004 #8

    cepheid

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    Whenever you integrate, you have to add the constant of integration so that your result is the most general antiderivative of the original function. Haven't you done that in calculus? In this case, the constant of integration is the initial velocity.
     
  10. Sep 19, 2004 #9
    Oh.

    Let's take your example:

    [itex]a = 4i +2j[/itex]

    [itex]v = \int{a}{\;dt} = (4t+c_{1})i + (2t+c_{2})j [/itex]

    The constants of integration [tex]c_{1}[/tex] & [tex]c_{2}[/tex] are the initial velocity of the particle.

    ([tex]c_{1} = c_{2}[/tex])
     
  11. Sep 19, 2004 #10
    No I have not done this in calc because I just started calc a couple days ago (i just started physics a couple days ago too) but we are doing this in physics. I read the chapter in my calc book about integrations but i have not really learned how to do them yet. That's why I'm soooo confused in physics now.
     
    Last edited: Sep 19, 2004
  12. Sep 19, 2004 #11
    How do you know what the constant of integration is?
     
  13. Sep 19, 2004 #12
    You substitute values in the new expression.

    For example, you'd be told that the particle's velocity is v=6i+8j when t=2.

    Edit:
    Errr... Actually, if you want to find the initial velocity, use t=0.
     
    Last edited: Sep 19, 2004
  14. Sep 19, 2004 #13
    but how did you know that you had to add the initial velocity? how did you know the constant of integration was initial velocity?
     
  15. Sep 19, 2004 #14
    Let's assume that after integrating you got:
    v = (2t+c)i + (4t+c)j, where c is the constant of integration.
    You should realise that at the initial velocity, t=0.
    Substituting this in the original equation gives:
    v = (2*0+c)i + (4*0+c)j = ci + cj, which is the initial velocity.
     
  16. Sep 19, 2004 #15
    ohhhhhhh i understand :smile: .

    When you take the integral of acceleration and you have the velocity of v = (2t+c)i + (4t+c)j, is this the average velocity or the final velocity?
     
  17. Sep 19, 2004 #16
    Is this correct?

    If you integrate:

    1.) acceleration: the constant constant of integration is initial velocity

    2.) velocity: the constant of integration is final velocity



    What do you get when you integrate displacement?
    Is there a constant when you find the derivative?
     
  18. Sep 19, 2004 #17

    HallsofIvy

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    (instantaneous) acceleration is defined as "the derivative of velocity" by the "fundamental theorem of calculus", integrating acceleration gives velocity.

    No, it is not correct to say that the constant of integration IS the initial velocity. IF setting t= 0 makes everything else in the integral 0, then it is true but if, for example, acceleration is given by et then the acceleration is et+ C. The initial velocity is 1+ C so in this case the constant of integration is the initial velocity minus 1. It is much better to understand finding the constant of integration by evaluating at some given time. (Yes, most of the time that is t= 0 and often (but not always, the initial velocity is C.)

    But that's not nearly as bad as saying "the constant of integration is final velocity"!
    What do you do if there is NO final velocity? The integral of velocity is "displacement". Again, you determine the constant, for a particular problem by evaluating the position (displacement) function at some given time (which is often t= 0). Often, but not always, the constant of integration is the initial displacement.

    Generally speaking, any quantity, that, as long as everything stays constant, is calculated by a division (speed = distance/time, density= mass/volume, pressure= force/area) is a derivative when dealing with variables and any thing that, as long as everything stays constant, is calculated by a product (distance= speed*time, mass= density*volume, force= pressure*area) requires and integral.
     
  19. Sep 19, 2004 #18
    :cry: :cry: :cry: :cry: :cry:
    So when I am doing a problem I have to figure out the constant of integration?
     
  20. Sep 19, 2004 #19
    could i use one of the kinematic equations?
     
  21. Sep 19, 2004 #20

    HallsofIvy

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    If you are given acceleration, then you will have to be given the velocity and position at some time to find the "constants of integration". If you are given the initial velocity and position you can certainly put those values into the equations.
     
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