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Defining derivates

  1. Mar 21, 2003 #1
    i've been oposing the traditional-modern physics as much as i can. unfortunately, no one is pating much attention to what i say. in one such effort my corespondant said dx/dt by definition is the momentary speed. we have a huge differences in interpreting the basics of both math and physics.

    (1)if you define dx/dt as V the do you neceserily agree that x is function of V and t?

    (2)if so then why do you derivate x partially only by dt?

    (3)what is dx/dV then?

    (4)how do you measure the momentary speed V if V=dx/dt?

    your physics needs to be redone. i mean it most sencerely.
     
  2. jcsd
  3. Mar 21, 2003 #2
    Actually, you do not need to know the velocity to determine the position at time t. Instantaneous position is a function of time only. It probably looks strange because you are used to Velocity = Distance / Time. This is used to define v = ds / dt, the equation for instantaneous velocity. v, too, is a function of time only. It does not seem to make sense, does it, that velocity is not a function of distance (actually displacement).
     
  4. Mar 21, 2003 #3

    Hurkyl

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    In certain cases, x may be written as a function of v and t, but in general that is not possible.

    You don't; you take the total derivative of x with respect to t.

    Taking total derivatives, dx/dv = v/a (where a is acceleration).
    The partial derivative of x with respect to v would depend upon the particular function you use to express x as a function of v and t.

    Hurkyl
     
    Last edited: Mar 21, 2003
  5. Mar 21, 2003 #4

    arivero

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    C'mon, move on. The concept of momentaneus velocity was defined by Newton and accepted on experimental grounds. These experimental grounds were nullified in the first decade of the XXth century. Pal, you are one century back!
     
  6. Mar 21, 2003 #5
    Instantaneous position is a function of time only
    this particular instance seems totally wrong because of counterparts incompatibility. look: t expresses in secounds, any function of t should be expressed in same function of secounds but on the other hand x or s (the position in the coresponding moment) expresses in meters. HALT ALL. as a logical consequence x=x(V,t) but then dx=Vdt+tdV. now if you compare with Vdt=dx comes that t or dV equals to zero(i.e. this is the starting moment or the velocity is constant)

    It probably looks strange because you are used to Velocity = Distance / Time
    not because of what i am use to but because of counterparts incompatibility as mentioned perviously.

    v, too, is a function of time only
    i've been thru that with integral(as his pervious nick was).look:
    case 1:
    x = t => dx = dt
    if Vdt = dx and dx = dt => V = 1
    go back in x = t => x = 1t = Vt.
    x = Vt

    case 2:
    x = tt => dx = 2tdt
    if V = dx/dt and dx = 2tdt then V = 2t
    go back in x = tt u'll have x = 2t(t/2) = Vt/2
    x = Vt/2

    how come this two cases have different forms then?

    should the law for instant (position, velocity and time) be valid no matter what/in any case?

    it is hard to be right when the goverment is wrong!!
     
  7. Mar 21, 2003 #6
    if only Newton was alive now i'd address him personally but.....

    back or forward in time we are about to see. the only certain thing is that i am one century away from you. i am sorry if i am boring you with all this but it is my highest priority to convice you that, and where you are wrong so you could coretc your self. i don't have a problem. even if i do i can deal with it my self. it is you who has a problem.
     
  8. Mar 21, 2003 #7

    arivero

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    Hmm? Hey, please note I agree with you if you are actually attacking the concept of momentaneous velocity.

    It is only that I can not understand why you are attacking on mathematical grounds when the concept has already been disproved just on physical grounds. Nowadays we have only averaged velocities.

    Even in the path-integral approach, which is the closest thing to classical trajectories we have, the distribution concentrates on path where the mathematical derivative can not be defined.
     
  9. Mar 21, 2003 #8

    Hurkyl

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    Don't forget that one is allowed to multiply and divide by dimensions as if they were ordinary numeric constants. For example:

    x = (3 meters / seconds^2) * t^2

    This is dimensionally consistent, and it expresses a functional dependance between x and t; no other variables are involved.


    You cannot conclude that x must be a function of v and t. Of course, as I've mentioned, in specific cases you can write x as some (not unique!) function of v and t. Using @ for the symbol for partial differentiation:

    if x = f(v, t)

    then dx = @f/@v * dv + @f/@t * dt

    Then if you "divide" by dt, one gets:

    dx/dt = @f/@v * dv/dt + @f/@t

    using the definition of velocity and acceleration:

    v = @f/@v * a + @f/@t


    Let's try this equation out on one of your examples.

    suppose x = t^2, then v = 2t and a = 2
    As you demonstrated, we can write x = vt / 2, so if we apply the above rule:

    x = vt / 2 => v = @(vt/2)/@v * a + @(vt/2)/@t
    => v = t/2 * a + v/2

    Then if we plug in our initial formula for v and a:

    2t = t/2 * 2 + (2t) / 2

    And, thankfully, this equation checks out.


    Because your two alternative forms depend strongly on the actual functional dependance between x and t. "x = vt / 2" is valid only when "x = t^2". There is no general functional dependance between x, v, and t, except for trivial dependancies where one of those 3 variables is absent (in the same sense that 0 is a function of t). Note that v=dx/dt is not functional dependance since it is a differential equation.


    Hurkyl
     
  10. Mar 21, 2003 #9
    Perhaps a graphical approach would work better.

    Consider a graph of position versus time. Let position in meters be on the vertical axis, and time in seconds be on the horizontal axis.

    We have som generic curve representing the behavior of a particle moving in time. We know that derivatives tell us the slope of the curve at any point (i.e. the tangent line to a curve at the point). Slope is also defined as rise over run. The dimensions then check out to be meters over seconds....or velocity. We now see the first derivative of a position-time function is the velocity of the particle at a particular time, t.

    Consider a velocity time function, in which velocity in meters per second is the vertical axis, and time in seconds is horizontal axis. Again we see the first derivative of that is none other than the acceleration of a particle at time t. Integrate the velocity time function, and by the fundamental theorem of calculus we see we wind up with the position time function...in other words the displacement of the particle is the area under the curve of a velocity time function.

    Not only is this mathematically consistent, it is also universally supported by experiments, of which you yourself can attempt as well with some equipment and time.
     
  11. Mar 23, 2003 #10
    stated this way x is a function of not only time but velocity too. otherwise meters per secounds (counterparts for velocity) would not appear in the equation.

    so in case x = vt/2 we have:
    v = t/2 * a + v/2 => v = a * t

    but if x = vt then
    v = @(vt)/@v * a + @(vt)/@t
    v = t * a + v => (a = 0) inclusive or (t = 0) and v is any

    conclusion: the results are different for different cases. since this is not behavior of any law your definition vdt = dx is not general enough to cover all the cases unlike it is the case with x = vt.

    In the pervious case (x = vt), the counerparts for velocity are meters per secound cause you can put x = 1m, t = 1s => V = 1m/s
    In this case (x = vt/2), would the counerparts for velocity be two meters per secound cause if x = 1m, t = 1s => V = 2m/s?
     
  12. Mar 23, 2003 #11
    I've found that when I get confused about what is a function of what, it helps to *always* write out the arguments of all your functions: eg say "x(t)=v(t)*t" not "x=vt."

    One reason we don't consider x to be a function of v and t -- that is, x( v(t), t ) -- is because this isn't even defined except for *one* particular value of v(t), in any given problem! You only have one position at any given moment in time, so it doesn't make a whole lot of sense to have a position function that can take on multiple values.
     
  13. Mar 23, 2003 #12
    I will point out that the equations you are using are actually differential equations

    x(t) = v(t)

    is

    x(t)= d(x(t))/dt

    Even then, when you are dealing with an equation of the form

    x = vt + a/2 *t2

    Velocity in this case is a constant as is the acceleration. The only variable is time. Yes, velocity is a function of time as well, but as stated it is a differential equation, so the actual notation for that equation is:

    x = d(x(t))/dt + 1/2 * d2(x(t))/dt2

    (if I'm wrong about that someone point it out please)

    Thus it is all based on the position function.

    So if you are given conditions say that start only with acceleration

    and that a is constant.

    a(t) = a.

    Let us integrate with respect to time:

    int (a(t) dt) = int (a dt)

    we have at + C = a(t)*t with C being some constant.

    Now, let is digress quickly and ask what a(t)*t is.

    If we say that a(t)*t = v(t) then we have

    v(t) = a*t + C.

    Let us integrate with respect to t again.

    int (v(t) dt) = int (a*t dt) + int (C dt)

    v(t)*t = a/2 *t2 + Ct + D where D is another constant.

    Furthermore, let us assume for the moment that v(t)*t = x(t)

    x(t) = a/2 *t2 +C*t + D.

    Now, let's go back to our assumption that a(t)*t = v(t). Let's call a(t) d2x(t) / dt2. We see then:

    d2x(t) / dt2 = a

    The dt then comes from the denominator:

    int (d2x(t) / dt) = int (adt)

    and thus the first is the integral of d(x(t))dx/dt.

    so it follows then:

    d(x(t))/dt = a*t +C. Above we defined a(t)*t = v(t). So v(t) = d(x(t))/dt.

    Integrating the bottom dt out gives:

    d(x(t)) = (a*t + C) dt

    Integrating both sides yields:

    x(t) = a/2 *t2 + C*t + D.

    C and D can be found from initial conditions. As it usually turns out, C = initial velocity due to initial conditions that can be known or found out, and tested through the second equation dealing with v(t) = a*t + C, and D = initial position usually.

    Overall differential equation:

    x(t) = d2x(t) *t2/ (2dt2) + C*t + D

    So there you have it..I believe that should prove that everyone else has been right.
     
  14. Mar 23, 2003 #13

    Hurkyl

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    Correct!

    If x = vt is valid, then acceleration must be zero at all times.


    To paraphrase Alice in Wonderland, do "I breathe when I sleep" and "I sleep when I breathe" mean the same thing?

    It is true that velocity is measured in meters per second (or other compatable units), but the converse is not true; something with the units meters per second is not always a velocity.


    And to make a minor correction, notice the units on my constant are "meters per second squared" not "meters per second", so it couldn't be a velocity anyways.


    If x = vt, and x = 1 m when t = 1 s, then v = 1 m / s at that time. As you proved, acceleration is zero in this case, so v = 1 m / s is true at all times.

    If x = vt/2 and x = 1m when t = 1 s, then v = 2 m / s at that time. In this case, though, velocity is not constant, and at later times will not be 2 m / s.

    What do you mean by counterpart?

    Hurkyl
     
  15. Mar 24, 2003 #14

    russ_watters

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    To me this is a metaphysical question and therefore pointless to argue on mathematical grounds. Can you have velocity without time? For the purpose of doing the math, yes. In reality? uhhhhhhhh.....
     
  16. Mar 24, 2003 #15
    this aint no metaphysical question but pure math. it is sort of presenting position as a function of two other variables. since time and position are different dimensions if you state that position x is proportional with time t i.e.
    x=vt
    then v is dimension - speed of x and position is also proportional with v. so it is not about having velocity without time but it is
    -you cannot present position only with time or in general
    -you cannot present any single dimension in this equation with only one other single dimension, but every single dimension apearing in the pervious equation is a function of two other dimensions.that is my argument is this thread.
    similar with the acceleration; if you want to present position x proportional with the square of time t^2 then position is proportional with acceleration a i.e.
    x=at^2
     
  17. Mar 24, 2003 #16
    In the equation x = vt, v is a constant. Watch this:

    Acceleration is constant:

    (1.) a(t) = a_0

    Integrating, we have velocity:

    (2.) v(t) = a_0 * t + v_0

    v_0 is the initial velocity. Integrating again, we have position:

    (3.) s(t) = (a_0/2) * t^2 + v_0 * t + s_0

    s_0 is the initial position. Note, if we let a_0 = 0, s_0 = 0, (3.) becomes this familiar expression:

    (4.) s(t) = v_0 * t

    This naturally implies that v(t) = v_0.


    One needs to distinguish between constants, various types of functions, and dependent and independent variables.
     
  18. Mar 24, 2003 #17
    acceleration a is not zero in this case but accel. a is
    a = x / t^2 = v / t
    furthermore, you can have constant velocity but variable accel at the same time. it's strange cause you are used to v being dx / dt and a being d(dx / dt) / dt.that's not true but

    x = vt and v = at

    as for trajectory i'm using the set of consequential events (x, v, t)
    starting from (x1, v1, t1) ending in (x2, v2, t2). in my case the trajectory is 3-dimensional while in your case you are working with its normal projection on v = 0.i've been thru your way and i decided on purpose to abandon it cause it's the wrong way.
    by couterparts i mean pair of dimension and quantity considered as singular.

    THE MAIN DIFFERENCE BETWEEN YOUR (OR TRADITIONAL) CONCEPT ON ONE SIDE AND MY CONCEPT ON THE OTHER SIDE FOR POSITION, TIME AND VELOCITY IS THAT YOU PRESUME THAT VELOCITY IS RATIO BETWEEN DISPLACEMENT AND ELAPSED TIME. UNFORUNATELY, I CAN PROVE, JUST AS I AM PROVING, STARTING FROM YOUR PRESUMPTIONS, THAT VELOCITY IS RATIO BETWEEN MOMENTARY POSITION OVER MOMENTARY TIME.

    I KNOW YOUR CONCEPT WELL.

    DO YOU KNOW MINE?


    here is a picture of what i'm talking about:
    http://members.fortunecity.com/dock0609/kinetics0089.JPG [Broken]

    you might want to visit this site of mine to get a wider picture:
    http://members.fortunecity.com/dock0609/kinetics.htm [Broken]
     
    Last edited by a moderator: May 1, 2017
  19. Mar 24, 2003 #18
    what should i see?

    if x = vt then
    case(1): v = 1m/s; x = 1m; t = 1s
    case(2): v = 2m/s; x = 2m; t = 1s
    case(2): v = 2m/s; x = 1m; t = 1/2s
    .
    .
    .
    csae(n): v = nm/s; x = nm; t = 1s
    of what constant you were talking about.
    can't you see what three-dimensional means.
    it means v and t are free to get any values and x has to aply x = vt.

    when v = dx / dt then v is const, cause for any dt in point t_0, dx / dt= const. v = dx / dt is straith line with equation:

    x - x_0 = (dx / dt)(t - t_0)

    this line is known as tangent in the same point.
     
  20. Mar 24, 2003 #19
    I thought that you might be doing it this way. Your diagram may be 3-D, but you are still thinking in 1-D. What if velocity and position are represented as 3-D vectors? The diagram would then be 7-D. Try plotting that!
     
  21. Mar 24, 2003 #20

    Hurkyl

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    No it's not. a = dv/dt is the definition of acceleration, not a = v / t.


    No you can't. Since acceleration is, by definition, the rate of change of velocity, the acceleration of an unchanging velocity must be zero.


    This is only possible when x is constant and both v and a are zero.


    And I did that too. The difference is I used the correct laws of multivariable differentation and you didn't.


    I haven't done any work on this thread where v = 0 must be true, except for my assertion above that x = vt & v = at can only be satisfied with both v and a are zero.


    If you can't even apply differentiation correctly, how can you say it's the wrong way?


    Care to be more precise? This definition makes no sense in the way you used the term.


    No, I presume that velocity is the derivative of displacement with respect to time. Average velocity is the ratio between displacement and elapsed time between two events.


    You can't prove that if you don't even know what my presumptions are, such as the laws of differentiation.


    (boldface mine)

    Your multiple case 2's demonstrate your misunderstanding of what x = vt means. x = vt means that for a given particle, at any point in time, position the product of velocity and time. x = vt does not mean that x, v, and t take on all possible combinations that satisfy x = v t.


    This is patently incorrect, trajectories only permit a single velocity and a single position for any given time.


    v is not a straight line, v is a number.

    Hurkyl
     
  22. Mar 24, 2003 #21
    Hey Hurkyl, how is the maths in my post? Does it follow all the rules ok? I'm fairly certain it does, but it never hurts to have it checked.
     
  23. Mar 25, 2003 #22
    now i'm pissed off.
    what the hack i've been talking all the time. every definition is from the start/unconditionally right. but between definitions about same thing there is more acurate and less acurate one. it depends on how general the definition is. the more general definition overrules the less general one. your definition 'adt=dv' is not general enough cause it covers only the case when da=0 inclusive or t=0. if you declare adt as dv then neceserily
    v=v(a,t)
    and then neceserily
    dv=(@v/@t)dt+(@v/@a)da
    now equalize the coeficients in front every d(something).
    you should get (dv=dv) and (@v/@t=a) and ((@v/@a=t=0) or (da=0))
    @v/@a=t=0 is a trivial case.
    so if adt=dv when da=0 then tda=dv when dt=0.
    when neither (dt=0 and da=0) then dv=adt+tda
    this is when my definition 'at=v'/'vt=x' overrules yours
    'adt=dv'/'vdt=dx'


    i meant dv=0 and presuming vdt=dx is same as presuming dv=0.
    i don't botter to apply differentiation at all. i think that there is more general something then doing that, like at=v and vt=x.
    meters, secounds, meters over secound are all dimensions, qualities. you get them answering the question that much of what?
    every real number like 1, 1.1, 0.4 is quantity. you get them answering how much/many of this quality?
    the pair of quantity and quality accepted as singular is counterpart.
    for example you can say counterparts for position or distance is
    1.4 meters
    then two counterparts are 2*counterpart=2*1.4meters=2.8meters
    i know your definition adt=dv or vdt=dx
    on the contrary it means just that. later the trajectory is subset of this general set.
    it can be also
    single position and time for any given velocity
    or
    single velocity and time for any given position
    i'm not talking about the number 'v' but about the equation 'v=dx/dt'
    if you define v as dx/dt and you integrate assuming the only possible (dv=0) you get
    X-x=v(T-t)=(dx/dt)(T-t)
    this equation indeed represents straith line called tangent in point (t, x).


    finally: I'm making fine and very clear point here for those skiled to follow me. I also know that no one will let my point be certified cause centuries of hard and wrong work are in stake. You'll rather be wrong then allow your self to admit the error.
    I'm thru with this thread. I won't reply any further on this subject.
    I'll just know what I know and keep silent cause you are making my effort worthless.
     
  24. Mar 25, 2003 #23

    Hurkyl

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    Maybe some examples will make it more clear.

    Take a particular solution for undamped 1-D harmonic motion about the origin:

    x = sin(t)
    v = cos(t)
    a = -sin(t)

    Let's try some things!

    Does v = dx / dt?

    Yes. the representation x(t) = sin(t) is a function of t alone, so we take the ordinary derivative yielding dx/dt = cos(t) = v

    What about the representation x(t, v) = sin(t)? This is a function of both t and v so we have to apply the chain rule:

    dx / dt = @x / @t * dt / dt + @x / @v * dv / dt
    = cos(t) * 1 + 0 * dv / dt = cos(t) = v

    It still works! What about a more complicated example that actually involves v in the expression?

    x(t, v) = tan(t) * v

    (of course when you plug in that v = cos(t) this equation reduces to x(t, v) = sin(t), so this equation is correct)

    Well, let's apply the chain rule here:

    dx / dt = @x / @t * dt / dt + @x / @v * dv / dt
    = (sec(t)^2 * v) * 1 + tan(t) * dv / dt

    This looks like something totally different! But it's not, remember that v = cos(t) and dv / dt = a by definition and a = -sin(t):

    dx / dt = (sec(t)^2 * v) * 1 + tan(t) * dv / dt
    = sec(t)^2 * cos(t) + tan(t) * (-sin(t))
    = sec(t) - tan(t) * sin(t)
    = (1 / cos(t)) - (sin(t) / cos(t)) * sin(t)
    = (1 - sin(t)^2) / cos(t) = cos(t)^2 / cos(t) = cos(t)

    Whaddya know, dx / dt = cos(t) = v yet again!

    What about "dv=(@v/@t)dt+(@v/@a)da" (which is the chain rule for v a function of t and a)?

    Well, using v(t, a) = cos(t) and a = -sin(t), this yields:

    dv = (-sin(t)) * dt+ 0 * da = -sin(t) * dt = a dt

    Wow, the equation worked! What about something more complicated? Try: v(t, x, a) = cos(t) + x + a

    dv = @v/@t dt + @v/@x dx + @v/@a da
    = -sin(t) dt + 1 dx + 1 da

    Remember that x(t) = sin(t), so dx = cos(t) dt.
    Also, da = -sin(t) so da = -cos(t) dt

    dv = -sin(t) dt + cos(t) dt - cos(t) dt = -sin(t) dt = a dt

    Whaddya know, it still works.


    You can't naively equalize the coefficients because dv, dt, and da are not independant. da depends on dt just like dv depends on dt. To give another example of why your operation here is absurd, suppose x = y. Then:

    2x = x + x = x + y

    therefore 2x = x + y

    equalizing coefficients yields (<------ this is the bad step)
    2x = x & y = 0

    therefore x = 0 and y = 0


    The flaw was assuming x and y are independant variables. They are not independant (in fact, x = y), so equalizing coefficients is not a valid operation.


    Ok that definition is clear. But counterparts are not constant things like you implied in a previous post.


    Yes, the trajectory is a subset of your general set. Your general set, however, is not a trajectory. And beyond that, only zero acceleration trajectories belong to your general set defined by x = v t.


    No it cannot. Trajectories can, for example, occupy the same position at different times, and can have the same velocity at different positions.


    That equation is indeed a tangent line. Not a trajectory, not a velocty, a tangent line.


    Right back at ya.

    Hurkyl
     
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