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Derivatives of a Parametric Equaiton for Motion

  1. Nov 25, 2011 #1
    I am exploring the motion of a particle who's path P(x,y) is given by two parametric equations:

    x = f(t) = a cos(bt) + cd (1)

    y = g(t) = a sin(bt) (2)

    To get a tangential velocity for this particle, I differentiated (1) and (2) then combined them using the form sqrt( m2 + n2 ):

    f'(t) = -ab sin(bt) + cd (3)

    g'(t) = ab cos(bt) (4)

    Tv(t) = sqrt ( (-ab sin(bt) + cd )2 + ( ab cos(bt) )2 ) (5)

    Next, I wanted to examine the tangential acceleration of the particle, so I differentiated (5):

    Ta(t) = -a2 b2 c d cos(bt) / sqrt ( (-ab sin(bt) + cd )2 + ( ab cos(bt) )2 ) (6)

    The results of everything so far seem to be as expected. Now I wanted to isolate the horizontal and vertical components of the tangential velocity and acceleration. So I use the following to isolate the horizontal velocity:

    Xv(t) = -sin(bt) sqrt( (-ab sin(bt) + cd )2 + ( ab cos(bt) )2 ) (7)

    Then differentiated that to get a horizontal acceleration function. This is where things started diverging from the expected results.

    Was my choice for the horizontal acceleration function wrong? Should have just multiplied the tangential acceleration function by same term that I used to multiply the tangential velocity by to get a horizontal acceleration function?

    Things got even worse when I applied this strategy to the vertical component.

    ***

    It just occurred to me that what I want is a vector for the tangential acceleration, then isolate the x and y components from that.
     
    Last edited: Nov 25, 2011
  2. jcsd
  3. Nov 25, 2011 #2
    Erm, I don't know exactly what you're trying to do, or by what method.

    x = acos(bt) + cd

    therefore (∂x/∂t) = -absin(bt)

    y = asin(bt)

    therefore (∂y/∂t) = abcos(bt)

    therefore (∂y/∂x) = (abcos(bt))/(-absin(bt))
    = -cot(bt)

    ...which I believe is your tangential velocity.

    Now, acceleration is the derivative of velocity with respect to time, yes?

    therefore:
    ∂ (∂y) = cosec(bt)
    ∂t(∂x)

    therefore (∂2y/∂x2) = (cosec(bt))/(-absin(bt))

    ...which again I believe is your tangential acceleration. Not really sure what you're doing about the horizontal and vertical components. But that's my two cents.
     
  4. Nov 25, 2011 #3
    Hey, thanks for the reply.

    Ordinarily that would be fhe case, although here it is just the slope of the position function.

    Pardon me for omitting this to begin with. I'm studying a particle, which can be considered point on a rotating disk, from a reference frame that the disk is moving through on a linear path. The tangential velocity and tangential acceleration of the particle are in regard to to that point as seen from the reference frame.

    This is involves a translation of coordinate systems. The particle motion is a combination of circular orbit, asin(bt), acos(bt) with reference to center of the disk, which is also moving parallel to the x axis of a second reference frame with velocity cd. So, when the slope of the path of the particle is vertical in the reference frame, the tangential velocity is actually only ab.

    I think I figured out the answer, this would be easier if I had a standard three semester calc book on hand. Anyway, The third derivative of position function is the slope of the acceleration curve.

    The arctangent of the slope of the acceleration curve is the angle of the tangent to the acceleration curve. Then, if I multiply the acceleration function by the cos of the arctan of the slope of the acceleration curve, that ought to give me the horizontal component of the acceleration curve, and the acceleration function times the sin of the arctan of the slope of the acceleration curve ought to give me the y component of the acceleration curve. All with respect to time of course.

    I tried deriving the third derivative of the position function with Axiom, although the answer it gives me seems off. It differs from the result I get from just differentiating the velocity function twice in SciDAVis.

    It's frustrating. Have I completely lost you?
     
    Last edited: Nov 26, 2011
  5. Nov 26, 2011 #4
    Let's start this over from the beginning. There was a mistake in the position function there anyway. I forgot a 't' after the 'cd'.

    I need three things from this to be able to integrate the tangential acceleration, as well at the acceleration parallel to the x axis and the acceleration parallel to the y axis over a period of time.

    I've also figured out from the calculus text that I do have, that a vector approach would be appropriate. So,

    p = ( a cos(bt) + cdt)i + a sin(bt)j

    s = dp/dt = (-ab sin(bt) + cd)i + ab cos(bt)j

    a = d2p/dt2 = -ab2cos(bt)i + -ab2sin(bt)j

    My focus is on acceleration here so I'll skip to that. The magnitude of the acceleration would be:

    |a| = sqrt ( (-ab2cos(bt))2 + (ab2sin(bt))2 )

    I got this far before without the vector approach. What I'd like to do now is to decompose the acceleration vector into it's x and y components, and this is where I started having trouble.

    The direction of the acceleration would be:

    a/|a| = -ab2cos(bt)i/|a| + -ab2sin(bt)j/|a|

    What I'm thinking now is if I can convert this direction vector in to an angle, θ, then I could multiply the magnitude of the acceleration by the sin and cos of θ to get the acceleration components that are parallel to the x and y axis.

    The only problem with this approach is that the denominator of the direction vector is going to be zero twice evey bt/2pi seconds, which might cause trouble when I try to integrate the acceleration curves over a time interval.
     
    Last edited: Nov 26, 2011
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