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Calculus Polar question

  1. Apr 19, 2012 #1
    a) I am having a hard time figuring this out.

    Im saying that x=rcos([itex]\phi[/itex]) and y = r sin ([itex]\phi[/itex])

    where r = a-vt.

    Im not sure how to work a DE into it.




    b) I believe it is [itex]\omega[/itex]a/v since [itex]\Delta[/itex][itex]\Theta[/itex]= [itex]\omega[/itex]t and [itex]\Delta[/itex]x=vt


    c) so we have the arc length formula would i use the x and y above for the dx and dy?
     

    Attached Files:

  2. jcsd
  3. Apr 19, 2012 #2
    a) r and theta, which I will call T, are each functions of time t, so you are looking for two functions: r(t) and T(t). You do not need to use Cartesian coordinates x and y for this problem in any way; it is simpler in polar coordinates. Just remember that r is the distance of the bug from the origin and T is the angle that the radius makes with a particular horizontal ray from the origin (which is usually chosen to be the positive x-axis when translating to a Cartesian coordinate system). The question says that the bug moves toward the center of the disc with constant velocity v. That implies that the change in the bug's radial coordinate r, which is the distance to the origin, with respect to time (dr/dt) is ? Next, it also states that the disc is spinning with constant angular velocity w. That implies that dT/dt = ?
    The path of the bug would be the solution to that pair of differential equations: P(t) = (r(t), T(t)).
     
    Last edited: Apr 19, 2012
  4. Apr 20, 2012 #3
    so dr/dt is -v? since r= a - vt.

    is dT/dt just [itex]\omega[/itex]?
     
  5. Apr 20, 2012 #4
    hmmm. For some reason I posted the same thing three times. Can a mod come and fix that for me... Thank you
     
  6. Apr 21, 2012 #5

    sharks

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    Gold Member

    You should be able to edit your posts and delete your own replies, within a certain amount of time after you've posted them. Use a good browser, like Firefox, to avoid these script issues.

    To get back to your problem, slider142 has explained it clearly. You need to find [itex]\frac{dr}{d\theta}[/itex]. Hint: Use chain rule.

    For part (a): [tex]\frac{dr}{d\theta}=\frac{dr}{dt} \times \frac{dt}{d\theta}[/tex]

    For part (b):

    Use the formula for periodic time: [itex]T=\frac{2\pi}{\omega}[/itex]
    In [itex]T[/itex] seconds -> the number of revolutions is [itex]\frac{2\pi}{\omega}[/itex]

    Now, how much time does the bug take to travel to the center of the disk?

    Let's say the bug takes [itex]T_1[/itex] seconds. Then, from [itex]v=\frac{r}{T_1}[/itex], you can find the time, [itex]T_1[/itex].

    Then, you can find the number of revolutions during that time, [itex]T_1[/itex].

    For part (c):
    Now, that you know the number of revolutions, you can find the total rotational distance travelled, since 1 rev. = [itex]2\pi r[/itex]
    Then, to get the total distance travelled, simply add to the radius, since the latter is the additional distance travelled by the bug to reach the centre of the disk.

    After some more thinking, i think the solution that i suggested in part (c) might be incorrect, as the bug is travelling all the time, while the disk is spinning. So, the radius changes continuously with respect to time, and hence the rotational distance travelled by the bug changes with respect to time (actually, it decreases at a constant rate).

    Using [itex]S=r\theta[/itex], we can find the total circumferential distance travelled, but since r is changing with respect to time, we can relate to:
    [tex]\frac{dS}{d\theta}=\frac{dS}{dr} \times \frac{dr}{d\theta}[/tex]

    We already know [itex]\frac{dr}{d\theta}[/itex] which we calculated in part (a). We need to solve for [itex]\frac{dS}{dr}[/itex].
     
    Last edited: Apr 21, 2012
  7. Apr 21, 2012 #6
    Ok I think I got this


    dr/dt= -v because derivative of r with respect to t is -v
    dt/dθ= 1/ω because θ= ωt

    from there we get

    dr/dθ= -v/ω



    b)

    The bug takes a/v to get to the center of the disk so

    ωa/v

    c)

    Wouldn't this have to be an integral?
     
  8. Apr 21, 2012 #7
    So would the arc length be given by

    ∫[itex]_{0}[/itex][itex]^{a/v}[/itex] [itex]\sqrt{
    (dx/dt)^2+(dy/dt)^2}[/itex] dt

    where dx is the derivative of x=(a -vt)cos(ωt) and dy = (a-vt)sin(ωt)
     
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