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Befuddled - Vector Analysis

  1. Mar 25, 2010 #1
    I'm stuck on both problems. Here's my work thus far. I have no idea he got rtt in (b).

    chp2.jpg

    chp2ab.jpg
     
  2. jcsd
  3. Mar 27, 2010 #2

    Mark44

    Staff: Mentor

    Use the hint. You have rt, so differentiate it again with respect to t to get rtt. You'll need to undo your substitution for u to get back into terms of theta. You'll need to use the product rule and then quotient rule. Also d/dt(dtheta/dt) = d^2(theta)/dt^2.

    One other thing: you'll need to use the equation r^2 d(theta)/dt = const ==> d(theta)/dt = const/r^2.

    I haven't worked it all the way through, but these are the things I would do.
     
  4. Mar 29, 2010 #3
    Is my rt correct?

    dr/d(theta) = [(a e sin(theta))/(1 + e cos(theta))^2 ] d(theta)/dt

    I differentiated with respect to t again and didn't get anything close to the rtt on the sheet. I'll scan it in a few minutes.
     
  5. Mar 29, 2010 #4
    Here's as far as I got.

    1b.jpg
     
  6. Mar 30, 2010 #5

    Mark44

    Staff: Mentor

    Yes, your rt looks fine. I'll look over the other part and get back to you.
     
  7. Mar 30, 2010 #6

    Mark44

    Staff: Mentor

    On second thought, even though your work was fine, I don't think what you're doing is the way to go about this problem. Let's look at the a part, since you said you were stuck on it, too. From Kepler's 2nd law, we have thetat = K/r2, where K is the constant referred to.

    Differentiate both sides with respect to t to get
    thetatt = (-2K/r3)*rt

    Using this and Kepler's 2nd law in the form above, you can show that the theta-component of the acceleration is zero.

    When you get to that point, you have shown that a = [rtt - r(thetat)2]ur and the b part is pretty straightforward after that.
     
  8. Mar 30, 2010 #7
    I don't see where to use thetatt = (-2K/r3)*rt.

    The only way I can see to get the theta-component form is if I do the following and divide by 1/r.

    d(r^2 * d(theta)/dt)/dt = r^2 * d^2(theta)/dt^2 + d(theta)/dt * d(r^2)/dt =
    r^2 * d^2(theta)/dt^2 + d(theta)/dt * (2r*d(r)/dt

    If I insert thetatt = (-2K/r3)*rt, I get (-2K/r)*d(r)/dt
     
    Last edited: Mar 30, 2010
  9. Mar 30, 2010 #8

    Mark44

    Staff: Mentor

    Look at the page you scanned. It's in the theta-component of acceleration.
     
  10. Mar 30, 2010 #9
    Yes. I plugged that in.

     
  11. Mar 30, 2010 #10
    Actually, I just showed that the theta-component

    = r^2 * d^2(theta)/dt^2 + d(theta)/dt * 2r * d(r)/dt

    Now, plugging in my d^2(theta)/dt^2 and d(theta)/dt from the law gives

    = r^2 * (-2K/r^3) + (K/r^2) * 2r * d(r)/dt

    = -2K/r * d(r)/dt + 2K/r * d(r)/dt = 0
     
  12. Mar 30, 2010 #11

    Mark44

    Staff: Mentor

    That's not quite right; you have an extra factor of r all the way through. Also, you didn't need to show it, since it is given right near the beginning of the page you scanned. From that page, the theta-component of acceleration is
    [tex]r \frac{d^2 \theta}{dt^2} + 2\frac{dr}{dt} \frac{d\theta}{dt}[/tex]

    I outlined what you can do to show that this expression is zero.
     
  13. Mar 30, 2010 #12
    Plugging in d^2(theta)/dt^2 and d(theta)/dt, I get r * (-2K/r^3) * (dr/dt) + 2 * (dr/dt) * (K/r^2) = 0.
     
  14. Mar 30, 2010 #13

    Mark44

    Staff: Mentor

    OK, that's what you needed to to. Did you understand my objection to your work in post 10?

    Anyway, that takes care of part a. Now you're ready to take on part b. Use the same substitution from Kepler's 2nd law - d(theta)/dt = K/r^2 - and work on the radial component of acceleration.
     
  15. Mar 30, 2010 #14
    I did not understand the objection. It was a sum of a term and its negative, which is zero. The last one I did is also the sum of a term and its negative, so I'm not quite following you.

    I understand the hint in (a), because the derivative of r^2 * (d(theta)/dt), which is a constant, is zero. So, we want to show that the theta-component is zero, and I guess I was missing the whole problem right off the bat by not using the second law.
     
  16. Mar 30, 2010 #15

    Mark44

    Staff: Mentor

    My objection was that the expression you showed was zero wasn't the right expression. Yours had an extra factor of r for some reason.
     
  17. Mar 30, 2010 #16
    Oh. But isn't that the d/dt of the second law as written?

    I just used the product rule and chain.
     
  18. Mar 30, 2010 #17

    Mark44

    Staff: Mentor

    OK, now I get it. We both came at this from slightly different directions.
     
  19. Mar 30, 2010 #18
    Yeah. Sorry about that. Your method was more direct and in tune with the directions.
    Thanks for the help so far. I'll post my (b) work in a little bit.
     
  20. Mar 30, 2010 #19
    I'm basically doing the same thing here.

    dr/dt = dr/d(theta) * d(theta)/dt

    d^2(r)/dt^2 = dr/d(theta) * d^2(theta)/dt^2 + d(theta)/dt * [dr/d(theta) * d(theta)/dt]

    = dr/d(theta) * d^2(theta)/dt^2 + (d(theta)/dt)^2 * dr/d(theta)

    = dr/d(theta) * (-2K/r^3 * dr/dt) + (K/r^2)^2 * dr/d(theta)
     
  21. Mar 31, 2010 #20

    Mark44

    Staff: Mentor

    You seem to be on the wrong track here. The goal for part b is to show that
    rtt - r([itex]\theta_t[/itex])2 ~ 1/r2

    Don't overthink this - it's much simpler than the first part.

    The subscript notation is a lot less cumbersome than the d/dt etc. notation you're using. I don't know if you're not understanding it, but they are using rtt to mean the same thing as d^2(r)/dt^2.
     
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