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Homework Help: One more particle question/s

  1. Jun 16, 2008 #1
    Really struggling with these particle questions, jsut can't grasp the concept

    A moving particle at time t ∈ [0, 10] (seconds) has position vector in metres from the origin (0, 0, 0) given by the vector function r(t) = (10 − t)i + (t 2 − 10t)j + sin tk.

    i. Describe the path of the particle, as seen from above (the positive k-direction), and
    also describe it in three dimensions.

    ii. Find the curvature of the path, at t = 2π ≈ 6.28 seconds.

    iii. Find the angle between the path (at start and end-points) and the k-direction.

    And this one

    A particle’s path, in two dimensions, is described by its position vector (in metres and
    time t ∈ [1, 2] seconds) relative to point (0, 0, 0) by r(t) = (2t + 1)i + (4 − t 2 )j.

    i. Sketch the path of the particle.

    ii. Find the value of t ∗ at which the particle has greatest distance from (0, 0, 0).
    Hint: optimising squared distance may be the simpler method here.

    iii. Show that at position r(t ∗ ), the particles velocity is not perpendicular to r(t∗ ).

    From a practice exam, for my final this week
  2. jcsd
  3. Jun 17, 2008 #2


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    You've posted two problems with a total of six parts. And you haven't given us any attempt to solve any of them. Read the forum rules. You have to start TRYING to solve one of these.
  4. Jun 17, 2008 #3
    sorry i have tried.

    i have so far

    x = 10 - t => t = 10 - x (1)

    y = t^2 - 10t (2)
    Substitute (1) into (2).
    which gives y=(10-x)^2 - 10(10-x) which describes the particles path?

    ii. and then i need to substitute 2*pi into the equation (not 100% sure which one though) to find the position

    iii. and to find the angle of the path i need to use the dot product but i can't work out which other vector to use to get the dot product.

    The second question:

    The distance between the particle and the origin |r| and its square |r|^2 will be minimum at the same time t*

    |r|^2 = (2t+1)^2+(4-t^2)^2
    this then needs to be expanded which i think i am doing wrong

    once the expansion is right i need to differentiate the equation?

    but part iii. i can't work out how to do

    sorry i'm a bit new here, i have definitely tried them just thought i might be on the wrong track
  5. Jun 17, 2008 #4


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    That's a start. Good. Now you'll hopefully get a lot of help. You have y=(10-x)^2-10(10-x). That's the path of the particle viewed from above, in the xy plane. It's a parabola, do you know why? To describe it in 3d you have to use words creatively, it's not any kind of a standard curve. Finding angles of the path and curvature requires calculus. For angles you need to take the dot product of tangent vectors. Can you find the tangent vector to the curve in the first question?
  6. Jun 17, 2008 #5
    the tangent vector is that r(t) / ||r(t)|| and then substitute in r(0)?
  7. Jun 17, 2008 #6


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    No. The tangent vector is r'(t). The normalized tangent vector is r'(t)/|r'(t)|. Notice the derivatives.
  8. Jun 17, 2008 #7
    so the derivative of r(t) would be:

    r(t) = (10 − t)i + (t 2 − 10t)j + sin tk.

    r'(t) = ti + 2tj + (-cos)tk ?

    also just to back track a little the 2 * pi gets subbed into y=(10-x)^2 - 10(10-x)?
  9. Jun 17, 2008 #8


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    You need a lot of work on derivatives. Start with the x component. What's the derivative of 10-t? On the backtrack you need to substitute 2pi for t, not x, and you need to substitute it into the 3d form. We haven't even started the curvature question, you'll need a second derivative for that. Can you find a formula? For iii), you should get one vector at t=0 and another at t=10. Use them to find the angles with the k direction.
  10. Jun 17, 2008 #9
    so pi would be substituted into

    r(t) = (10 − t)i + (t 2 − 10t)j + sin tk
    giving (10-2*pi) + (2*pi^2 - 10(2*pi))+sin(2*pi)

    and would the equation for r'(t) be:
    i + 2tj + -cosk?
  11. Jun 17, 2008 #10


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    Don't even try to substitute 2pi into anything until you have at least the tangent right. And you don't. I asked you what is d/dt of (10-t) and you got that one wrong. Next I'll ask you what is d/dt of t^2-10t. What's your guess? Then if you want to practice while I sleep, what is d/dt sin(t). I'll give you a hint. It's NOT -cos(t).
  12. Jun 17, 2008 #11
    d/dt of 10-t is 1
    and the other equation is
    d/dt 2t -10

    is that right?

    giving i + (2t -10)j + cos(t)?
  13. Jun 17, 2008 #12


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    No. Still not. d/dt(10-t) is NOT 1. Try again. The other parts are ok.
  14. Jun 17, 2008 #13
    is it -1i + (2t-10)j + cos(t)?
  15. Jun 17, 2008 #14


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    Yessss. That's the tangent vector for the curve in the first problem. Can you use that to answer part iii)? What's the angle between that vector and k at t=0 and t=10
  16. Jun 17, 2008 #15
    so i put 2*pi into that equation and then for part iii:

    theta= cos^-1 (u.v/ |u||v|)

    with r(t) being u and k being v
  17. Jun 17, 2008 #16


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    The only mention of 2pi is in the curvature question, part ii. The third part says for the start and end points. Since the curve is defined on [0,10], I think the endpoints are t=0 and t=10. Yes, you have the right dot product relation to use.
  18. Jun 17, 2008 #17
    thanks very much for all your help and patience dick, i think i finally understand it
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