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Homework Help: Vector Kinematics

  1. Sep 16, 2007 #1
    An object is moving in the xy plane with the position as a function of time given by r(vector) = x(t)i + y(t)j
    Point O is at r(vector) = 0 . The object's distance to O is decreasing when

    A) Vx > 0 , Vy > 0
    B) Vx < 0 , Vy < 0
    C) xVx + yVy < 0
    D) xVx + yVy > 0

    i figure i can take the v=r/t

    take the derivative

    v=dr/dt = d(x(t)i + y(t)j ) / t

    I think the answer should be B, since Vx and Vy should both be in the negative direction to return to one, assuming there can't be negative distance.
    I'm pretty lost of how to actually prove any of this.

    Any suggestions?

    Thanks very much
  2. jcsd
  3. Sep 16, 2007 #2


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    What is the distance of the object from 0? What is the derivative of the distance with respect to time?
  4. Sep 16, 2007 #3
    there is no stated distance
    we are just given position as a function of time

    would the derivative of distance with respect to time be
    r(t) = v*dt
    x(t)i + y(t)j = v*dt
    xi + yj = v
  5. Sep 16, 2007 #4


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    get the formula for distance in terms of x(t) and y(t), then take its derivative.
  6. Sep 16, 2007 #5
    r(t) = x(t)i + y(t)j
    i'm not sure what i can derive from that

    thanks for your help
  7. Sep 16, 2007 #6


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    Suppose the position vector of an object is r = 3i + 4j, what is the distance from the origin of this object?
  8. Sep 16, 2007 #7
    r = sqrt((3i)^2 + (4j)^2)

    would I want:
    r(t) = sqrt[(x(t)i)^2 + (y(t)j)^2]
  9. Sep 16, 2007 #8


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    exactly... but you wouldn't worry about the i and j (i and j are just unit vectors...)

    so r = sqrt((3)^2 + (4)^2)

    and for your question

    r(t) = sqrt[(x(t))^2 + (y(t))^2]

    what do you get for dr/dt ?
  10. Sep 16, 2007 #9

    v(t)=dr(t)/dt= d(sqrt[(x(t))^2 + (y(t))^2]) / dt

    im not sure how to proceed from here if thats what you meant by dr/dt
  11. Sep 16, 2007 #10


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    sorry I shouldn't use r(t) since that refers to the position...

    instead call it s(t) = sqrt[(x(t))^2 + (y(t))^2] (so s(t) is the distance from the origin)

    we are trying to find out when the distance is decreasing... ie when ds(t)/dt is negative.

    take the derivative ds(t)/dt. Do you know the chain rule in calculus?
  12. Sep 16, 2007 #11
    no i dont know the chain rule
    it's concept was briefly explained to us
    but i havent taken calculus before, and my physics class is going faster than my calc class which is causing my problems with questions like this

    would the derivative ds(t)/dt be
    ds(t)/dt = d(sqrt[(x(t))^2 + (y(t))^2]) / dt
    or is there something else i can put in?
  13. Sep 16, 2007 #12


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    Hmmm... the chain rule is a concept that takes some practice to get used to...

    As far as I can see you need the chain rule to solve this problem... It isn't that hard... I highly recommend just looking through your calculus book with regards to the chain rule... they'll explain it much better than I could. Then come back and look at this problem.

    Once you're able to write that derivative ds(t)/dt in terms of dx(t)/dt and dy(t)/dt... then you'll see the answer to the question...
  14. Sep 16, 2007 #13


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    I'm going to chime in for a moment because your situation sounds like that of a number of students I've worked with who got poor advising and were told you can take first semester physics and first semester calculus at the same time. (Not really true -- the physics course assumes you've seen all the necessary math and will *not* take time to review it.)

    It sounds like you are in a calculus-based physics course, so you are going to run into a few occasions where your physics course uses something in calculus that you aren't going to see until weeks later. You will need to acquaint yourself with the basic rules of differentiation (derivatives of elementary functions, Product Rule, Quotient Rule, Chain Rule) and basic integration (integrals of elementary functions and definite integration).

    You will probably be able to get by in the physics course without the mathematical theory (pretty much), but you'll at least need to know how to use the "tools". See if you can get a little help with that at your end (it shouldn't take too much time -- you just need to know how to do the calculations for now) or maybe ask questions in the appropriate forum here. You should look into doing this *soon*.
  15. Sep 17, 2007 #14
    I've read over the chain rule, aswell as some chapters on derivatives from my calculus textbook, it clears up some of the things I have been doing.

    if I want ds(t)/dt in terms of x(t)/dt and y(t)/dt

    would i do

    s(t) = sqrt[(x(t))^2 + (y(t))^2]

    x(t) = xo + Vxt
    y(t) = yo + Vyt

    s(t) = sqrt[(xo + Vxt)^2 + (yo + Vyt)^2]
  16. Sep 17, 2007 #15


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    You are assuming here that the velocity components are constant, which I think is not implied in the problem statement. However, this is a good example to start from. You are now ready to differentiate s(t) with respect to time using the Chain Rule...
  17. Sep 17, 2007 #16
    I thought of that as I was writing it, I don't know why I just assumed the velocities would be constant, let me try again.

    ax(t) = Vx/t
    ay(t) = Vx/t

    Vx(t) = Vxo + ax(t)t
    Vy(t) = Vyo + ay(t)t

    x(t) = xo + Vxo + ax(t)t
    y(t) = yo + Vyo + ay(t)t

    s(t) = sqrt[(xo + Vxo + ax(t)t)^2 + (yo + Vyo + ay(t)t)^2]
  18. Sep 17, 2007 #17


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    I think it is best to start from here:

    s(t) = sqrt[(x(t))^2 + (y(t))^2]

    Now without assuming anything about x(t) and y(t)... try to differential s(t) just as it is...
  19. Sep 18, 2007 #18
    do I want to be replacing the x(t) and y(t) from s(t) = sqrt[(x(t))^2 + (y(t))^2]
  20. Sep 18, 2007 #19


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    well, let's start here. let h = (x(t))^2 + (y(t))^2... I substitute that into the s equation (i'm not going to write s(t), just s).

    s = sqrt(h)

    now what is ds/dh.
  21. Sep 18, 2007 #20
    let h = (x(t))^2 + (y(t))^2
    s = sqrt(h)

    ds/dh = d sqrt(h) / d (x(t))^2 + (y(t))^2
    = d sqrt((x(t))^2 + (y(t))^2) / d (x(t))^2 + (y(t))^2

    im starting to get lost here
    what would ds/dh
    if s is the point away from the start
    what is h for?
  22. Sep 18, 2007 #21


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    I'm just making a substitution to illustrate the chain rule... don't worry about x(t) and y(t) for the moment.

    s = sqrt(h), ie s = h^(1/2). what's the derivative of s with respect to h... don't substitute anything. just take the derivative.

    or what's the derivative of y = x^(1/2) with respect to x.
  23. Sep 18, 2007 #22
    ds/dh = d (h^1/2) / dh = (1/2)h^(-1/2)

    i did that following the power rule d x^n / dx = nx^n-1
  24. Sep 18, 2007 #23


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    Now the chain rule says that ds/dt = (ds/dh)*(dh/dt)

    That's why I substituted that h = (x(t))^2 + (y(t))^2 into the s equation. The chain rule helps me calculate the derivative ds/dt, by calculating two easier ones.

    Now, try taking the derivative of h with respect to t. ie try to calculate dh/dt. using: h = (x(t))^2 + (y(t))^2
  25. Sep 18, 2007 #24
    ds/dt = (ds/dh)*(dh/dt)

    dh/dt = d (x(t))^2 + (y(t))^2) / dt = 2x(t) + 2y(t)
  26. Sep 18, 2007 #25


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    Here actually you need to use the chain rule also... I'm just going to call it x, instead of x(t) and y instead of y(t)

    I'll show the steps here:

    dh/dt = d(x^2 + y^2)/dt = d(x^2)/dt + d(y^2)/dt = [d(x^2)/dx]*(dx/dt) + [d(y^2)/dy]*(dy/dt) (take note of how I used the chain rule in this last step)

    So then we get: 2x*(dx/dt) + 2y*(dy/dt)

    So ds/dt = (ds/dh)*(dh/dt) = [(1/2)h^(-1/2)]*(2x*(dx/dt) + 2y*(dy/dt))

    Hope this makes at least a little sense... takes a while getting used to the chain rule.

    Here's a simpler example of the chain rule:

    taking the derivative of: (x^2 + 3)^5 with respect to x. I get [5(x^2+3)^4]*(2x)
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