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Rate of distance between two objects

  1. Nov 15, 2009 #1
    1. The problem statement, all variables and given/known data
    An object A moves along the positive horizontal axis, and object B along the graph of f(x) =
    -sqrt(3)x for x <= 0. At a certain time, A is at the point (5,0) and moving with the speed 3 units/sec and B is at a distance of 3 units from the origin and moving with speed 4 units/ sec. At what rate is the distance between A and B changing?

    2. Relevant equations

    line equation

    3. The attempt at a solution

    don't I only have to draw a line from A to B at that point and take the slope? Probably not, since I;'m given so much information... anybody want to help me out? thank you
  2. jcsd
  3. Nov 16, 2009 #2
    What are
    [tex] \frac{d A_x}{dt}, \frac{d A_y}{dt}, \frac{d B_x}{dt} \mathrm{and} \frac{d B_y}{dt} [/tex]?
    What is the rate at which the distance changes in terms of these?
  4. Nov 16, 2009 #3
    thank you for the reply, I realized that I will have to have a distance function dependent on A and B... I don't know how to get a distance function.. thank you
  5. Nov 16, 2009 #4

    well Ay/dt and By/dt are given , 3 and 4 units per second right
  6. Nov 16, 2009 #5
    please help, I'm stuck here
  7. Nov 16, 2009 #6
    What you are given is the velocity, which is [tex] \left| \frac{d \mathbf{A}}{dt} \right| [/tex].
  8. Nov 16, 2009 #7
    okay, I have the following:
    d(t) = distance with respect to time
    d(t)^2 = (xA - xB)^2 + (yA-yB)^2

    d(t)^2/dt ...

    2d(t)d'(t) = 2 (5 - 3/2)(3-dxB/dt) + 2(0-(-sqrt(3)(3/2)))(0 - dyB/dt)
    so now I solve for dxB/dt and dyB/dt?

    thank you
  9. Nov 17, 2009 #8
    not sure how to find dxB/dt and dyB/dt.. anybody want to help? thank you
  10. Nov 17, 2009 #9
    Use Pythagoras. You know what shape curve you are doing so just start by

    [tex] \sqrt{(dx/dt)^2 + f'(x)^2} = 4 [/tex] and from there, solve for [tex]dx/dt.[/tex]
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