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Vector problem ?

  1. Sep 26, 2006 #1
    1. The points S and T are midpoints of the sides AB and AD respectively of a parallelogram ABCD .The line CS and CT cut the diagonal BD at the points U and V respectively .
    Show that BU = x BC + x CD and also BU = (1-y)BC +1/2 y CD , where x and y are constant.Hence show that BU=1/3 BD
    Deduce that the lines CS and CT trisect the diagonal BD .

    I can do the first two part but I don't know how to deduce that the lines CS and CT trisect the diagonal BD .Please help me .

    2. A canal of width 2a has parallel straight banks and the water flows due north. The points A and B are on opposite banks and B is due east of A , with point O as the midpoint of AB . The x-axis and y-axis are taken in the east and north directions respectively with O as the origin. The speed of the current in the canal ,Vc is given by
    Vc=Vo(1-x^2/a^2) where Vo is the speed of the current in the middle of the canal and x is the distance eastwards from the middle of the canal.A swimmer swims from A towards east at speed Vr relative to the current in the canal . Taking y to denote the distance northwards travelled by the sewimmer ,show that
    dy/dx = Vo/Vr (1-x^2/a^2) .
    I have no idea with this question . Is it want to find the velocity of the swimmer relative to ground ? But I cannot find the answer also if I find the velocity of the swimmer relative to ground . How to do this ?

    Please help me . Thanks .:!!)
     
  2. jcsd
  3. Sep 26, 2006 #2

    HallsofIvy

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    1. You are able to show that BU= 1/3 BD? That says immediately that U is 1/3 the distance from B to D. You should by exactly the same argument, using D instead of B, be able to show that DV= 1/3 BD. That is, U and V cut BD into thirds- they trisect BD.

    2. The speed of the swimmer, relative to the water, is Vr i and the speed of the current is V0(1-x^2/a^2)j so the speed of the swimmer relative to the ground is the sum of those vectors, Vr i + V0(1-x^2/a^2)j.
    That is [itex]\frac{dx}{dt}= V_r[/itex] also
    [tex]\frac{dy}{dt}= V_0(1- \frac{x^2}{a^2})[/tex]
    Now, just use the chain rule:
    [tex]\frac{dy}{dx}= \frac{\frac{dy}{dt}}{\frac{dx}{dt}}[/tex]
     
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