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A Swimmer's Problem

  1. Aug 30, 2012 #1
    I found an answer, but I am skeptical.

    What must the swimmer's speed, vs, be in order that he drifts only 1 mile downstream as he crosses the river?

    The width of the river = 2 miles
    The velocity of the river, vr = 3(1-x^2)

    Relevant equations:

    tanα=vr/vs
    dy/dx = tanα
    vr = v0(1-x^2/a^2)

    Image:
    9K15G.png


    So I started off with:

    tanα=vr/vs
    plugged in dy/dx for tanα, and vr = 3(1-x^2)

    dy/dx = (3/vs)(1-x^2)
    I assumed that vs was a constant.
    ∫dy/dx = (3/vs)∫(1-x^2)dx

    y = (3/vs)(x - 1/3 x^3) + C

    plugged in initial values y(-1) = 0,

    0 = (3/vs)((-1) - 1/3(-1)^3) + C
    ((-1) - 1/3(-1)^3) = -2/3
    0 = (3/vs)(-2/3) + C
    C = 2/vs

    now I have the equation all with 1 unknown:

    y = (3/vs)(x - 1/3 x^3) + 2/vs

    here is the part where I just improvised not really knowing what to do...
    I figured that if I used the initial values, I wouldn't get my vs, so I used the final values:
    y(1)=1

    1 = (3/vs)(1 - 1/3) + 2/vs
    and for this one I get that:
    vs = 4 mi/h
     
  2. jcsd
  3. Aug 30, 2012 #2

    LCKurtz

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    Science Advisor
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    Gold Member

    Looks OK to me. The ##a## in the picture is evidently 1.
     
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