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    The population x, in thousands, of red squirrels is modelled by the equation
    x = a/1+kt
    where t is the time in years, and a and k are constants. When ,

    Show that
    dx/dt = -kx^2/a

    can anyone do this step by step cos i'm clearly missing something simple, I'm using the quotient rule
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    (Original post by londongirl)
    The population x, in thousands, of red squirrels is modelled by the equation
    x = a/1+kt
    where t is the time in years, and a and k are constants. When ,
    Is that 'When ,' meant to continue...?
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    (Original post by londongirl)
    The population x, in thousands, of red squirrels is modelled by the equation
    x = a/1+kt
    where t is the time in years, and a and k are constants. When ,

    Show that
    dx/dt = -kx^2/a

    can anyone do this step by step cos i'm clearly missing something simple, I'm using the quotient rule
    x = a/(1+kt)
    .. = a(1+kt)^(-1)
    dx/dt = -ak(1+kt)^(-2) = -ak/(1+kt)^2

    dx/dt = -ak/(1+kt)^2

    but OMG! (1+kt) = a/x, so dx/dt = -ak/(a/x)^2 = -k/(a/x^2) = -kx^2/a
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    (Original post by MrTrig)
    Is that 'When ,' meant to continue...?
    Not necessary.
    x = a/1+kt = a(1+kt)^-1
    using chain rule (u = 1+kt):
    dx/dt = -ak (1+kt)^-2
    Now x/a = 1/(1+kt)
    so dx/dt = -ak(x/a)^2 = (-kx^2)/a
    QED
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    a(1+kt)^-1
    Goes to
    -ak(1+kt)^-2
    From the first thing:
    1+kt=a/x
    -ak(a/x)^-2

    And that rearranges to your thingy. I think xD


    [edit] Okay okay. You win this time.
    But I still take the bronze. >>;;
 
 
 
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