Calc

A patient receives a 3 ml dose of medicine. The amount, y, of medicine in the body decreases at a rate of 15% per hour (dy/dt=-.15y). When the medicine in the patient is down to 1 ml, the nurse administers a second 3 ml dose and plans to give the patient a third dose when the medicine in the patient is again down to 1 ml. To the nearest hour, the patient will receive that third dose ___ hours after the initial dose.

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asked by Anon
  1. y' = -.15y
    dy/y = -.15 dt
    lny = -.15t+c
    y = ce^-.15t

    with y(0) = 3, c=3
    with y(0) = 4, c=4

    with c=3,
    3e^-.15t = 1
    e^-.15t = 1/3
    -.15t = ln(1/3)
    t = ln(1/3)/-.15
    t = 7.32

    with c=4,
    t = 9.24

    So, the patient gets the 3rd dose after 7.32+9.24 = 16.60 hours.

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    posted by Steve
  2. dy/dt=-0.15y
    dy/y = -0.15t
    ln(y)=-(0.15/2)t²+C
    f(t)=y=Ce-0.075t^sup2;
    At t=0, y=3 ml.
    f(0)=C=3
    so, let t=time for second dose,
    f(t)=3e-0.075t^sup2;
    solve for f(t)=1 ml.
    -0.075t²=ln(1/3)
    t=sqrt(-ln(1/3)/-0.075)
    =3.8273 (approx.)

    Time for third dose (after second dose),
    when medication remaining=1+3 ml=4 ml
    so
    f(t)=1=4e-0.075t^sup2;
    t=sqrt(-ln(1/4)/-0.075)
    =4.2993 hours approx.
    Time for third dose after first
    =3.8273+4.2993
    =8.1266 hous
    =8 hours approx

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  3. Oops, I got the function wrong, got an extra t on the right hand side.
    Go with Steve's solution.

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