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.

Answer 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.

Answer 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

Answer 3

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