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.

Question

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.

Answer 4

To find the time it takes for the medicine in the patient to decrease to 1 ml after the initial dose, we need to solve the differential equation dy/dt = -0.15y.

The differential equation represents exponential decay, where the rate of change of the medicine amount (dy/dt) is proportional to the amount of medicine in the body (y) with a proportionality constant of -0.15.

Let's solve this differential equation to find the solution.

First, divide both sides of the equation by y to separate variables:
dy/y = -0.15 dt

Then, integrate both sides:
∫(1/y) dy = ∫-0.15 dt

The integral of (1/y) with respect to y is ln|y| (natural logarithm of the absolute value of y) and the integral of -0.15 with respect to t is -0.15t.

So, we have:
ln|y| = -0.15t + C

Where C is the constant of integration.

To determine the value of C, we need to use the initial condition, given that when t = 0, y = 3 ml (the initial dose):
ln|3| = -0.15(0) + C
ln|3| = C

We can rewrite the equation as:
ln|y| = -0.15t + ln|3|

Now, let's consider the second dose. When the medicine in the patient is down to 1 ml, the nurse administers the second 3 ml dose. So, at that time, y = 1 ml. Now we can find the time it takes for y to become 1 ml.

ln|1| = -0.15t + ln|3|

Since the natural logarithm of 1 is 0:
0 = -0.15t + ln|3|

Solving for t:
0.15t = ln|3|
t = ln|3| / 0.15

Using a calculator, we can find that ln|3| ≈ 1.099.

So, t = 1.099 / 0.15 ≈ 7.326.

Therefore, the patient will receive the third dose approximately 7 hours and 19 minutes after the initial dose. Rounding to the nearest hour, the patient will receive the third dose around 7 hours after the initial dose.