A 0.2 ml dose of a drug is injected into a patient steadily for 0.75 seconds. At the end of this time, the quantity, Q, of the drug in the body starts to decay exponentially at a continuous rate of 0.2 percent per second. Using formulas, express Q as a continuous function of time, t, in seconds
Q(t) = ______ if 0 ¡Ü t ¡Ü ______
and Q(t) = ______ if _____ ¡Ü t ¡Ü ¡Þ
To express Q as a continuous function of time, t, we can break it down into two intervals:
1) For 0 ≤ t ≤ 0.75 seconds:
During this interval, the drug is being injected steadily into the patient. The quantity of the drug in the body is increasing at a constant rate. The formula for this type of situation is:
Q(t) = Q₀ + R * t
Where:
Q₀ is the initial quantity of the drug in the body (at t = 0)
R is the rate of injection (ml/second)
t is the time in seconds (0 ≤ t ≤ 0.75)
In this case, Q₀ is 0 (as no drug is present initially), and R is 0.2 ml/0.75 seconds. Therefore, the formula becomes:
Q(t) = 0 + (0.2 ml/0.75 sec) * t
= (0.2/0.75) * t
= (4/15) * t
So, for 0 ≤ t ≤ 0.75 seconds, the continuous function for Q is:
Q(t) = (4/15) * t
2) For 0.75 ≤ t ≤ ∞ (infinity seconds):
After 0.75 seconds, the drug starts to decay exponentially at a continuous rate of 0.2 percent per second. Exponential decay can be modeled using the formula:
Q(t) = Q₀ * e^(rt)
Where:
Q₀ is the initial quantity of the drug in the body (at t = 0.75 seconds)
r is the continuous decay rate (0.2 percent per second)
t is the time in seconds (0.75 ≤ t ≤ ∞)
In this case, Q₀ is the value of Q(0.75), which we can calculate using the previous interval's formula:
Q₀ = Q(0.75)
= (4/15) * 0.75
= 0.2 ml
Now we can rewrite the formula for the second interval:
Q(t) = (0.2 ml) * e^(0.002t)
So, for 0.75 ≤ t ≤ ∞ (infinity seconds), the continuous function for Q is:
Q(t) = (0.2 ml) * e^(0.002t)
Putting it all together:
Q(t) = (4/15) * t if 0 ≤ t ≤ 0.75
(0.2 ml) * e^(0.002t) if 0.75 ≤ t ≤ ∞