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
(If you write your function without breaking it up, use an endpoint of infinity and enter none for the remaining answer blanks.)
What is "0 ¡Ü t ¡Ü" ?
What is "¡Ü t ¡Ü ¡Þ"?
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To express Q as a continuous function of time, we need to break down the problem into two parts: the injection period (0 ≤ t ≤ 0.75) and the decay period (0.75 ≤ t ≤ ∞).
1. Injection period (0 ≤ t ≤ 0.75):
During this period, the drug is being injected into the patient steadily, so the quantity Q increases linearly with time. We know that a 0.2 ml dose is injected steadily over 0.75 seconds, so the rate of increase is given by:
Rate of increase = 0.2 ml / 0.75 s = 0.2667 ml/s
Therefore, during the injection period, Q can be expressed as a linear function:
Q(t) = 0.2667t if 0 ≤ t ≤ 0.75
2. Decay period (0.75 ≤ t ≤ ∞):
After 0.75 seconds, the quantity of the drug starts to decay exponentially at a continuous rate of 0.2% per second. To express this as a function, we can use the formula for exponential decay:
Q(t) = Q₀ * e^(-rt)
Where Q₀ is the initial quantity, e is the base of the natural logarithm, r is the decay rate, and t is the time.
In our case, the initial quantity Q₀ is the amount of the drug at the end of the injection period, which is Q(0.75). The decay rate r is given as 0.2% per second, so we convert it to a decimal: r = 0.002.
Q₀ = Q(0.75) = 0.2667 * 0.75 = 0.2 ml
r = 0.002
Therefore, during the decay period, Q can be expressed as an exponential function:
Q(t) = 0.2 * e^(-0.002t) if t ≥ 0.75
Now we have the two expressions for Q(t) for different time ranges:
Q(t) = 0.2667t if 0 ≤ t ≤ 0.75
Q(t) = 0.2 * e^(-0.002t) if t ≥ 0.75