A 0.9 ml dose of a drug is injected into a patient steadily for half a second. At the end of this time, the quantity, Q, of the drug in the body starts to decay exponentially at a continuous rate of 0.4% per second. Using formulas, express Q as a continuous function of time, t in seconds.
For 0 less than or equal to t which is less than or equal to 0.5
And for 0.5 < t
Q=
Q=
after t seconds,
Q(t) = 9*0.996^t
Now just solve for what you need.
To express Q as a continuous function of time, we need to consider two different scenarios: one for the time interval from 0 to 0.5 seconds and another for the time interval from 0.5 seconds onwards.
For 0 ≤ t ≤ 0.5 seconds:
During this time interval, the drug is being injected steadily into the patient's body. Given that the dose is 0.9 ml and the injection takes half a second, the rate of drug accumulation in the body is constant at 0.9 ml / 0.5 s = 1.8 ml/s.
Therefore, the quantity of the drug in the body as a function of time, Q(t), can be expressed as:
Q(t) = 1.8t
For 0.5 < t seconds:
After half a second, the drug injection stops, and the quantity of drug in the body starts to decay exponentially at a continuous rate of 0.4% per second.
The exponential decay formula is given by:
Q(t) = Q0 * e^(kt)
Where:
- Q(t) is the quantity of the drug at time t
- Q0 is the initial quantity of the drug at t = 0
- e is the base of natural logarithms (approximately 2.71828)
- k is the decay constant.
Since the initial dose is 0.9 ml, we have Q0 = 0.9 ml. The decay constant, k, is given as 0.4% per second, which can be written as 0.004.
Therefore, for t > 0.5 seconds, the continuous function that expresses the quantity of the drug in the body is:
Q(t) = 0.9 * e^(0.004t)
Combining the two scenarios, the continuous function for the quantity of the drug in the body, Q(t), is as follows:
Q(t) = 1.8t, for 0 ≤ t ≤ 0.5
Q(t) = 0.9 * e^(0.004t), for t > 0.5