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