To express the quantity of the drug in the patient's body, $Q(t)$, as a function of time, we need to break the problem into two parts: the injection phase and the decay phase.

1. Injection Phase (0 ≤ t ≤ 0.6 seconds)

During the injection, the drug is administered steadily at a constant rate of 0.2 ml over 0.6 seconds. So, the quantity of the drug in the body increases linearly during this time.

The rate of injection is:

$\text{Rate of injection} = \frac{0.2 \, \text{ml}}{0.6 \, \text{seconds}} = \frac{1}{3} \, \text{ml/second}$

Thus, for $0 \le t \le 0.6$, the drug is added continuously to the body at a rate of $\frac{1}{3}$ ml per second. The function for the quantity of the drug, $Q(t)$, during this time can be expressed as:

$Q(t) = \frac{1}{3} \cdot t \quad \text{for} \quad 0 \le t \le 0.6$

At $t = 0.6$, the total amount of the drug in the body is:

$Q(0.6) = \frac{1}{3} \cdot 0.6 = 0.2 \, \text{ml}$

2. Decay Phase (t > 0.6 seconds)

Once the injection is complete (at $t = 0.6$), the drug starts to decay exponentially at a continuous rate of 0.45% per second. The formula for exponential decay is:

$Q(t) = Q_0 \cdot e^{-kt}$

where:

• $Q_0$ is the initial quantity of the drug at $t = 0.6$, which is 0.2 ml.
• $k$ is the decay rate, which is 0.0045 per second (since 0.45% = 0.0045 in decimal form).
• $t$ is the time in seconds.

Thus, for $t > 0.6$, the quantity of the drug decays as:

$Q(t) = 0.2 \cdot e^{-0.0045(t - 0.6)} \quad \text{for} \quad t > 0.6$

This expression represents the amount of drug in the patient's body after $t$ seconds, taking into account both the initial steady injection and the subsequent decay.

Final Answer:

\begin{cases} \frac{1}{3} \cdot t & \text{if} \, 0 \le t \le 0.6 \\ 0.2 \cdot e^{-0.0045(t - 0.6)} & \text{if} \, t > 0.6 \end{cases}$$ Let me know if you would like further clarification or additional details!