Question: A 0.4 ml dose of a drug is injected into a patient steadily for $0.61$ 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 $0.8\%$ as a function of time, $t$, in seconds. Then, $$Q(t) = \begin{cases} \text{(blank)} & \text{if } 0 \leq t \leq 0.61 \\ 0.4e^{-0.008(t-0.61)} & \text{if } 0.61 \leq t < \infty \end{cases}$$ Graph shape: piecewise function with a blank first segment for $0 \leq t \leq 0.61$ and an exponential decay curve for $t \geq 0.61$; position_hint = center
1. **State the problem:**
We have a drug dose of $0.4$ ml injected steadily over $0.61$ seconds. After $t=0.61$ seconds, the drug quantity $Q(t)$ decays exponentially at a continuous rate of $0.8\%$ per second.
2. **Understand the function:**
The piecewise function is given by:
$$Q(t) = \begin{cases} \text{blank} & 0 \leq t \leq 0.61 \\ 0.4e^{-0.008(t-0.61)} & t \geq 0.61 \end{cases}$$
The decay rate $0.8\%$ per second means the decay constant $k = 0.008$ (since $0.8\% = 0.008$ in decimal).
3. **Interpret the blank segment:**
For $0 \leq t \leq 0.61$, the drug is being injected steadily, so the quantity increases from $0$ to $0.4$ ml. The problem states the dose is injected steadily, so the quantity increases linearly from $0$ to $0.4$ ml over $0.61$ seconds.
Thus, for $0 \leq t \leq 0.61$:
$$Q(t) = \frac{0.4}{0.61} t$$
4. **Final piecewise function:**
$$Q(t) = \begin{cases} \frac{0.4}{0.61} t & 0 \leq t \leq 0.61 \\ 0.4 e^{-0.008(t-0.61)} & t \geq 0.61 \end{cases}$$
5. **Explanation:**
- From $0$ to $0.61$ seconds, the drug amount increases linearly because the dose is injected steadily.
- After $0.61$ seconds, the drug amount decays exponentially at a continuous rate of $0.8\%$ per second.
This matches the problem description and the given decay formula.
**Final answer:**
$$Q(t) = \begin{cases} \frac{0.4}{0.61} t & 0 \leq t \leq 0.61 \\ 0.4 e^{-0.008(t-0.61)} & t \geq 0.61 \end{cases}$$