Let's break down the problem step by step:
Part A: The rate of decay
The problem states that approximately 6% of the caffeine in the body is metabolized and eliminated every hour. Consequently, the rate of decay of caffeine is \( 6% \).
Part B: The initial amount of caffeine consumed
The initial amount of caffeine in a typical cup of coffee is given as \( 100 , \text{mg} \).
Part C: Write C(t) as a function of t
The amount of caffeine in the body can be described by the exponential decay formula:
\[ C(t) = C_0 \cdot (1 - r)^t \]
Where:
- \( C_0 \) is the initial amount of caffeine,
- \( r \) is the decay rate (as a decimal),
- \( t \) is the time in hours.
In this scenario:
- \( C_0 = 100 \) mg,
- \( r = 0.06 \).
Thus, the function becomes:
\[ C(t) = 100 \cdot (1 - 0.06)^t \]
\[ C(t) = 100 \cdot (0.94)^t \]
Part D: Fill out the table
Now, we need to calculate \( C(t) \) for the given values of \( t \).
- For \( t = 0 \):
\[ C(0) = 100 \cdot (0.94)^0 = 100 \cdot 1 = 100 , \text{mg} \]
- For \( t = 1 \):
\[ C(1) = 100 \cdot (0.94)^1 = 100 \cdot 0.94 = 94.00 , \text{mg} \]
- For \( t = 3 \):
\[ C(3) = 100 \cdot (0.94)^3 \approx 100 \cdot 0.830584 = 83.06 , \text{mg} \quad (\text{rounded to the nearest hundredth}) \]
- For \( t = 5 \):
\[ C(5) = 100 \cdot (0.94)^5 \approx 100 \cdot 0.735091 = 73.51 , \text{mg} \quad (\text{rounded to the nearest hundredth}) \]
Now let's summarize the results in the table:
| t | C(t) | |-----|--------------| | 0 | 100.00 mg | | 1 | 94.00 mg | | 3 | 83.06 mg | | 5 | 73.51 mg |
Summary of Answers
- Part A: 6%
- Part B: 100 mg
- Part C: \( C(t) = 100 \cdot (0.94)^t \)
- Part D:
- \( C(0) = 100.00 , \text{mg} \)
- \( C(1) = 94.00 , \text{mg} \)
- \( C(3) = 83.06 , \text{mg} \)
- \( C(5) = 73.51 , \text{mg} \)