Question

A typical cup of coffee contains about 100 milligrams (mg) of caffeine.

Every hour, approximately 6%
of the amount of caffeine in the body is metabolized and eliminated.



Part A: The rate of decay is
%

Part B: The initial amount of caffeine consumed in this scenario is
mg.

Part C: Let C
represent the amount of caffeine in the body, in mg, and t
represents the number of hours since a cup of coffee was consumed. Write C
as a function of t
.

C(t) =

Part D: Fill out the table below. Round your answer to the nearest hundredth.

t
0 1 3 6
C(t)


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Answer 1

Let's go through each part of the question step by step.

### Part A
The rate of decay is the percentage of the substance that is lost per hour. According to the problem, 6% of caffeine is metabolized and eliminated every hour. Therefore, the rate of decay is:

**Rate of Decay = 6%**

### Part B
The initial amount of caffeine consumed is stated in the problem. A typical cup of coffee contains about 100 milligrams (mg) of caffeine.

**Initial amount of caffeine = 100 mg**

### Part C
We need to express the amount of caffeine \( C \) as a function of time \( t \) in hours. Since 6% is eliminated every hour, 94% remains.

The decay function can be expressed as:
\[
C(t) = C_0 \cdot (1 - r)^t
\]
Where:
- \( C_0 \) is the initial amount of caffeine (100 mg),
- \( r \) is the decay rate in decimal form (6% is 0.06),
- \( t \) is time in hours.

So, the function is:
\[
C(t) = 100 \cdot (0.94)^t
\]

### Part D
Now, we will calculate \( C(t) \) for the specified values of \( t \): 0, 1, 3, and 6.

1. For \( t = 0 \):
\[
C(0) = 100 \cdot (0.94)^0 = 100 \cdot 1 = 100
\]

2. For \( t = 1 \):
\[
C(1) = 100 \cdot (0.94)^1 = 100 \cdot 0.94 = 94
\]

3. For \( t = 3 \):
\[
C(3) = 100 \cdot (0.94)^3 \approx 100 \cdot 0.830584 = 83.06
\]

4. For \( t = 6 \):
\[
C(6) = 100 \cdot (0.94)^6 \approx 100 \cdot 0.528987 = 52.90
\]

### Here’s the completed table:

| t | 0 | 1 | 3 | 6 |
|----|-----|-----|------|------|
| C(t) | 100 | 94 | 83.06 | 52.90 |

Feel free to ask if you have any other questions!