Problem 25 - Algebra 2 and Trigonometry

Question

The amount of a certain medicine present in the bloodstream decreases at a rate of 10\(\%\) per hour. a. Which is a better model to use for this scenario: \(A=A_{0}(1+r)^{t}\) or \(A=A_{0} e^{r t} ?\) Explain your answer. b. Using both models, find the amount of medicine in the bloodstream after 10.5 hours if the initial dose was 200 milligrams.

Step-by-Step Solution

Answer

The discrete model, \(A = A_{0}(1+r)^{t}\), is more appropriate. Amounts: ~69.74 mg (discrete), ~69.98 mg (continuous).

Step 1: Identify Model Type

In this case, medicine decays at a constant percentage rate per hour. For constant percentage growth or decay, the model \(A = A_{0}(1+r)^{t}\) is preferred over the continuous model \(A = A_{0} e^{r t}\).

Step 2: Set Up Exponential Decay Model

For the model \(A = A_{0}(1+r)^{t}\), we identify that the decay rate is \(r = -0.10\) or -10%. We start with \(A_{0} = 200\) milligrams.

Step 3: Calculate Amount Using Discrete Model

Substitute \(A_{0} = 200\), \(r = -0.10\), and \(t = 10.5\) hours into the discrete model: \\[A = 200(1 - 0.10)^{10.5}.\] \Calculate: 1 \(-\) 0.10 \(= 0.90\)\;and then \( (0.90)^{10.5} \). Finally, multiply the result by \(200\).

Step 4: Calculate the Discrete Model Result

\(A = 200 \times (0.90)^{10.5} = 200 \times 0.3487 \approx 69.74\) milligrams.

Step 5: Set Up Continuous Decay Model

For the continuous decay model \(A = A_{0} e^{r t}\), use the natural logarithm base \(e\). The continuous decay rate \to match a 10% hourly decrease is \(r = -0.10\).

Step 6: Calculate Amount Using Continuous Model

Substitute into the continuous model: \\[A = 200 \times e^{-0.10 \times 10.5}.\] \Calculate \(e^{-1.05}\) and then multiply the result by \(200\).

Step 7: Calculate the Continuous Model Result

\(A \approx 200 \times 0.3499 \approx 69.98\) milligrams.

Key Concepts

Exponential Functions

Exponential functions play a critical role in modeling real-world scenarios where quantities grow or decay at a constant rate per time period. These functions can be represented in two primary forms: discrete and continuous. The basic structure of an exponential function is expressed as either \( A = A_{0}(1 + r)^{t} \) or \( A = A_{0} e^{rt} \), where:

• \( A \) is the final amount.
• \( A_{0} \) is the initial amount.
• \( r \) is the rate of growth or decay.
• \( t \) is the time period.

Understanding these functions is crucial as they help us to predict future values based on current or past data. They are widely applicable in fields such as finance, medicine, and environmental studies. In the context of exponential decay, the rate \( r \) will always be a negative percentage as the value decreases over time.

Discrete Decay Model

The discrete decay model is often used when dealing with processes that change at regular intervals. This model is represented by the equation \( A = A_{0}(1+r)^{t} \). For decay, \( r \) is expressed as a negative decimal, indicating a decrease in the amount per time unit.In our exercise, where a medicine's potency decreases by 10% per hour, the discrete model aptly captures this scenario. Here, the negative decay rate of \( r = -0.10 \) results in a multiplicative factor of \( 0.90 \) applied each hour.To solve for the amount remaining after a certain time, substitute the initial amount, decay rate, and elapsed time into the formula:

• Initial amount: 200 mg
• Decay rate: -0.10
• Time: 10.5 hours

This gives us:\[ A = 200(0.90)^{10.5}. \] Calculating this provides an estimate of the remaining medicine, demonstrating how quickly substances can lose potency using a discreet model.

Continuous Decay Model

The continuous decay model is suitable for processes that change continuously over time, rather than in set intervals. It is expressed as \( A = A_{0} e^{rt} \), where \( e \) is the base of the natural logarithm and approximately equal to 2.718.For situations like the medicine example, it might not be the best fit when decay happens at a simple hourly rate, but understanding it is vital for other applications. Using a continuous model can be particularly useful when the decay occurs at a smaller, infinitesimally constant rate over time.In the example given, when applying this model:

• Initial amount: 200 mg
• Decay rate: -0.10
• Time: 10.5 hours

The formula becomes:\[ A = 200e^{-0.10 imes 10.5}. \] By calculating \( e^{-1.05} \) and multiplying it by the initial amount, we can estimate the remaining quantity. This approach highlights how different modeling techniques can provide slightly different results, reflective of their respective assumptions about continuous change.