Problem 30 - Applied Calculus

Question

Put the functions in the form \(P=P_{0} e^{k t}\). $$P=4(0.55)^{t}$$

Step-by-Step Solution

Answer

The function is \(P = 4e^{-0.59784t}\).

Step 1: Identify Original Form

The given function is in the form \(P = 4 (0.55)^t\). We need to transform this into the exponential growth or decay form \(P = P_0 e^{kt}\).

Step 2: Determine Initial Value \(P_0\)

In the general form \(P = P_0 (b)^t\), \(P_0\) corresponds to the initial value of the function. Here, \(P_0 = 4\).

Step 3: Express as Exponential Base \(e\)

The given base, 0.55, must be expressed as an exponential function with base \(e\). We do this by setting \(0.55 = e^{k}\) and solving for \(k\).

Step 4: Solve for \(k\) using Natural Logarithms

Take the natural logarithm of both sides of the equation: \(\ln(0.55) = k \ln(e)\). Simplifying, we have \(k = \ln(0.55) \approx -0.59784\).

Step 5: Write in Final Form

Substitute \(P_0\), \(k\), and \(t\) back into the exponential form to get \(P = 4e^{-0.59784t}\).

Key Concepts

Exponential Growth and Decay

Exponential functions appear in the form of \( P = P_0 e^{kt} \), and they describe how quantities grow or shrink over time.
They are common in fields like biology, economics, and physics for modeling populations, investments, or radioactive decay.
- **Exponential Growth:** This occurs when \( k > 0 \). The quantity increases rapidly as time goes on.
- **Exponential Decay:** Conversely, when \( k < 0 \), we have exponential decay, and the quantity decreases over time.For example, the function \( P = 4(0.55)^t \) is an illustration of exponential decay.
In this case, every time \( t \) increases, the quantity \( P \) decreases because the base \( 0.55 \) is less than 1.
This pattern is typical of processes like radioactive decay or cooling processes in physics.To convert this function to the form \( P = P_0 e^{kt} \), we must identify that \( k = \ln(0.55) \).
This tells us how fast the decay happens, giving a deeper understanding of the process.

Natural Logarithms

Natural logarithms are an essential mathematical tool when dealing with exponential functions.
Represented as \( \ln(x) \), they are logarithms with base \( e \), where \( e \approx 2.71828 \), the natural constant.
- **Why Use Natural Logarithms?** Natural logarithms are used to simplify or solve expressions involving the exponential constant \( e \).
They help in rearranging equations, such as finding out the \( k \) value in exponential functions.
- **Conversion Example:** If you have \( a = e^b \), taking the natural logarithm of both sides results in \( \ln(a) = b \).
This conversion is crucial when transforming a base other than \( e \) into the \( e \) base.In the exercise, knowing \( 0.55 = e^k \) allows finding \( k \) by using \( k = \ln(0.55) \).
This successfully transforms the function to use the base \( e \), which is often easier to interpret or compare.

Transforming Functions

Transforming functions involves changing a given function into another form.
This can make them easier to analyze or interpret depending on the context.
In exponential functions, the transformation may involve changing bases or rearranging the formula.
- **Why Transform?** It helps in applying more advanced mathematical tools or incorporating the function into larger models.
Transforming to an \( e \) base specifically allows the use of calculus techniques, providing insights into growth or decay rates.- **Transformation Process:** As seen in the exercise, the original function \( P = 4(0.55)^t \) was transformed.
The base \( 0.55 \) was converted to an \( e \) base, resulting in \( P = 4e^{-0.59784t} \).
Through such transformations, we aim to make the equation easier to manage or more suitable for mathematical analysis.
This process shows the flexibility of mathematical expressions and how they can adapt to different needs or audiences.