Problem 30 - Applied Calculus

Question

In Problems \(29-30,\) a quantity \(P\) is an exponential function of time \(t .\) Use the given information about the function \(P=P_{0} e^{k t}\) to: (a) Find values for the parameters \(k\) and \(P_{0}\). (b) State the initial quantity and the continuous percent rate of growth or decay. \(P=40\) when \(t=4\) and \(P=50\) when \(t=3\)

Step-by-Step Solution

Answer

\( k \approx -0.2231 \), \( P_0 \approx 78.125 \); Initial quantity is 78.125, decay rate is 22.31\%.

Step 1: Write down the given exponential function equations

We know the function is given by \( P = P_0 e^{kt} \). We have two data points \( P = 40 \) at \( t = 4 \), and \( P = 50 \) at \( t = 3 \). This gives us two equations: 1. \( 40 = P_0 e^{4k} \) 2. \( 50 = P_0 e^{3k} \).

Step 2: Divide the equations to eliminate \( P_0 \)

To find \( k \), divide the first equation by the second equation: \[ \frac{P_0 e^{4k}}{P_0 e^{3k}} = \frac{40}{50} \]This simplifies to: \[ e^{k} = \frac{4}{5} \]

Step 3: Solve for \( k \)

Take the natural logarithm of both sides to find \( k \): \[ k = rac{\ln(4/5)}{1} \]Thus, \( k \approx -0.2231 \).

Step 4: Use one equation to solve for \( P_0 \)

Use the original equation \( 50 = P_0 e^{3k} \) to solve for \( P_0 \): \[ P_0 = \frac{50}{e^{3k}} \]Substitute \( k \approx -0.2231 \) to find:\[ P_0 \approx \frac{50}{(4/5)^3} \approx 78.125 \].

Step 5: Identify the initial quantity and the continuous rate

The initial quantity, when \( t = 0 \), is \( P_0 \approx 78.125 \). The continuous rate of decay is given by the rate \( k \), which is approximately \( -22.31\% \) because \( k \approx -0.2231 \). Negative \( k \) indicates a decay, hence \( 22.31\% \) decay per unit time.

Key Concepts

Growth and Decay

Exponential functions are essential in representing various real-world scenarios, such as population growth or radioactive decay. The general form of these functions is given by

1. \( P = P_0 e^{kt} \)

where \( P \) is the quantity at time \( t \), \( P_0 \) is the initial quantity, and \( k \) is the rate constant. Growth occurs when \( k > 0 \), and decay happens when \( k < 0 \). In this instance, the quantity \( P \) is an exponential function showing decay since we found a negative \( k \) value.
Decay implies that the initial quantity decreases over time. The rate of decay is indicated by the negative value of \( k \). Each unit of time, the quantity shrinks by a certain percentage, which can be interpreted through the equation as continuous decay.

• If \( k \) is negative, it tells us how quickly the decay occurs.
• The larger the absolute value of \( k \), the faster the decay rate.

Understanding whether a quantity is experiencing growth or decay is pivotal, as it directly impacts predictions and decision-making in practical applications like economics, biology, and physics.

Natural Logarithm

Natural logarithms play a crucial role in solving exponential equations. They allow us to "undo" the effect of exponentials and find unknown parameters.
In this problem, solving for the rate constant \( k \) required converting an exponential equation into a linear form using logarithms. The natural logarithm, denoted as \( \ln \), is used when dealing with the base \( e \), the natural exponential base.

To isolate \( k \), we apply the natural logarithm to both sides of the equation:

• \( e^k = \frac{4}{5} \)
• Apply \( \ln \): \( \ln(e^k) = \ln\left(\frac{4}{5}\right) \)
• Since \( \ln(e^k) = k \), it follows that \( k = \ln\left(\frac{4}{5}\right) \).

Using the natural logarithm, we found that \( k \approx -0.2231 \). This transformation simplifies computations by leveraging properties of logarithms, turning multiplicative relationships into additive ones.

Parameter Estimation

Parameter estimation involves finding unknown values such as \( k \) and \( P_0 \) in the function \( P = P_0 e^{kt} \). Begin with what you know:

• Two known points: \( P = 40 \) at \( t = 4 \) and \( P = 50 \) at \( t = 3 \).
• Write corresponding equations: \( 40 = P_0 e^{4k} \) and \( 50 = P_0 e^{3k} \).
• Use algebraic manipulation to solve for \( k \) without needing \( P_0 \).
• Substitute the found \( k \) to solve for \( P_0 \).

By dividing the two equations, we isolate \( e^k \) and solve using logarithms to determine \( k \). Once \( k \) is known, substituting it back to find \( P_0 \) completes the process. This method is effective and flexible, adapting to various types of growth and decay problems.