Problem 22 - Applied Calculus

Question

A quantity \(P\) is an exponential function of time \(t .\) Use the given information about the function \(P=P_{b} a^{t}\) to: (a) Find values for the parameters \(a\) and \(P_{0}\). (b) State the initial quantity and the percent rate of growth or decay. \(P=320\) when \(t=5\) and \(P=500\) when \(t=3\)

Step-by-Step Solution

Answer

The initial quantity is approximately 976.56, and the decay rate is 20%.

Step 1: Write down the equations

As given, the exponential function is \( P = P_0 a^t \). We have two conditions: \( P = 320 \) when \( t = 5 \), and \( P = 500 \) when \( t = 3 \). So we have two equations, \( 320 = P_0 a^5 \) and \( 500 = P_0 a^3 \).

Step 2: Set up a system of equations

We will solve for \( a \) and \( P_0 \). We have:1. \( 320 = P_0 a^5 \) 2. \( 500 = P_0 a^3 \). We can divide the second equation by the first equation to find \( a \).

Step 3: Divide the equations to find 'a'

Divide Equation 2 by Equation 1: \( \frac{500}{320} = \frac{P_0 a^3}{P_0 a^5} \) leads to \( \frac{5}{3.2} = a^{-2} \). Simplifying gives \( a^2 = \frac{3.2}{5} \). Solve for \( a \) by taking the square root on both sides to get \( a = \sqrt{\frac{3.2}{5}} \).

Step 4: Calculate the value of 'a'

Calculate \( a \): \( a = \sqrt{\frac{3.2}{5}} = \sqrt{0.64} = 0.8 \). So \( a \) is 0.8.

Step 5: Substitute 'a' to find 'P_0'

Substitute \( a = 0.8 \) back into one of the original equations, say Equation 2: \( 500 = P_0 (0.8)^3 \). This gives \( P_0 = \frac{500}{0.8^3} = \frac{500}{0.512} \).

Step 6: Calculate the value of 'P_0'

Calculate \( P_0 \):\( P_0 = \frac{500}{0.512} \approx 976.56 \). Thus, \( P_0 \) is approximately 976.56.

Step 7: State the initial quantity and rate

The initial quantity \( P_0 \) is approximately 976.56. Since \( a = 0.8 \) which is less than 1, this implies a decay. The percent rate of decay is found using \( 1-a \), which is \( 1 - 0.8 = 0.2 \), or 20% decay per time unit.

Key Concepts

Exponential Growth and Decay

Exponential functions describe how quantities grow or decay over time by a constant factor. When the base of an exponential function, represented by the parameter \(a\), is greater than 1, it signifies growth. Conversely, if \(a\) is between 0 and 1, it indicates decay.
For example, in this exercise, we used the function \( P = P_0 a^t \) to model how a quantity changes over time. Here, \(a\) was determined to be 0.8, which is less than 1, indicating the scenario is an example of exponential decay rather than growth.
The percent rate of decay is calculated as \( (1-a) \times 100 \% \). In our case, it means a 20% decrease per time unit since \(a = 0.8\) gives us a decay rate of 20%. Understanding whether a situation represents growth or decay is crucial, as it helps us interpret and predict the behavior of variables modeled exponentially.

Initial Value Problems

Initial value problems abound in scenarios where you need to determine the starting point of a dynamic process. In exponential functions, the initial quantity or initial value \( P_0 \) represents the amount we start with when \( t = 0 \).
In the exercise, after determining \(a\), the parameter \(P_0\) was calculated as approximately 976.56. This value highlights the starting quantity, which is a crucial step when setting up models for processes involving growth or decay.
To find \( P_0 \), substitute the values of the function back into the known equations. Solving these initial value equations gives insight into the baseline from which changes over time can be measured, assisting in both understanding and predicting the evolution of the system under study.

Solving Equations

Solving exponential equations often involves finding unknown parameters like \( a \) and \( P_0 \). In this exercise, the steps were methodical and systematic, showcasing the importance of creating a clear path from known values to unknowns.
We began by setting up two equations based on the given conditions: \( P = 320 \) at \( t = 5 \) and \( P = 500 \) at \( t = 3 \). By solving for \( a \) first, we managed part of the equation using division to simplify and isolate \( a \).
Once \( a \) was determined as 0.8, substitution back into the equation yielded the initial value \( P_0 \). Understanding how to manipulate these equations step-by-step is key to solving more complex problems where exponential parameters need uncovering or verifying.