Problem 13 - Applied Calculus

Question

The net worth, \(f(t)\), of a company is growing at a rate of \(f^{\prime}(t)=2000-12 t^{2}\) dollars per year, where \(t\) is in years since \(2005 .\) How is the net worth of the company expected to change between 2005 and \(2015 ?\) If the company is worth \(\$ 40.000\) in \(2005,\) what is it worth in \(2015 ?\)

Step-by-Step Solution

Answer

The net worth increases by $16,000; in 2015, it is $56,000.

Step 1: Define the Problem

We need to determine how the net worth of the company changes from 2005 to 2015 given the rate of change function \( f'(t) = 2000 - 12t^2 \). Moreover, we have an initial net worth of $40,000 in 2005.

Step 2: Find the Net Worth Function

To find the net worth function \( f(t) \), we need to integrate the rate of change function \( f'(t) \). It is given by:\[ f(t) = \int (2000 - 12t^2) \, dt = 2000t - 4t^3 + C \]where \( C \) is the constant of integration.

Step 3: Determine the Constant of Integration

Using the initial condition, \( f(0) = 40000 \), substitute \( t = 0 \) into the net worth function:\[ 40000 = 2000(0) - 4(0)^3 + C \]Thus, \( C = 40000 \). The net worth function is:\[ f(t) = 2000t - 4t^3 + 40000 \]

Step 4: Calculate the Net Worth Change from 2005 to 2015

Calculate \( f(10) - f(0) \) to find the net worth change over 10 years:\[ f(10) = 2000(10) - 4(10)^3 + 40000 = 20000 - 4000 + 40000 = 56000 \]Thus, the change in net worth is \( 56000 - 40000 = 16000 \). The net worth increases by $16,000.

Step 5: Conclude with Final Value in 2015

Since the initial net worth was \(40,000 in 2005 and the net worth increased by \)16,000, the net worth in 2015 is:\( \text{Net Worth in 2015} = \text{Net Worth in 2005} + \text{Change in Net Worth} = 40000 + 16000 = 56000. \)

Key Concepts

Integration

Integration is a mathematical process used to find the accumulated value from a rate of change, which is crucial in this exercise. Here, we are given the rate of change of a company’s net worth, expressed as a differential function \(f'(t) = 2000 - 12t^2\). To find the actual net worth function \(f(t)\), we perform integration.

The integration of the rate of change function would look like this:

• We integrate \(f'(t)\) over \(t\) to find \(f(t)\).
• The indefinite integral results in \(f(t) = 2000t - 4t^3 + C\), where \(C\) is the constant of integration.

This function \(f(t)\) helps us determine the net worth at any given time \(t\), provided we know the initial condition. Integration connects us to the broader picture from a single rate of change, offering insights into how quantities accumulate over time.

Rate of Change

The rate of change essentially describes how a function, such as net worth in this case, varies with respect to time. Here, the rate of change is given by the function \(f'(t) = 2000 - 12t^2\), which represents the net change in dollars per year.

This rate of change is calculated by observing:

• A constant rate component \(2000\) which suggests a stable increase.
• A variable rate \(-12t^2\) indicating that as time increases, the reduction in growth rate becomes more significant.

The negative quadratic term \(-12t^2\) means that while the net worth initially grows at a fast rate, this growth slows down significantly as \(t\) progresses. Understanding this rate helps predict future values under changing conditions.

Initial Conditions

Initial conditions serve as the starting point when solving for the integration constant in our net worth function. In this exercise, the company’s net worth at the year 2005 is given as $40,000.

When solving for our integrated function \(f(t)\), setting \(t = 0\) allows us to apply this initial condition in the equation:

• \(f(0) = 2000(0) - 4(0)^3 + C = 40000\)

This results in determining \(C = 40000\), anchoring our net worth function to realistic values. Initial conditions are vital because they help transform a general solution into a specific one that's applicable to a real-world scenario.

Calculus Applications

Calculus offers several practical applications, especially when analyzing changes over time, like the net worth of a company. By employing calculus techniques, such as differentiation and integration, we understand not just the current state but project future financial landscapes.

For instance, through integration, we get a functional form of net worth over time, revealing how investments may pan out between two points. The ability to apply both differentiation (for rates of change) and integration (for accumulated values) allows us to:

• Predict future trends and values.
• Make informed financial decisions.
• Understand underlying economic behavior.

In this case, calculus completely describes the company’s financial journey from a rate of growth to its accumulated worth ten years later, demonstrating its power in quantitative analysis.