Problem 31 - Applied Calculus

Question

A mutual fund is currently valued at \(\$ 80\) per share and its value per share is increasing at a rate of \(\$ 0.50\) a day. Let \(V=f(t)\) be the value of the share \(t\) days from now. (a) Express the information given about the mutual fund in term of \(f\) and \(f^{\prime}\). (b) Assuming that the rate of growth stays constant, estimate and interpret \(f(10)\).

Step-by-Step Solution

Answer

The mutual fund's value per share will be $85 in 10 days.

Step 1: Define the Initial Conditions

At day 0, the value of the mutual fund is \( V = f(t) = 80 \). This serves as our initial condition to form the function \( f(t) \).

Step 2: Interpret the Rate of Change

The rate at which the value of the mutual fund is increasing is \$0.50 per day. Therefore, \( f'(t) = 0.50 \), indicating the derivative of \( f(t) \) with respect to \( t \) is 0.50.

Step 3: Formulate the Function f(t)

Since \( f(t) = 80 \) at \( t = 0 \) and \( f'(t) = 0.50 \), we use the formula for a linear function: \( f(t) = f(0) + f'(t) \, t \). Therefore, \( f(t) = 80 + 0.50t \).

Step 4: Calculate f(10)

Estimate \( f(10) \) by substituting \( t = 10 \) into the expression for \( f(t) \): \[ f(10) = 80 + 0.50 \times 10 = 80 + 5 = 85. \]

Step 5: Interpret f(10)

The interpretation of \( f(10) = 85 \) is that in 10 days, the value of the mutual fund share is projected to be \$85 if the daily increase remains constant.

Key Concepts

Linear Functions

Linear functions form a backbone in calculus, especially when modeling situations where change is constant. A linear function features a straight-line graph, typically symbolized as \( f(t) = mt + b \), where \( m \) represents the slope, and \( b \) is the y-intercept. In our context, this means that the value of the mutual fund is increasing in a predictable way.

For the mutual fund problem where \( f(t) = 80 + 0.50t \):

• \( 80 \) is the starting value of the function, known as the y-intercept. It represents the share value at day zero.
• \( 0.50 \) is the slope of the function, indicating the rate at which the stock's value increases per day.

Linear functions like this are powerful tools in predicting future values, providing us with an easy-to-calculate formula for any given day \( t \). If the rate of change remains constant, linear functions assure us of consistency in our predictions.

Rate of Change

The rate of change tells us how one quantity changes over time. In the mutual fund example, the rate of change is the amount by which the mutual fund value increases each day, which is \\(0.50.

This means for every day that passes, the value of the mutual fund goes up by \\)0.50.

Analyzing the rate of change helps us understand the dynamics of a scenario:

• Steady: With a constant rate of change, the growth of the mutual fund is predictable.
• Real-world application: In economics and finance, the rate of change is crucial for making predictions and forming strategies.

Understanding the concept of rate of change allows us to easily calculate how the mutual fund value changes with time and provides a mathematical structure for these predictions.

Derivatives

Derivatives are a fundamental concept in calculus that measure how a function changes as its input changes. In simpler terms, derivatives tell us the rate at which one quantity changes with respect to another.

For our mutual fund scenario, the derivative, \( f'(t) = 0.50 \), indicates the rate of change of the mutual fund’s value over time.

Derivatives have several key applications:

• Understanding change: They tell us how quickly quantities grow or shrink.
• Predictive power: In finance, they help anticipate changes in stock prices or asset values.
• Optimization: Derivatives are used in finding maximum profits or minimum costs.

Through derivatives, we gain a clearer understanding of the mutual fund's growth, reaffirming that it increases by \$0.50 daily. These insights are invaluable in various fields ranging from physics to economics, where change and optimization are crucial.