Problem 100 - Calculus and its applications

Question

The value, \(V(t),\) in dollars, of a share of Cypress Mills stock \(t\) months after it is purchased is modeled by $$ V(t)=58\left(1-e^{-1.1 t}\right)+20$$. a) Find \(V(1)\) and \(V(12)\). b) Find \(V^{\prime}(t)\). c) After how many months will the value of a share of the stock first reach \(\$ 75 ?\) d) Find \(\lim _{t \rightarrow \infty} V(t)\). Discuss the value of a share over a long period of time. Is this trend typical?

Step-by-Step Solution

Answer

a) \(V(1) \approx \$58.69\), \(V(12) = \$78\); b) \(V'(t) = 63.8e^{-1.1t}\); c) \(t \approx 2.69\) months; d) \(\lim_{t \to \infty} V(t) = 78\).

Step 1: Substitute to find V(1)

To find the value of a share after 1 month, we substitute \( t = 1 \) into the function \( V(t) = 58\left(1-e^{-1.1 t}\right)+20 \). This gives: \[ V(1) = 58\left(1-e^{-1.1 \times 1}\right) + 20 \]. Calculate this expression.

Step 2: Compute V(1)

Calculate \( e^{-1.1} \) and substitute this value into the equation from Step 1: \[ V(1) = 58(1 - e^{-1.1}) + 20 \]. Use a calculator to find \( e^{-1.1} \approx 0.3329 \). Thus, \[ V(1) = 58(1 - 0.3329) + 20 \approx 58(0.6671) + 20 = 38.692 + 20 = 58.692 \].

Step 3: Substitute to find V(12)

Substitute \( t = 12 \) into the function: \[ V(12) = 58\left(1-e^{-1.1 \times 12}\right) + 20 \].

Step 4: Compute V(12)

Calculate \( e^{-13.2} \), which is very close to zero. Therefore, \( V(12) = 58(1 - 0) + 20 = 58 + 20 = 78. \) So, \( V(12) = 78 \).

Step 5: Find the derivative V'(t)

To find \( V'(t) \), differentiate \( V(t) = 58(1-e^{-1.1 t}) + 20 \) with respect to \( t \): \[ V'(t) = 58 \cdot (-1) \cdot (-1.1) \cdot e^{-1.1 t} = 63.8 e^{-1.1 t} \].

Step 6: Solve for t when V(t) = 75

Set \( V(t) = 75 \) and solve for \( t \): \[ 75 = 58(1-e^{-1.1 t}) + 20 \]. Simplify to \[ 55 = 58(1-e^{-1.1 t}) \], so \[ \frac{55}{58} = 1 - e^{-1.1 t} \]. This gives \[ e^{-1.1 t} = 1 - \frac{55}{58} = 0.0517 \].

Step 7: Solve for t

Take the natural log of both sides: \[ -1.1 t = \ln(0.0517) \] which gives \[ t = \frac{\ln(0.0517)}{-1.1} \]. Calculate \( \ln(0.0517) \approx -2.964 \), so \[ t \approx \frac{2.964}{1.1} \approx 2.69 \]. Thus, the stock reaches \$75 in approximately 2.69 months.

Step 8: Analyze limit as t approaches infinity

Calculate the limit \( \lim_{t \to \infty} V(t) \). As \( t \to \infty \), \( e^{-1.1 t} \to 0 \), so \[ V(t) = 58(1-0) + 20 = 78 \]. Thus, \( \lim_{t \to \infty} V(t) = 78 \).

Step 9: Discuss long-term trend and typicality

The value of the share approaches \\(78 as \( t \to \infty \), indicating a horizontal asymptote at \\)78. This behavior suggests that the stock value stabilizes over time, a trend that is common in many financial investments as they reach their intrinsic value.

Key Concepts

Function Evaluation

Function evaluation is the process of finding the output of a particular function for given input values. In the context of stock market modeling, this often involves calculating the value of a stock at a specific time using an exponential growth or decay model.

In the problem, the function given is: \[V(t) = 58\left(1-e^{-1.1 t}\right)+20\] This equation models the value of Cypress Mills stock over time. Evaluating a function like this requires substituting a specific value for \(t\), which is the time in months.

Let's understand the steps:

• For \(t = 1\), substitute \(1\) into the function to find \(V(1)\), which indicates the stock's value after one month.
• For \(t = 12\), substituting \(12\) helps find the stock's value after a year.

Function evaluation helps us understand the short-term versus long-term changes in the stock’s value.

Derivative of Exponential Functions

The derivative of a function provides the rate at which the function's value changes with respect to changes in the input, which in this case is time, \(t\).

For exponential functions, differentiation involves using the chain rule. Given the function:
\[V(t) = 58\left(1-e^{-1.1 t}\right)+20\]
The derivative, \(V'(t)\), is found by differentiating the exponential component.

• The exponential term, \(e^{-1.1t}\), behaves uniquely. Its derivative is naturally negative due to the negative exponent, leading to derivative \(V'(t) = 63.8e^{-1.1t}\).
• This indicates the measure of change in the stock's value over time, highly important for predicting stock trends and making investment decisions.

Understanding the derivative helps financial analysts measure how quickly stock values might rise or fall.

Limit of a Function

Limits describe the behavior of a function as its argument approaches a certain point, often infinity. In financial models, determining the limit as \(t \to \infty\) helps predict long-term behavior.

For the function:
\[V(t) = 58\left(1-e^{-1.1 t}\right)+20\]
Taking the limit as \(t\) approaches infinity involves examining the behavior of \(e^{-1.1t}\).

• As \(t\) increases, \(e^{-1.1t}\) approaches zero.
• This results in \(V(t)\) stabilizing to \(78\), reflecting a plateau in stock growth.

This stabilization is typical for exponential decay models, indicating that the stock will reach a certain value and remain constant as time progresses.

Stock Market Modeling

Modeling the stock market involves predicting future stock values using mathematical equations. These models can be linear or exponential, with exponential growth or decay being vital in capturing how real-world stock prices rise or fall over time.

In this example, we use:
\[V(t) = 58\left(1-e^{-1.1 t}\right)+20\]
This model considers how stock values rise rapidly initially before leveling off.

• The exponential component \(e^{-1.1t}\) signifies quick changes initially, with gradual stabilization.
• The limit at \(t \to \infty\) provides long-term value expectations, which is critical for investors planning their strategies.

Stock market modeling thus aids investors and analysts in understanding both the immediate dynamics and future expectations of stock behaviors.