Problem 33 - Applied Calculus

Question

Soybean production, in millions of tons $$\begin{array}{c|c|c|c|c|c|c} \hline \text { Year } & 2000 & 2001 & 2002 & 2003 & 2004 & 2005 \\ \hline \text { Production } & 161.0 & 170.3 & 180.2 & 190.7 & 201.8 & 213.5 \\ \hline \end{array}$$ (a) Niki invested \(\$ 10,000\) in the stock market. The investment was a loser, declining in value \(10 \%\) per year each year for 10 years. How much was the investment worth after 10 years? (b) After 10 years, the stock began to gain value at \(10 \%\) per year. After how long will the investment regain its initial value \((\$ 10,000) ?\)

Step-by-Step Solution

Answer

After 10 years, the investment was worth \( \$3,486.78 \). It will regain its initial value in approximately 12 years of growth.

Step 1: Calculate Investment Decline After 10 Years

Niki's investment decreases by 10% each year. Initially, the investment is worth \( \\(10,000 \), and after each year, it is worth 90% of the previous year's amount due to a 10% decrease. This can be modeled by the formula \( V = P (1 - r)^t \), where \( V \) is the final amount, \( P \) is the initial principal \( \\)10,000 \), \( r \) is the rate of decline (0.1), and \( t \) is the time in years (10 years). Substitute the values and solve: \[ V = 10,000 \times (0.9)^{10} \] Calculating yields: \( V \approx \$3,486.78 \).

Step 2: Determine Years to Regain Initial Investment

Now the investment starts gaining 10% per year. We use the formula \( F = P (1 + r)^t \), with \( F = 10,000 \), \( P = 3,486.78 \), and \( r = 0.1 \). We need to solve for \( t \):\[ 10,000 = 3,486.78 \times (1.1)^t \] Divide both sides by 3,486.78: \[ (1.1)^t = \frac{10,000}{3,486.78} \approx 2.8679 \] Take the logarithm of both sides:\[ t \log(1.1) = \log(2.8679) \]\[ t = \frac{\log(2.8679)}{\log(1.1)} \] Calculating gives \( t \approx 11.27 \).

Step 3: Final Interpretation

It will take just over 11 years, specifically 11.27 years, for Niki's investment to regain its original value of \( \$10,000 \) after beginning to grow again at a rate of 10% per year. Since we typically consider complete years, Niki will regain her initial investment value after approximately 12 complete years of growth.

Key Concepts

Investment Value Calculation

Calculating the future value of an investment is crucial to understanding how money can grow or shrink over time. In this scenario, Niki's investment starts at \\(10,000 but decreases by 10% each year for 10 years. This situation is modeled using the exponential decay formula:

• \( V = P(1 - r)^t \)

Here,

• \( V \) is the future value,
• \( P \) is the principal (initial amount),
• \( r \) is the decline rate (0.1 or 10%),
• \( t \) is time in years.

Substituting the known values: \[ V = 10,000 \times (0.9)^{10} \]After calculating, the investment value after 10 years is approximately \\)3,486.78.
This example demonstrates how quickly investments can lose value, stressing the importance of understanding negative growth impact.

Logarithmic Functions

Logarithmic functions play a vital role in solving equations involving exponential growth. After the investment stops declining and starts gaining value at 10% per year, we need to find how long it will take for Niki's investment to reach \$10,000 again. The equation is set up as:

• \( F = P(1 + r)^t \)

Substitute the known values:\[ 10,000 = 3,486.78 \times (1.1)^t \]First, divide by the principal:\[ (1.1)^t = \frac{10,000}{3,486.78} \approx 2.8679 \]To solve for \( t \), take the logarithm of both sides:\[ t \log(1.1) = \log(2.8679) \]The solution becomes:\[ t = \frac{\log(2.8679)}{\log(1.1)} \]Computing this gives \( t \approx 11.27 \).
Logarithms simplify solving exponentials by converting multiplication into addition, making complex calculations manageable.

Mathematical Modeling

Mathematical modeling provides a framework to simulate real-world scenarios. Here, two distinct phases of Niki's investment lifecycle, exponential decay and exponential growth, are modeled mathematically. These models help predict future outcomes based on present data. ### Exponential Decay in Investments When investments decrease, the decay is exponential, as shown when the value dropped for 10 years. This model takes into account frequent proportional reductions. ### Exponential Growth in Investments The growth model is similar, used when the investment rebounds, allowing us to calculate when a desired future value will be achieved. The key to effective mathematical modeling lies in accurately incorporating rates and time frames, enabling practical financial planning. Thus, understanding such models equips investors to anticipate gains or losses over periods, optimizing decision-making processes.