Problem 28 - Applied Calculus

Question

(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

(a) $3,487; (b) just over 11 years.

Step 1: Understanding the Problem

We need to calculate how much the investment is worth after declining by 10% each year for 10 years, and then determine how long it will take for it to regain the original value if it gains 10% per year.

Step 2: Calculating the Value After 10 Years of Decline

Starting with the initial investment of \(\\)10,000\(, we need to apply a decline of 10% per year over 10 years. The formula to find the value after depreciation is: \[ V = P \times (1 - r)^n \]where \)P\( is the principal amount (\)\\(10,000\)), \(r\) is the rate of decline (0.10), and \(n\) is the number of years (10). Substituting these values in, we have:\[ V = 10,000 \times (1 - 0.10)^{10} \]

Step 3: Solving for the Investment's Value After Decline

Calculate:\[ V = 10,000 \times (0.90)^{10} \]First, compute the power: \[ (0.90)^{10} = 0.3487 \]Then multiply by the initial investment:\[ V = 10,000 \times 0.3487 = 3487 \]After 10 years, the investment is worth \(\\)3,487$.

Step 4: Calculating Time to Regain Initial Value

Now, we need to calculate how long it will take for the investment to grow from \(\\)3,487\( back to \)\\(10,000\) assuming a 10% growth per year. The formula for compound interest is:\[ P_f = P_i \times (1 + r)^n \]where \(P_f\) is the final amount (\(\\)10,000\(), \)P_i\( is the new principal (\)\\(3,487\)), and we need to solve for \(n\). Thus:\[ 10,000 = 3,487 \times (1.10)^n \]

Step 5: Solving for the Number of Years to Regain Initial Value

To isolate \(n\), divide both sides by 3,487:\[ (1.10)^n = \frac{10,000}{3,487} \]Calculate the right side:\[ 2.868 \approx \frac{10,000}{3,487} \]Now we use logarithms to solve for \(n\):\[ n \times \log(1.10) = \log(2.868) \]Divide both sides by \(\log(1.10)\):\[ n = \frac{\log(2.868)}{\log(1.10)} \approx 11.072 \]Thus, it will take slightly over 11 years for the investment to regain its initial value.

Key Concepts

Investment Depreciation

When you invest in something, it’s possible for its value to decrease over time. This decrease is known as _investment depreciation_. Depreciation simply means losing value at a certain rate each year. In the given problem, Niki's investment depreciates by 10% each year over a span of 10 years. To calculate the depreciated value, we use the formula for exponential decay:

• \[ V = P \times (1 - r)^n \]
• Where:
  • \(V\) is the value of the investment after depreciation,
  • \(P\) is the initial principal amount,
  • \(r\) is the rate of depreciation,
  • \(n\) is the number of years.

For Niki’s investment, we substitute to get \(10,000 \times (0.90)^{10}\), which results in an investment worth \$3,487 after 10 years.Think of depreciation like a balloon losing air each year, slowly getting smaller. It shows that investments can sometimes lose value before potentially gaining it back.

Compound Interest

When an investment starts to gain value, we calculate this increase using _compound interest_. Compound interest means that you add to the original amount not just the percentage increase from the initial amount each time, but also on the interest that has been added over the previous periods. In the problem, after 10 years of depreciation, the value of the investment begins to increase by 10% each year. Here is where compound interest comes into play:

• The formula used is:\[ P_f = P_i \times (1 + r)^n \]
• Where:
  • \(P_f\) is the final amount that we want to achieve—in this case, it is the original amount of \\(10,000,
  • \(P_i\) is the current principal, which has depreciated to \\)3,487,
  • \(r\) is the annual growth rate of 10%,
  • \(n\) is the number of years needed to return to the original value.

By calculating, we find that it takes a bit more than 11 years for the value to return to \$10,000, thanks to the power of compounding.

Logarithmic Equations

Solving problems involving growth and decay often requires the use of _logarithmic equations_. In Niki's example, logarithms help us find the exact number of years needed for the investment to regain its original value.After setting up the equation for compound interest: \(10,000 = 3,487 \times (1.10)^n\), we solve for \(n\) by taking logarithms:

• First, isolate the growth factor raised to a power.\[ (1.10)^n = \frac{10,000}{3,487} \]
• Next, take the logarithm of both sides:\[ n \times \log(1.10) = \log(2.868) \]
• Finally, divide by \(\log(1.10)\) to solve for \(n\):\[ n = \frac{\log(2.868)}{\log(1.10)} \approx 11.072 \]

This calculation shows that using logarithms efficiently helps find the time period involved when the rate and initial values are given.Logarithmic equations are useful tools for untangling the exponent in equations, showing us the time period over which changes occurred.