Problem 9 - Calculus and its applications

Question

If an amount \(P_{0}\) is invested in the Mandelbrot Bond Fund and interest is compounded continuously at \(5.9 \%\) per year, the balance \(P\) grows at the rate given by \(\frac{d P}{d t}=0.059 P\) a) Find the function that satisfies the equation. Write it in terms of \(P_{0}\) and 0.059 b) Suppose \(\$ 1000\) is invested. What is the balance after I yr? After 2 yr? c) When will an investment of \(\$ 1000\) double itself?

Step-by-Step Solution

Answer

a) The function is \( P(t) = P_0 e^{0.059t} \). b) After 1 year: \$1059.66, after 2 years: \$1122.14. c) Investment doubles in approximately 11.75 years.

Step 1: Solve the differential equation

The given differential equation is \( \frac{dP}{dt} = 0.059P \). This is a separable differential equation which can be solved by separating variables and integrating. Divide both sides by \( P \) and multiply both sides by \( dt \) to obtain \( \frac{1}{P} dP = 0.059 dt \). Integrate both sides to get \( \ln |P| = 0.059t + C \), where \( C \) is a constant of integration.

Step 2: Exponentiate to solve for P(t)

To solve for \( P \), exponentiate both sides to eliminate the natural logarithm: \( |P| = e^{0.059t + C} = e^{C}e^{0.059t} \). Let \( C' = e^{C} \), a new constant, so \( |P| = C'e^{0.059t} \). Since \( P \) is always positive in this context, we can drop the absolute value to get \( P = C'e^{0.059t} \).

Step 3: Determine the constant using initial condition

The initial condition is \( P(0) = P_0 \). Substituting \( t = 0 \) into the equation \( P = C'e^{0.059t} \), we have \( P_0 = C'e^{0} = C' \). Thus, the constant \( C' = P_0 \), and the function is \( P(t) = P_0 e^{0.059t} \).

Step 4: Calculate the balance after 1 year for $1000 initial investment

With \( P_0 = 1000 \), substitute into the function: \( P(1) = 1000 e^{0.059 \times 1} \). Simplifying gives \( P(1) = 1000 e^{0.059} \). Calculate this to find the balance after 1 year: \( P(1) \approx 1059.66 \).

Step 5: Calculate the balance after 2 years for $1000 initial investment

Using the function \( P(t) = 1000 e^{0.059t} \), substitute \( t = 2 \): \( P(2) = 1000 e^{0.059 \times 2} \). Simplify and compute \( P(2) = 1000 e^{0.118} \), which calculates to approximately \( 1122.14 \).

Step 6: Determine when the investment doubles with $1000 initial investment

The investment is doubled when \( P(t) = 2000 \). Using the equation \( 2000 = 1000 e^{0.059t} \), divide both sides by 1000 to get \( 2 = e^{0.059t} \). Take the natural logarithm of both sides to find \( \ln(2) = 0.059t \). Solve for \( t \): \( t = \frac{\ln(2)}{0.059} \), which approximately equals 11.75 years.

Key Concepts

Exponential Growth

Exponential growth is a fascinating concept where the value of anything increases by a constant percentage over a period of time. When we're dealing with continuous compounding, like in this exercise, the change isn't in simple linear steps. Instead, the principal grows at an ever-accelerating pace. This is because the interest gets continually reinvested, so the principal amount—including interest earned—is ever-growing.Exponential growth can be visualized with the formula:

• \[ P(t) = P_0 e^{rt} \]

Here, \( P(t) \) represents the future value, \( P_0 \) is the original principal, \( r \) is the rate of interest, and \( t \) is time.In our example, this means:\[ P(t) = P_0 e^{0.059t} \]indicating that with an initial investment of \( P_0 \) at an annual interest rate of 5.9%, the investment grows exponentially over time. The natural base \( e \) (approximately 2.718) is critical because it represents growth that is continuously compounded.

Differential Equations

The backbone of modeling growth and decay in continuous systems often lies within differential equations. In this exercise, the differential equation takes the form:

• \[ \frac{dP}{dt} = 0.059P \]

This tells us how the investment changes with time. The term on the left, \( \frac{dP}{dt} \), represents the rate of change of the investment over time, while the right side shows that this rate is proportional to the current amount \( P \). This is why it's called a first-order linear differential equation with constant coefficients. To find a solution, the process is straightforward but requires some integration steps. We rearrange it into:

• \[ \frac{1}{P} dP = 0.059 \, dt \]

Integrating both sides helps us form a solution, leading us to find the exponential growth function, which describes how our initial amount grows continuously at 5.9% per annum.

Investment Doubling Time

One of the compelling aspects of exponential growth is determining how quickly an investment can double. In mathematical terms, this is finding when:

• \( P(t) = 2P_0 \)

For this specific case with continuous compounding, it mandates the equation:

• \[ 2 = e^{0.059t} \]

Solving for \( t \) involves taking the natural logarithm of both sides:

• \[ \ln(2) = 0.059t \]
• \[ t = \frac{\ln(2)}{0.059} \]

This calculates to approximately 11.75 years. This means that under these conditions, it will take around 11.75 years for the initial investment to double in value. Methods like these highlight the mesmerizing acceleration of exponential growth over time.