Problem 10 - Calculus and its applications

Question

If an amount \(P_{0}\) is invested in a savings account and interest is compounded continuously at \(4.3 \%\) per year, the balance \(P\) grows at the rate given by \(\frac{d p}{d t}=0.043 P\) a) Find the function that satisfies the equation. Write it in terms of \(P_{0}\) and \(0.043 .\) b) Suppose \(\$ 20,000\) is invested. What is the balance after 1 yr? After 2 yr? c) When will an investment of \(\$ 20,000\) double itself?

Step-by-Step Solution

Answer

The balance function is \( P(t) = P_0 e^{0.043t} \). For \$20,000, after 1 year \(\approx\) \$20,878, after 2 years \(\approx\) \$21,796. Investment doubles in \(\approx\) 16.13 years.

Step 1: Understand the Continuous Compounding Formula

Continuous compounding means that the interest is added to the initial amount at an infinite number of times per year, leveraging the formula for continuous growth: \( P(t) = P_0 e^{rt} \). Here, \( r \) is the growth rate, which in this case is \( 0.043 \) or \(4.3\%\).

Step 2: Solve Differential Equation for Part (a)

The differential equation given is \( \frac{dP}{dt} = 0.043P \). This is a standard exponential growth problem. Integrating both sides, we use the fact that \( \int \frac{1}{P} dP = \int 0.043 dt \), leading to \( \ln |P| = 0.043t + C \). Solving for \( P \), we exponentiate to get \( P = e^{0.043t+C} = e^{C}e^{0.043t} = P_0 e^{0.043t} \). Thus, \( P(t) = P_0 e^{0.043t} \).

Step 3: Part (b): Calculate Balance After 1 and 2 Years

Using the function \( P(t) = 20000 e^{0.043t} \):- After 1 year: \( P(1) = 20000 e^{0.043 \cdot 1} \approx 20000 \cdot 1.0439 \approx 20878 \).- After 2 years: \( P(2) = 20000 e^{0.043 \cdot 2} \approx 20000 \cdot 1.0898 \approx 21796 \).

Step 4: Part (c): Determine Doubling Time

To find when the investment doubles, set \( P(t) = 2P_0 \). Using \( P(t) = 20000 e^{0.043t} \), set \( 40000 = 20000 e^{0.043t} \). Simplifying, equate \( 2 = e^{0.043t} \). Taking the natural logarithm of both sides gives \( \ln 2 = 0.043t \). Solving for \( t \), \( t = \frac{\ln 2}{0.043} \approx 16.1265 \). This means the investment doubles in approximately 16.13 years.

Key Concepts

Exponential Growth

Exponential growth is a fascinating concept that describes how a quantity increases rapidly in proportion to its current value. This type of growth is often seen in populations, finance, and physics. The key feature of exponential growth is that the rate of increase is proportional to the current value.

In the context of finance and interest, exponential growth is closely tied to continuous compounding. When interest is compounded continuously, the value of an investment grows according to the function:\[ P(t) = P_0 e^{rt} \]Here, \( P_0 \) is the initial investment, \( r \) is the rate of growth or interest rate, and \( t \) is the time. The function shows that as time progresses, the investment grows exponentially because of the constant compounding of interest.

By using the exponential growth model, you can predict the future value of investments more accurately. This helps both investors and financial analysts to make better decisions by understanding how investments can grow over time.

Differential Equation

A differential equation is a mathematical equation that involves the derivatives of a function. These equations are used to describe various natural and physical phenomena, including growth processes. The differential equation presented in this exercise is:\[ \frac{dP}{dt} = 0.043P \]This particular equation is a great representation of exponential growth, where the rate of change of the investment \( P \) with respect to time \( t \) depends proportionally on the current amount \( P \).

The solution to a differential equation often involves integration. In this case, by integrating both sides, you determine the general solution for \( P(t) \). Through integration, it was shown that:\[ P(t) = P_0 e^{0.043t} \]The solution indicates how an investment grows over time with continuous compounding. Understanding such equations can be crucial in fields like physics, biology, and economics, where predicting future behavior based on current rates and states is necessary.

Interest Calculation

Interest calculation is a fundamental concept in finance, where it determines how much extra money can be earned over time on an initial amount of investment. There are different types of interest calculations; one of the most impactful is continuous compounding.

In continuous compounding, the interest accumulates constantly rather than at discrete intervals (like annually, semi-annually, etc.). This results in higher returns because the investment balance is compounded on an ongoing basis. The formula used for continuous compounding is:\[ P(t) = P_0 e^{rt} \]This exercise specifically considered an initial investment of $20,000 with a 4.3% annual interest rate. By substituting these values into the formula:- After 1 year: The balance was calculated as \( 20878 \) by plugging \( t = 1 \) into the formula.- After 2 years: The balance increased to \( 21796 \).- Doubling time: The problem also explored when the investment would double. Using logarithms, it was found to double around 16.13 years.

Understanding interest calculations through continuous compounding allows investors to better appreciate how their investments grow over time, benefiting from every moment of compounded growth.