Problem 7 - Applied Calculus

Question

A deposit of \(\$ 5000\) is made to a bank account paying \(1.5 \%\) annual interest, compounded continuously. (a) Write a differential equation for the balance in the account, \(B\), as a function of time, \(t\), in years. (b) Solve the differential equation. (c) How much money is in the account in 10 years?

Step-by-Step Solution

Answer

(a) \( \frac{dB}{dt} = 0.015B \); (b) \( B(t) = 5000 \cdot e^{0.015t} \); (c) \(\$ 5809\).

Step 1: Express the Differential Equation

To express the differential equation for the balance \(B\) as a function of time \(t\) for continuous compounding interest, we use the formula \( \frac{dB}{dt} = rB \). Here, \(r\) is the interest rate expressed as a decimal. With an interest rate of \(1.5\%\), we have \(r=0.015\). Therefore, the differential equation is \( \frac{dB}{dt} = 0.015B \).

Step 2: Solve the Differential Equation

The differential equation \( \frac{dB}{dt} = 0.015B \) is a separable differential equation, which can be solved by integrating both sides. Separating the variables gives \( \frac{dB}{B} = 0.015 \, dt \). Integrating both sides, we get \( \int \frac{1}{B} \, dB = \int 0.015 \, dt \), which results in \( \ln |B| = 0.015t + C \), where \(C\) is the constant of integration.

Step 3: Solve for B(t)

To find \(B(t)\), we exponentiate both sides of \( \ln |B| = 0.015t + C \). This gives \( B = e^{0.015t + C} = e^C \cdot e^{0.015t} \). Let \(A = e^C\), a new constant representing the initial account balance. Therefore, \(B(t) = A \cdot e^{0.015t} \).

Step 4: Determine the Initial Condition

We are given an initial deposit of \( \$5000 \), so at \(t=0\), \(B(0) = 5000\). Substituting into \(B(t) = A \cdot e^{0.015t}\) gives \(5000 = A \cdot e^0 = A\). Hence, \(A = 5000\). So, \(B(t) = 5000 \cdot e^{0.015t}\).

Step 5: Compute the Balance After 10 Years

Substitute \(t = 10\) into \(B(t) = 5000 \cdot e^{0.015t}\) to find the balance after 10 years: \(B(10) = 5000 \cdot e^{0.015 \times 10}\). Calculate \(e^{0.15} \approx 1.1618\), so \(B(10) = 5000 \cdot 1.1618 \approx 5809\). Therefore, the balance in the account after 10 years is approximately \(\$5809\).

Key Concepts

Continuous Compounding

Continuous compounding is a concept in finance where interest is added to an account balance at every possible instant. Unlike compounding annually or monthly, continuous compounding results in the most frequent application of interest. This means:

• Your money grows at a faster rate.
• The formula used to describe this process is an exponential function.

The formula for continuous compounding can be expressed as:\[ B(t) = P \cdot e^{rt} \]where:

• \( B(t) \) is the balance after time \( t \)
• \( P \) is the principal, or initial amount
• \( r \) is the annual interest rate expressed as a decimal
• \( t \) is the time in years
• \( e \) is Euler's number (approximately 2.71828)

For instance, with a principal of \(\$5000\) and a rate of \(1.5\%\), your balance grows over time via the expression \( B(t) = 5000 \cdot e^{0.015\,t} \). This represents the power of continuous compounding in maximizing your investment's growth.

Separable Differential Equations

Separable differential equations are a class of differential equations that can be broken down, or separated, into two individual integrals. The goal here is to isolate the dependent variable from the independent variable to solve the equation. When dealing with an equation such as \[ \frac{dB}{dt} = rB \]we can reformulate it by separating variables. Essentially, we get the form:\[ \frac{dB}{B} = r\,dt \]This separation allows:

• The left side to include terms related to \(B\).
• The right side to feature terms dependent on \(t\).
• Integration of both sides to solve the equation.

Once separated, the solution involves:

• Calculating the integral of \( \frac{1}{B} \, dB \), which results in \( \ln|B| \).
• Integrating \( r \, dt \) to obtain \( rt + C \), with \(C\) as the integration constant.
• Exponentiating to solve for \( B(t) \), leading to the general solution \( B(t) = A \cdot e^{rt} \).

This methodology is applied in scenarios like continuous compounding to determine how balances evolve over time.

Exponential Growth

Exponential growth is a pattern of data that shows greater increases with passing time, in proportion to the growing total number or size. This growth type plays a crucial role in various fields such as finance, biology, and physics. For instance, in a bank account where interest compounds continuously, the account balance does not grow linearly over time:

• Instead, initial amounts grow slowly at first.
• As time progresses, the balance increases at an accelerating rate.

The mathematical representation of exponential growth is an exponential function, such as:\[ B(t) = A \cdot e^{rt} \]In this function:

• \( A \) denotes the initial amount or balance.
• \( r \) is the growth rate.
• \( t \) is time.

With a starting balance of \(\$5000\) at an interest rate of \(1.5\%\), the function becomes \( 5000 \cdot e^{0.015t} \). Doubling, tripling, and eventually tenfold increases are smoother because of the continuous application of this growth, effectively illustrating the concept's core characteristic: the potential for rapid increases over time.