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Magazine Chapter 11: Problem 16
A bank account earns \(2 \%\) annual interest, compounded continuously. Money is deposited in a continuous cash flow at a rate of \(\$ 1200\) per year into the account. (a) Write a differential equation that describes the rate at which the balance \(B=f(t)\) is changing. (b) Solve the differential equation given an initial balance \(B_{0}=0.\) (c) Find the balance after 5 years.
Short Answer
Expert verified
The balance after 5 years is approximately $5712.
Step by step solution
01
Establish the Differential Equation
To describe the rate at which the balance \( B(t) \) is changing, we need to consider two contributions to the change in balance: the interest earned and the continuous deposit. The interest earned at any time \( t \) is \( 0.02B(t) \), since the interest rate is \( 2\% \) compounded continuously. The continuous cash flow adds \( 1200 \) per year. Therefore, the differential equation is given by: \[ \frac{dB}{dt} = 0.02B + 1200. \]
02
Solve the Differential Equation
The given differential equation is a first-order linear differential equation with the form \( \frac{dB}{dt} = aB + b \). We use the integrating factor method to solve it. The integrating factor \( \mu(t) \) is \[ e^{\int 0.02 \, dt} = e^{0.02t}. \]Multiply the entire differential equation by this factor:\[ e^{0.02t} \frac{dB}{dt} = 0.02 e^{0.02t} B + 1200 e^{0.02t}. \]Rewriting, we find:\[ \frac{d}{dt}(e^{0.02t} B) = 1200 e^{0.02t}. \]Integrate both sides:\[ e^{0.02t} B = \frac{1200}{0.02} e^{0.02t} + C = 60000 e^{0.02t} + C. \] Solve for \( B(t) \):\[ B(t) = 60000 + Ce^{-0.02t}. \]
03
Apply Initial Conditions
Using the initial condition \( B(0) = 0 \), substitute into the equation:\[ 0 = 60000 + C e^{0}. \]Thus, \( C = -60000 \). Therefore, the particular solution for \( B(t) \) is:\[ B(t) = 60000 - 60000 e^{-0.02t}. \]
04
Calculate the Balance After 5 Years
Substitute \( t = 5 \) into the expression for \( B(t) \):\[ B(5) = 60000 - 60000 e^{-0.1}. \]Calculate \( e^{-0.1} \) which is approximately \( 0.9048 \). Thus,\[ B(5) = 60000 - 60000 \times 0.9048 = 60000 - 54288 = 5712. \] The balance after 5 years is approximately \( 5712 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Continuous Compounding
Continuous compounding is a concept where interest on an account is calculated and added back to the principal continuously, thus earning interest on interest at an infinite rate. This means that every tiny fraction of time, your balance grows by a small amount.
Unlike traditional compounding, which occurs monthly or annually, continuous compounding offers a way to maximize the effect of growth. It uses the natural exponential function with the formula:
\[A = Pe^{rt}\]where:
Unlike traditional compounding, which occurs monthly or annually, continuous compounding offers a way to maximize the effect of growth. It uses the natural exponential function with the formula:
\[A = Pe^{rt}\]where:
- A is the amount,
- P is the principal amount,
- r is the annual interest rate,
- t is the time in years.
Continuous Cash Flow
Continuous cash flow means that money is being continually deposited into an account, rather than at discrete intervals. In this scenario, a rate, such as \( \$1200 \) per year, is continuously contributing to the balance.
This can be visualized as a faucet that pours a continuous stream of cash into your account! Unlike lump sum deposits, continuous cash flow changes the balance gradually, allowing for more consistent growth.
In calculations, this continuous addition simplifies to a constant in the differential equation, representing a steady rate of inflow.
This can be visualized as a faucet that pours a continuous stream of cash into your account! Unlike lump sum deposits, continuous cash flow changes the balance gradually, allowing for more consistent growth.
In calculations, this continuous addition simplifies to a constant in the differential equation, representing a steady rate of inflow.
Initial Conditions
Initial conditions are crucial in solving differential equations, as they allow us to find a particular solution to the equation that fits a specific scenario.
In our problem, the initial condition is the balance at time zero, expressed as \( B(0) = 0 \).
This tells us that the account starts with no initial money. By incorporating this into our differential equation, we are able to find the constant term that completes the specific equation for our scenario.
Initial conditions ensure that the mathematical model reflects the real-world situation accurately.
In our problem, the initial condition is the balance at time zero, expressed as \( B(0) = 0 \).
This tells us that the account starts with no initial money. By incorporating this into our differential equation, we are able to find the constant term that completes the specific equation for our scenario.
Initial conditions ensure that the mathematical model reflects the real-world situation accurately.
Integrating Factor
The integrating factor is a method used to solve first-order linear differential equations. It transforms the original equation into a simpler form that can be easily integrated.
For our differential equation,\[\frac{dB}{dt} = 0.02B + 1200\]we identify the integrating factor \( \mu(t) \) as:
\[\mu(t) = e^{\int 0.02 \, dt} = e^{0.02t}\]Multiplying through by this factor allows us to reframe the differential equation, ultimately leading to an expression that can be directly integrated.
This process helps simplify the solution path for complex equations like the one in this exercise, facilitating a precise solution.
For our differential equation,\[\frac{dB}{dt} = 0.02B + 1200\]we identify the integrating factor \( \mu(t) \) as:
\[\mu(t) = e^{\int 0.02 \, dt} = e^{0.02t}\]Multiplying through by this factor allows us to reframe the differential equation, ultimately leading to an expression that can be directly integrated.
This process helps simplify the solution path for complex equations like the one in this exercise, facilitating a precise solution.
Interest Rate
The interest rate is a percentage that indicates how much the principal will grow each year due to earned interest.
In continuous compounding, the interest rate provides constant growth influence over the balance. In our example, the interest rate is \(2\%\), meaning that annually the balance grows by that percentage.
In mathematical terms, it forms part of the exponential function in the integrating factor and across the differential equation, representing the impact of interest.
This rate is crucial, as it dictates the overall growth potential and, along with cash inflows, determines how quickly the account balance will increase over time.
In continuous compounding, the interest rate provides constant growth influence over the balance. In our example, the interest rate is \(2\%\), meaning that annually the balance grows by that percentage.
In mathematical terms, it forms part of the exponential function in the integrating factor and across the differential equation, representing the impact of interest.
This rate is crucial, as it dictates the overall growth potential and, along with cash inflows, determines how quickly the account balance will increase over time.
Balance Calculation
Balance calculation refers to solving for the total amount in the bank account after a certain period, considering initial conditions, continuous compounding, interest rates, and continuous cash flow.
In this problem, once the differential equation was solved, we substitute specific values (like \( t = 5 \)) to determine the account's balance.
The formula derived for the problem is:
\[B(t) = 60000 - 60000 e^{-0.02t}\]Plugging in \( t = 5 \) allowed us to calculate the balance at five years:
\[B(5) = 5712\]This computed value gives insight into how the account grows under the defined parameters, providing a complete understanding of the balance after the specified period.
In this problem, once the differential equation was solved, we substitute specific values (like \( t = 5 \)) to determine the account's balance.
The formula derived for the problem is:
\[B(t) = 60000 - 60000 e^{-0.02t}\]Plugging in \( t = 5 \) allowed us to calculate the balance at five years:
\[B(5) = 5712\]This computed value gives insight into how the account grows under the defined parameters, providing a complete understanding of the balance after the specified period.
