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A person planning for her retirement arranges to make continuous deposits into a savings account at the rate of $$\$ 3600$$ per year. The savings account earns \(5 \%\) interest compounded continuously. (a) Set up a differential equation that is satisfied by \(f(t)\), the amount of money in the account at time \(t\). (b) Solve the differential equation in part (a), assuming that \(f(0)=0\), and determine how much money will be in the account at the end of 25 years.

Short Answer

Expert verified
The differential equation is \( \frac{df(t)}{dt} = 3600 + 0.05f(t) \). The amount of money in the account at the end of 25 years is approximately $$51,372$$.

Step by step solution

01

Define the differential equation for the balance

Let the amount of money in the account at time \(t\) be \(f(t)\). The rate of change of the money in the account consists of two parts: the continuous deposits at the rate of \(3600\) per year and the interest at the rate of \(5\%\) per year compounded continuously. Thus, the differential equation can be written as: \[ \frac{df(t)}{dt} = 3600 + 0.05f(t) \]
02

Solve the differential equation

Rewrite the differential equation in a standard linear form: \[ \frac{df(t)}{dt} - 0.05f(t) = 3600 \] To solve this, we use an integrating factor \(\mu(t)\), which is given by \[ \mu(t) = e^{-\int 0.05 dt} = e^{-0.05t}. \] Multiply both sides of the differential equation by \(e^{-0.05t}\) to get: \[ e^{-0.05t} \frac{df(t)}{dt} - 0.05 e^{-0.05t} f(t) = 3600 e^{-0.05t} \] This simplifies to: \[ \frac{d}{dt}[f(t)e^{-0.05t}] = 3600e^{-0.05t}. \]
03

Integrate both sides

Integrate both sides of the equation: \[ \int \frac{d}{dt}[f(t)e^{-0.05t}] dt = \int 3600e^{-0.05t} dt \] This gives: \[ f(t)e^{-0.05t} = -3600 \left( \frac{e^{-0.05t}}{0.05} \right) + C \] Simplify the right side to obtain: \[ f(t)e^{-0.05t} = -72000 e^{-0.05t} + C \]
04

Solve for the constant of integration

Given that \(f(0) = 0\), substitute \(t = 0\) into the equation: \[ 0 = -72000 e^{0} + C \] Therefore, \[ C = 72000. \] The solution to the differential equation is then: \[ f(t)e^{-0.05t} = -72000 e^{-0.05t} + 72000. \]
05

Solve for \(f(t)\)

Multiply both sides by \(e^{0.05t}\) to isolate \(f(t)\): \[ f(t) = 72000 (1 - e^{-0.05t}). \]
06

Determine the amount of money in the account at \(t = 25\) years

Substitute \(t = 25\) into the solution for \(f(t)\): \[ f(25) = 72000 (1 - e^{-0.05 \cdot 25}) = 72000 (1 - e^{-1.25}). \] Using a calculator, \[ e^{-1.25} \approx 0.2865. \] Therefore, \[ f(25) = 72000 (1 - 0.2865) = 72000 \times 0.7135 = 51372. \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

retirement savings
When planning for your retirement, one of the key elements to consider is how much money you will need to save to ensure a comfortable retirement. There are a variety of strategies to save for retirement, but one consistent method is to make continuous deposits into a savings account over the years. This approach means regularly putting aside a fixed amount of money to grow over time.

Think of retirement savings like planting a tree. You water it regularly (continuous deposits) and over time, it grows (with interest) to give you shade when you need it (retirement).

Taking the example from the exercise, suppose an individual decides to deposit $$3600$$ per year into a savings account. Understanding the growth of this savings over time can be grounded in the principles of differential equations and continuous compound interest.
compound interest
Compound interest is a powerful concept that significantly impacts long-term savings. It refers to the interest earned not only on the initial principal but also on the accumulated interest from previous periods.

In the retirement exercise, the account earns interest at a continuous rate of 5% per year. This continuous compounding can be described by a differential equation. The continuous nature means that interest is constantly calculated and added to the account balance, creating a scenario where money grows exponentially over time.

Mathematically, this continuous compounding is expressed as part of the differential equation \(\frac{df(t)}{dt} = 3600 + 0.05f(t)\). Here, \(\frac{df(t)}{dt}\) represents the rate of change of the account balance at time \( t \). The term 0.05\(f(t)\) shows how the existing balance grows at an annual rate of 5% due to interest. By solving this equation, one can predict the future value of the retirement savings.
continuous deposits
Continuous deposits refer to the regular and systematic addition of funds into an account over time. In the context of retirement planning, making continuous deposits helps build a substantial amount of savings by retirement.

The example provided involves depositing $$3600$$ every year consistently. The continuity of these deposits, combined with compound interest, ensures an exponential growth of savings. By setting up a differential equation \( \frac{df(t)}{dt} = 3600 + 0.05f(t) \), one can capture the impact of these continuous deposits.

Solving this equation involves several steps, including finding the integrating factor and applying boundary conditions like \( f(0) = 0 \) to arrive at the solution \( f(t) = 72000 (1 - e^{-0.05t}) \). This solution models how the amount of money in the account grows over time.

After understanding the equation and the continuous deposit model, it's possible to compute the final amount in the savings account. For instance, at the end of 25 years, the amount can be calculated using \( f(25) = 72000 (1 - e^{-1.25}) \), which gives sturdy evidence of how continuous deposits coupled with compound interest work in synergetic growth.

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Most popular questions from this chapter

The National Automobile Dealers Association reported that the average retail selling price of a new vehicle was $$\$ 30,303$$ in 2012 . A person purchased a new car at the average price and financed the entire amount. Suppose that the person can only afford to pay $$\$ 500$$ per month. Assume that the payments are made at a continuous annual rate and that interest is compounded continuously at the rate of \(3.5 \%\). (Source: The National Automobile Dealers Association, www.nada.com.) (a) Set up a differential equation that is satisfied by the amount \(f(t)\) of money owed on the car loan at time \(t\). (b) How long will it take to pay off the car loan?

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Solve the following differential equations with the given initial conditions. $$ \frac{d y}{d x}=\frac{\ln x}{\sqrt{x y}}, y(1)=4 $$

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