91Ó°ÊÓ

Solving the differential equations that arise from modeling may require using integration by parts. [See formula (1).] After depositing an initial amount of \(\$ 10,000\) in a savings account that earns \(4 \%\) interest compounded continuously, a person continued to make deposits for a certain period of time and then started to make withdrawals from the account. The annual rate of deposits was given by \(3000-500 t\) dollars per year, \(t\) years from the time the account was opened. (Here, negative rates of deposits correspond to withdrawals.) (a) How many years did the person contribute to the account before starting to withdraw money from it? (b) Let \(P(t)\) denote the amount of money in the account, \(t\) years after the initial deposit. Find an initial-value problem satisfied by \(P(t)\). (Assume that the deposits and withdrawals were made continuously.)

Step by step solution

01

Find the Time When Deposits End

The person contributes to the account as long as the annual rate of deposit is non-negative. The annual rate of deposits is given by the equation: Set the annual rate to zero and solve for t:3000 - 500t = 0

02

Solve for t

Solve the equation 3000 - 500t = 0 to find the time when the deposits end:\[ 3000 - 500t = 0 \]\[ 500t = 3000 \]\[ t = \frac{3000}{500} \]\[ t = 6 \]Thus, the person contributed to the account for 6 years before starting to withdraw money from it.

03

Set Up The Differential Equation

The amount of money in the account, P(t), is affected by the continuous compounding and the deposits. This can be modeled by the differential equation: \[ \frac{dP}{dt} = 0.04P(t) + R(t) \] where R(t) is the annual rate of deposits.R(t) = 3000 - 500t for 0 ≤ t ≤ 6 and 0 for t > 6.

04

Setup the Initial Condition

The initial amount deposited in the account is given as \( P(0) = 10,000 \).

05

Construct the Initial Value Problem

The initial value problem can be summarized as follows:For 0 ≤ t ≤ 6, \[ \frac{dP}{dt} = 0.04P(t) + 3000 - 500t \] with \( P(0) = 10,000 \). For t > 6, the rate of deposit R(t) is zero.For t > 6, \[ \frac{dP}{dt} = 0.04P(t) \] with initial value from the solution at t = 6.Thus, the initial value problem captures the amount in the account both during the deposit and withdrawal phases.

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Key Concepts

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

Continuous Compounding Interest

Interest that is compounded continuously is a powerful concept in finance. It means that the interest is added to the principal balance exponentially and constantly, rather than at discrete intervals. This results in a higher amount of interest being earned over time compared to compound interest that is calculated periodically. The formula used is:
\[ P(t) = P_0 e^{rt} \] where:

• \( P(t) \) is the amount of money after time \( t \)
• \( P_0 \) is the initial principal or deposit amount
• \( r \) is the annual interest rate
• \( t \) is the time in years

With continuous compounding, even small interest rates can lead to significant growth over long periods.

Initial-Value Problem

An initial-value problem (IVP) in differential equations involves finding a function that satisfies a given differential equation and meets an initial condition specified at some point. For example, in financial modeling, we often want to find the amount \( P(t) \) in an account given by a differential equation, along with the initial amount deposited.
The steps are:

• Identify the differential equation that models the problem.
• Provide an initial value, such as the amount deposited at \( t = 0 \).
• Solve the differential equation to find the function \( P(t) \).

In our specific problem, we set up the initial-value problem for the amount in an account with an initial deposit of $10,000 and additional deposit/withdrawal rates given by \( 3000 - 500t \).

Integration by Parts

Integration by parts is a technique used to evaluate integrals where the standard methods of integration are difficult to apply. It is particularly useful when dealing with the product of two functions. The integration by parts formula is:

• \[ \int u \,dv = uv - \int v \,du \]

where

• \( u \) and \( dv \) are differentiable functions of \( x \)
• \( du \) and \( v \) are their respective derivatives and antiderivatives

In the context of our financial modeling exercise, integration by parts might be employed to solve the differential equations that describe the growth of money in the account, especially when the deposit and withdrawal rates need to be integrated over time. Understanding how to apply this technique is crucial for solving such problems. Make sure to choose \( u \) and \( dv \) wisely to simplify the integral.