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A certain college graduate borrows 8000 dollar to buy a car. The lender charges interest at an annual rate of 10 dollar % Assuming that interest is compounded continuously and that the borrower makes payments continuously at a constant annual rate \(k\), determine the payment rate \(k\) that is required to pay off the loan in 3 years. Also determine how much interest is paid during the 3 -year period.

Short Answer

Expert verified
Answer: The payment rate required to pay off the loan in 3 years is approximately 3545.49 dollars per year, and the total interest paid during this period is approximately 2636.46 dollars.

Step by step solution

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01

Calculate the total amount due after 3 years

To calculate the total amount due after 3 years, we will use the formula for continuous compounding: A = P * e^(r*t) Where: - A is the amount due after 3 years - P is the initial borrowed amount (8000 dollars) - r is the annual interest rate (10% = 0.10) - t is the time in years (3 years) - e is the base of the natural logarithm, approximately equal to 2.71828 A = 8000 * e^(0.10 * 3) Calculate the amount after 3 years: A ≈ 8000 * e^(0.3) ≈ 8000 * 2.71828^(0.3) ≈ 10636.46
02

Determine payment rate k

Now that we know the total amount due after 3 years, we can determine the payment rate k. We will set up the following equation to find k: Total amount / Total time = k 10636.46 / 3 = k Calculate the payment rate k: k ≈ 10636.46 / 3 ≈ 3545.49 dollars per year
03

Calculate the total interest paid

To calculate the total interest paid, we will subtract the initial loan amount from the total amount due after 3 years: Total interest = Total amount - Initial loan amount Total interest ≈ 10636.46 - 8000 ≈ 2636.46 dollars So, the payment rate required to pay off the loan in 3 years is approximately 3545.49 dollars per year, and the total interest paid during this period is approximately 2636.46 dollars.

Key Concepts

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

Differential Equations
Differential equations are at the heart of modeling real-world phenomena in various scientific disciplines, including economics, physics, and biology. Essentially, a differential equation is a mathematical equation that relates some function with its derivatives. In the context of continuously compounded interest, a differential equation helps us describe how the balance of the loan changes over time.

For example, if we denote the balance of the loan at any time as a function, say, \( B(t) \), where \( t \) is time, then its rate of change (derivative), \( B'(t) \), is proportional to the balance itself, following the natural exponential growth pattern, with the interest rate as the proportionality constant. This can be written as \( B'(t) = r \times B(t) \), which is a simple form of a differential equation. When we solve this equation, we get the function that describes how the balance grows over time due to interest alone, without considering payments being made.

The significance of understanding how to formulate and solve these equations cannot be overstated. They enable students to predict future values based on current trends and to calculate quantities like payoff time or interest paid, as seen in the exercise problem.
Exponential Growth
Exponential growth denotes the increase of a quantity at a rate that is proportional to its current value. This type of growth is characterized by the presence of a constant base raised to a variable exponent, as shown by the formula \( A = Pe^{rt} \), defining continuous compounding interest in finance. The base of the natural logarithm, \( e \), is vital for these calculations.

This concept explains why the amount owed on a loan can grow so quickly over time when the interest is compounded continuously. The exponential nature of the growth means small changes in the interest rate or time can have large effects on the total amount owed. In the exercise, the use of exponential growth allows us to determine the future balance of a loan, accounting for the continuous accumulation of interest, which leads to calculating the payments needed to pay off the loan over a specific period. The simplicity and power of exponential growth models make them incredibly useful for planning and predicting financial scenarios.
Natural Logarithm
The natural logarithm, commonly denoted as \( ln \), is the logarithm to the base \( e \), where \( e \) (approximately 2.71828) is an irrational and transcendental number known as Euler's number. The natural logarithm is the inverse operation of taking the power of \( e \), which means that if \( e^y = x \), then \( ln(x) = y \).

It plays a pivotal role when working with exponential growth and continuous compounding interest. Understanding the properties of the natural logarithm allows one to manipulate equations involving exponential terms in a highly effective way. For instance, when we are dealing with the formula for continuously compounded interest, we sometimes need to solve for the time \( t \) or the rate \( r \), which requires taking the natural logarithm of both sides of the equation.

In the context of our loan repayment problem, the natural logarithm does not appear directly, but the understanding of its properties is implicit in managing the compound interest formula. This knowledge could also be used to improve the search for an exact payment rate or in adjustments to the payment plan, to correspond with changes in interest rates or repayment periods.

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Most popular questions from this chapter

Show that the equations are not exact, but become exact when multiplied by the given integrating factor. Then solve the equations. $$ (x+2) \sin y d x+x \cos y d y=0, \quad \mu(x, y)=x e^{x} $$

Suppose that a certain population has a growth rate that varies with time and that this population satisfies the differential equation $$ d y / d t=(0.5+\sin t) y / 5 $$ $$ \begin{array}{l}{\text { (a) If } y(0)=1, \text { find (or estimate) the time } \tau \text { at which the population has doubled. Choose }} \\ {\text { other initial conditions and determine whether the doubling time } \tau \text { depends on the initial }} \\ {\text { population. }} \\ {\text { (b) Suppose that the growth rate is replaced by its average value } 1 / 10 . \text { Determine the }} \\ {\text { doubling time } \tau \text { in this case. }}\end{array} $$ $$ \begin{array}{l}{\text { (c) Suppose that the term sin } t \text { in the differential equation is replaced by } \sin 2 \pi t \text { ; that is, }} \\\ {\text { the variation in the growth rate has a substantially higher frequency. What effect does this }} \\ {\text { have on the doubling time } t ?} \\ {\text { (d) Plot the solutions obtained in parts (a), (b), and (c) on a single set of axes. }}\end{array} $$

we indicate how to prove that the sequence \(\left\\{\phi_{n}(t)\right\\},\) defined by Eqs. (4) through (7), converges. Note that $$ \phi_{n}(t)=\phi_{1}(t)+\left[\phi_{2}(t)-\phi_{1}(t)\right]+\cdots+\left[\phi_{x}(t)-\phi_{n-1}(t)\right] $$ (a) Show that $$ \left|\phi_{n}(t)\right| \leq\left|\phi_{1}(t)\right|+\left|\phi_{2}(t)-\phi_{1}(t)\right|+\cdots+\left|\phi_{n}(t)-\phi_{n-1}(t)\right| $$ (b) Use the results of Problem 17 to show that $$ \left|\phi_{n}(t)\right| \leq \frac{M}{K}\left[K h+\frac{(K h)^{2}}{2 !}+\cdots+\frac{(K h)^{n}}{n !}\right] $$ (c) Show that the sum in part (b) converges as \(n \rightarrow \infty\) and, hence, the sum in part (a) also converges as \(n \rightarrow \infty\). Conclude therefore that the sequence \(\left\\{\phi_{n}(t)\right\\}\) converges since it is the sequence of partial sums of a convergent infinite series.

Harvesting a Renewable Resource. Suppose that the population \(y\) of a certain species of fish (for example, tuna or halibut) in a given area of the ocean is described by the logistic equation $$ d y / d t=r(1-y / K) y . $$ While it is desirable to utilize this source of food, it is intuitively clear that if too many fish are caught, then the fish population may be reduced below a useful level, and possibly even driven to extinction. Problems 20 and 21 explore some of the questions involved in formulating a rational strategy for managing the fishery. At a given level of effort, it is reasonable to assume that the rate at which fish are caught depends on the population \(y:\) The more fish there are, the easier it is to catch them. Thus we assume that the rate at which fish are caught is given by \(E y,\) where \(E\) is a positive constant, with units of \(1 /\) time, that measures the total effort made to harvest the given species of fish. To include this effect, the logistic equation is replaced by $$ d y / d t=r(1-y / K) y-E y $$ This equation is known as the Schaefer model after the biologist, M. B. Schaefer, who applied it to fish populations. (a) Show that if \(E0 .\) (b) Show that \(y=y_{1}\) is unstable and \(y=y_{2}\) is asymptotically stable. (c) A sustainable yield \(Y\) of the fishery is a rate at which fish can be caught indefinitely. It is the product of the effort \(E\) and the asymptotically stable population \(y_{2} .\) Find \(Y\) as a function of the effort \(E ;\) the graph of this function is known as the yield-effort curve. (d) Determine \(E\) so as to maximize \(Y\) and thereby find the maximum sustainable yield \(Y_{m}\).

Draw a direction field for the given differential equation and state whether you think that the solutions are converging or diverging. $$ y^{\prime}=-t y+0.1 y^{3} $$

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