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A better investment: You open an account by investing \(\$ 250\) with a financial institution that advertises an APR of \(5.75 \%\), with continuous compounding. a. Find an exponential formula for the balance in your account as a function of time. In your answer, give both the standard form and the alternative form for an exponential function. b. What account balance would you expect 5 years after your initial investment? Answer this question using both of the forms you found in part a. Which do you think gives a more accurate answer? Why?

Short Answer

Expert verified
The exponential formula is \( A(t) = 250 e^{0.0575t} \). After 5 years, the balance is \$333.25, same for both forms; both forms give equally accurate answers.

Step by step solution

01

Understand the Continuous Compounding Formula

The formula for the balance with continuous compounding interest is given by \[ A(t) = P e^{rt} \]where \( A(t) \) is the amount after time \( t \), \( P \) is the principal amount (initial investment), \( r \) is the annual interest rate (as a decimal), and \( t \) is the time in years. Here, \( P = 250 \), \( r = 0.0575 \), and \( t \) is variable.
02

Write the Exponential Formula (Standard Form)

Substitute the given values into the continuous compounding formula: \[ A(t) = 250 e^{0.0575t} \]This is the exponential formula in its standard form, with \( e^{0.0575t} \) representing the growth factor as a function of time.
03

Write the Exponential Formula (Alternative Form)

The alternative form of an exponential function can be expressed using the property \( e^{x} = (1+x/n)^n \) as \( n \to \infty \). For continuous compounding, the alternative form mirrors the standard form: \[ A(t) = 250 e^{(0.0575 imes t)} \]Both forms are effectively identical for continuous growth.
04

Calculate the Account Balance after 5 Years using Standard Form

For the standard form formula, substitute \( t = 5 \) into the equation: \[ A(5) = 250 e^{0.0575 imes 5} \]Using \( e^{0.2875} \), we calculate \[ A(5) \approx 250 imes 1.333 \approx 333.25 \]Thus, the balance after 5 years is approximately \$333.25 using the standard form.
05

Calculate the Account Balance after 5 Years using Alternative Form

For the alternative form, the calculation remains the same as the standard form: \[ A(5) = 250 e^{0.0575 imes 5} \approx 333.25 \]The alternative form supports the same mathematical operation for a continuous compounding scenario, confirming the result is \$333.25.
06

Determine Which Form Gives a More Accurate Answer

Both the standard and alternative forms yield the same result for continuous compounding. In cases of continuous growth, there is no difference between these forms; hence, they are both equally accurate and precise.

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Key Concepts

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

Exponential Growth
Exponential growth describes a process that increases rapidly over time. It is characterized by the quantity growing at a rate proportional to its current value. In the context of investments, exponential growth means that the more money you have, the more it will grow. This is modeled by mathematical functions where the rate of increase becomes faster as time goes on.

The equation for exponential growth is typically written as \( A(t) = A_0 e^{rt} \), where:
  • \( A(t) \) is the amount of growth after time \( t \).
  • \( A_0 \) is the initial amount or principal.
  • \( r \) is the growth rate.
  • \( t \) is the time period.
This equation shows how exponential growth reflects any situation where quantities increase quickly over time.
Compound Interest
Compound interest is a powerful financial concept that allows your money to grow faster than simple interest. Instead of earning interest only on the initial investment, compound interest ensures that the interest earned is added to the principal. This way, you earn interest on both the original amount and the accumulated interest.

When interest is compounded, the balance is calculated using the principal plus the accumulated interest. The formula for continuous compounding is a special case where interest is compounded infinitely throughout the time period, maximizing the growth potential.

This creates a snowball effect, where earnings themselves generate more earnings. It's a fundamental principle for understanding how investments grow over time.
APR (Annual Percentage Rate)
The Annual Percentage Rate (APR) is a crucial concept that represents the annualized credit cost or investment gain expressed as a percentage. It includes any fees or additional costs associated with the financial product. In investments, the APR is often referred to as the annual interest rate.

APR gives you an idea of the potential yearly earning on an investment. However, it doesn't tell the whole story when it comes to continuous compounding as it assumes compounding occurs at discrete intervals. In continuous compounding, the effective annual rate effectively becomes higher than the stated APR due to the constant compounding.

Thus, while APR provides a baseline, it's important to understand how compounding can lead to different effective returns.
Continuous Growth Formula
The continuous growth formula is fundamental when discussing continuous compounding. Unlike traditional compounding that occurs at regular intervals, continuous compounding assumes the reinvestment of earnings constantly at every possible instant over time.

The formula is expressed as \( A(t) = P e^{rt} \). Here's the breakdown:
  • \( P \) is the principal, or initial amount invested.
  • \( e \) is the base of the natural logarithm, an irrational constant approximately equal to 2.71828.
  • \( r \) is the annual interest rate.
  • \( t \) is the time the money is invested for.
By using continuous compounding, the formula offers the maximum possible return from an investment, illustrating the purest form of exponential growth in financial contexts. This approach is especially useful for understanding how investments can grow consistently and rapidly over time without gaps in the compounding process.

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Most popular questions from this chapter

Making up a story about a car trip: You begin from home on a car trip. Initially your velocity is a small positive number. Shortly after you leave, velocity decreases momentarily to zero. Then it increases rapidly to a large positive number and remains constant for this part of the trip. After a time, velocity decreases to zero and then changes to a large negative number. a. Make a graph of velocity for this trip. b. Discuss your distance from home during this driving event, and make a graph. c. Make up a driving story that matches this description.

Equation of change for logistic growth: The logistic growth formula \(N=6.21 /\left(0.035+0.45^{t}\right)\) that we used in Chapter 2 for deer on the George Reserve actually came from the following equation of change: $$ \frac{d N}{d t}=0.8 N\left(1-\frac{N}{177}\right) \text {. } $$ a. Plot the graph of \(\frac{d N}{d t}\) versus \(N\), and use it to find the equilibrium solutions. Explain their physical significance. b. In one plot, sketch the equilibrium solutions and graphs of \(N\) versus \(t\) in each of the two cases \(N(0)=10\) and \(N(0)=225 .\) c. To what starting value for \(N\) does the solution \(N=6.21 /\left(0.035+0.45^{t}\right)\) correspond?

Hiking: You are hiking in a hilly region, and \(E=\) \(E(t)\) is your elevation at time \(t\). a. Explain the meaning of \(\frac{d E}{d t}\) in practical terms. b. Where might you be when \(\frac{d E}{d t}\) is a large positive number? c. You reach a point where \(\frac{d E}{d t}\) is briefly zero. Where might you be? d. Where might you be when \(\frac{d E}{d t}\) is a large negative number?

Experimental determination of the drag coefficient: When retardation due to air resistance is proportional to downward velocity \(V\), in feet per second, falling objects obey the equation of change $$ \frac{d V}{d t}=32-r V $$ where \(r\) is known as the drag coefficient. One way to measure the drag coefficient is to measure and record terminal velocity. a. We know that an average-size man has a terminal velocity of 176 feet per second. Use this to show that the value of the drag coefficient is \(r=0.1818\) per second. (Hint: To say that the terminal velocity is 176 feet per second means that when the velocity \(V\) is 176 , velocity will not change. That is, \(\frac{d V}{d t}=0\). Put these bits of information into the equation of change and solve for \(r\).) b. An ordinary coffee filter has a terminal velocity of about 4 feet per second. What is the drag coefficient for a coffee filter?

Looking up: The constant \(g=32\) feet per second per second is the downward acceleration due to gravity near the surface of the Earth. If we stand on the surface of the Earth and locate objects using their distance up from the ground, then the positive direction is up, so down is the negative direction. With this perspective, the equation of change in velocity for a freely falling object would be expressed as $$ \frac{d V}{d t}=-g $$ (We measure upward velocity \(V\) in feet per second and time \(t\) in seconds.) Consider a rock tossed up- ward from the surface of the Earth with an initial velocity of 40 feet per second upward. a. Use a formula to express the velocity \(V=V(t)\) as a linear function. (Hint: You get the slope of \(V\) from the equation of change. The vertical intercept is the initial value.) b. How many seconds after the toss does the rock reach the peak of its flight? (Hint: What is the velocity of the rock when it reaches its peak?) c. How many seconds after the toss does the rock strike the ground? (Hint: How does the time it takes for the rock to rise to its peak compare with the time it takes for it to fall back to the ground?)

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