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Chapter 5: Problem 13

If an investment triples in 15 years, what yearly interest rate (compounded continuously) does the investment earn?

Short Answer

Expert verified
The yearly interest rate is approximately 7.32%.

Step by step solution

01

Understand the Continuous Compounding Formula

The formula for continuous compounding is given by \( A = P e^{rt} \), where \(A\) is the amount of money accumulated after n years, including interest. \(P\) is the principal amount (the initial amount of money), \(r\) is the annual interest rate (as a decimal), \(t\) is the time the money is invested for in years, and \(e\) is the base of the natural logarithm.
02

Set up the Equation with Given Information

Here, the investment triples in 15 years. This means \(A = 3P\) (since it triples) and \(t = 15\) years. Plugging these values into the formula \(A = P e^{rt}\), we get \( 3P = P e^{15r} \).
03

Simplify the Equation

Divide both sides of the equation by \(P\): \( 3 = e^{15r} \).
04

Solve for the Interest Rate

To solve for \(r\), first take the natural logarithm of both sides: \( \ln 3 = 15r \), which simplifies to \( r = \frac{\ln 3}{15} \).
05

Calculate the Interest Rate

Use a calculator to find \( \ln 3 \) (approximately 1.0986), and then divide by 15: \( r = \frac{1.0986}{15} \approx 0.0732 \). Thus, the yearly interest rate is approximately 0.0732, or 7.32%.

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Key Concepts

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

Exponential Growth
Exponential growth is a process where the quantity grows at a rate proportional to its current size. In our problem, the investment grows continuously over time, rather than at regular intervals.
This kind of growth is represented by the exponential function, specifically using the base of the natural logarithm, denoted by 'e'.
The formula for continuous compounding is: \[A = P e^{rt}\]
  • \(A\) is the future value of the investment.
  • \(P\) is the principal or initial amount.
  • \(r\) is the annual interest rate, as a decimal.
  • \(t\) is the time period in years.
In our case, the investment triples in 15 years, so we solve for the interest rate using this formula.
Natural Logarithm
The natural logarithm, represented as \( \ln \) , is the logarithm to the base 'e'. It is used to solve for the exponent in equations involving exponential growth.
In our exercise, we reach the equation: \[ 3 = e^{15r} \]To isolate the variable \(r\) , we take the natural logarithm of both sides. This is the inverse operation of the exponential function.
So, taking the natural logarithm on both sides, we get: \[ \ln 3 = \ln(e^{15r}) = 15r \]This simplifies to: \[ r = \frac{\ln 3}{15} \]Using the property that \( \ln(e^x) = x \), we can solve for \(r\) effectively.
Calculus
Calculus helps us understand change and motion, and it is instrumental in solving problems involving continuous growth.
When dealing with continuous compounding interest, calculus concepts like exponential functions and logarithms play a pivotal role.
Here's a step-by-step outline of how calculus applies to this problem:
  • Identify the continuous growth formula: \[ A = P e^{rt} \]
  • Set up the equation with given values: \[ 3P = P e^{15r} \]
  • Divide both sides by \(P\) to simplify: \[ 3 = e^{15r} \]
  • Take the natural logarithm of both sides to isolate \(r\) : \[ \ln 3 = 15r \]
  • Finally, solve for \(r\): \[ r = \frac{\ln 3}{15} \]
By understanding these calculus principles, solving for the interest rate becomes systematic and approachable. Calculus also ensures accurate and precise solutions for problems involving continuous change.

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Most popular questions from this chapter

A person is given an injection of 300 milligrams of penicillin at time \(t=0 .\) Let \(f(t)\) be the amount (in milligrams) of penicillin present in the person's bloodstream \(t\) hours after the injection. Then, the amount of penicillin decays exponentially, and a typical formula is \(f(t)=300 e^{-0.6 t}\) (a) Give the differential equation satisfied by \(f(t)\) (b) How much will remain at time \(t=5\) hours? (c) What is the biological half-life of the penicillin (that is, the time required for half of a given amount to decompose) in this case?

Physiologists usually describe the continuous intravenous infusion of glucose in terms of the excess concentration of glucose, \(C(t)=A(t) / V\) where \(V\) is the total volume of blood in the patient. In this case, the rate of increase in the concentration of glucose due to the continuous injection is \(r / V .\) Find a differential equation that gives a model for the rate of change of the excess concentration of glucose.

A model incorporating growth restrictions for the number of bacteria in a culture after \(t\) days is given by \(f(t)=5000\left(20+t e^{-0.04 t}\right)\) (a) Graph \(f^{\prime}(t)\) and \(f^{\prime \prime}(t)\) in the window [0,100] by [-700,300] (b) How fast is the culture changing after 100 days? (c) Approximately when is the culture growing at the rate of 76.6 bacteria per day? (d) When is the size of the culture greatest? (e) When is the size of the culture decreasing the fastest?

A population is growing exponentially with growth constant .05. In how many years will the current population triple?

A small amount of money is deposited in a savings account with interest compounded continuously. Let \(A(t)\) be the balance in the account after \(t\) years. Match each of the following answers with its corresponding question. Answers. a. \(P e^{r t}\) b. \(A(3)\) c. \(A(0)\) \(\mathbf{d} . A^{\prime}(3)\) e. Solve \(A^{\prime}(t)=3\) for \(t\) f. Solve \(A(t)=3\) for \(t\) g. \(y^{\prime}=r y\) h. Solve \(A(t)=3 A(0)\) for \(t\) Questions A. How fast will the balance be growing in 3 years? B. Give the general form of the function \(A(t)\) C. How long will it take for the initial deposit to triple? D. Find the balance after 3 years. E. When will the balance be 3 dollars? F. When will the balance be growing at the rate of 3 dollars per year? G. What was the principal amount? H. Give a differential equation satisfied by \(A(t)\)
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