Problem 99 - College Algebra and Calculus: An Applied Approach

Question

Find the time required for a \(\$ 1000\) investment to double at interest rate \(r\), compounded continuously.\(r=0.0625\)

Step-by-Step Solution

Answer

The time required for the investment to double at an interest rate of 0.0625 compounded continuously is approximately 11.09 years.

Step 1: Identify given values

In this case, the given values are: Principal, \(P = \$ 1000\), Final amount, \(A = 2 * P = \$ 2000\), and Interest rate, \(r = 0.0625\).

Step 2: Apply the continuous compounding formula

The formula is \( A = P e^{rt} \). Substituting the known values, we get: \( \$2000 = \$1000 e^{0.0625t} \)

Step 3: Simplify the equation

Dividing both sides of equation by \$1000 gives: \(2 = e^{0.0625t}\), then apply natural logarithm (ln) on both sides to get: \(\ln{2} = \ln{(e^{0.0625t})}\), which simplifies to \(\ln{2} = 0.0625t\).

Step 4: Solve for time, \(t\)

Finally, solve for \(t\) by dividing both sides by 0.0625: \(t = \frac{\ln{2}}{0.0625}\).

Key Concepts

Investment Growth

When we talk about investment growth, especially in the context of continuous compounding, we are looking at how money increases in value over time. This is a vital concept because it shows how even a modest investment can grow significantly. Continuous compounding is a powerful tool in finance that helps investors maximize their returns.

• **Continuous Compounding:** This is when interest is added to the principal continuously, causing your investment to grow exponentially over time. Instead of compounding monthly or annually, the interest in continuous compounding is calculated and added at every possible moment.


• **Doubling Your Investment:** The exercise showed us how to calculate the time required for an investment to double at a given interest rate. By analyzing such growth over time, investors can plan effectively and set realistic financial goals.


• **Principal and Final Amount:** The process begins with setting your principal (the initial amount you invest) and looking at what you want it to grow into (final amount). Knowing these helps you use mathematical models to predict growth, like using the exponential growth formula.

Exponential Functions

Exponential functions are mathematical expressions that involve variables raised to a power. In the context of continuous compounding, they model how investments grow over time.

• **The Exponential Growth Formula:** In finance, the exponential formula used is: \[ A = P e^{rt} \] where \(A\) is the final amount, \(P\) is the principal, \(r\) is the rate of interest, and \(t\) is the time. This formula showcases the exponential increase in the initial investment.


• **Significance:** Exponential functions are vital because they accurately portray the nature of compounding growth, which is not linear but rather accelerates over time. This acceleration is due to the compounding effect, where interest earns its own interest.


• **Understanding \(e\):** In the formula, \(e\) stands for Euler's number, approximately 2.718, and it plays a vital role in continuously compounding interest. It is the base of the natural logarithm and is pivotal in calculating exponential growth smoothly and continuously.

Natural Logarithm

The natural logarithm, denoted as \( \ln \), is used to reversely interpret exponential functions, particularly in solving for the variable of time, \(t\), in growth equations.

• **What is a Natural Logarithm?** It's the logarithm to the base of \(e\), where \(e\) is Euler's number. It helps in simplifying expressions and solving equations where the unknown variable is an exponent.


• **Application in the Exercise:** In solving this exercise, we used the natural logarithm to solve for \(t\), the time required for the investment to double. This is crucial because it allows us to rearrange the equation: \[ 2 = e^{0.0625t} \]into a form where \(t\) can be isolated: \[ t = \frac{\ln{2}}{0.0625} \]


• **Rearranging using \(\ln\):** Using \(\ln\) allows us to transform exponential equations to linear ones, which are mathematically easier to handle. This transformation helps in clearly understanding the relationship between different variables like time and growth.