Problem 29
Question
Shannon has a trust fund that will yield \(\$ 80,000\) in 13 yr. A CPA is preparing a financial statement for Shannon and wants to take into account the present value of the trust fund in computing her net worth. Interest is compounded continuously at \(4.8 \% .\) What is the present value of the trust fund?
Step-by-Step Solution
Verified Answer
The present value of the trust fund is approximately $42,824.
1Step 1: Understand the Problem
In this problem, we need to find the present value of a future sum of money, which will be available in 13 years. The future amount is $80,000, and the interest is compounded continuously at an annual rate of 4.8%. The present value is the amount that should be invested today to grow to $80,000 in 13 years at the given interest rate.
2Step 2: Use the Present Value Formula for Continuous Compounding
The formula for the present value with continuous compounding is given by \( PV = FV \times e^{-rt} \), where \( PV \) is the present value, \( FV \) is the future value, \( r \) is the interest rate (as a decimal), and \( t \) is the time in years. Here, \( FV = 80,000 \), \( r = 0.048 \) (since 4.8% is 0.048 as a decimal), and \( t = 13 \) years.
3Step 3: Substitute Values Into the Formula
Substitute the given values into the formula: \[ PV = 80,000 \times e^{-0.048 \times 13} \].
4Step 4: Calculate the Exponent
Calculate the exponent \( -0.048 \times 13 \). This equals \( -0.624 \). So the formula becomes \[ PV = 80,000 \times e^{-0.624} \].
5Step 5: Calculate Using the Exponential Function
Calculate \( e^{-0.624} \) using a calculator. The value of \( e^{-0.624} \) is approximately 0.5353.
6Step 6: Find the Present Value
Now calculate the present value by multiplying: \[ PV = 80,000 \times 0.5353 \]. This equals approximately \( 42,824 \).
Key Concepts
Continuous CompoundingPresent ValueExponential Growth
Continuous Compounding
Continuous compounding is a concept in calculus where interest is calculated and added to the principal balance continuously, rather than on a monthly, quarterly, or yearly basis. This concept is a hypothetical scenario that illustrates the maximum impact of compounding interest. In real life, interest is typically compounded periodically, but continuous compounding offers a useful model for understanding limits and growth over time.The formula to calculate future value with continuous compounding is:\[ FV = PV imes e^{rt} \]where:
- \( FV \) is the future value.
- \( PV \) is the present value.
- \( e \) is the base of the natural logarithm, approximately equal to 2.71828.
- \( r \) is the annual interest rate in decimal form.
- \( t \) is the time in years.
Present Value
Present value is a concept that represents the current worth of a sum of money that you are expected to receive in the future. When using continuous compounding, it helps to determine how much an amount of money to be received in the future is worth today.The present value formula with continuous compounding is given by:\[ PV = FV \times e^{-rt} \]where:
- \( PV \) is the present value.
- \( FV \) is the future value, or the amount of money in the future.
- \( r \) is the interest rate (as a decimal).
- \( t \) is the time in years until the money is received.
Exponential Growth
Exponential growth is a concept where quantities increase at a constant proportional rate over time, resulting in growth that accelerates rapidly. This is commonly observed in populations, finance, and physical processes where a continuous process causes the amount to grow exponentially.In the context of finance and continuous compounding, the exponential growth equation is:\[ A(t) = A_0 e^{rt} \]where:
- \( A(t) \) is the amount of money at time \( t \).
- \( A_0 \) is the initial principal amount (what you start with).
- \( e^{rt} \) represents the exponential growth factor.
- \( r \) is the growth rate, often an interest rate for investments or savings.
- \( t \) is time.
Other exercises in this chapter
Problem 28
Differentiate. $$ F(x)=\left(\log _{9} x\right)^{7} $$
View solution Problem 28
Given \(\log _{b} 3=1.099\) and \(\log _{b} 5=1.609\), find each value. $$ \log _{b} 15 $$
View solution Problem 29
Differentiate. $$ G(x)=x^{3}-5 e^{2 x} $$
View solution Problem 29
Differentiate. $$ f(x)=\frac{6^{x}}{5 x-1} $$
View solution