Problem 30 - Precalculus

Question

A new Hyundai Triburon has a book value of \(\$ 22,000\), and after 2 years a book value of \(\$ 14,000 .\) What is the car's value in 4 years? Apply the formula \(N=N_{0} e^{-n t},\) where \(N\) represents the value of the car. Round to the nearest hundred.

Step-by-Step Solution

Answer

The car's value in 4 years is approximately \(\$8900\).

Step 1: Understand the Formula

The given formula is \( N = N_0 e^{-nt} \), where \( N_0 \) is the initial value of the car, \( N \) is the value of the car after \( t \) years, \( t \) is the time in years, and \( n \) is the depreciation rate. Our goal is to find \( N \) when \( t = 4 \).

Step 2: Calculate the Depreciation Rate

We know the car's value decreased from \( \\(22,000 \) to \( \\)14,000 \) over 2 years. Using the formula \( N = N_0 e^{-nt} \), plug in these numbers: \( 14000 = 22000 e^{-2n} \). Solve for \( n \): \[ e^{-2n} = \frac{14000}{22000}. \] Simplify the fraction: \[ e^{-2n} = \frac{7}{11}. \] Take the natural logarithm of both sides: \[ -2n = \ln \left( \frac{7}{11} \right). \] Solve for \( n \): \[ n = -\frac{1}{2} \ln \left( \frac{7}{11} \right). \]

Step 3: Calculate the Value After 4 Years

Substitute \( n \) into the formula to find the value after 4 years. \( N_0 = 22000 \), so \[ N = 22000 e^{-4n}. \] We found \( n \) in Step 2 as \(-\frac{1}{2} \ln \left( \frac{7}{11} \right)\). Calculate \( N \): \[ N = 22000 \left( \frac{7}{11} \right)^2. \] \[ N = 22000 \times \frac{49}{121}. \] \[ N \approx 8917.36. \] Round to the nearest hundred: \( N \approx \$8900. \)

Key Concepts

Depreciation Rate

The depreciation rate is a crucial element when it comes to understanding how the value of an asset decreases over time. In the context of cars, this rate helps estimate how quickly a car's value drops as it ages. Given the equation \( N = N_0 e^{-nt} \), the depreciation rate is represented by \( n \). It signifies the rate at which an asset loses its value annually.
To find the depreciation rate \( n \), we can use the values provided for the car's current and initial values. From the equation:

• \( N_0 = 22000 \)
• \( N = 14000 \)
• \( t = 2 \)

By plugging these into the equation, and solving for \( n \), we derive that \( n = -\frac{1}{2} \ln \left( \frac{7}{11} \right) \). This rate is negative because it represents decay or decrease in value rather than growth. Understanding the depreciation rate provides valuable insight into the financial trajectory of long-term investments like automobile ownership.

Natural Logarithm

Natural logarithm is a fundamental mathematical concept often used in problems involving exponential decay or growth, like in this car depreciation scenario. The natural logarithm, denoted as \( \ln \), is the power to which the base \( e \) (approximately 2.718) must be raised to yield a given number.
When calculating the depreciation rate, the natural logarithm is used to simplify the expression and solve for \( n \). Specifically, after rewriting the equation \( e^{-2n} = \frac{7}{11} \), we took the natural logarithm of both sides to obtain \( -2n = \ln \left( \frac{7}{11} \right) \).

• This transformation helps convert an exponential equation into a linear form.
• \( \ln \) functions are especially handy in solving equations involving the constant \( e \).

Understanding natural logarithms enriches problem-solving strategies, especially where exponential functions are concerned. They serve as a pivotal tool in dissecting the relationships embedded within growth and decay.

Value Function

The value function represents the changing value of an asset, such as a car, over time. In this scenario, the function \( N = N_0 e^{-nt} \) is used to calculate the car's value after any given period \( t \).
The parameters of this function include:

• \( N_0 \): The initial value of the asset before depreciation, which was \( 22000 \) for the car.
• \( N \): The value of the asset after \( t \) years.
• \( -nt \): The exponent that signifies the total accumulated depreciation after \( t \) years.

In application, once the depreciation rate \( n \) is known, the value function can predict future asset values. For instance, after 4 years, substituting in the values and previously calculated \( n \), we find \( N = 22000 e^{-4n} \). This illustrates the power of exponential decay represented in the value function, specifically showcasing how rapidly an asset's value can diminish under exponential rules.