Problem 37 - Calculus and its applications

Question

An office machine is purchased for \(\$ 5200 .\) Under certain assumptions, its salvage value, \(V\), in dollars, is depreciated according to a method called double declining balance, by basically \(80 \%\) each year, and is given by \(V(t)=5200(0.80)^{t}\), where \(t\) is the time, in years, after purchase. a) Find \(V^{\prime}(t)\). b) Interpret the meaning of \(V^{\prime}(t)\).

Step-by-Step Solution

Answer

a) \( V'(t) = 5200 \ln(0.80) (0.80)^{t} \). b) It shows the rate at which the machine's value decreases annually.

Step 1: Differentiate the Function

To find the derivative, we need to differentiate the function given: \( V(t) = 5200(0.80)^{t} \). Use the power rule and chain rule for differentiation. Recall that the derivative of a constant times a function is the constant times the derivative of the function. Therefore, we acknowledge that the derivative of \((0.80)^{t}\) is \((0.80)^{t} \ln(0.80)\). So, \( V'(t) = 5200 \times (0.80)^{t} \times \ln(0.80) \). This gives us our derivative: \( V'(t) = 5200 \ln(0.80) (0.80)^{t} \).

Step 2: Simplifying the Derivative

We know that \( \ln(0.80) \) is a negative number (because \(0.80 < 1\)). Thus, \( V'(t) = 5200 \ln(0.80) (0.80)^{t}\) will also be negative, indicating a decreasing function. Calculating \( \ln(0.80) \approx -0.2231 \), so \( V'(t) = 5200 \times -0.2231 \times (0.80)^{t} \approx -1160.12(0.80)^{t} \).

Step 3: Interpret the Derivative

The derivative \( V'(t) \approx -1160.12(0.80)^{t} \) represents the rate of change of the machine’s salvage value with respect to time. Since the derivative is negative, it shows how the value of the machine decreases per year. The magnitude tells us how quickly the value is decreasing; specifically, it's decreasing at a rate proportional to \(1160.12(0.80)^{t} \) dollars per year.

Key Concepts

Double Declining Balance

The double declining balance is a method of depreciation where an asset loses value over time at a rate that is faster than traditional straight-line depreciation methods. In this method, an asset's value decreases by the same percentage each year, leading to a rapid initial depreciation that slows over time.
This technique reflects how some assets lose value quickly in their early life, like cars or office machines. For example, in our exercise, the machine depreciates by \(80\%\) per year, meaning it retains only \(20\%\) of its value after each year.

• The formula for double declining balance is \(V(t) = \, V_0 \, (0.80)^t\).
• In this exercise, \(V_0\) is \(\$5200\), which reduces by \(80\%\) annually.
• The salvage value is given as \(V(t) = 5200(0.80)^t\).

This helps businesses quickly recover the cost of assets that lose usefulness quickly, enhancing tax efficiency.

Derivative Interpretation

Derivatives are a cornerstone in calculus and are used to understand how functions change. The interpretation of a derivative provides insights into the rate of change of a dependent variable, like an asset's value, with respect to an independent variable, such as time.

• For our exercise, the derivative \(V'(t) = 5200 \, \ln(0.80) \, (0.80)^t\) represents how fast the salvage value of the machine declines each year.
• A negative derivative, like \(-1160.12(0.80)^t\), shows the value is decreasing over time.
• The magnitude of the derivative tells us how quickly this decrease occurs, which is crucial for understanding depreciation.

In summary, derivative interpretation helps in comprehending the rate at which economic activities, like asset valuation, change over time, aiding in strategic decision making.

Chain Rule

The chain rule is an essential tool in calculus that allows us to differentiate composite functions - functions of functions. It's pivotal in our exercise where the function \( V(t) = 5200 (0.80)^t \) involves both an exponential and a linear component.

• The chain rule states: If you have a composite function \( f(g(x)) \), its derivative is \( f'(g(x)) \cdot g'(x) \).
• In our case, the function is \( (0.80)^t \), where the inner function is \( g(t) = t \) and the outer function is \( f(t) = (0.80)^t \).
• We differentiate \( (0.80)^t \) as \( (0.80)^t \cdot \ln(0.80) \) exploiting the chain rule.

This approach simplifies finding derivatives of complicated functions, making it indispensable for tackling problems involving exponential growth or decay, as seen in exponential depreciation.

Exponential Depreciation

Exponential depreciation describes the reduction in an asset’s value over time in an exponentially decreasing fashion, often used for assets that quickly lose value. It differs from linear depreciation by accelerating the decrease early in an asset's life.
In our exercise, exponential depreciation is captured by \(V(t) = 5200(0.80)^t\):

• The exponential component, \((0.80)^t\), dictates the rapid decline postulating \(80\%\) depreciation each year.
• This exponential model is suitable for financial analysis where initial rapid depreciation benefits tax savings.
• Unlike linear models, exponential depreciation captures the real-life decay of value for rapidly depreciating assets, forecasting more accurately their salvage value.

This method provides a realistic approach to valuing assets and preparing financial plans, balancing between taxation benefits and asset valuation.