Problem 113 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Let \(l\) be the purchase value of an equipment and \(V(t)\) be the value of equipment after it has been used for \(t\) years. The value \(V(t)\) depreciates at a rate given by the differential equation \(\frac{d V(t)}{d t}=k(T-t)\), where \(k>0\) is a constant and \(T\) is the total life in years of the equipment. Then, the scrap value \(V(T)\) of the equipment is (A) \(l-\frac{k T^{2}}{2}\) (B) \(l-\frac{k(T-t)^{2}}{2}\) (C) \(e^{-k T}\) (D) \(T^{2}-\frac{l}{k}\)

Step-by-Step Solution

Answer

The correct option is (A) \( l - \frac{k T^2}{2} \).

Step 1: Understand Given Information

We are given a differential equation \( \frac{d V(t)}{d t} = k(T-t) \), which describes the rate at which the value of the equipment depreciates over time. Here, \( V(t) \) is the value of the equipment at time \( t \), \( k \) is a positive constant, and \( T \) is the total life of the equipment.

Step 2: Set Up the Integral Equation

To find \( V(t) \), we need to integrate the given differential equation. We have \( \int \frac{d V}{d t} \, dt = \int k(T-t) \, dt \), which when integrated will give us \( V(t) \).

Step 3: Perform the Integration

Integrate the right-hand side: \( \int k(T-t) \, dt = k \left( \int T \, dt - \int t \, dt \right) = k(Tt - \frac{t^2}{2}) + C \), where \( C \) is the integration constant.

Step 4: Determine the Integration Constant

Using the initial condition \( V(0) = l \) (equipment value at \( t = 0 \)), solve for \( C \):\[ V(0) = k(T \cdot 0 - \frac{0^2}{2}) + C = l \]Thus, \( C = l \).

Step 5: Find General Solution for \( V(t) \)

Substituting the value of \( C \) into our integrated function gives:\( V(t) = k(Tt - \frac{t^2}{2}) + l \).

Step 6: Compute the Scrap Value \( V(T) \)

Substitute \( t = T \) (total life of equipment) into the equation for \( V(t) \):\[ V(T) = k \left(TT - \frac{T^2}{2}\right) + l = l - \frac{kT^2}{2} \].

Step 7: Select the Correct Answer

The expression for \( V(T) \) matches option (A): \( V(T) = l - \frac{k T^{2}}{2} \).

Key Concepts

Depreciation

Depreciation is a concept that describes how the value of an asset, such as equipment, decreases over time. This loss in value occurs due to several factors, including wear and tear, age, and technological obsolescence. In this specific exercise, depreciation is studied in the context of a differential equation. The rate at which the equipment's value declines is given by \( \frac{d V(t)}{d t} = k(T-t) \), where \( V(t) \) stands for the equipment's value at a given time \( t \), \( k \) is a positive constant highlighting how fast depreciation occurs, and \( T \) signifies the equipment's expected lifespan.

Key points about depreciation in this context:

• The value decreases linearly over the equipment's lifespan from purchase to scrap.
• The rate of depreciation is not constant; it decreases as \( t \) approaches \( T \).

Understanding how depreciation works is crucial, as it helps companies plan for the future, ensuring assets are replaced timely, and also helps in financial reporting and tax calculations.

Integration

Integration is a mathematical technique used to calculate the accumulation of quantities, which can be thought of as the area under a curve. When it comes to differential equations, integration is often used to determine the original function from its rate of change. In this exercise, the integration is required to solve the differential equation and find \( V(t) \), the value of the equipment as it depreciates over time.

The process is as follows:

• Set up the integral from the differential equation, \( \int \frac{d V}{d t} \, dt = \int k(T-t) \, dt \).
• Integrate the expression \( k(T-t) \, dt \), leading to \( k(Tt - \frac{t^2}{2}) + C \), where \( C \) is the constant of integration.

Integration is an essential skill in calculus, providing a way to transition from the rate of change back to the actual function, helping us understand processes like depreciation.

Scrap Value

The scrap value of an asset refers to its value at the end of its useful life, the point when it can no longer serve its intended purpose. In the exercise, the scrap value is given by \( V(T) = l - \frac{kT^2}{2} \). This formula shows us what remains of the initial investment after accounting for the depreciation over time.

Important factors regarding scrap value:

• The scrap value might differ from the book value and should be considered when calculating depreciation.
• Typically, businesses aim to recover some cost by selling the asset's materials or parts.

In practice, accurate estimation of an asset's scrap value is vital for future financial planning and investment decisions.

Equipment Value

Equipment value, represented as \( V(t) \) in the problem, embodies the worth of the equipment at any given time \( t \) during its lifespan. Initially, the equipment is valued at \( l \), which is the purchase price. As time progresses, the equipment's value decreases due to depreciation.

The differential equation provides a way to model this decrease in value:

• At the start \( t=0 \), \( V(0) = l \).
• Throughout the life \( t \), the value reduces according to the derived equation \( V(t) = k(Tt - \frac{t^2}{2}) + l \).

Understanding how equipment value changes over time is essential for managing asset resources efficiently and helps predict when equipment should be upgraded or replaced.