Problem 34 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Let \(I\) be the purchase value of an equipment and \(V(t)\) be the value after it has been used for \(t\) years. The value \(V(t)\) depreciates at a rate given by 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) \(e^{-k T}\) (B) \(T^{2}-\frac{I}{k}\) (C) \(I-\frac{k T^{2}}{2}\) (D) \(I-\frac{k(T-t)^{2}}{2}\)

Step-by-Step Solution

Answer

Option C: \(I-\frac{kT^2}{2}\).

Step 1: Understand the problem

The problem involves a differential equation for the depreciation of equipment over time. We need to find the scrap value, which is the value of the equipment after its total life, denoted by \( V(T) \).

Step 2: Set up the differential equation

We have the differential equation \( \frac{dV(t)}{dt} = -k(T-t) \). We need to solve this equation to find \( V(t) \).

Step 3: Integrate the differential equation

Integrate the equation \( \frac{dV(t)}{dt} = -k(T-t) \) with respect to \( t \). This yields \( V(t) = \int -k(T-t) \, dt \).

Step 4: Perform the integration

Carry out the integration: \( V(t) = -k \int (T-t) \, dt = -k \left[ Tt - \frac{t^2}{2} \right] + C \), where \( C \) is the integration constant.

Step 5: Determine the integration constant

We know the initial condition: when \( t = 0 \), \( V(0) = I \). Using this, we can find \( C \): \[ I = -k(T \cdot 0 - \frac{0^2}{2}) + C \] \[ I = C \]. Hence, \( C = I \).

Step 6: Substitute back to find V(t)

Substitute \( C = I \) into the equation: \( V(t) = -k \left[ Tt - \frac{t^2}{2} \right] + I \). Simplify: \( V(t) = I - k(Tt - \frac{t^2}{2}) \).

Step 7: Calculate the scrap value V(T)

Substitute \( t = T \) into \( V(t) \) to find the scrap value: \[ V(T) = I - k\left(T \cdot T - \frac{T^2}{2}\right) \] Simplify: \[ V(T) = I - \frac{kT^2}{2} \].

Step 8: Match to given options

Compare the expression \( V(T) = I - \frac{kT^2}{2} \) to the provided options to find that it matches option (C).

Key Concepts

Depreciation

Depreciation refers to the decline in value of an asset over time. In our context, we are looking at how the value of equipment decreases yearly as it is used. This reduction in value is crucial in accounting as it affects the reported worth of assets in financial statements. Understanding depreciation is important because:

• It reflects the real-world dropping in value of assets over time.
• Helps in planning for future expenses, like maintenance or replacement of equipment.
• Aids in tax calculations since depreciation can be used as a deductible expense.

In this exercise, depreciation is modeled by a differential equation that represents how the rate of change in value of the equipment depends on time.

Integration

Integration is a fundamental concept in calculus that allows us to find functions from their derivatives. This is particularly useful when working with differential equations. In our example, integration is used to solve for the function that gives us the equipment's value over time, denoted by \(V(t)\).
To solve the differential equation \(\frac{dV(t)}{dt} = -k(T-t)\), we need to integrate it with respect to \(t\). This gives us:

• The integral \( V(t) = \int -k(T-t) \, dt \) provides a way to express \(V(t)\) in terms of known variables and constants.

Once integrated, the equation helps us ascertain the value of the asset at any given time \(t\), provided we know the initial value and other required parameters.

Initial Condition

An initial condition, in the context of differential equations, is a value that is used to find specific solutions to those equations. It effectively anchors the solution of the differential equation at a certain point.In this problem, we are given the initial condition when the equipment is first purchased, denoting its initial value as \(V(0) = I\).

• This initial condition allows us to calculate the integration constant \(C\), necessary to completely resolve the equation for \(V(t)\).
• With \(I = C\), we ensure our solution doesn't just describe a generic scenario but is applicable to the specific case outlined in the problem.

This use of an initial condition is essential for tailoring the solution to actual situations.

Scrap Value

Scrap value is the amount an asset is worth at the end of its useful life. It is crucial for determining residual value for accounting and tax purposes. In this exercise, after solving the differential equation, we find the scrap value as \(V(T) = I - \frac{kT^2}{2}\), where \(T\) is the total lifespan of the equipment:

• Scrap value helps businesses plan whether to continue using an old asset or consider replacing it.
• It also affects the computation of depreciation when using methods like straight-line or reducing balance.

Understanding scrap value not only aids in maintaining accurate financial books, but it is also key for making informed business decisions about asset management.