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

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

The correct answer is (C) \(I - \frac{kT^2}{2}\).

Step 1: Analyze the Given Differential Equation

The differential equation provided is \(\frac{dV(t)}{dt} = -k(T-t)\). This equation implies that the rate of change of the value \(V(t)\) with respect to time \(t\) is proportional to \(T-t\), the remaining time until the end of the equipment's life. Here, \(k\) is a positive constant.

Step 2: Integrate the Differential Equation

To find \(V(t)\), integrate the differential equation. Start with:\[\frac{dV(t)}{dt} = -k(T-t)\]Integrate with respect to \(t\):\[\int dV(t) = -k \int (T - t) \, dt\]This results in:\[ V(t) = -k \left[Tt - \frac{t^2}{2}\right] + C\]where \(C\) is the integration constant.

Step 3: Determine the Integration Constant

At \(t = 0\), the equipment is new, so \(V(0) = I\). Substitute these values into the equation to find \(C\):\[I = -k \left[T \times 0 - \frac{0^2}{2}\right] + C \]Hence, \(C = I\).

Step 4: Compute the Value of the Equipment After Time \(t\)

Substitute the value of \(C\) back into the integrated equation:\[V(t) = -k \left(Tt - \frac{t^2}{2}\right) + I\]Thus, the value \(V(t)\) simplifies to:\[V(t) = I - k \left(Tt - \frac{t^2}{2}\right)\]

Step 5: Calculate the Scrap Value

The scrap value is the value of the equipment when \(t = T\). Set \(t = T\) in the equation for \(V(t)\):\[V(T) = I - k \left(T \times T - \frac{T^2}{2}\right)\]This simplifies to:\[V(T) = I - k \left(\frac{T^2}{2}\right)\]Thus, the scrap value of the equipment is \(I - \frac{kT^2}{2}\).

Step 6: Select the Correct Option

Compare the derived expression for scrap value \(I - \frac{kT^2}{2}\) with the options given. Option C is \(I - \frac{kT^2}{2}\), which matches our derived expression.

Key Concepts

Understanding Depreciation

Depreciation is an accounting method used to allocate the cost of an asset over its useful life. It represents the decrease in value of the equipment as time progresses. In the context of differential equations, depreciation can be described mathematically to help predict how much an asset will be worth at any future point in time.

• Straight-Line Depreciation: This assumes a constant rate of depreciation each year.
• Declining Balance Method: This accounts for higher depreciation initially, which decreases over time.

In this problem, we look at a unique way of modeling depreciation through a differential equation. It expresses the rate of depreciation in terms of time and helps to compute the value of an asset at any given point. By understanding the relationship between time and value decrease, we can make informed financial decisions regarding equipment lifecycle management.

Integration in Differential Equations

Integration is a fundamental tool in calculus used to find the accumulated change of a quantity. When we have a differential equation, like \(-k(T-t)\), it can describe how a quantity, such as the value of an asset, changes over time. To solve for the value function \(V(t)\), we perform integration on \(-k(T-t)\), which means finding a function whose derivative gives back the original differential expression. In this case, after integrating, the resulting expression is: \[V(t) = -k \left[Tt - \frac{t^2}{2}\right] + C\]

• Definite vs. Indefinite Integration: Indefinite integration signifies finding the function's antiderivative. Definite integration, on the other hand, computes the value over a specific interval.
• Integration Constant: After integrating, we introduce an integration constant \(C\) because differentiation of a constant gives zero. We determine its value using initial conditions or known values at specific points.

Integration is powerful, aiding in resolving complex problems by reconstructing functional relationships from rates, especially in finance and engineering scenarios.

Initial Value Problem and Its Solution

An initial value problem is a differential equation accompanied by a specific value at a start point. It's crucial in determining a unique solution to the problem since different starting conditions can lead to different solutions. In the exercise, the given differential equation \(\frac{dV(t)}{dt} = -k(T-t)\) required an initial condition to solve completely. For this problem, the initial condition is at time \(t = 0\), where the equipment's value is known as \(V(0) = I\). Using this, we determine the constant of integration \(C\), ensuring our solution fits the initial scenario. This initial condition leads to:\[C = I\] Thus, the full solution for \(V(t)\) becomes:\[V(t) = I - k \left(Tt - \frac{t^2}{2}\right)\]

• Importance: Initial conditions ensure that solutions of differential equations are not just talking points but applicable real-world scenarios.

These problems help enhance conceptual understanding by reinforcing how initial conditions and differential equations work together harmoniously to solve real-life problems.