Let I be the purchase value of an equipment and V(t) be the value after it has been used for t years

Q: Wait, why is the rate of depreciation dependent on $(T-t)$? Shouldn't it just be a constant rate?

That is a brilliant observation! In many simple models, we assume linear depreciation, but here, the problem defines a more sophisticated reality. By making the rate proportional to $(T-t)$, the problem implies that as the equipment nears the end of its life ($t \to T$), the rate of depreciation actually slows down. It is a mathematical way of modeling how assets lose value rapidly at first and then stabilize as they become 'scrap'.

Q: I got $V(t) = -\frac{k(T-t)^2}{2} + C$. Why is my sign different from the solution?

Ah, the classic sign trap! When you integrate $\int -k(T-t) dt$, you are integrating with respect to $t$. The derivative of $(T-t)$ is $-1$. Therefore, when you use the power rule, you must divide by the derivative of the inner function, which is $-1$. So, the two negatives cancel out to become positive. Always double-check your chain rule during integration!

Q: Why do we use $V(0) = I$ to find $C$? Is there any other way?

You could use a definite integral from $0$ to $t$, which naturally incorporates the initial condition. However, finding the constant $C$ is a powerful habit because it gives you the 'General Equation of Value' for any time $t$. It allows you to see the entire life cycle of the equipment, not just the final scrap value.

Q: Is it possible for the scrap value to be negative?

Mathematically, yes, if $I < \frac{kT^2}{2}$. But physically, in the context of this JEE problem, we are looking for the value of the asset. If the math leads to a negative value, it might imply the cost of disposal exceeds the value of the materials. However, for this specific problem, we follow the algebraic derivation to reach the intended expression.

Q: Why does the term $(T-t)^2$ vanish at $t=T$?

Because at $t=T$, the equipment has reached the end of its life. The term $(T-t)$ becomes $(T-T) = 0$. Physically, this means the depreciation process has completed its cycle. It is a beautiful moment of mathematical symmetry where the complexity of the integral simplifies perfectly at the boundary condition.

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 $\frac{dV(t)}{dt} = -k(T - t)$, 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

Options:

(A) $I - \frac{kT^2}{2}$

(B) $I - \frac{k(T-t)^2}{2}$

(C) $e^{-kT}$

(D) $T^2 - \frac{1}{k}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are an engineer tasked with managing the lifecycle of a high-precision industrial machine. You purchase this equipment for an initial value $I$.

As the years roll by, the machine undergoes wear and tear. Its value follows a specific, elegant mathematical path defined by the differential equation:

$$
\frac{dV(t)}{dt} = -k(T - t)
$$

The term $-k(T-t)$ tells us that the rate of value loss is tied to the remaining lifespan of the machine. As $t$ approaches $T$, the rate of depreciation slows down. Let us embark on the journey to find the final scrap value, $V(T)$.

### The Art of Separation

To understand the value of our machine at any given time $t$, we must solve this differential equation. We begin by separating the variables to isolate $V$ and $t$:

$$
dV = -k(T - t) dt
$$

This is the moment of truth. We are essentially summing up all the tiny, infinitesimal losses in value over the entire duration of the machine's life. To do this, we integrate both sides:

$$
\int dV = \int -k(T - t) dt
$$

### The Integration Challenge

Now, we face the integral. Many students stumble here because of the negative sign inside the parenthesis. Remember, when we integrate a function of the form $(ax+b)^n$, we must account for the coefficient $a$.

Here, our 'inner function' is $(T-t)$, and its derivative with respect to $t$ is $-1$. Applying the power rule, we get:

$$
V(t) = -k \left[ \frac{(T - t)^2}{2 	imes (-1)} ight] + C
$$

Notice how the two negative signs—one from the original equation and one from the chain rule—cancel out beautifully to yield:

$$
V(t) = \frac{k(T - t)^2}{2} + C
$$

This is our general solution. It describes the value of the machine at any time $t$, provided we can determine the constant $C$.

### Anchoring the Reality

Mathematics is a language of constraints. We know that at the very beginning, when $t = 0$, the value of the machine is its purchase price $I$. This is our boundary condition: $V(0) = I$.

By substituting this into our general solution, we find:

$$
I = \frac{k(T - 0)^2}{2} + C
$$

Solving for $C$, we get:

$$
C = I - \frac{kT^2}{2}
$$

This constant $C$ is the 'anchor' of our equation. It represents the baseline value that persists through the depreciation process. Now, our specific value function is complete:

$$
V(t) = \frac{k(T - t)^2}{2} + I - \frac{kT^2}{2}
$$

### The Final Reveal

We have reached the end of the machine's life. We want to find the scrap value $V(T)$. We simply substitute $t = T$ into our refined equation:

$$
V(T) = \frac{k(T - T)^2}{2} + I - \frac{kT^2}{2}
$$

Since $(T - T) = 0$, the first term vanishes entirely. We are left with the elegant result:

$$
**V(T) = I - \frac{kT^2}{2}**
$$

Look at this result. It is not just an answer; it is a testament to the power of calculus. We started with a rate of change and, through the process of integration and boundary matching, we arrived at a precise value for the end of the machine's life.

Select Answer:

Visualized Solution

Visualizing Depreciation

The Rate of Depreciation

Separating Variables

Setting up Integration

Executing the Integration

Finding Constant $C$

Solving for $C$

The Specific Value Function

Calculating Scrap Value

Final Result

The Sigma Insight: Variable Separable Method

Analyzing the Setup

The Art of Separation

The Integration Challenge

Anchoring the Reality

The Final Reveal

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