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 `(d)/(dt) V(t)= -k(T-t), k gt 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. `T^(2)-(1)/(k)`
B. `I -(kT^(2))/(2)`
C. `I-(k(T-t)^(2))/(2)`
D. `e^(-kT)`

Answer

To solve the problem, we start with the given differential equation that describes the depreciation of the value of the equipment:

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

where \( V(t) \) is the value of the equipment after \( t \) years, \( k \) is a positive constant, and \( T \) is the total life of the equipment in years.

### Step 1: Separate the Variables
We can rearrange the equation to separate the variables \( V \) and \( t \):

\[ dV = -k(T - t) dt \]

### Step 2: Integrate Both Sides
Now we integrate both sides. The left side integrates with respect to \( V \), and the right side integrates with respect to \( t \):

\[ \int dV = \int -k(T - t) dt \]

The left side gives us \( V \), and for the right side, we can expand and integrate:

\[ V = -k \left( Tt - \frac{t^2}{2} \right) + C \]

where \( C \) is the constant of integration.

### Step 3: Simplify the Expression
Rearranging gives us:

\[ V = -kTt + \frac{kt^2}{2} + C \]

### Step 4: Determine the Constant of Integration
To find the constant \( C \), we use the initial condition. When \( t = 0 \), the value \( V(0) \) is equal to the purchase value \( I \):

\[ V(0) = I \implies I = -kT(0) + \frac{k(0)^2}{2} + C \implies C = I \]

### Step 5: Substitute \( C \) Back into the Equation
Substituting \( C \) back into the equation gives us:

\[ V(t) = -kTt + \frac{kt^2}{2} + I \]

### Step 6: Find the Scrap Value \( V(T) \)
To find the scrap value \( V(T) \), we substitute \( t = T \):

\[ V(T) = -kT(T) + \frac{kT^2}{2} + I \]

This simplifies to:

\[ V(T) = -kT^2 + \frac{kT^2}{2} + I \]

Combining the terms gives:

\[ V(T) = I - \frac{kT^2}{2} \]

### Final Result
Thus, the scrap value \( V(T) \) of the equipment is:

\[ V(T) = I - \frac{kT^2}{2} \]