Question
Let $$I$$ be the purchase value of an equipment and $$V\left( t \right)$$ be the value after it has been used for $$t$$ years. The value $$V\left( t \right)$$ depreciates at a rate given by differential equation $$\frac{{dV\left( t \right)}}{{dt}} = - k\left( {T - t} \right),$$ where $$k > 0$$ is a constant and $$T$$ is the total life in years of the equipment. Then the scrap value $$V\left( T \right)$$ of the equipment is-
A.
$$I - \frac{{k{T^2}}}{2}$$
B.
$$I - \frac{{k{{\left( {T - t} \right)}^2}}}{2}$$
C.
$${e^{ - \,kT}}$$
D.
$${T^2} - \frac{1}{k}$$
Answer :
$$I - \frac{{k{T^2}}}{2}$$
Solution :
$$\eqalign{
& \frac{{dV\left( t \right)}}{{dt}} = - k\left( {T - t} \right)\,\,\,\,\, \Rightarrow \int {dVt} = - k\int {\left( {T - t} \right)dt} \cr
& V\left( t \right) = \frac{{k{{\left( {T - t} \right)}^2}}}{2} + c \cr
& V\left( 0 \right) = I\,\, \Rightarrow I = \frac{{K{T^2}}}{2} + C\,\,\, \Rightarrow C = I - \frac{{K{T^2}}}{2} \cr
& \therefore \,V\left( T \right) = 0 + C = I - \frac{{K{T^2}}}{2} \cr} $$