Practicing Success
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 the differential equation $\frac{d V}{d t}=-k(T-t)$, 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: |
$T^2-\frac{I}{k}$ $I-\frac{k T^2}{2}$ $I-\frac{k(T-t)^2}{2}$ $e^{-k T}$ |
$I-\frac{k T^2}{2}$ |
We have, $\frac{d V}{d t} =-k(T-t)$ $\Rightarrow d V =-k(T-t) d t$ On integrating, we get $\int d V=-k \int(T-t) d t$ $\Rightarrow V(t)=k \frac{(T-t)^2}{2}+C$ ....(i) Initially i.e. at t = 0, we have V(t) = I. Putting t = 0 and V(t) = I in (i), we get ∴ $I=\frac{k T^2}{2}+C \Rightarrow C=I-\frac{k T^2}{2}$ Putting this value of $C$ in (i), we get $V(t)=\frac{k(T-t)^2}{2}+I-\frac{k T^2}{2}$ At $t=T$, we get $V(T)=I-\frac{k T^2}{2}$ |