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

• A. $I-\frac{k(T-t{)}^2}{2}$
• B. $\frac{k{(T-t)}^2}{2}+I$
• C. $T^2-\frac{I}{k}$
• D. $I-\frac{k{T}^2}{2}$

Answer 1

The correct option is B $\frac{k{(T-t)}^2}{2}+I$

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

On integrating we get,

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

Using condition that at $t=T,V\left(t\right)=I$

we get , $V\left(t\right)=\frac{k{(T-t)}^2}{2}+I$