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 V(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 : (1) `T^2-1/k` (2) `I-(k T^2)/2` (3) `I-(k(T-t)^2)/2` (4) `e^(-k T)`

Options

A. `T^(2)-(I)/(k)`
B. `I-(kT^(2))/(2)`
C. `I-((T-t)^(2))/(2)`
D. `e^(-kT)`

Answer

D

Explanation

We have,
`(dV)/(dt)=-k(T-t)`
`rArr" "dV=-k(T-t)dt`
On integrating, we get
`intdV=-k int (T-t) dt`
`rArr" "V(t)=k((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
`therefore" "I=(kT^(2))/(2)+CrArr C=I-(kT^(2))/(2)`
Putting this value of C in (i), we get
`V(t)=(k(T-t)^(2))/(2)+I-(kT^(2))/(2)`
At t = T, we get
`V(T)=I-(kT^(2))/(2)`