Question

Let I be the purchase value of 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 `(dV)/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 ______ 

Options

• `T^2 - I/k`

• `I - (kT^2)/2`

• `I - (k(T - 2)^2)/2`

• `e^{-kT}`

Answer 1

Let I be the purchase value of 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 `(dV)/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 `underline(I - (kT^2)/2)`.

Explanation:

`(dV)/dt = -k(T - t)` 

⇒ dV = -k(T - t)dt

Integrating on both sides, we get

`intdV = -kint(T - t)dt + c`

⇒ V(t) = `(k(T - t)^2)/2 + c` ...........(i)

Initially, when t = 0, V(t) = I

∴ I = `(kT^2)/2 +  c` ⇒ c = `I - (kT^2)/2`

∴ V(t) = `(k(T - t)^2)/2 + I - (kT^2)/2` ..........[From (i)]

When t = T,

V(T) = `I - (kT^2)/2`