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)/d t)=-k(T-t), where k0 is a constant and T is the total life in years of the equipment. Then the scrap value V(T) of the equipment is

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 (d V(t)/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 (A) e-k T (B) T2-(1/k) (C) I-(k T2/2) (D) I-(k(T-t)2/2)

Tardigrade Admin · Accepted Answer

∫ limitslV(T) d V(t)=∫t=0T-k(T-t) d t ⇒ V(T)-I=k[((T-t)2/2)]0T ⇒ V(T)-I=-k[(T2/2)] ⇒ V(T)=I-(k T2/2)

$e^{- k T}$

$T^{2} - \frac{1}{k}$

$I - \frac{k T ^{2}}{2}$

$I - \frac{k ( T - t ) ^{2}}{2}$

$l \int V (T) d V (t) = \int_{t = 0}^{T} - k (T - t) d t$
$\Rightarrow V (T) - I = k [\frac{( T - t ) ^{2}}{2}]_{0}^{T}$
$\Rightarrow V (T) - I = - k [\frac{T ^{2}}{2}]$
$\Rightarrow V (T) = I - \frac{k T ^{2}}{2}$

e−kT

T2−k1​

I−2kT2​

I−2k(T−t)2​

l∫V(T)​dV(t)=∫t=0T​−k(T−t)dt ⇒V(T)−I=k[2(T−t)2​]0T​ ⇒V(T)−I=−k[2T2​] ⇒V(T)=I−2kT2​

$e^{- k T}$

$T^{2} - \frac{1}{k}$

$I - \frac{k T ^{2}}{2}$

$I - \frac{k ( T - t ) ^{2}}{2}$

$l \int V (T) d V (t) = \int_{t = 0}^{T} - k (T - t) d t$
$\Rightarrow V (T) - I = k [\frac{( T - t ) ^{2}}{2}]_{0}^{T}$
$\Rightarrow V (T) - I = - k [\frac{T ^{2}}{2}]$
$\Rightarrow V (T) = I - \frac{k T ^{2}}{2}$