JEE Main · 2011 · 10 Apr 2011 · Offline · Differential Equations

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{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

Answer: (A)

Solution

Integrating

\frac{d V}{d t} = - k (T - t)

gives

V = \frac{k ( T - t ) ^{2}}{2} + c

. At

t = 0

the value is the purchase price

I

, so

c = I - \frac{k T ^{2}}{2}

. The scrap value is

V

t = T

, where the squared term vanishes, leaving

V (T) = I - \frac{k T ^{2}}{2}

JEE Main · 2011 · 10 Apr 2011 · Offline · Single correct option

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