The rate of depreciation `(dV)/ dt` of a machine is inversely proportional to the square of t + 1, where V is the value of the machine t years after it was purchased. The initial value of the machine was ₹ 8,00,000 and its value decreased ₹1,00,000 in the first year. Find its value after 6 years.

#### Solution

According to the given condition,

`(dV)/dt ∝ 1/(t+1)^2`

∴ `(dV)/dt = (-k)/(t+1)^2` …[Negative sign indicates disintegration]

∴ `dV = (-kdt)/(t+1)^2`

Integrating on both sides, we get

`int dV = - k int dt/(t+1)^2`

∴ `V = k/(t+1) + c`

when t = 0, V = 8,00,000

∴ `8,00,000 = k/((0+1)) +c`

∴ 8,00,000 = k + c …(i)

when t = 1, V = 7,00,000

∴ `7,00,000 = k /((1 +1)) + c`

∴ `7,00,000 = k/ 2 + c` …(ii)

From (i) – (ii), we get

`1,00,000 = k /2`

∴ k = 2,00,000 …(iii)

Substituting (iii) in (i), we get

c = 6,00,000 …(iv)

when t = 6, we get

`V = k/((6+ 1)) + c`

=`(2,00,000 )/7 + 6,00,000`

= 6,28,571.4286

≈6,28,571

∴ Value of the machine after 6 years is ₹ 6,28,571.