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Question
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.
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