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Question
Solve the following:
A bank pays interest by continuous compounding, that is by treating the interest rate as the instantaneous rate of change of principal. A man invests ₹ 1,00,000 in the bank deposit which accrues interest, 8% per year compounded continuously. How much will he get after 10 years? (e0.8 = 2.2255)
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Solution
Let P(t) denotes the amount of money in the account at time t.
Then the differential equation governing the growth of money is
`"dp"/"dt" = 8/100 "p"` = 0.08 p
⇒ `"dp"/"p"` = 0.08 dt
Integrating on both sides
`int "dp"/"p" = int 0.08 "dt"`
loge P = 0.08 t + c
P = `"e"^(0.08"t") + "c"`
P = `"e"^(0.08"t")* "e"^"c"`
P = `"C"_1 "e"^(0.08"t")` .........(1)
when t = 0, P = ₹ 1,00,000
Equation (1)
⇒ 1,00,000 = C1 e°
C1 = 1,00,000
∴ P = `100000 "e"^(0.08"t")`
At t = 10
P = `1,00,000 * "e"^(0.08(10))`
= 1,00,000 e0.8 .......{∵ e0.8 = 2.2255}
= 100000 (2.2255)
p = ₹ 2,25,550
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