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? (e^0.8 = 2.2255)

Answer 1

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"`

log_e 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 e^0.8  .......{∵ e^0.8 = 2.2255}

= 100000 (2.2255)

p = ₹ 2,25,550