Question

If the population of a town increases at a rate proportional to the population at that time. If the population increases from 40 thousand to 60 thousand in 40 years, what will be the population in another 20 years? `("Given" sqrt(3/2) = 1.2247)`

Answer 1

Let 'x' be the population at time 't'.

∴ `("d"x)/("d"t) ∞ x`

∴ `("d"x)/("d"t)` = kx,

where k is the constant of proportionality.

`("d"x)/"dt"` = k dt

Integrating on both sides, we get

`int ("d"x)/x = "k" int 1 "dt"`

∴ log x = kt + c

∴ x = a.e^kt,     ......(i)

where a = e^c 

When t = 0, x = 40,000

∴ 40000 = a.e⁰ 

∴ a = 40000

∴ x = 40000.e^kt    ......(ii)  ......[From (i)]

When t = 40, P = 60000

∴ 60000 = 40000.e^40k

∴ e^40k = `60000/40000 = 3/2`    ......(iii)

Now, we have to find x when t = 40 + 20

= 60 years

∴ P = 40000.e^60k    .....[From (iii)]

= `40000 ("e"^(40"k"))^(3/2)`

= `40000 (3/2)^(3/2)`   ......[From (iii)]

= `40000 (3/2) sqrt(3/2)`

= 60000(1.2247)

= 73482

∴ The required population will be 73482.