Question

the world population was 2560 million in 1950 and 6,080 million in 2000. assume the growth rate of the population P'(t) is proportional to the size of the population P(t): P'(t)=kP(t)

Answer 1

Let
`P(t)` denote the population at time
`t` in millions.

The population's growth rate is modeled by the differential equation


`P'(t)=kP(t)`

This equation is separable, and we can solve easily:


`(\mathrm dP(t))/(\mathrm dt)=kP(t)\iff(\mathrm dP(t))/(P(t))=k\,\mathrm dt`

`\implies\displaystyle\int\frac{\mathrm dP}P=k\int\mathrm dt`

`\implies \ln|P(t)|=kt+C`

`\implies P(t)=e^(kt+C)=e^(kt)e^C=Ce^(kt)`

Let's assume 1950 corresponds to year
`t=0`, so that 2000 would correspond to
`t=50`. Then
`P(0)=2560`, which means we have


`2560=Ce^(0k)=C`

and
`P(50)=6080`, which means


`6080=2560e^(50k)\implies \frac{19}8=e^(50k)`

`\implies 50k=\ln\frac{19}8`

`\implies k=\frac1{50}\ln\frac{19}8\approx0.0173`

So the population is modeled by


`P(t)=2560e^(0.0173t)`

(assuming
`t=0` refers to the year 1950)