Question

Since January 1, 1960, the population of Slim Chance has been described by the formula P = 40000(0.94)', where P is the population of the city & years after the start of 1960. At what rate was the popularion changingon January 1, 1985?numerical rate of change= ___ people/yr

Answer 1

The Solution:

Given that the population of Slim Chance is described by the exponential function below:

`P(t)=40000(0.94)^t\ldots\text{eqn}(1)`

We are required to find the rate at which the population is changing, and also give a numerical rate of change in the population.

By formula, the exponential function is:

`\begin{gathered} P(t)=p_0(1-r)^t\ldots eqn(2) \\ \text{where r=rate (\%)} \end{gathered}`

Comparing eqn(1) and eqn(2), we have that:

`1-r=0.94`

Solving for r in the above equation, we get

`\begin{gathered} r=1-0.94 \\ r=0.06\text{ } \\ \text{ So, to convert to percentage, we multiply by 100 to get,} \\ r=0.06*100=6\% \end{gathered}`

To find the numerical rate of change in the population, we get

`\text{ Rate of change in the population = 6\% of 40000}=0.06*40000=2400`

Therefore, the population decreases by 2400 people every year.