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Question

In April 2015 in the US, there was one birth every 8 seconds, one death every 12 seconds, and one new international migrant every 32 seconds. (a) Let f(t) be the population of the US, where t is time in seconds measured from the start of April 2015. Find f'(0). Give units. (b) To the nearest second, how long did it take for the US population to add one person in April 2015?

Solution

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In April 2015 in the US, there was one birth every 8 seconds, one death every 12 seconds, and one new international migrant every 32 seconds.

(a)

$f(t)$ is the population of the US, where $t$ is time in seconds measured from the start of April 2015. Let $b$, $d$ and $m$ is the birth, death and migration rates form the start of April 2015. Therefore,

\begin{align*} b & = \dfrac{1}{8}\ \mathrm{people/sec} \\ d & = -\dfrac{1}{12}\ \mathrm{people/sec} \\ m & = \dfrac{1}{32}\ \mathrm{people/sec} \end{align*}

Therefore, the total rate of the change of the population is the addition of all of these rates.

\begin{align*} f'(0) & = b + d+ m \\ f'(0) & = \dfrac{1}{8} - \dfrac{1}{12} + \dfrac{1}{32}\ \mathrm{people/sec} \\ f'(0) & = \dfrac{7}{96}\ \mathrm{people/sec} \\ f'(0) & = 0.073\ \mathrm{people/sec} \\ &\hspace*{-9mm}\boxed{f'(0)= 0.073\ \mathrm{people/sec} } \end{align*}

The population is increasing at 0.073 people/sec.

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