# The population of a city is growing at a rate of 3 thousand people per year. The city starts with...

## Question:

The population of a city is growing at a rate of 3 thousand people per year. The city starts with a population of 250 thousand people, and we assume that the population grows exponentially. Write the initial value problem for the city's population and solve the initial value problem.

## Modeling Exponential Growth and decay:

A process is said to undergo exponential change if the relative rate of change of the process is modeled by a constant. In other words,

{eq}\hspace{30mm} \displaystyle{ \dfrac{f'(t)}{f(t)} = k \\ \dfrac{df}{dt} = k f(t) } {/eq}

If {eq}k {/eq} is positive, the exponential change is called growth, and if {eq}k {/eq} is negative, then exponential change is called decay.

## Answer and Explanation:

Let {eq}P(t) {/eq} denote the population of city at time {eq}t {/eq} year. And the rate of growth of the population is {eq}3000 {/eq} thousand per year. Therefore the exponential model of the given problem is:

{eq}\hspace{30mm} \displaystyle{ \dfrac{dP}{dt} = 3 P } {/eq}

And it is given that the city starts with a population of {eq}250 {/eq} thousand people. Therefore

{eq}\hspace{30mm} \displaystyle{ P(0) = 250 } {/eq}

Hence the initial value problem is:

{eq}\hspace{30mm} \displaystyle{ \dfrac{dP}{dt} = 3 P \ , \qquad \ \ \ \ P(0) = 250 \\ \dfrac{dP}{P} = 3 dt \\ \ln P = 3t + \ln C \\ P(t) = C e^{3t } } {/eq}

Now use the initial condition:

{eq}\hspace{30mm} \displaystyle{ P(0) = 250 \\ C = 250 } {/eq}

Therefore the solution of the initial value problem is

{eq}\hspace{30mm} \displaystyle{ P(t) = 250 e^{3t } } {/eq}

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from High School Algebra I: Help and Review

Chapter 6 / Lesson 10