Copyright

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}



Learn more about this topic:

Loading...
Exponential Growth: Definition & Examples

from High School Algebra I: Help and Review

Chapter 6 / Lesson 10
105K

Related to this Question

Explore our homework questions and answers library