Copyright

The population of a town is growing at a rate given by \displaystyle\frac { d P } { d t } = 132 -...

Question:

The population of a town is growing at a rate given by {eq}\displaystyle\frac { d P } { d t } = 132 - 12 t ^ { \frac { 2 } { 3 } } {/eq} people per year. Find a function to describe the population {eq}t {/eq} years from now if the present population is 9800 people.

Differential Equation:

Integration of a function f(x) over an interval x= a to x=b is defined as {eq}\displaystyle \in_{a}^{b}f(x) dx {/eq}

Formulae used:

{eq}\displaystyle * \int (x^n) dx= \frac{x^{n+1}}{n+1} {/eq}

We have used an assumption of p(0)=9800 since population after 't' years from is asked

Answer and Explanation:

Given, The population of a town is growing at a rate given by

{eq}\displaystyle \frac{ d P }{ d t } = 132 - 12 t ^ { \frac{ 2 }{ 3 } }......(1) {/eq}people per year.

Popoulation after 't' years is obtained by integrating (1)

{eq}\displaystyle \begin{align} \frac{ d P }{ d t } &= 132 - 12 t ^ { \frac{ 2 }{ 3 } } \\ \int dp &=\int (132 - 12 t ^ { \frac{ 2 }{ 3 } }) dt \\ p &= 132t - 12 \frac{t ^ {( \frac{ 2 }{ 3 }+1) }}{(\frac{2}{3}+1)} +c \quad \left[ \int (x^n) dx= \frac{x^{n+1}}{n+1}\right] \\ &= 132t -(12)\frac{3}{5}t ^ { \frac{ 5 }{ 3 } } +c\\ &= 132t -\frac{36}{5}t ^ { \frac{ 5 }{ 3 } }+c \\ p &= 132t -\frac{36}{5}t ^ { \frac{ 5 }{ 3 } }.+c....(2) \\ \end{align} {/eq}

But given present population of the town {eq}\displaystyle p(0) = 9800 {/eq}

Substituting in (2)

{eq}\displaystyle \begin{align} p &= 132t -\frac{36}{5}t ^ { \frac{ 5 }{ 3 } } \\ 9800 &= 132(0)-\frac{36}{5} (0)+c \\ \rightarrow c &= 9800 \end{align} {/eq}

Population equation of the town is {eq}\displaystyle p(t) = 132t -\frac{36}{5}t ^ { \frac{ 5}{ 3 } }+9800 {/eq}


Learn more about this topic:

Loading...
Separable Differential Equation: Definition & Examples

from GRE Math: Study Guide & Test Prep

Chapter 16 / Lesson 1
3.5K

Related to this Question

Explore our homework questions and answers library