Researchers use formula f (t) = P e^{r t} to model the population of fruit flies growing...

Question:

Researchers use formula {eq}f (t) = P e^{r t} {/eq} to model the population of fruit flies growing exponentially after t days. Measurements show that {eq}f (0) = 30 {/eq} and {eq}f (5) = 1620 {/eq}.

a) Find a function of the form {eq}f(t) = P e^{rt} {/eq} that models the fruit fly population after {eq}t {/eq} days.

b) When the function f is expressed in the {eq}f (x) = Pa^x {/eq}, the constant {eq}a {/eq} is the growth factor. Find the growth factor.

c) When will there be {eq}18,000 {/eq} fruit flies?

Exponential Growth

Exponential growth describes the increase of a quantity over time, wherein the derivative of the quantity is directly proportional to the quantity itself. It is commonly used to model the growth behavior of microorganisms and insects.

Answer and Explanation:

a) To determine the corresponding function, we must calculate the value of P and r using the given boundary conditions.

{eq}f (0) = 30\\ f (5) = 1620\\ {/eq}

{eq}f(0) =P e^{r(0)}=30\\ Pe^0=30\\ P=30\\ {/eq}

{eq}f(5)=30e^{r(5)}=1620\\ 30e^{5r}=1620\\ e^{5r}=\dfrac{1620}{30}\\ \ln \mid e^{5r} \mid=\ln \mid 54 \mid\\ 5r\approx 3.99\\ r\approx \dfrac{3.99}{5}\\ r\approx 0.8 {/eq}

Therefore, the exponential equation would be given by

{eq}\color{red}{f (t) = 30 e^{0.8t}}\\ {/eq}

b) Using the same set of boundary conditions, we can the determine the exponential equation in the following form

{eq}f (x) = Pa^x {/eq}

{eq}f(0)=Pa^(0)=30\\ P=30\\ {/eq}

{eq}f(5)=30a^5=1620\\ a^5=\dfrac{1620}{30}\\ a= (54)^{1/5}\\ \color{blue}{a\approx 2.22}\\ {/eq}

The equation becomes

{eq}\color{red}{f(x)=(30)(2.22)^{x}}\\ {/eq}

c) We solve for the corresponding time for P = 18,000 using both equations.

{eq}30 e^{0.8t}=18000\\ e^{0.8t}=\dfrac{18000}{30}\\ \ln \mid e^{0.8t} \mid =\ln \mid 600 \mid\\ 0.8t= 6.40 \\ t= \dfrac{6.4}{0.8}\\ \color{blue}{t = 8\,\rm days}\\[0.2cm] {/eq}

{eq}(30)(2.22)^{x}=18000\\ 2.22^x=\dfrac{18000}{30}\\ \ln \mid2.22^x \mid= \ln \mid 600 \mid\\ x\ln \mid2.22 \mid= 6.4\\ x= \dfrac{6.4}{\ln \mid2.22 \mid}\\ \color{blue}{x= 8\,\rm days} {/eq}


Learn more about this topic:

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Exponential Growth: Definition & Examples

from High School Algebra I: Help and Review

Chapter 6 / Lesson 10
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