A fish population is approximated by P(t) = 10e^{0.6t}, where t is in months. Calculate and use...


A fish population is approximated by {eq}P(t) = 10e^{\large 0.6t} {/eq}, where t is in months. Calculate and use units to explain what each of the following tells us about the population:

(a) P(12)

(b) P'(12)

Population modeling

Population modeling is a technique in biology where the population of organisms is described by a function. The most common functions used in modeling exponential functions and logistic functions of time. An exponential growth function for population is given by {eq}P(t) = Ae^{kt} {/eq}.

Answer and Explanation:

The population of fish is given by the function

{eq}P(t) = 10e^{0.6 t} {/eq}

where P(t) is the population or the number of fish and t is the time in months.

a.) The expression P(12) is the value of the function at t = 12 months. This means that the value P(12) is the number of fish in 12 months.

Solving for P(12).

{eq}\displaystyle \begin{align*} P(12) &= 10e^{0.6(12)} \\ &= 13394\ \rm{fishes} \end{align*} {/eq}

This means that the population of fish in 12 months is at 13394 fish.

b.) The expression, P'(12) is the derivative of the function with respect to time at t = 12. We can write P'(t) explicitly as

{eq}\displaystyle P'(t) = \frac{dP}{dt}. {/eq}

The units of the derivative is {eq}\displaystyle \frac{\Delta \text{ Population}}{\Delta \text{time}} \to \frac{\text{number of fish}}{\text{month}} {/eq}.

The derivative with respect to time refers to the rate of change in the population of fish. Therefore the term P'(12) is the rate of change in population of the fish in 12 months.

Calculating P'(12).

{eq}\displaystyle \begin{align*} P'(t) &= \frac{dP}{dt} \\ &= \frac{dP}{dt} 10e^{0.6t} \\ &= 0.6( 10e^{0.6t})\\ P'(t) &= 6e^{0.6t}\\ P'(12) &= 6 e^{0.6(12)}\\ P'(12) &= 8037\ \rm{fish\ per\ month} \end{align*} {/eq}

This means that in 12 months the rate of change in the population of fish is 8037 fish per month. The positive rate of change means that the population is increasing at a rate of 8037 fish per month.

Learn more about this topic:

Exponential Growth: Definition & Examples

from High School Algebra I: Help and Review

Chapter 6 / Lesson 10

Related to this Question

Explore our homework questions and answers library