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A major corporation is building a 4325-acre complex of homes, offices, stores, schools, and churches in the rural community of Glen Cove. As a result of this development, the planners have estimated that Glen Cove's population (in thousands) \(t\) yr from now will be given by $$ P(t)=\frac{25 t^{2}+125 t+200}{t^{2}+5 t+40} $$ a. Find the rate at which Glen Cove's population is changing with respect to time. b. What will be the population after 10 yr? At what rate will the population be increasing when \(t=10\) ?

Short Answer

Expert verified
a. The rate at which Glen Cove's population is changing with respect to time is given by the derivative of the population function: $$ \frac{dP}{dt}=\frac{250t-625}{(t^2 + 5t + 40)^2} $$ b. After 10 years, the population of Glen Cove will be \(\frac{125}{2}\) thousands. At this time, the rate at which the population is increasing is \(\frac{1750}{20^4}\) people per year.

Step by step solution

01

a. Differentiate the population function with respect to time t

To find the rate at which Glen Cove's population is changing with respect to time, we need to differentiate the given population function \(P(t)\) with respect to time \(t\). Here, $$ P(t) = \frac{25t^2 + 125t + 200}{t^2 + 5t + 40} $$ Differentiate \(P(t)\) with respect to \(t\): $$ \frac{dP}{dt} = \frac{(50t + 125)(t^2 + 5t + 40) - (25t^2 + 125t + 200)(2t + 5)}{(t^2 + 5t + 40)^2} $$ Now simplify the expression: $$ \frac{dP}{dt} = \frac{50t^3 + 375t^2 + 750t + 6250 - (50t^3 + 375t^2 + 1000t^2 + 625t)}{(t^2 + 5t + 40)^2} $$ Combining the terms, we get: $$ \frac{dP}{dt}=\frac{250t-625}{(t^2 + 5t + 40)^2} $$ Now, we have the rate at which the Glen Cove's population is changing with respect to time.
02

b. Find the population after 10 years and the rate of increase at that time

To find the population after 10 years, we have to plug in the value of \(t=10\) into the population function \(P(t)\): $$ P(10) = \frac{25(10)^2 + 125(10) + 200}{(10)^2 + 5(10) + 40} = \frac{2500}{200} = \frac{125}{2} $$ So, after 10 years, the population of Glen Cove will be \(\frac{125}{2}\) thousands. To find the rate at which the population is increasing when \(t=10\), we have to plug in the value of \(t=10\) into the derivative of population function: $$ \frac{dP}{dt}(10)= \frac{250(10)-625}{(10^2 + 5(10) + 40)^2} = \frac{1750}{20^4} $$ So, when \(t=10\), the population will be increasing at \(\frac{1750}{20^4}\) people per year.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Population Growth
Population growth refers to the increase in the number of individuals in a population over time. This concept is crucial in various fields such as ecology, urban planning, and economics. In the context of Differential Calculus, we can model population growth using mathematical equations that help predict future population size based on current data.
Understanding population growth involves appreciating how various factors contribute to changes in population size. For Glen Cove, as more residential and commercial facilities like homes and offices are developed, the population is expected to grow. This is because people are attracted to the amenities and job opportunities in the area.
Additionally, a population model, like the one given in the exercise, allows planners to estimate how the population might grow over time. This model uses variables such as time, represented by "\( t \)," to predict future population values. Such predictions are essential for resource planning, infrastructure development, and community services.
Rate of Change
The rate of change is a critical concept in calculus and describes how a quantity increases or decreases over time. For population growth, the rate of change tells us how quickly the population is expected to grow or shrink.
When we differentiate a population function with respect to time, we obtain the rate of change of the population. In simpler terms, it is the derivative of the population with respect to time, denoted as \( \frac{dP}{dt} \). In Glen Cove's case, this tells planners how rapidly the population is expected to increase as infrastructure is developed.
Calculating \( \frac{dP}{dt} \) involves applying differentiation rules (like the quotient rule, if the function is a fraction) to the population function \( P(t) \). This will result in a new function that describes the velocity of population growth at any point in time. For instance, at \( t = 10 \), we found the growth rate using this derivative, indicating how many people are expected to be added to the population per year at that specific time.
Mathematical Modeling
Mathematical modeling is the process of using mathematical expressions to represent real-world phenomena. It's a powerful tool in understanding and predicting outcomes in various fields like physics, biology, and economics.
In the context of population growth, a mathematical model such as \( P(t) = \frac{25 t^{2}+125 t+200}{t^{2}+5 t+40} \) creates a structure that helps us predict how Glen Cove's population will evolve over time. Such models rely on historical data and assumptions which help in forming the function that describes how variables interact.
Modeling involves making assumptions about initial conditions and external factors that might influence the system. By analyzing this model, urban planners can effectively anticipate changes brought about by new developments. This method supports decision-making processes by providing a simulated forecasting scenario of future population sizes and growth rates. Mathematical models thus offer a predictive insight that can lead to better planning and resource allocation.
  • Models simplify complex real-world processes into manageable computations.
  • They help test different "what if" scenarios to see potential outcomes.
  • Quantitative models often require validation and periodic updates to remain accurate over time.

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