Problem 91 - Calculus and its applications

Question

Population growth rate. In \(t\) years, the population of Kingsville grows from 100,000 to a size \(P\) given by \(P(t)=100,000+2000 t^{2}\) a) Find the growth rate, \(d P / d t\). b) Find the population after 10 yr. c) Find the growth rate at \(t=10\). d) Explain the meaning of your answer to part (c).

Step-by-Step Solution

Answer

a) \(\frac{dP}{dt} = 4000t\). b) 300,000 people. c) 40,000 people/year at \(t=10\). d) At year 10, the population grows by 40,000 annually.

Step 1: Find the growth rate formula

To find the growth rate, we take the derivative of the population equation with respect to time, \(t\). Given, \(P(t) = 100,000 + 2000t^2\), the derivative \(\frac{dP}{dt}\) will give us the growth rate.Thus, \(\frac{dP}{dt} = \frac{d}{dt}(100,000 + 2000t^2) = 0 + 4000t = 4000t\).

Step 2: Calculate the population after 10 years

Insert \(t = 10\) into the population formula \(P(t) = 100,000 + 2000t^2\).\[ P(10) = 100,000 + 2000(10)^2 = 100,000 + 2000 \times 100 = 100,000 + 200,000 = 300,000 \]So, the population after 10 years is 300,000.

Step 3: Calculate the growth rate at 10 years

Insert \(t = 10\) into the growth rate formula \(\frac{dP}{dt} = 4000t\).\(\frac{dP}{dt}\bigg|_{t=10} = 4000 \times 10 = 40,000\)Thus, the growth rate at \(t = 10\) is 40,000.

Step 4: Interpret the growth rate at 10 years

The growth rate of 40,000 at \(t = 10\) means that at year 10, the population of Kingsville is increasing by 40,000 people per year. This indicates the speed at which the population is growing at that specific time.

Key Concepts

Understanding the Derivative

The concept of the derivative is fundamental in calculus and helps us understand how functions change. In the context of population dynamics, the derivative allows us to calculate how quickly or slowly a population is growing over time. Using a simple function like the one given for Kingsville, which is \( P(t) = 100,000 + 2000t^2 \), the derivative \( \frac{dP}{dt} \) can provide an accurate measure of the rate of population change.

• The derivative of a constant is zero. Here, the constant is 100,000, which does not change, and thus its contribution to the derivative is zero.
• The derivative of a term like \(2000t^2\) is calculated by bringing down the exponent 2 and multiplying it with the coefficient 2000, resulting in \(4000t\).

By finding the derivative of the population equation, we get \( \frac{dP}{dt} = 4000t \). This expression now represents a growth rate function that varies with time \(t\). You can plug different values of \(t\) into this function to find out the growth rate at those specific times.

Growth Rate Calculation - A Practical Approach

Calculating the growth rate helps us evaluate how a population size changes over time. This is particularly useful for predicting future populations or understanding trends. In our example of Kingsville, the derivative \( \frac{dP}{dt} = 4000t \) serves as the growth rate formula.

• To find the population's growth rate at any given time \( t \), substitute \( t \) into the derivative equation. For example, if \( t = 10 \), the growth rate would be \( 4000 \times 10 = 40,000 \).
• This result means that at year 10, the population is increasing by 40,000 people per year, reflecting a dynamic pace of change over time.

Having this formula allows you to determine how various factors might affect population changes. It's a versatile tool for various scenarios—whether planning urban developments or resource management. Interpreting these results, you get a clear sense of how rapidly a population grows at particular instances.

Exploring Population Dynamics

Population dynamics is the study of how and why populations change over time. The mathematical models, like the ones used for Kingsville, reveal insights into the nature of these changes.

• For Kingsville, the initial population is 100,000, and it increases as a function of time with a quadratic term influencing the speed of growth.
• After 10 years, the formula \( P(t) \) predicts a population of 300,000, highlighting an increase influenced by the quadratic nature of \( 2000t^2 \).

Such models are essential for ecologists, urban planners, and policy developers, providing data to draft strategies that accommodate or guide population growth. The growth rate interpretation, like how 40,000 people per year at \( t = 10 \) reflects the town's rapid growth phase, is essential for understanding wider population dynamics. They show us not just the numbers but the implications for real-world scenarios—how infrastructure, housing, and resource demands will evolve.