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The population of a city has increased since 1985 at a rate of \(1.5+0.3 \sqrt{t}+0.006 t^{2}\) thousand people per year, where \(t\) is the number of years after 1985 . Assuming that this rate continues and that the population was 50,000 in \(1985,\) estimate the population in \(1994 .\)

Short Answer

Expert verified
The estimated population in 1994 is 70,358.

Step by step solution

01

Understand the Growth Rate Function

The city's population growth rate is given by the function \( R(t) = 1.5 + 0.3\sqrt{t} + 0.006 t^2 \), where \( t \) is the number of years after 1985. This rate tells us how many thousands of people are added per year, starting from the population in 1985.
02

Determine the Time Interval

We need to find the population in 1994. Since \( t \) is the number of years after 1985, we calculate \( t = 1994 - 1985 = 9 \). So, we are interested in the population change from 1985 to 1994, which means \( t = 9 \) years.
03

Integrate the Growth Rate Function

To find the total change in population, integrate the function \( R(t) \) from \( t = 0 \) to \( t = 9 \). This integral gives us the total number of thousands of people added to the population over 9 years. \[ \int_{0}^{9} \left( 1.5 + 0.3\sqrt{t} + 0.006 t^2 \right) \ dt \]
04

Calculate the Definite Integral

First, compute the integral of each term separately:- \( \int 1.5 \ dt = 1.5t \) - \( \int 0.3\sqrt{t} \ dt = \frac{0.3}{1.5} t^{1.5} = 0.2t^{1.5} \)- \( \int 0.006 t^2 \ dt = \frac{0.006}{3} t^3 = 0.002 t^3 \)Evaluate the integral from \( t = 0 \) to \( t = 9 \):\[ \left[ 1.5t + 0.2t^{1.5} + 0.002t^3 \right]_{0}^{9} \]
05

Evaluate the Expression at Upper and Lower Limits

Compute the expression at \( t = 9 \):\( 1.5(9) + 0.2(9^{1.5}) + 0.002(9^3) = 13.5 + 0.2(27) + 0.002(729) = 13.5 + 5.4 + 1.458 \)This equals 20.358. For \( t = 0 \), the value is 0. Thus, the total increase in population is 20.358 thousand people.
06

Calculate the Population in 1994

The population increased by 20.358 thousand people from 1985 to 1994. The initial population in 1985 was 50,000. Therefore, the population in 1994 is:\( 50000 + 20358 = 70358 \)

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Key Concepts

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

Differential Equations
Differential equations play a crucial role in modeling various real-world phenomena, and population growth is one such area. When we talk about differential equations, we're referring to equations that involve functions and their derivatives. These equations model how a particular quantity changes over time.
The growth of a population can often be represented as a rate of change, which is where differential equations come in handy. In the context of our exercise, the given growth rate function, \( R(t) = 1.5 + 0.3\sqrt{t} + 0.006 t^2 \), is in differential equation form. The function \( R(t) \) tells us the rate at which the population is increasing every year, expressed in thousands of people.
Solving a differential equation allows us to predict future values of a quantity—in this case, the future population. By integrating the growth rate function, the differential equation helps us determine the total change in population over a specified period.
Definite Integral
The definite integral is a powerful tool in calculus, used to compute the total accumulation of a quantity over a certain interval. In our exercise, the definite integral helps determine the total population increase from 1985 to 1994.
By integrating the growth rate function \( R(t) = 1.5 + 0.3\sqrt{t} + 0.006 t^2 \) over the interval from \( t = 0 \) to \( t = 9 \), we can calculate the total number of thousands of people added to the city's population during these 9 years. Mathematically, this is expressed as finding \[ \int_{0}^{9} (1.5 + 0.3\sqrt{t} + 0.006 t^2) \, dt \].
The process involves calculating the integral of each component of \( R(t) \), which when evaluated from \( t = 0 \) to \( t = 9 \), gives the total change in the population. This method effectively accumulates the yearly changes in population over the given time period.
Growth Rate Function
Understanding the growth rate function is key to modeling population dynamics. This function characterizes how the population changes over time and is essential for making predictions about future populations.
The given function, \( R(t) = 1.5 + 0.3\sqrt{t} + 0.006 t^2 \), provides the rate of population increase "per year" and is described in thousands of people. It incorporates both constant and variable components:
  • A constant term (\(1.5\)), which contributes a steady increase annually.
  • A square root term (\(0.3\sqrt{t}\)), which implies a diminishing growth effect as time passes, suggesting early acceleration.
  • A quadratic term (\(0.006 t^2\)), indicating increasing growth impact over time, representing long-term acceleration.
Analyzing such a growth rate function allows us to understand how different factors contribute to population growth and how these factors combine to influence the total growth trajectory over time.

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