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The population \(y\) (in thousands) of a colony of bacteria after \(t\) hr is given by $$ y=(6 t+12) /(2 t+1) $$ where \(t \geq 0\) (a) Find the initial population and the long-term population. Which is larger? (b) Use a graphing utility to graph the population function. Is the function increasing or decreasing? Check that your response here is consistent with your answers in part (a).

Short Answer

Expert verified
The initial population is 12; the long-term population is 3. The function decreases.

Step by step solution

01

Find the Initial Population

The initial population occurs when \( t = 0 \). Substitute \( t = 0 \) into the function \( y(t) = \frac{6t + 12}{2t + 1} \). This gives:\[ y(0) = \frac{6(0) + 12}{2(0) + 1} = \frac{12}{1} = 12 \]So, the initial population is 12 (in thousands of bacteria).
02

Determine the Long-Term Population

The long-term population is found by determining the horizontal asymptote as \( t \to \infty \). For the function \( y(t) = \frac{6t + 12}{2t + 1} \), we consider the coefficients of the highest power term in the numerator and denominator:\[ y(t) \approx \frac{6t}{2t} = 3 \]Therefore, the long-term population as \( t \to \infty \) is 3 (in thousands of bacteria).
03

Compare Initial and Long-Term Population

The initial population of 12 (thousand bacteria) is larger than the long-term population of 3 (thousand bacteria). This indicates that the population decreases over time.
04

Graph the Population Function

Using a graphing utility, graph the function \( y(t) = \frac{6t + 12}{2t + 1} \). As \( t \) increases, the graph shows that the population decreases, approaching the horizontal asymptote at \( y = 3 \).
05

Verify Increasing or Decreasing Trend

The graph supports our earlier finding from Step 3: the function is decreasing over time, which is consistent with the initial population being larger than the long-term population.

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Key Concepts

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

Initial Population
The initial population refers to the number of individuals at the very beginning of our observation, which is usually when time, denoted as \( t \), is zero. For the given problem, the population function is defined by the equation:
\[ y = \frac{6t + 12}{2t + 1} \]
To find the initial population, we substitute \( t = 0 \) into the function. This allows us to determine what the population quantity is when time has not yet started passing:
- Substitute \( t = 0 \) into the formula: \[ y(0) = \frac{6(0) + 12}{2(0) + 1} = \frac{12}{1} = 12 \]
From this process, it becomes clear that the initial population is 12 thousand bacteria. This number gives us a starting point, from which we can observe how the population changes overtime under the given conditions.
Long-Term Population
To understand the long-term population, we look at the behavior of the function as time goes to infinity (\( t \to \infty \)). This involves finding what is known as the horizontal asymptote of the function. A horizontal asymptote helps us predict the population when enough time has passed so that transient dynamics have settled.
For the function \( y(t) = \frac{6t + 12}{2t + 1} \), as \( t \) becomes very large, the behavior is dominated by the highest degree terms in both the numerator and the denominator:
  • The numerator's highest degree term is \( 6t \).
  • The denominator's highest degree term is \( 2t \).
When simplifying the function for large \( t \), we find:
\[ y(t) \approx \frac{6t}{2t} = 3 \]
Thus, as \( t \to \infty \), the long-term population approaches 3 thousand bacteria. This asymptotic analysis tells us that over a long period, the bacterial population will stabilize at this lower figure compared to its initial state.
Graphing Utility
A graphing utility can visually represent the changes in population by plotting the function \( y(t) = \frac{6t + 12}{2t + 1} \) over a suitable range of time \( t \). This practical tool helps verify theoretical calculations and provides a clear picture of how the population evolves over time.
When using a graphing utility, input the function and set a range for \( t \). As the graph is drawn, several key details emerge:
  • At \( t = 0 \), the population starts at 12, confirming our initial population finding.
  • As \( t \) increases, the curve approaching a horizontal line at \( y = 3 \) indicates the long-term population.
  • The trend shows a decreasing function, from 12 to 3, matching our calculations regarding the population’s decline over time.
Graphing functions like this one make abstract concepts more concrete and allow for a better understanding of trends and behaviors in mathematical models.

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Most popular questions from this chapter

A line is drawn from the origin \(O\) to a point \(P(x, y)\) in the first quadrant on the graph of \(y=1 / x .\) From point \(P,\) a line is drawn perpendicular to the \(x\) -axis, meeting the \(x\) -axis at \(B\) (a) Draw a figure of the situation described. (b) Express the perimeter of \(\triangle O P B\) as a function of \(x\).(c) Try to express the area of \(\triangle O P B\) as a function of \(x\) What happens?

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