/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 4 A herd of deer has an initial po... [FREE SOLUTION] | 91Ó°ÊÓ

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A herd of deer has an initial population of 10 at time \(t=0\), with \(t\) in years. a. If the size of the herd increases by 8 per year, find the formula for the population of deer, \(P(t),\) over time. b. If the size of the herd increases by a factor of 1.8 each year, find the formula for the deer population, \(Q(t),\) over time. c. For each model create a table of values for the deer population for a 10-year period. d. Using the table, estimate when the two models predict the same population size.

Short Answer

Expert verified
a. \(P(t) = 10 + 8t\), b. \(Q(t) = 10 \times 1.8^t\), c. Tables needed, d. Compare to find intersection.

Step by step solution

01

- Understand the Linear Growth

The first part of the problem involves linear growth. The population increases by a fixed amount each year.
02

- Define Linear Growth Formula

To find the population of deer over time for linear growth, use the formula: \(P(t) = P_0 + rt\). Here, \(P_0\) is the initial population (10 deer), and \(r\) is the growth rate per year (8 deer/year).
03

- Substitute Values into Linear Formula

Substitute the given values: \(P_0 = 10\) and \(r = 8\). So the formula becomes \(P(t) = 10 + 8t\).
04

- Understand Exponential Growth

The second part of the problem involves exponential growth where the population increases by a fixed factor each year.
05

- Define Exponential Growth Formula

To find the population of deer over time for exponential growth, use the formula: \(Q(t) = Q_0 \times a^t\). Here, \(Q_0\) is the initial population (10 deer), and \(a\) is the growth factor per year (1.8).
06

- Substitute Values into Exponential Formula

Substitute the given values: \(Q_0 = 10\) and \(a = 1.8\). So the formula becomes \(Q(t) = 10 \times 1.8^t\).
07

- Create Table for Linear Model

Calculate the population for each year from \(t = 0\) to \(t = 10\) using \(P(t) = 10 + 8t\).
08

- Create Table for Exponential Model

Calculate the population for each year from \(t = 0\) to \(t = 10\) using \(Q(t) = 10 \times 1.8^t\).
09

- Compare Tables

Compare the values in both tables to find the year when both models predict the same population size.

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Key Concepts

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

linear growth
Linear growth is a type of progression where the population increases consistently by the same amount every period. In our exercise, the deer population grows by 8 deer each year. This uniform addition to the population can be modeled by the formula: ewline ewline \[P(t) = P_0 + rt\] ewline Let's break down the formula: ewline
  • \(P(t)\) represents the population at time \(t\).
  • \(P_0\) is the initial population, which is 10 deer.
  • \(r\) is the rate of growth per year, which is 8 deer.
By substituting the given values, we get:ewline \[P(t) = 10 + 8t\] This tells us that every year, the population will increase by 8. For instance, after 1 year (\(t=1\)), the population will be 18; after 2 years (\(t=2\)), it will be 26, and so on.
exponential growth
Exponential growth occurs when populations grow by a fixed percentage each year, resulting in a much faster increase over time compared to linear growth. In the exercise, the deer population grows by a factor of 1.8 each year. This can be modeled by the formula:ewline \[Q(t) = Q_0 \times a^t\] Similar to linear growth, let's break this formula down:
  • \(Q(t)\) represents the population at time \(t\).
  • \(Q_0\) is the initial population, which is 10 deer.
  • \(a\) is the growth factor, which is 1.8.
By substituting the given values, we get: \[ Q(t) = 10 \times 1.8^t\] This function tells us that the population is multiplied by 1.8 each year. For example, after 1 year (\(t=1\)), the population will be 18; after 2 years (\(t=2\)), it will be approximately 32.4, and so forth. Compared to linear growth, exponential growth produces a steeper rise over time.
population dynamics
Population dynamics study how and why populations change over time. In the context of the deer herd, we analyzed two types of growth: linear and exponential. Each type has different implications.ewlineLinear growth is steady and predictable, with the population increasing by a fixed number each year. In our scenario, the population grows by 8 deer annually, leading to a linear progression. This can be excellent for short-term projections but might not hold for long-term predictions as environmental limits come into play.ewlineExponential growth represents more rapid population increases, reflecting real biological scenarios where resources are initially abundant. The deer population growing by a factor of 1.8 annually results in a quickly rising population. This model might be more accurate for populations with plenty of resources but can also lead to issues like overpopulation and resource depletion in the long run.ewlineUnderstanding these growth patterns helps predict population trends and manage wildlife conservation effectively.
algebraic functions
Algebraic functions are essential tools in mathematical modeling. They provide ways to represent relationships and changes, like the population growth in our deer herd problem. Two specific types of algebraic functions we used are linear and exponential functions:ewline
  • Linear functions like \(P(t) = 10 + 8t\) are straightforward, denoting a constant rate of change over time.
  • Exponential functions like \(Q(t) = 10 \times 1.8^t\) show how quantities grow by a consistent percentage, leading to faster increases over time.
Using these functions, we can create models and predict population sizes. By calculating and comparing values for each year, we see how the population changes under different growth conditions.ewlineFor instance, after 3 years, the linear model predicts a population of 34 deer, while the exponential model predicts about 58.32 deer. Algebraic functions give us the framework to make these calculations and understand different scenarios in population dynamics.

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