/*! 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 50 Homeostasis refers to a state in... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Homeostasis refers to a state in which the nutrient content of a consumer is independent of the nutrient content of its food. In the absence of homeostasis, a model proposed by Sterner and Elser is given by $$ \frac{d y}{d x}=\frac{1}{\theta} \frac{y}{x} $$ where \(x\) and \(y\) represent the nutrient content of the food and the consumer, respectively, and \(\theta\) is a constant with \(\theta \geqslant 1\) $$\begin{array}{l}{\text { (a) Solve the differential equation. }} \\ {\text { (b) What happens when } \theta=1 ? \text { What happens when }} \\ {\quad \theta \rightarrow \infty ?}\end{array}$$

Short Answer

Expert verified
(a) The solution is \( y = C' x^{1/\theta} \). (b) With \( \theta = 1 \), \( y \) is directly proportional to \( x \). As \( \theta \to \infty \), \( y \) approaches a constant, indicating homeostasis.

Step by step solution

01

Understand the given differential equation

The given differential equation \( \frac{d y}{d x}=\frac{1}{\theta} \frac{y}{x} \) is a first-order linear differential equation which involves the variables \(x\) and \(y\), and \(\theta\) is a constant. Our task is to find the expression for \(y\) in terms of \(x\).
02

Separate the variables

The next step is to separate the variables. Rearrange the equation \(\frac{d y}{d x}=\frac{1}{\theta} \frac{y}{x}\) to separate \(y\) on one side and \(x\) on the other side:\[ \frac{dy}{y} = \frac{1}{\theta} \frac{dx}{x} \]
03

Integrate both sides

Integrate both sides of the equation \( \frac{dy}{y} = \frac{1}{\theta} \frac{dx}{x} \):\[ \int \frac{dy}{y} = \int \frac{1}{\theta} \frac{dx}{x} \]The left-hand side gives \( \ln|y| \) and the right-hand side gives \( \frac{1}{\theta} \ln|x| + C \) where \( C \) is the integration constant.
04

Solve for y

To solve for \( y \), exponentiate both sides of the equation:\[ y = C' x^{1/\theta} \]where \( C' = e^C \) is a new constant.
05

Answer part (a)

The solution to the differential equation is:\[ y = C' x^{1/\theta} \]where \( C' \) is an arbitrary constant.
06

Analyze \( \theta = 1 \)

If \( \theta = 1 \), \( y = C'x \). This indicates a direct proportionality between \( y \) and \( x \), meaning the nutrient content of the consumer mirrors that of the food.
07

Analyze \( \theta \rightarrow \infty \)

As \( \theta \to \infty \), \( y \approx C' \), suggesting that the nutrient content of the consumer becomes nearly independent of \( x \), representing the concept of homeostasis.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Homeostasis
Homeostasis is a fundamental concept in biology, referring to the ability of organisms to maintain a stable internal environment despite changes in the external surroundings. This stability is crucial for the survival and proper functioning of living organisms.
In the context of nutrient content, homeostasis implies that the consumer (be it an organism or a cell) can regulate its internal nutrient levels regardless of the fluctuating nutrient content in its food supply.
This means that instead of having a direct relationship between the nutrients available in their environment and their internal nutrient content, organisms can remain balanced. This balance helps them survive through varying external nutrient conditions.
To put it simply:
  • Homeostasis helps maintain internal stability.
  • It allows organisms to survive in diverse environments.
  • It involves complex feedback mechanisms to adjust to changes.
Understanding homeostasis provides insights into how organisms thrive and reproduce in ever-changing ecosystems.
Nutrient Content Modeling
Nutrient content modeling refers to the mathematical representation of the nutrient dynamics within ecological systems. In simpler terms, it's about using equations to understand how nutrients move from one part of an ecosystem, like food, to another, like a consumer.
  • This type of modeling allows biologists and ecologists to predict how an organism's nutrient levels change over time.
  • It provides insights into how nutrient availability affects growth, health, and reproductive success.
  • Such models can simulate different scenarios, helping scientists evaluate the impact of changing environments or human activities on ecosystems.
The Sterner and Elser model is a classic example of nutrient content modeling used to explore what happens when there is no homeostatic regulation. It reveals that without regulation, an organism's nutrient content would depend directly on its food. The model uses differential equations to describe these processes accurately.
Through modeling, we gain a deeper understanding of the delicate balance organisms need to maintain in order to sustain their populations over generations.
First-Order Linear Differential Equations
First-order linear differential equations are foundational tools in both mathematics and biological modeling. These equations involve derivatives and represent rates of change in various contexts.
  • A first-order equation means only the first derivative of the unknown function (here, nutrient content, represented as \( y \)) is present.
  • Linear implies that the unknown function and its derivatives appear to the power of one, meaning they are not multiplied or divided among themselves.
  • These equations are invaluable in predicting how a system evolves over time based on initial conditions.
The given exercise presents the differential equation \( \frac{d y}{d x}=\frac{1}{\theta} \frac{y}{x} \). It is particularly insightful as it demonstrates the interaction between nutrient content in food and in a consumer.
Solving such differential equations involves techniques like separation of variables and integration, leading to solutions that describe the relationship between nutrient content in consumers and their food under different conditions of \( \theta \).
Understanding these equations is essential for students of biology and ecology, as they form the mathematical underpinning of many models depicting real-world biological processes.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Find all equilibria and use the local stability criterion to determine if each is locally stable or unstable. Then construct the phase plot. (a) $$y^{\prime}=5 y\left(2 e^{-y}-1\right)$$ (b) $$y^{\prime}=y^{5 / 3}-2 y \quad (assume y \geqslant 0 )$$

A system of differential equations is given. (a) Construct the phase plane, plotting all nullclines, labeling all equilibria, and indicating the direction of motion. (b) Obtain an expression for each equilibrium. \(p^{\prime}=-p^{2}+q-1, \quad q^{\prime}=q(2-p-q)\)

Use a computer algebra system to draw a direction field for the given differential equation. Get a printout and sketch on it the solution curve that passes through \((0,1) .\) Then use the CAS to draw the solution curve and compare it with your sketch. \(y^{\prime}=x\left(y^{2}-4\right)\)

(a) Use a computer algebra system to draw a direction field for the differential equation. Get a printout and use it to sketch some solution curves without solving the differential equation. (b) Solve the differential equation. (c) Use the CAS to draw several members of the family of solutions obtained in part (b). Compare with the curves from part (a) \(y ^ { \prime } = x y\)

Modeling yeast populations In Section 7.1 we introduced the following differential equation to describe the dynamics of an experimental yeast population: $$\frac{d N}{d t}=(0.55-0.0026 N) N$$ where \(N(t)\) is the population size (in millions of individuals per mL) at time \(t\) (in hours). Sketch the direction field of the differential equation for values of \(N\) between 0 and \(250 .\)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.