/*! 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 38 Denote the size of a population ... [FREE SOLUTION] | 91Ó°ÊÓ

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Denote the size of a population by \(N(t)\), and assume that \(N(t)\) satisfies $$ \frac{d N}{d t}=N\left(1-\frac{N}{K}\right)-N \ln N $$ where \(K\) is a positive constant. (a) Show that if \(K>1\), then there exists a nontrivial equilibrium \(N^{*}>0\) that satisfies $$ 1-\frac{N^{*}}{K}=\ln N^{*} $$ (b) Assume now that the nontrivial equilibrium \(N^{*}\) is a function of the parameter \(K\). Use implicit differentiation to show that \(N^{*}\) is an increasing function of \(K\).

Short Answer

Expert verified
If \(K>1\), a nontrivial equilibrium \(N^*>0\) exists. \(N^*\) increases with \(K\).

Step by step solution

01

Understanding Equilibrium Points

An equilibrium point is where the population size does not change over time. Mathematically, this means solving \(\frac{dN}{dt} = 0\). Given the equation \(\frac{dN}{dt}=N\left(1-\frac{N}{K}\right)-N \ln N\), we set it to zero: \(N\left(1-\frac{N}{K}\right)-N \ln N = 0\). Simplifying, we divide both sides by \(N\) (assuming \(N eq 0\)): \(1-\frac{N}{K}= \ln N\). This defines the condition for a nontrivial equilibrium \(N^*>0\).
02

Showing Existence of Nontrivial Equilibrium

Consider the function \(f(N) = 1 - \frac{N}{K} - \ln N\). We seek \(N^* > 0\) such that \(f(N^*) = 0\). At \(N = 1\), \(f(1) = 1 - \frac{1}{K}\). Since \(K > 1\), \(f(1) > 0\). As \(N \rightarrow 0^+\) or \(N \rightarrow \infty\), \(f(N)\) becomes negative. By the Intermediate Value Theorem, there exists some \(N^* > 0\) such that \(f(N^*) = 0\).
03

Implicit Differentiation

Given \(1 - \frac{N^*}{K} = \ln N^*\), we want to find \(\frac{dN^*}{dK}\). Differentiating both sides with respect to \(K\): \(-\frac{dN^*}{dK} \cdot \frac{1}{K} + \frac{N^*}{K^2} = \frac{1}{N^*} \cdot \frac{dN^*}{dK}\). Simplifying, we get \(-\frac{1}{K} \cdot \frac{dN^*}{dK} = \frac{1}{N^*} \cdot \frac{dN^*}{dK} - \frac{N^*}{K^2}\).
04

Solving for Derivative

Rearrange the differentiated equation to find \(\frac{dN^*}{dK}\): \(-\frac{1}{K} \cdot \frac{dN^*}{dK} - \frac{1}{N^*} \cdot \frac{dN^*}{dK} = -\frac{N^*}{K^2}\) which simplifies to \((\frac{1}{K} + \frac{1}{N^*}) \cdot \frac{dN^*}{dK} = \frac{N^*}{K^2}\). Thus, \(\frac{dN^*}{dK} = \frac{N^*}{K^2 (\frac{1}{K} + \frac{1}{N^*})}\).

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Key Concepts

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

Equilibrium Points
In population dynamics, equilibrium points represent scenarios where the population size remains constant over time. For a given population size function, specifically here noted as \(N(t)\), an equilibrium point is obtained when the rate of change of the population, \(\frac{dN}{dt}\), is zero. This effectively signifies a balance where births equal deaths – the population neither grows nor shrinks.

In the equation \(\frac{dN}{dt}=N\left(1-\frac{N}{K}\right)-N \ln N\), setting \(\frac{dN}{dt} = 0\) leads us to the condition \(1-\frac{N}{K}= \ln N\). To ensure this situation describes a population, we focus on a positive solution, meaning \(N^* > 0\), which defines a nontrivial equilibrium point, particularly meaningful when \(K > 1\). Such a point is the size at which the population stays stable without tending toward zero or infinite growth.

The trick here is realizing that this balance requires solving a transcendental equation, often best approached using numerical methods or further mathematical insights like the Intermediate Value Theorem (IVT), ensuring solutions exist under specific conditions.
Implicit Differentiation
Implicit differentiation is a powerful tool in calculus used when a function is not given explicitly in terms of one independent variable. In our context, we tackle the equation for equilibrium \(1 - \frac{N^*}{K} = \ln N^*\) with \(N^*\) depending implicitly on \(K\) (the carrying capacity).

By differentiating both sides with respect to \(K\), we can find out how the equilibrium points \(N^*\) change as the carrying capacity \(K\) varies. This involves taking the derivative while respecting each variable's relationship to the other, essentially treating \(N^*\) as a function of \(K\).
  • Start by differentiating both sides. On the left, the chain rule applies, resulting in \(-\frac{1}{K} \frac{dN^*}{dK} + \frac{N^*}{K^2}\).
  • On the right, differentiate with respect to \(K\) using properties of natural logarithms, yielding \(\frac{1}{N^*} \frac{dN^*}{dK}\).
By re-arranging the resulting expression, you can isolate \(\frac{dN^*}{dK}\), solving the relationship between \(N^*\) and \(K\). This process helps reveal whether increases in \(K\) correspond to increases in the stable population size \(N^*\), a typical demand when managing sustainable populations.
Intermediate Value Theorem
The Intermediate Value Theorem (IVT) is an essential concept in calculus that supports finding solutions to equations within specific intervals. This theorem states that if a continuous function, say \(f(x)\), changes signs over an interval \([a, b]\), then there exists at least one point \(c\) within that interval where \(f(c) = 0\).

Applying IVT to the problem involves examining the function \(f(N) = 1 - \frac{N}{K} - \ln N\). With \(K > 1\), we observed that at \(N = 1\), \(f(1) = 1 - \frac{1}{K} > 0\), indicating the function is positive. Meanwhile, as \(N\) approaches zero or infinity, \(f(N)\) becomes negative. The change from positive to negative values as \(N\) varies confirms there is at least one \(N^* > 0\) where \(f(N^*) = 0\).

Thus, IVT guarantees the existence of nontrivial equilibrium points for the population within the domain, particularly highlighting the presence of a solution without explicitly solving the complex equation. This theorem acts as a bridge in proving existence, critical for validating models in theoretical ecology and real-world applications.

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Most popular questions from this chapter

Let $$ f(x)=\frac{x}{x-1}, \quad x \neq 1 $$ (a) Show that $$ \lim _{x \rightarrow-\infty} f(x)=1 $$ and $$ \lim _{x \rightarrow+\infty} f(x)=1 $$ That is, show that \(y=1\) is a horizontal asymptote of the curve \(y=\frac{x}{x-1}\) (b) Show that $$ \lim _{x \rightarrow 1^{-}} f(x)=-\infty $$ and $$ \lim _{x \rightarrow 1^{+}} f(x)=+\infty $$ That is, show that \(x=1\) is a vertical asymptote of the curve \(y=\frac{x}{x-1}\) (c) Determine where \(f(x)\) is increasing and where \(i t\) is decreasing. Does \(f(x)\) have local extrema? (d) Determine where \(f(x)\) is concave up and where it is concave down. Does \(f(x)\) have inflection points? (e) Sketch the graph of \(f(x)\) together with its asymptotes.

Determine whether the functions have absolute maxima and minima, and, if so, find their coordinates. Find inflection points. Find the intervals on which the function is increasing, on which it is decreasing, on which it is concave up, and on which it is concave down. Sketch the graph of each function. $$ y=x^{4}-2 x^{2}, x \in \mathbf{R} $$

Determine whether each function has absolute maxima and minima and find their coordinates. For each function, find the intervals on which it is increasing and the intervals on which it is decreasing. $$ y=\left|16-x^{2}\right|,-5 \leq x \leq 8 $$

use l'Hospital's rule to find $$ \lim _{x \rightarrow 0} \frac{a^{x}-1}{b^{x}-1} $$ where \(a, b>0\).

Determine whether each function has absolute maxima and minima and find their coordinates. For each function, find the intervals on which it is increasing and the intervals on which it is decreasing. $$ y=\sqrt{1+x^{2}}, x \in \mathbf{R} $$

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