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Chapter 10: Problem 2

You are given a logistic equation with one or more initial conditions. (a) Determine the carrying capacity and intrinsic rate. (b) Sketch the graph of \(\frac{d N}{d t}\) versus \(N\) in an \(N z\) -plane. (c) In the \(t N\) -plane, plot the constant solutions and place a dashed line where the concavity of certain solutions may change. (d) Sketch the solution curve corresponding to each given initial condition. $$ d N / d t=.3 N(100-N), N(0)=25 $$

Short Answer

Expert verified
Carrying capacity is 100, intrinsic rate is 0.3; equilibrium points at N=0, N=100; inflection point at N=50; solution curve for N(0)=25 is sigmoidal.

Step by step solution

01

Identify the logistic equation components

Given logistic equation: \[ \frac{d N}{d t}=0.3N(100-N) \] The standard form of the logistic equation is \[\frac{d N}{d t} = rN(1-\frac{N}{K}) \] Comparing both forms, we identify the intrinsic rate (\r) and carrying capacity (\rK).
02

Determine the carrying capacity and intrinsic rate

From the comparison, we find: \[\r=0.3,\rK=100 \] Therefore, the intrinsic rate (\rr) is 0.3 and the carrying capacity (\rK) is 100.
03

Sketch the graph of \frac{dN}{dt} versus \rN

The equation \(\r\frac{dN}{dt} = 0.3N(100-N)\r\) implies that \(\r\frac{dN}{dt} = 0\r\) when \(\r\rN = 0\r\) or \(\r\rN = 100\r\), showing two equilibrium points. Sketch \(\r\frac{dN}{dt}\r\) against \(\rN\r\), noting the shape will be a parabola that opens downwards, intersecting the \(\rN\r\) axis at \(\rN=0\r\) and \(\rN=100\r\).
04

Plot the constant solutions in the \r(t,N) plane

In the \r(t,N) plane, constant solutions are \(\rN=0\r\) and \(\rN=100\r\). These lines are horizontal, corresponding to the equilibrium solutions derived earlier.
05

Identify points where concavity changes

The derivative of \[ \frac{d N}{d t}=0.3N(100-N) \] with respect to \[ N \] is \[ \frac{d^2N}{d t^2} =0.3(100-2N) \] Set \[ \frac{d^2N}{d t^2} =0 \] to find inflection points. Solve \[100-2N=0, N=50\ \] The line \(\rN=50\r\) is where the concavity changes.
06

Sketch solution curves for given initial conditions

Given \(N(0)=25\r\), the solution starts at \(\rN=25\r\) and will follow a sigmoidal (S-shaped) curve approaching \(N=100\r\). Sketch this curve in the \r(t, N) plane.

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Key Concepts

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

Carrying Capacity
To fully understand the logistic equation, you need to grasp the concept of carrying capacity. This is the maximum population size that an environment can sustain indefinitely. In the given equation \[ \frac{d N}{d t}=0.3N(100-N) \], the term '100' represents the carrying capacity. Next, let鈥檚 break down why this is the case:
  • The logistic equation generally looks like \[ \frac{d N}{d t}= rN\left(1-\frac{N}{K}\right) \], where the K is the carrying capacity.
  • By comparing parameters, we see K (carrying capacity) = 100.
So, for any population N, when N is 100, the rate of growth will be zero, meaning the population will stabilize. This makes carrying capacity essential for understanding the population limits imposed by environmental factors.
Intrinsic Rate
Another crucial part of the logistic equation is the intrinsic rate, which is the rate at which the population grows when there are no constraints. In our equation \( \frac{d N}{d t}=0.3N(100-N) \), the intrinsic rate, represented by 'r,' is 0.3. Here鈥檚 why it matters:
  • Intrinsic rate (r) influences how quickly the population grows.
  • An r value of 0.3 means the population grows at 30% of the current population per unit of time when it starts growing.
A higher intrinsic rate results in a faster-growing population, assuming all other factors are constant. So, understanding the intrinsic rate helps you predict how quickly the population can grow in ideal conditions.
Solution Curve
Finally, we have the solution curve, which visually represents how the population changes over time given initial conditions. For the equation \( \frac{d N}{d t}=0.3N(100-N)\) with \(N(0)=25\), here鈥檚 what you should know:
  • The initial condition tells us that when t=0, the population is 25.
  • The solution curve follows a sigmoidal (S-shaped) trajectory, gradually accelerating and then decelerating as it approaches the carrying capacity.
To sketch the solution curve:
  • Start at the point (0, 25) on the tN plane.
  • Follow the S-shaped trajectory towards the horizontal line at \(N=100\)\
This demonstrates how the population stabilizes over time, provided there are no drastic changes in the intrinsic rate or carrying capacity. So, the curve helps visualize real-world population dynamics in a constrained environment.

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Most popular questions from this chapter

One or more initial conditions are given for each differential equation in the following exercises. Use the qualitative theory of autonomous differential equations to sketch the graphs of the corresponding solutions. Include a \(y z\) -graph if one is not already provided. Always indicate the constant solutions on the \(t y\) -graph whether they are mentioned or not. \(y^{\prime}=5 y-y^{2}, y(0)=1, y(0)=7\)

Find an integrating factor for each equation. Take \(t>0\). $$ y^{\prime}+t y=6 t $$

A person planning for her retirement arranges to make continuous deposits into a savings account at the rate of $$\$ 3600$$ per year. The savings account earns \(5 \%\) interest compounded continuously. (a) Set up a differential equation that is satisfied by \(f(t)\), the amount of money in the account at time \(t\). (b) Solve the differential equation in part (a), assuming that \(f(0)=0\), and determine how much money will be in the account at the end of 25 years.

Let \(f(t)\) be the solution of \(y^{\prime}=y(2 t-1), y(0)=8\). Use Euler's method with \(n=4\) to estimate \(f(1)\).

In an autocatalytic reaction, one substance is converted into a second substance in such a way that the second substance catalyzes its own formation. This is the process by which trypsinogen is converted into the enzyme trypsin. The reaction starts only in the presence of some trypsin, and each molecule of trypsinogen yields one molecule of trypsin. The rate of formation of trypsin is proportional to the product of the amounts of the two substances present. Set up the differential equation that is satisfied by \(y=f(t)\), the amount (number of molecules) of trypsin present at time \(t .\) Sketch the solution. For what value of \(y\) is the reaction proceeding the fastest? [Note: Letting \(M\) be the total amount of the two substances, the amount of trypsinogen present at time \(t\) is \(M-f(t) .]\)
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