Logout Open In App
Open In App
Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Chapter 9: Problem 4

sketch the trajectory corresponding to the solution satisfying the specified initial conditions, and indicate the direction of motion for increasing t. $$ d x / d t=a y, \quad d y / d t=-b x, \quad a>0, \quad b>0 ; \quad x(0)=\sqrt{a}, \quad y(0)=0 $$

Short Answer

Expert verified
Question: Sketch the trajectory of a particle moving in a plane with initial conditions x(0) = √a, y(0) = 0. The motion is described by the following system of first order differential equations: dx/dt = ay, dy/dt = -bx, where a > 0 and b > 0. Answer: The trajectory of the particle can be represented by the equation y = -1/6 * ab^2 * y^3 - (b√a)/2 * y^2. The initial point of the trajectory is at (x, y) = (√a, 0), and the direction of motion for increasing t will be in the direction of a decreasing y-coordinate. To sketch the trajectory, plot a curve starting from the initial point that moves downward and to the left, reflecting the negative slope for increasing y values. The exact shape of the trajectory will depend on the specific values of coefficients a and b.

Step by step solution

Achieve better grades quicker with Premium

  • Unlimited AI interaction
  • Study offline
  • Say goodbye to ads
  • Export flashcards
Start your free trial Skip

Over 22 million students worldwide already upgrade their learning with Vaia!

01

Rewrite the given system of differential equations

Given the system of first order differential equations: $$ \frac{dx}{dt} = ay, \quad \frac{dy}{dt} = -bx, \quad a > 0, \quad b > 0; \quad x(0) = \sqrt{a}, \quad y(0) = 0 $$
02

Eliminate one variable (y) by integrating dx/dt

First, integrate the equation \(\frac{dx}{dt} = ay\) with respect to t: $$ \int \frac{dx}{dt} dt = \int ay dt \implies x = \frac{1}{2}a y^2 + C_1 $$ Now, apply the initial condition x(0) = \(\sqrt{a}\) and y(0) = 0: $$ \sqrt{a} = \frac{1}{2}a (0)^2 + C_1 \implies C_1 = \sqrt{a} $$ Thus, the equation becomes: $$ x = \frac{1}{2}a y^2 + \sqrt{a} $$
03

Eliminate x from the second equation (dy/dt)

Replace x from Step 2 in the second equation \(\frac{dy}{dt} = -bx\): $$ \frac{dy}{dt} = -b \left(\frac{1}{2}a y^2 + \sqrt{a}\right) $$
04

Derive the equation of the trajectory

Integrate the equation \(\frac{dy}{dt} = -b \left(\frac{1}{2}a y^2 + \sqrt{a}\right)\) with respect to y: $$ \int \frac{dy}{dt} dt = \int -b \left(\frac{1}{2}a y^2 + \sqrt{a}\right) dy \implies y = -\frac{1}{6}ab^2 y^3 - \frac{b\sqrt{a}}{2}y^2 + C_2 $$ Now, apply the initial condition y(0) = 0: $$ 0 = -\frac{1}{6}ab^2 (0)^3 - \frac{b\sqrt{a}}{2}(0)^2 + C_2 \implies C_2 = 0 $$ Thus, the equation of the trajectory becomes: $$ y = -\frac{1}{6}ab^2 y^3 - \frac{b\sqrt{a}}{2}y^2 $$
05

Identify the direction of motion for increasing t

We are now ready to analyze the direction of motion for increasing t. Starting from the equation of the trajectory: $$ y = -\frac{1}{6}ab^2 y^3 - \frac{b\sqrt{a}}{2}y^2 $$ Notice that for increasing t, the term with \(y^3\) grows faster than the term with \(y^2\). Moreover, since a and b are both positive, the trajectory will have a negative slope for increasing y values. Therefore, the direction of motion for increasing t will be in the direction of a decreasing y-coordinate.
06

Sketch the trajectory

To sketch the trajectory, consider the equation: $$ y = -\frac{1}{6}ab^2 y^3 - \frac{b\sqrt{a}}{2}y^2 $$ 1. The initial point is \((\sqrt{a}, 0)\). 2. For increasing t, y decreases (i.e., motion is downwards). Plot the trajectory starting from the initial point, showing a curve that initially moves downward and to the left in the direction of motion for increasing t. Remember that the parabolic shape is influenced by the coefficients a and b.

Key Concepts

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

First Order Differential Equations
First order differential equations are fundamental in mathematics and applications like physics or engineering. They describe how one variable changes with respect to another. The general form is \( \frac{dx}{dt} = f(t, x) \) where \( x \) is a function of \( t \) and \( f(t, x) \) is some known function. In our exercise, we have a system of two first order differential equations where the rate of change of both \( x \) and \( y \) is given in terms of each other and the constants \( a \) and \( b \). These types of equations are solved by integration, which yields the functions \( x(t) \) and \( y(t) \) that describe the trajectory of the system over time.

Understanding first order differential equations is crucial because they often represent real-world phenomena such as growth rates, speed, or electrical currents. In the given problem, we integrate these equations with respect to time to find the trajectories defined by the initial conditions.
Phase Plane Analysis
Phase plane analysis is a graphical method to study the behavior of systems of first order differential equations. It involves plotting the variables of the system against one another on a plane called the phase plane. Each point on this plane represents a state of the system, and the trajectory of points as time progresses shows how the system evolves. In our case, plotting \( x \) and \( y \) against each other with the initial conditions \( x(0) = \sqrt{a} \) and \( y(0) = 0 \) gives us insight into the motion of the system.

The trajectory in the phase plane provides a visual representation of the solution to the system of differential equations. By analyzing the curves and their direction, we can predict the long term behavior of the system without solving the equations completely. This method is particularly powerful in systems that are difficult or impossible to solve analytically.
Initial Conditions
Initial conditions are specific values that determine the starting point of a system described by differential equations. They are necessary to find a unique solution to a differential equation since these equations often have infinitely many solutions. By specifying \( x(0) \) and \( y(0) \) in our problem, we confine the system to a single trajectory out of the many possibilities.

The initial conditions \( x(0) = \sqrt{a} \) and \( y(0) = 0 \) anchor our solution to a particular point in time, ensuring that the subsequent behaviour of the system follows a path consistent with these values. As we integrate to find \( x(t) \) and \( y(t) \), we apply these conditions to solve for constants of integration, which ultimately enables us to sketch the specific trajectory of the system.
Integration of Differential Equations
Integration is the process used to solve differential equations, taking us from the rate of change to the total change over time. By integrating the system of equations given in the problem, we find expressions for \( x(t) \) and \( y(t) \) in terms of \( a \) and \( b \). This step is done by finding an antiderivative for each side of the differential equation, leading us to the general solution.

In the problem, integrating \( \frac{dx}{dt} = ay \) and \( \frac{dy}{dt} = -bx \) with respect to \( t \) yields the underlying relationship between \( x \) and \( y \) without time as a variable, which is a key aspect of sketching the trajectory. After integration, we use the initial conditions to solve for the constants of integration, allowing us to plot the system's trajectory distinctly.
Trajectory Sketching
Trajectory sketching is a method to visualize the path that the system follows over time in the phase plane, using the equations derived from integrating the differential equations. In our exercise's context, the trajectory represents the combination of \( x(t) \) and \( y(t) \) charted on a graph. It gives us a picture of how, starting from the initial conditions, the system's states evolve with increasing \( t \).

In the final step of our solution, after determining the relationship between \( y \) and the constants \( a \) and \( b \) via integration, we use the initial conditions to plot the starting point. From there, we note that as \( t \) increases, the value of \( y \) decreases since the coefficients are positive, indicating a downward trajectory. We then sketch the path based on these determinations, carefully noting the influence of \( a \) and \( b \) on the curvature of the trajectory.

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

The equation of motion of an undamped pendulum is \(d^{2} \theta / d t^{2}+\omega^{2} \sin \theta=0,\) where \(\omega^{2}=g / L .\) Let \(x=\theta, y=d \theta / d t\) to obtain the system of equations $$ d x / d t=y, \quad d y / d t=-\omega^{2} \sin x $$ (a) Show that the critical points are \((\pm n \pi, 0), n=0,1,2, \ldots,\) and that the system is almost lincar in the neighborhood of cach critical point. (b) Show that the critical point \((0,0)\) is a (stable) center of the corresponding linear system. Using Theorem 9.3.2 what can be said about the nonlinear system? The situation is similar at the critical points \((\pm 2 n \pi, 0), n=1,2,3, \ldots\) What is the physical interpretation of these critical points? (c) Show that the critical point \((\pi, 0)\) is an (unstable) saddle point of the corresponding linear system. What conclusion can you draw about the nonlinear system? The situation is similar at the critical points \([\pm(2 n-1) \pi, 0], n=1,2,3, \ldots\) What is the physical interpretation of these critical points? (d) Choose a value for \(\omega^{2}\) and plot a few trajectories of the nonlinear system in the neighborhood of the origin. Can you now draw any further conclusion about the nature of the critical point at \((0,0)\) for the nonlinear system? (e) Using the value of \(\omega^{2}\) from part (d) draw a phase portrait for the pendulum. Compare your plot with Figure 9.3 .5 for the damped pendulum.

Two species of fish that compete with each other for food, but do not prey on each other, are bluegill and redear. Suppose that a pond is stocked with bluegill and redear, and let \(x\) and \(y\) be the populations of bluegill and redear, respectively, at time \(t\). Suppose further that the competition is modeled by the equations $$\frac{d x}{d t}=x\left(\epsilon_{1}-\sigma_{1} x-\alpha_{1} y\right), \frac{d y}{d t}=y\left(\epsilon_{2}-\sigma_{2} y-\alpha_{2} x\right)$$ a. If \(\epsilon_{2} / \alpha_{2}>\epsilon_{1} / \sigma_{1}\) and \(\epsilon_{2} / \sigma_{2}>\epsilon_{1} / \alpha_{1},\) show that the only equilibrium populations in the pond are no fish, no redear, or no bluegill. What will happen for large \(t ?\) b. If \(\epsilon_{1} / \sigma_{1}>\epsilon_{2} / \alpha_{2}\) and \(\epsilon_{1} / \alpha_{1}>\epsilon_{2} / \sigma_{2}\), show that the only equilibrium populations in the pond are no fish, no redear, or no bluegill. What will happen for large \(t ?\)

Can be interpreted as describing the interaction of two species with population densities \(x\) and \(y .\) In each of these problems carry out the following steps. (a) Draw a direction field and describe how solutions seem to behave. (b) Find the critical points. (c) For each critical point find the corresponding linear system. Find the eigenvalues and eigenvectors of the linear system; classify each critical point as to type, and determine whether it is asymptotically stable, or unstable. (d) Sketch the trajectories in the neighborhood of each critical point. (e) Draw a phase portrait for the system. (f) Determine the limiting behavior of \(x\) and \(y\) as \(t \rightarrow \infty\) and interpret the results in terms of the populations of the two species. $$ \begin{array}{l}{d x / d t=x(1.5-0.5 y)} \\ {d y / d t=y(-0.5+x)}\end{array} $$

Prove that if a trajectory starts at a noncritical point of the system $$ d x / d t=F(x, y), \quad d y / d t=G(x, y) $$ then it cannot reach a critical point \(\left(x_{0}, y_{0}\right)\) in a finite length of time. Hint: Assume the contrary, that is, assume that the solution \(x=\phi(t), y=\psi(t)\) satisfies \(\phi(a)=x_{0}, \psi(a)=y_{0}\). Then use the fact that \(x=x_{0}, y=y_{0}\) is a solution of the given system satisfying the initial condition \(x=x_{0}, y=y_{0}\) at \(t=a\).

(a) Find an equation of the form \(H(x, y)=c\) satisfied by the trajectories. (b) Plot several level curves of the function \(H\). These are trajectories of the given system. Indicate the direction of motion on each trajectory. $$ d x / d t=2 y, \quad d y / d t=-8 x $$
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Statistics

Read Explanation

Logic and Functions

Read Explanation

Probability and Statistics

Read Explanation

Decision Maths

Read Explanation

Calculus

Read Explanation
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.

Sign-up for free