/*! 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 1 sketch the trajectory correspond... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 1

sketch the trajectory corresponding to the solution satisfying the specified initial conditions, and indicate the direction of motion for increasing t. $$ d x / d t=-x, \quad d y / d t=-2 y ; \quad x(0)=4, \quad y(0)=2 $$

Short Answer

Expert verified
The trajectory for this motion is described by the equation: $$x = 4 e^{-\frac{2x}{y}}$$ Starting at the point (4,2) and moving towards the origin (0,0) as t increases. The direction of motion is towards the origin, since both x(t) and y(t) decrease with increasing t.

Step by step solution

01

Solve the differential equations independently

We have two independent differential equations: 1. $$\frac{dx}{dt} = -x$$ 2. $$\frac{dy}{dt} = -2y$$ We can solve each of these equations by using the method of separation of variables. For the first equation, we get: $$\frac{dx}{x} = -dt$$ Now, integrate both sides: $$\int\frac{dx}{x} = -\int dt$$ Which gives us: $$\ln|x| = -t + C_1$$ Where C_1 is the integration constant. Solve for x: $$x(t) = e^{-t+C_1} = e^{-t}e^{C_1}$$ Similarly, for the second equation, we get: $$\frac{dy}{y} = -2 dt$$ And after integrating both sides: $$\ln|y| = -2t + C_2$$ Where C_2 is another integration constant. Now, solve for y: $$y(t) = e^{-2t+C_2} = e^{-2t}e^{C_2}$$ Now we have the general solutions for both equations. Next, we will apply the initial conditions to obtain the particular solutions.
02

Apply the initial conditions

Given the initial conditions, we have: $$x(0)=4 \Rightarrow e^{-0+C_1}e^0 = 4 \Rightarrow e^{C_1}=4$$ And: $$y(0)=2 \Rightarrow e^{-2(0)+C_2}e^0 = 2 \Rightarrow e^{C_2}=2$$ So, the particular solutions for x(t) and y(t) are: $$x(t) = 4e^{-t}$$ $$y(t) = 2e^{-2t}$$
03

Sketch the trajectory and indicate the direction of motion for increasing t

The equations for x(t) and y(t) describe the trajectory of the motion. From these equations, we can see that as t increases, both x(t) and y(t) decrease: For x(t): $$\frac{d}{dt}x(t) = \frac{d}{dt}(4e^{-t}) = -4e^{-t}=-x(t)<0$$ For y(t): $$\frac{d}{dt}y(t) = \frac{d}{dt}(2e^{-2t}) = -4e^{-2t}=-2y(t)<0$$ To sketch the trajectory, consider the parametric equations: $$x = 4e^{-t}$$ $$y = 2e^{-2t}$$ Here, we can eliminate t by dividing both equations: $$\frac{y}{x} = \frac{2e^{-2t}}{4e^{-t}}$$ Simplifying, we get: $$\frac{y}{x} = \frac{1}{2}e^{-t}$$ Solve for $$e^t$$: $$e^t = \frac{2x}{y}$$ Now substitute this expression back into $$x(t)$$: $$x = 4 e^{-\frac{2x}{y}}$$ So, the trajectory for this motion will be described by: $$x = 4 e^{-\frac{2x}{y}}$$ This is the trajectory of an exponential decay curve starting at the point (4,2) and moving towards the origin (0,0) as t increases. The direction of motion will be towards the origin, as both x(t) and y(t) decrease with increasing t.

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 91影视!

Key Concepts

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

Separation of Variables
The method of separation of variables is a common technique used to solve ordinary differential equations (ODEs). It involves rearranging an equation to isolate the variables on different sides of the equation, hence 'separating' them. Let's break down how this works using a simple example.

For the differential equation \(\frac{dx}{dt} = -x\), we can separate the variables by moving all terms involving \(x\) to one side and all terms involving \(t\) to the other side, yielding \(\frac{dx}{x} = -dt\). What follows is the integration of both sides, leading to a function in terms of \(x\) and a function in terms of \(t\), coupled with an integration constant. This process converts the differential equation into an equation that is easier to handle and solve. It is widely used because it simplifies the equation to the point where the solution can be easily obtained by integration.
Integration Constants
When solving differential equations by integrating, we often encounter integration constants such as \(C_1\) and \(C_2\). These constants appear because when we integrate a function, we get a family of functions that differ by a constant. These are the 'plus C' terms you may remember from basic calculus.

In the context of differential equations, these constants play a vital role as they allow us to tailor our general solution to meet specific initial conditions or boundary values. For instance, in the given problem, after integrating, we obtained natural logarithms of the absolute values of \(x\) and \(y\) equal to the product of a negative time and the integration constants. By applying initial conditions, \(x(0) = 4\) and \(y(0) = 2\), we were able to solve for these constants and thus generate the particular solutions tailored to the problem's specifics. The presence of these constants helps transform a general solution into one that fits specific scenarios.
Exponential Decay
Exponential decay refers to a process where a quantity decreases at a rate proportional to its current value, a concept that's beautifully illustrated by the solutions to the given problem's differential equations. The expressions \(x(t) = 4e^{-t}\) and \(y(t) = 2e^{-2t}\) both suggest that as time \(t\) increases, the values of \(x\) and \(y\) diminish exponentially.

The negative exponents indicate the decay nature; the variable being multiplied by \(e\) to the power of a negative constant times \(t\) shows us that as \(t\) becomes larger, the overall expression shrinks rapidly. Exponential decay models are pervasive in natural phenomena, where they describe everything from the cooling of a hot object to radioactive decay. Understanding the behavior of these models is important not only in math but also in various scientific and engineering disciplines.
Parametric Equations
Parametric equations are a set of equations that express a set of quantities as explicit functions of one or more independent variables known as parameters. In the problem given, we have a system of parametric equations with \(t\) as the parameter: \(x = 4e^{-t}\) and \(y = 2e^{-2t}\).

The beauty of parametric equations lies in their ability to represent curves in the plane that may be difficult to describe using a single equation in just \(x\) and \(y\). By eliminating the parameter (in our case, \(t\)), we can find a relationship between \(x\) and \(y\) that defines the trajectory independently of time. This is particularly useful when we wish to visualize the path traced by a moving object, as it accounts for both the horizontal and vertical changes simultaneously, giving us a complete picture of the object's movement in space.

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

For certain \(r\) intervals, or windows, the Lorenz equations exhibit a period- doubling property similar to that of the logistic difference equation discussed in Section \(2.9 .\) Careful calculations may reveal this phenomenon. (a) One period-doubling window contains the value \(r=100 .\) Let \(r=100\) and plot the trajectory starting at \((5,5,5)\) or some other initial point of your choice. Does the solution appear to be periodic? What is the period? (b) Repeat the calculation in part (a) for slightly smaller values of \(r .\) When \(r \cong 99.98\), you may be able to observe that the period of the solution doubles. Try to observe this result by performing calculations with nearby values of \(r\). (c) As \(r\) decreases further, the period of the solution doubles repeatedly. The next period doubling occurs at about \(r=99.629 .\) Try to observe this by plotting trajectories for nearby values of \(r .\)

Using Theorem \(9.7 .2,\) show that the linear autonomous system $$ d x / d t=a_{11} x+a_{12} y, \quad d y / d t=a_{21} x+a_{22} y $$ does not have a periodic solution (other than \(x=0, y=0\) ) if \(a_{11}+a_{22} \neq 0\)

(a) Find the eigenvalues and eigenvectors. (b) Classify the critical point \((0,0)\) as to type and determine whether it is stable, asymptotically stable, or unstable. (c) Sketch several trajectories in the phase plane and also sketch some typical graphs of \(x_{1}\) versus \(t .\) (d) Use a computer to plot accurately the curves requested in part (c). \(\frac{d \mathbf{x}}{d t}=\left(\begin{array}{rr}{1} & {2} \\ {-5} & {-1}\end{array}\right) \mathbf{x}\)

(a) Find all the critical points (equilibrium solutions). (b) Use a computer to draw a direction field and portrait for the system. (c) From the plot(s) in part (b) determine whether each critical point is asymptotically stable, stable, or unstable, and classify it as to type. $$ d x / d t=(2+x)(y-x), \quad d y / d t=(4-x)(y+x) $$

If \(x=r \cos \theta, y=r \sin \theta,\) show that \(y(d x / d t)-x(d y / d t)=-r^{2}(d \theta / d t)\)
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Decision Maths

Read Explanation

Probability and Statistics

Read Explanation

Mechanics Maths

Read Explanation

Logic and Functions

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.