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Find the orthogonal trajectories of the family of curves.Use a graphing device to draw several members of each family on a common screen. $$y=\frac{x}{1+k x}$$

Short Answer

Expert verified
The orthogonal trajectories satisfy \( y - \frac{1}{2}ky^2 + x + \frac{1}{2}kx^2 = C \), subject to parameter \( k \).

Step by step solution

01

- Identify the Family of Curves

The given family of curves is described by the equation \( y = \frac{x}{1+kx} \), where \( k \) is a parameter. We need to find the orthogonal trajectories for this family.
02

- Express the Family with Implicit Differentiation

First, we differentiate the given equation with respect to \( x \), treating \( y \) as a function of \( x \). Begin by rewriting the equation in implicit form: \( y(1+kx) = x \). Differentiate this with respect to \( x \) to obtain \( y'(1+kx) + yk = 1 \).
03

- Solve for the Slope \( y' \)

From \( y'(1+kx) + yk = 1 \), solve for \( y' \). This gives us \( y' = \frac{1 - yk}{1+kx} \). This represents the slope of the original family of curves.
04

- Use Orthogonality Condition

Orthogonal trajectories will have slopes that are the negative reciprocals of the original family. Thus, the slope \( y'_\perp \) for the orthogonal trajectory is \( y'_\perp = -\frac{1+kx}{1 - yk} \).
05

- Set Up and Solve Differential Equation for Orthogonal Trajectories

Set up the differential equation for the orthogonal trajectories: \( \frac{dy}{dx} = -\frac{1+kx}{1 - yk} \). We need to solve this differential equation to find the orthogonal trajectories.
06

- Simplify and Solve using Separation of Variables

Rearrange the differential equation to use separation of variables, giving \( (1 - yk)dy = -(1+kx)dx \). Integrate both sides: \( \int(1 - yk) dy = \int -(1+kx) dx \). This results in \( y - \frac{1}{2}ky^2 = -x - \frac{1}{2}kx^2 + C \), where \( C \) is the constant of integration.
07

- Rearrange and Substitute

The above integration gives us an implicit form of the equation for the orthogonal trajectories. As we solved using an implicit form, the novel orthogonal trajectory family is: \( y - \frac{1}{2}ky^2 + x + \frac{1}{2}kx^2 = C \). Substitute for different values of \( k \) to illustrate several members on a common graphing device.

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Key Concepts

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

Implicit Differentiation
Implicit differentiation is a powerful mathematical tool used when dealing with equations that define y implicitly as a function of x, rather than explicitly. Instead of having y directly expressed, you often encounter equations where y appears mixed with other terms. In such cases, you need to differentiate both sides of the equation with respect to x, treating y as a dependent variable and applying the chain rule where necessary.

In the given exercise, the family of curves is represented by the equation \( y = \frac{x}{1+kx} \). To use implicit differentiation, we first rewrite this equation in an implicit form as \( y(1+kx) = x \). Then, we differentiate with respect to x. As y is a function of x, applying implicit differentiation and treating y as dependent means including \( y' \) (the derivative of y with respect to x) in our differentiation result. This gives us the differentiated form \( y'(1+kx) + yk = 1 \).

This process allows us to find the derivative of y with respect to x without needing to solve for y first, making it very handy when dealing with complex or combined expressions.
Family of Curves
A family of curves is a set of curves described by an equation with one or more parameters that can vary. Each different parameter value gives us a new member of the curve family. For the given equation \( y = \frac{x}{1+kx} \), \( k \) serves as the parameter representing the family member. Each value of \( k \) gives a different curve, but they all share the same form.

Understanding a family of curves helps us identify patterns and relationships between different curves and provides a clearer view of a spectrum of possibilities that arise from modifying the parameter(s). This is crucial when determining orthogonal trajectories, as you need the whole family to explore intersecting paths that are perpendicular to each member of the given family.

Graphically, plotting several members of a family on the same graph allows you to visually assess their orientation and distribution relative to each other. This helps in predicting or analyzing how another family of orthogonal trajectories would interact with them.
Differential Equations
Differential equations are equations that involve an unknown function and its derivatives. They are a central tool in modeling and solving problems involving rates of change. In context, finding orthogonal trajectories requires setting up a differential equation whose solution gives the equation of the trajectory curve.

For the problem at hand, through the process of differentiating and applying the orthogonality condition, we formed a differential equation: \( \frac{dy}{dx} = -\frac{1+kx}{1 - yk} \). This equation represents the orthogonal condition necessary for the trajectories to intersect each member of the original family of curves at right angles (90 degrees).

Solving this differential equation requires further techniques, such as separation of variables, which allows us to obtain an implicit equation representing the orthogonal trajectories.
Separation of Variables
Separation of variables is a method used to solve certain differential equations by separating the variables (often denoted as x and y) on opposite sides of the equation, allowing us to integrate each side independently. It works effectively when the differential equation can be reorganized such that each side of the equation contains only one of the variables.

In our exercise, from the orthogonal differential equation \( \frac{dy}{dx} = -\frac{1+kx}{1 - yk} \), we rearrange it to separate variables: \( (1 - yk)dy = -(1+kx)dx \). By isolating \( y \) and \( x \) with their respective differentials, we can then integrate both sides: \( \int(1 - yk) dy = \int -(1+kx) dx \).

After integration, we obtain an implicit equation: \( y - \frac{1}{2}ky^2 = -x - \frac{1}{2}kx^2 + C \), where \( C \) is the constant of integration. This implicit form characterizes the family of orthogonal trajectories, shedding light on their layout relative to the original family of curves.

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Most popular questions from this chapter

Use a computer algebra system to draw a direction field for the differential equation \(y^{\prime}=y^{3}-4 y .\) Get a printout and sketch on it solutions that satisfy the initial condition \(y(0)=c\) for various values of \(c .\) For what values of \(c\) does $$\lim _{t \rightarrow \infty} y(t)$$ exist? What are the possible values for this limit?

The population of the world was about 5.3 billion in \(1990 .\) Birth rates in the 1990 s ranged from 35 to 40 million per year and death rates ranged from 15 to 20 million per year. Let's assume that the carrying capacity for world population is 100 billion. (a) Write the logistic differential equation for these data. (Because the initial population is small compared to the carrying capacity, you can take \(k\) to be an estimate of the initial relative growth rate.) (b) Use the logistic model to estimate the world population in the year 2000 and compare with the actual population of 6.1 billion. (c) Use the logistic model to predict the world population in the years 2100 and \(2500 .\) (d) What are your predictions if the carrying capacity is 50 billion?

Populations of aphids and ladybugs are modeled by the equations $$\begin{aligned} \frac{d A}{d t} &=2 A-0.01 A L \\ \frac{d L}{d t} &=-0.5 L+0.0001 A L \end{aligned}$$ $$\begin{array}{l}{\text { (a) Find the equilibrium solutions and explain their }} \\ {\text { significance. }} \\ {\text { (b) Find an expression for } d L / d A} \\ {\text { (c) The direction field for the differential equation in part (b) }} \\ {\text { is shown. Use it to sketch a phase portrait. What do the }} \\ {\text { phase trajectories have in common? }}\\\\{\text { (d) Suppose that at time } t=0 \text { there are } 1000 \text { aphids and }} \\ {200 \text { ladybugs. Draw the corresponding phase trajectory }} \\ {\text { and use it to describe how both populations change. }} \\ {\text { (e) Use part (d) to make rough sketches of the aphid and }} \\\ {\text { ladybug populations as functions of } t \text { . How are the graphs }} \\ {\text { related to each other? }}\end{array}$$

\(11-14\) Sketch the direction field of the differential equation. Then use it to sketch a solution curve that passes through the given point. $$y^{\prime}=y-2 x, \quad(1,0)$$

Find an equation of the curve that passes through the point \((0,1)\) and whose slope at \((x, y)\) is \(x y .\)

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