/*! 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 9 Find the orthogonal trajectories... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find the orthogonal trajectories of each given family of curves. In each case sketch several members of the family and several of the orthogonal trajectories on the same set of axes. $$ x=\frac{y^{2}}{4}+\frac{c}{y^{2}} $$

Short Answer

Expert verified
In summary, to find the orthogonal trajectories of the given family of curves \(x = \frac{y^2}{4} + \frac{c}{y^2}\), we computed the derivative of \(x\) with respect to \(y\) and found \(\frac{dx}{dy} = \frac{1}{2}y -2cy^{-3}\). Then, we replaced the slope with its negative reciprocal, obtaining a new differential equation: \(\frac{dX}{dY} = \frac{2}{y(1-2cY)}\). After solving this differential equation, we found the general equation for the orthogonal trajectories to be \(-\frac{c}{2} \ln|1 - 2cY| = \ln|y| + C\). To sketch the family of curves and the orthogonal trajectories, choose different values of c and C and plot them on the same set of axes using graphing software or by hand.

Step by step solution

01

Find the derivative of x with respect to y.

Differentiate the given equation with respect to y. We have: $$ x = \frac{y^2}{4} + \frac{c}{y^2} $$ Differentiate both sides with respect to y: $$ \frac{dx}{dy} = \frac{d}{dy}\left(\frac{y^2}{4}\right) + \frac{d}{dy}\left(\frac{c}{y^2}\right) $$ Use the power rule and the chain rule: $$ \frac{dx}{dy} = \frac{1}{2}y -2cy^{-3} $$
02

Replace the slope with the negative reciprocal.

To find the equation for the orthogonal trajectories, replace the slope in the differential equation with the negative reciprocal of the slope: $$ -\frac{1}{\frac{dx}{dy}} = -\frac{1}{\frac{1}{2}y - 2cy^{-3}} $$ Now, let's denote the new slope by \(\frac{dX}{dY}\): $$ \frac{dX}{dY} = \frac{2}{y(1-2cY)} $$
03

Solve for the orthogonal trajectories.

We now have a new differential equation for the orthogonal trajectories: $$ \frac{dX}{dY} = \frac{2}{y(1-2cY)} $$ To find these trajectories, we should solve this differential equation. Start by separating the variables: $$ \frac{dX}{2} = \frac{dy}{y(1 - 2cY)} $$ Now, integrate both sides: $$ \int \frac{dX}{2} = \int \frac{dy}{y(1 - 2cY)} $$ Let \(u = 1 - 2cY\), so \(du = -2cdy\): $$ \int \frac{1}{u}(-\frac{cdu}{2y}) = \int \frac{dy}{y} $$ Now integrate: $$ -\frac{c}{2} \int \frac{1}{u} du = \int \frac{1}{y} dy $$ $$ -\frac{c}{2} \ln|u| = \ln|y| + C $$ Substitute back \(u = 1 - 2cY\): $$ -\frac{c}{2} \ln|1 - 2cY| = \ln|y| + C $$ Now we have the general equation for the orthogonal trajectories: $$ -\frac{c}{2} \ln|1 - 2cY| = \ln|y| + C $$
04

Sketch the family of curves and the orthogonal trajectories.

To sketch the given family of curves and orthogonal trajectories, first, find specific members of both families by choosing different values of c and C and plot them on the same set of axes. For this task, you may use graphing software or do it by hand.

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 91Ó°ÊÓ!

Key Concepts

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

Differential Equations
Differential equations are mathematical equations that describe the relationship between a function and its derivatives. They serve as a tool to model various physical phenomena, such as motion, heat flow, and population dynamics. A differential equation involves an unknown function and one or more of its derivatives.

For example, the equation given in our exercise, \( x = \frac{y^2}{4} + \frac{c}{y^2} \), can be differentiated with respect to \( y \) to obtain a relationship between \( x \) and its derivative. This relationship is represented by the derivative \( \frac{dx}{dy} \), which tells us how \( x \) changes as \( y \) changes, holding the constant \( c \) fixed. By understanding how to manipulate and solve differential equations, we can predict and comprehend complex systems and their behavior over time.

In the field of orthogonal trajectories, finding the derivative is the first essential step to determining the family of curves that intersect a given family of curves perpendicularly.
Slope Fields
Slope fields, also known as direction fields, are visual representations of a differential equation that depict the slope or direction of the tangent line to the solutions at various points in the plane. These fields provide a graphical method to predict the behavior of solutions without actually solving the differential equation.

To create a slope field for an equation like \( \frac{dx}{dy} = \frac{1}{2}y - 2cy^{-3} \), we plot short line segments at a variety of points that have slopes equal to the value of \( \frac{dx}{dy} \) at those points. When sketching orthogonal trajectories, the slope field must be adjusted so that at each point, the new direction is perpendicular to the original slope. This involves using negative reciprocals, which effectively rotate the original slopes by 90 degrees.

The usefulness of slope fields lies in their ability to show us how the solutions of differential equations behave, and they can be particularly valuable for understanding the relationships between families of curves, such as orthogonal trajectories.
Separable Equations
Separable equations are a type of differential equation in which the variables can be separated on either side of the equation, allowing us to integrate and find the solution. In essence, a separable equation has the form \( \frac{dy}{dx} = g(x)h(y) \), where the function on the right-hand side can be written as a product of two functions, one in terms of \( x \) and the other in terms of \( y \).

The process of solving a separable equation typically involves rearranging the equation so that all terms involving \( x \) are on one side and all terms involving \( y \) are on the other. Then, both sides are integrated separately. In our exercise, the equation \( \frac{dX}{dY} = \frac{2}{y(1 - 2cY)} \) is made separable and solved by separating the variables \( X \) and \( Y \) and then integrating. This technique is a foundational tool for solving many types of differential equations and is particularly effective for finding orthogonal trajectories.
Integrating Factors
Integrating factors are a method used to solve certain types of differential equations, particularly linear non-separable equations. An integrating factor is a function that, when multiplied by an equation, allows the equation to be rewritten in a form that is easier to integrate. The goal is to convert the equation into an exact equation, which can be integrated directly with respect to one of the variables.

While the example from our exercise does not directly require the use of an integrating factor (since it deals with a separable equation), integrating factors are essential when dealing with non-separable equations where the standard separation of variables does not apply. By multiplying both sides of these more complex equations by an appropriately chosen integrating factor, the resultant equation can often be simplified to allow for straightforward integration of both sides, thus leading to a solution. For students tackling various differential equations, understanding integrating factors is a crucial step towards solving a wider range of problems.

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

A tank initially contains 100 gal of brine in which there is dissolved \(20 \mathrm{lb}\) of salt. Starting at time \(t=0\), brine containing \(3 \mathrm{Ib}\) of dissolved salt per gallon flows into the tank at the rate of \(4 \mathrm{gal} / \mathrm{min}\). The mixture is kept uniform by stirring and the well-stirred mixture simultaneously flows out of the tank at the same rate. (a) How much salt is in the tank at the end of \(10 \mathrm{~min} ?\) (b) When is there \(160 \mathrm{lb}\) of salt in the tank?

Assume that the population of a certain city increases at a rate proportional to the number of inhabitants at any time. If the population doubles in 40 years, in how many years will it triple?

A girl on her sled has just slid down a hill onto a level field of ice and is starting to slow down. At the instant when their speed is \(5 \mathrm{ft} / \mathrm{sec}\), the girl's father runs up and begins to push the sled forward, exerting a constant force of \(15 \mathrm{lb}\) in the direction of motion. The combined weight of the girl and the sled is \(96 \mathrm{lb}\), the air resistance (in pounds) is numerically equal to one-half the velocity (in feet per second), and the coefficient of friction of the runners on the ice is \(0.05 .\) How fast is the sled moving 10 sec after the father begins pushing?

An object weighing \(12 \mathrm{lb}\) is placed beneath the surface of a calm lake. The buoyancy of the object is \(30 \mathrm{lb}\); because of this the object begins to rise. If the resistance of the water (in pounds) is numerically equal to the square of the velocity (in feet per second) and the object surfaces in \(5 \mathrm{sec}\), find the velocity of the object at the instant when it reaches the surface.

A bullet weighing 1 oz is fired vertically downward from a stationary helicopter with a muzzle velocity of \(1200 \mathrm{ft} / \mathrm{sec}\). The air resistance (in pounds) is numerically equal to \(16^{-5} v^{2}\), where \(v\) is the velocity (in feet per second). Find the velocity of the bullet as a function of the time.

See all solutions

Recommended explanations on Math Textbooks

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.