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In each exercise, find the orthogonal trajectories of the given family of curves. Draw a few representative curves of each family whenever a figure is requested. \(y=c_{1}(\sec x+\tan x)\)

Short Answer

Expert verified
The orthogonal trajectories are given by \(y^3 = -3x + 3c_2\).

Step by step solution

01

Identify the Family of Curves

The given family of curves is represented by the equation \(y = c_{1} (\sec x + \tan x)\), where \(c_1\) is a constant. The task is to find the orthogonal trajectories to these curves.
02

Find the Differential Equation of the Family

Differentiate the given family of curves with respect to \(x\) to obtain the differential equation. Starting with:\[d(y) = c_1 \cdot d(\sec x + \tan x)\]Using the derivatives \(\frac{d}{dx}(\sec x) = \sec x \tan x\) and \(\frac{d}{dx}(\tan x) = \sec^2 x\), we have:\[\frac{dy}{dx} = c_1(\sec x \tan x + \sec^2 x) = c_1 \sec x (\tan x + \sec x)\]Given \(y = c_1(\sec x + \tan x)\), replace \(c_1\) by \(\frac{y}{\sec x + \tan x}\) to get:\[\frac{dy}{dx} = \frac{y \sec x (\tan x + \sec x)}{\sec x + \tan x}\]
03

Simplify the Differential Equation

Notice that the right-hand expression can be simplified and adjusted to match the given family equation. This simplifies to:\[\frac{dy}{dx} = y \sec x \]
04

Solve the Differential Equation for Orthogonal Trajectories

To find the orthogonal trajectories, replace \(\frac{dy}{dx}\) by the negative reciprocal \(-\frac{1}{y \sec x}\) and solve:\[\frac{dy}{dx} = -\frac{1}{y \sec x}\]Thus,\(y \, d(y) = -\frac{1}{\sec x} \, d(x)\) leads to:\[y^2 \, dy = - d(x)\]
05

Integrate to Find Orthogonal Trajectories

Integrate both sides:\[\int y^2 \, dy = -\int dx\]Solving, we get:\[\frac{y^3}{3} = -x + c_2\]This represents the orthogonal trajectories as:\[y^3 = -3x + 3c_2\]

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Key Concepts

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

Orthogonal Trajectories
Orthogonal trajectories offer a fascinating look into how curves can relate to each other in geometry, particularly in analytical contexts. When two families of curves intersect, the angles at which they meet can be of great interest. Orthogonal trajectories are specifically concerned with families of curves that intersect each other at right angles, or are perpendicular to each other at every point of intersection.

Finding orthogonal trajectories involves identifying a family of curves that maintain this perpendicularity with a given family. This process requires converting an original differential equation, derived from the given family, by utilizing the negative reciprocal of its derivative. The resulting differential equation then describes the new family of curves, which are the orthogonal trajectories.

In practical terms, such a process is not only mathematical but also visual and can be pivotal in fields such as physics or engineering, where understanding how different force lines or wave patterns intersect can be crucial.
Family of Curves
A family of curves consists of multiple curves that can be described using a single equation containing one or more arbitrary constants. These constants can be adjusted to generate different members within the family, effectively weaving through a space with seamless continuity.

In the context of the exercise, the family is described by the equation \(y = c_{1} (\sec x + \tan x)\). Here, \(c_1\) is the arbitrary constant that allows for a variety of curves to be generated. By differentiating with respect to \(x\), you isolate a common trajectory or pattern that is inherent to every member within this family. Differentiating these equations with respect to their parameters helps in achieving a unifying differential equation.

These curve families can represent physical phenomena, like wavefronts or propagation lines, showing how deeply interlinked mathematical constructs are with real-world applications. By understanding how these curves are interrelated, one can gain insights into broader patterns and dynamics of a system.
Differential Equation Solution
At the heart of finding orthogonal trajectories is the manipulation and solving of differential equations. The solution often involves several distinct procedural steps: deriving the differential equation, finding the negative reciprocal, and integrating to obtain the solution.

In this exercise, starting with the derivative \(\frac{dy}{dx} = y \sec x \) of the given family of curves, the quest for orthogonal trajectories leads to replacing this with its negative reciprocal \(-\frac{1}{y \sec x}\). This action fundamentally changes the slope characteristic of the curves, allowing for the new set of orthogonal curves to be formed.

Integration then transforms this expression into an explicit equation \(y^3 = -3x + 3c_2\). Solving differential equations in this manner is crucial, as it uncovers the underlying curve structures and dynamics. These solutions not only describe trajectories mathematically but often translate into usable models or patterns in practical applications.

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Most popular questions from this chapter

Two substances, \(A\) and \(B\), are being converted into a single compound \(C .\) In the laboratory it has been shown that, for these substances, the following law of conversion holds: the time rate of change of the amount \(x\) of compound \(C\) is proportional to the product of the amounts of unconverted substances \(A\) and \(B\). Assume the units of measure so chosen that one unit of compound \(C\) is formed from the combination of one unit of \(A\) with one unit of \(B\). If at time \(t=0\) there are \(a\) units of substance \(A, b\) units of substance \(B,\) and none of compound \(C\) present, show that the law of conversion may be expressed by the equation $$\frac{d x}{d t}=k(a-x)(b-x)$$ Solve this equation with the given initial condition.

In each exercise, find the orthogonal trajectories of the given family of curves. Draw a few representative curves of each family whenever a figure is requested. The trisectrices of Maclaurin, \((a+x) y^{2}=x^{2}(3 a-x)\).

In each exercise, find the orthogonal trajectories of the given family of curves. Draw a few representative curves of each family whenever a figure is requested. \(x^{3}=3(y-c) .\) Draw the figure.

In each exercise, find the orthogonal trajectories of the given family of curves. Draw a few representative curves of each family whenever a figure is requested. \(y^{2}=a x^{2}(1-c x) ;\) with \(a\) held fixed.

The supply of food for a certain population is subject to a seasonal change that affects the growth rate of the population. The differential equation \(d x / d t=c x(t) \cos t\) where \(c\) is a positive constant, provides a simple model for the seasonal growth of the population. Solve the differential equation in terms of an initial population \(x_{0}\) and the constant \(c .\) Determine the maximum and the minimum populations and the time interval between maxima.

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