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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. \(a x^{2}+b y^{2}=c ;\) with \(a\) and \(b\) held fixed.

Short Answer

Expert verified
The orthogonal trajectories are given by \(y = K x^{\frac{b}{a}}\), where \(K\) is a constant.

Step by step solution

01

Understand the Family of Curves

The given family of curves is described by the equation \(a x^{2}+b y^{2}=c\). This is the equation of an ellipse (when \(a\) and \(b\) are positive) or a hyperbola (when \(a\) or \(b\) is negative) in standard form. Here, \(a\) and \(b\) are constants, and \(c\) is a parameter representing different members of the family.
02

Determine the Implicit Form

To find the orthogonal trajectories, we first express the given family in implicit form. The implicit differentiation of \(a x^{2}+b y^{2}=c\) with respect to \(x\) is needed. Differentiating both sides with respect to \(x\) gives: \(2ax + 2by\frac{dy}{dx} = 0\).
03

Solve for the Slope \(\frac{dy}{dx}\)

From the differentiated equation \(2ax + 2by\frac{dy}{dx} = 0\), solve for \(\frac{dy}{dx}\):\[\frac{dy}{dx} = - \frac{ax}{by}.\] This represents the slope of the tangent to a curve in the given family at any point \((x, y)\).
04

Determine the Orthogonal Slope

Orthogonal curves have negative reciprocal slopes. Therefore, the slope \(m\) of the orthogonal trajectories is:\[m = \frac{by}{ax}.\] This equation describes the slope at any point \((x, y)\) for the orthogonal trajectory.
05

Set up Differential Equation for Orthogonal Trajectories

To find the orthogonal trajectories, setup a differential equation:\[\frac{dy}{dx} = \frac{by}{ax}.\]
06

Separate Variables and Integrate

We separate the variables and integrate:\[\frac{dy}{y} = \frac{b}{a} \frac{dx}{x}.\]Integrating both sides gives:\[\ln|y| = \frac{b}{a} \ln|x| + C,\] or equivalently:\[ y = K x^{\frac{b}{a}},\] where \(K = e^C\) is a constant.
07

Finalize the Orthogonal Trajectories

The equation \(y = K x^{\frac{b}{a}}\) represents the family of orthogonal trajectories to the given ellipses or hyperbolas. For different values of \(K\), we have different members of the orthogonal family.

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Key Concepts

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

Ellipses and Hyperbolas
Ellipses and hyperbolas are fascinating curves that have many applications in fields such as physics and engineering. They are both conic sections, which means they arise from the intersection of a cone with a plane.

For ellipses, the sum of the distances from any point on the ellipse to two fixed points called foci is constant. An equation of an ellipse takes the form \(ax^2 + by^2 = c\) where both \(a\) and \(b\) are positive constants. This results in an elongated circular shape.

Hyperbolas, on the other hand, have the property that the difference of the distances from any point on the hyperbola to two fixed points (also called foci) is constant. The equation \(ax^2 + by^2 = c\) represents a hyperbola when one of the coefficients \(a\) or \(b\) is negative. This gives a shape that opens in two directions, like two mirrored curves.

Understanding the distinction between ellipses and hyperbolas is crucial when solving problems related to their properties and their orthogonal trajectories.
Implicit Differentiation
Implicit differentiation is a method used when it's challenging or impossible to solve for one variable in terms of another explicitly. This technique is handy when differentiating equations that describe curves, like our ellipse or hyperbola family in the form \(ax^2 + by^2 = c\).

To perform implicit differentiation, apply derivatives to each term while treating the dependent variable as an implicit function of the independent variable. In our problem, we treat \(y\) as a function of \(x\) and differentiate both sides of the equation \(ax^2 + by^2 = c\) with respect to \(x\).

This gives us:\[2ax + 2by\frac{dy}{dx} = 0.\]

Remember, when differentiating \(y^2\), you use the chain rule, resulting in \(2y \cdot \frac{dy}{dx}\). Implicit differentiation is a powerful tool when dealing with complex curves that cannot be described by simple functions.
Differential Equations
Differential equations involve functions and their derivatives and are essential for describing the behavior of complex systems. In this task, we use differential equations to find orthogonal trajectories, which are paths intersecting given curves at specific angles.

After computing the slope of the tangent from the implicit differentiation step \(\frac{dy}{dx} = -\frac{ax}{by}\), we seek curves that intersect these original curves orthogonally. This involves finding the negative reciprocal of this slope, \(m = \frac{by}{ax}\).

The idea is to create a new differential equation \(\frac{dy}{dx} = \frac{by}{ax}\) to describe these orthogonal trajectories. Solving this equation produces a family of curves that taps into the beauty of mathematics by producing precisely oriented intersections with the original family.
Curve Families
Curve families refer to sets of curves that share a common characteristic or equation. In our current context, we explore two families: the original set of curves (ellipses or hyperbolas) and their orthogonal trajectories.

The original curve family is described by \(ax^2 + by^2 = c\) where different parameters \(c\) change the specific curve within the family. Each value of \(c\) gives an individual curve belonging to this family.

Orthogonal trajectories form another family, each member crossing the original curve family at right angles. They are represented by equations like \(y = Kx^{\frac{b}{a}}\), where \(K\) alters which specific curve within the orthogonal family we're considering.

Exploring how these families relate provides a richer understanding of geometry. It exemplifies how different mathematical approaches can elegantly describe complex relationships. This can help deepen insights into not just math itself, but also into natural and physical phenomena modeled by these concepts.

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Most popular questions from this chapter

A bacterial population \(B\) is known to have a rate of growth proportional to \(B\) itself. If between noon and 2 P.M. the population triples, at what time, no controls being exerted, should \(B\) become 100 times what it was at noon?

Suppose that the human body dissipates a drug at a rate proportional to the amount \(y\) of drug present in the bloodstream at time \(t\). At time \(t=0\) a first injection of \(y_{0}\) grams of the drug is made into a body that was free from that drug prior to that time. (a) Find the amount of residual drug in the bloodstream at the end of \(T\) hours. (b) If at time \(T\) a second injection of \(y_{0}\) grams is made, find the residual amount of drug at the end of \(2 T\) hours. (c) If at the end of each time period of length \(T,\) an injection of \(y_{0}\) grams is made, find the residual amount of drug at the end of \(n T\) hours. (d) Find the limiting value of the answer to part (c) as \(n\) approaches infinity.

For the family \(x^{2}+3 y^{2}=c y,\) find that member of the orthogonal trajectories which passes through (1,2)

A certain population is known to be growing at a rate given by the logistic equation \(d x / d t=x(b-a x) .\) Show that the maximum rate of growth will occur when the population is equal to half its equilibrium size, that is, when the population is \(b / 2 a\).

Glucose is being fed intravenously into the bloodstream of a patient at a constant rate \(c\) grams per minute. At the same time, the patient's body converts the glucose and removes it from the bloodstream at a rate proportional to the amount of glucose present. If the constant of proportionality is \(k\), show that as time increases, the amount of glucose in the bloodstream approaches an equilibrium value of \(c / k\).

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