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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. $$y=\frac{k}{x}$$

Short Answer

Expert verified
Orthogonal trajectories are given by \( y^2 - x^2 = C \).

Step by step solution

01

Understand the Problem

We're tasked with finding the orthogonal trajectories for the family of curves described by the equation \( y = \frac{k}{x} \). Orthogonal trajectories are curves that intersect the given family of curves at right angles.
02

Differentiate the Given Family

Differentiating \( y = \frac{k}{x} \) implicitly with respect to \( x \), we get \( \frac{dy}{dx} = -\frac{k}{x^2} \). This represents the slope of the family curves for each \( k \).
03

Find the Negative Reciprocal Slope

For orthogonal trajectories, slopes of intersecting paths are negative reciprocals. So, if \( \frac{dy}{dx} = -\frac{k}{x^2} \), then the orthogonal trajectory has a slope \( \frac{dy}{dx} = \frac{x^2}{k} \).
04

Set Up the Differential Equation for Orthogonal Trajectories

Substitute \( y \) back into the equation. Since \( k = xy \), \( \frac{dy}{dx} = \frac{x^2}{xy} = \frac{x}{y} \). Thus, the differential equation is \( y \frac{dy}{dx} = x \).
05

Solve the Differential Equation

Separate variables: \( y \, dy = x \, dx \). Integrate both sides: \( \int y \, dy = \int x \, dx \). This gives \( \frac{y^2}{2} = \frac{x^2}{2} + C \). Simplifying, we find \( y^2 - x^2 = C \).
06

Graph the Family of Curves and Their Trajectories

Graph the original family of hyperbolas \( y = \frac{k}{x} \) and the orthogonal trajectories \( y^2 - x^2 = C \) using a graphing device. The original family should appear as a set of hyperbolas through the origin, while the orthogonal trajectories will form hyperbolas interlacing them.

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Key Concepts

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

Differentiation
Differentiation is a fundamental concept in calculus that involves finding the rate at which a function is changing at any given point. In the context of finding orthogonal trajectories, we first differentiate the original family of curves. These curves are expressed as \( y = \frac{k}{x} \).

The process of differentiation gives us the slope of the tangent line to the curve at any point. When we differentiate \( y = \frac{k}{x} \) with respect to \( x \), using implicit differentiation, we get \( \frac{dy}{dx} = -\frac{k}{x^2} \). This equation lets us find the slope of the original curves and serves as the starting point for determining the orthogonal trajectories.

Ortho means right or perpendicular, and hence, to find orthogonal trajectories, finding the negative reciprocal of this slope is necessary.
Differential Equations
Differential equations involve equations that involve an unknown function and its derivatives. Solving differential equations is crucial in finding orthogonal trajectories since it helps determine new curves that intersect the given family of curves at right angles.

When we talk about differential equations in this context, we aim to replace variables to find a new slope for the orthogonal trajectories. We set the slope, \( \frac{dy}{dx} = \frac{x^2}{k} \), and rewrite it using the relationship \( k = xy \), giving us \( \frac{dy}{dx} = \frac{x}{y} \).

Hence, we arrive at the differential equation \( y \frac{dy}{dx} = x \). Solving this equation enables us to compute the orthogonal trajectories, which are expressed as \( y^2 - x^2 = C \), where \( C \) is a constant. The process revolves around integrating both sides separately, enabling us to find an explicit form for the orthogonal family of curves.
Implicit Differentiation
Implicit differentiation is a technique used when differentiation is needed for functions not explicitly solved for one of the variables, like \( y = f(x) \). It is particularly useful here because in the given equation \( y = \frac{k}{x} \), "\( y \)" is not isolated.

To apply implicit differentiation, differentiate both sides with respect to \( x \). You'll remember that the derivative of \( y \) with respect to \( x \) introduces \( \frac{dy}{dx} \), which we solve for. Thus, differentiating \( y = \frac{k}{x} \) gives \( \frac{dy}{dx} = -\frac{k}{x^2} \).

The role of implicit differentiation is crucial as it allows us to track how one variable changes with another, even when they are interlinked in non-straightforward ways. It is an invaluable tool in scenarios like these where finding orthogonal trajectories means differentiating an equation not evenly solvable for a single variable.

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Most popular questions from this chapter

In a seasonal-growth model, a periodic function of time is introduced to account for seasonal variations in the rate of growth. Such variations could, for example, be caused by seasonal changes in the availability of food. (a) Find the solution of the seasonal-growth model $$\frac{d P}{d t}=k P \cos (r t-\phi) \quad P(0)=P_{0}$$ where \(k, r,\) and \(\phi\) are positive constants. (b) By graphing the solution for several values of \(k, r,\) and \(\phi\) explain how the values of \(k, r,\) and \(\phi\) affect the solution. What can you say about lim \(_{t \rightarrow \infty} P(t) ?\)

$$\begin{array}{l}{\text { (a) Solve the differential equation } y^{\prime}=2 x \sqrt{1-y^{2}}} \\ {\text { (b) Solve the initial-value problem } y^{\prime}=2 x \sqrt{1-y^{2}}} \\ {y(0)=0, \text { and graph the solution. }} \\\ {\text { (c) Does the initial-value problem } y^{\prime}=2 x \sqrt{1-y^{2}}} \\ {y(0)=2, \text { have a solution? Explain. }}\end{array}$$

A sphere with radius 1 \(\mathrm{m}\) has temperature \(15^{\circ} \mathrm{C}\) . It lies inside a concentric sphere with radius 2 \(\mathrm{m}\) and temperature \(25^{\circ} \mathrm{C}\) . The temperature \(T(r)\) at a distance \(r\) from the common center of the spheres satisfies the differential equation $$\frac{d^{2} T}{d r^{2}}+\frac{2}{r} \frac{d T}{d r}=0$$ If we let \(S=d T / d r,\) then \(S\) satisfies a first-order differential equation. Solve it to find an expression for the temperature \(T(r)\) between the spheres.

In an elementary chemical reaction, single molecules of two reactants \(A\) and \(B\) form a molecule of the product \(C\) : \(A+B \rightarrow C\) . The law of mass action states that the rate of reaction is proportional to the product of the concentrations of \(A\) and \(B :\) $$\frac{d[\mathrm{C}]}{d t}=k[\mathrm{A}][\mathrm{B}]$$ (See Example 4 in Section \(3.7 . )\) Thus, if the initial concentrations are \([\mathrm{A}]=a\) moles \(/ \mathrm{L}\) and \([\mathrm{B}]=b\) moles \(/ \mathrm{L}\) and we write \(x=[\mathrm{C}],\) then we have $$\frac{d x}{d t}=k(a-x)(b-x)$$ $$\begin{array}{l}{\text { (a) Assuming that } a \neq b, \text { find } x \text { as a function of } t \text { . Use the }} \\ {\text { fact that the initial concentration of } C \text { is } 0 \text { . }} \\ {\text { (b) Find } x(t) \text { assuming that } a=b . \text { How does this expres- }} \\\ {\text { sion for } x(t) \text { simplify if it is known that }[C]=\frac{1}{2} a \text { after }} \\ {20 \text { seconds? }}\end{array}$$

\(9-10\) Sketch a direction field for the differential equation. Then use it to sketch three solution curves. $$y^{\prime}=1+y$$

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