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Find the orthogonal trajectories of the family of ellipses having center at the origin, a focus at the point \((c, 0)\), and semimajor axis of length \(2 c\).

Short Answer

Expert verified
The orthogonal trajectories of the family of ellipses with center at the origin, a focus at the point \((c, 0)\), and semimajor axis of length \(2c\) is given by the equation: \[y^4 = 8x^4\]

Step by step solution

01

Find the equation of the family of ellipses

To find the equation for the given family of ellipses, we will use the general equation of an ellipse centered at the origin: \[\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\] Since the semimajor axis has a length of \(2c\), we have \(a = 2c\). To find the length of semiminor axis \(b\), we will use the distance between the center of the ellipse and a focus, which is \(c\). The relationship between \(a\), \(b\), and \(c\) is given by the following equation: \[c^2 = a^2 - b^2\] Substitute \(a = 2c\) into the equation: \[c^2 = (2c)^2 - b^2\] Solve for \(b\): \[b^2 = (2c)^2 - c^2 = 3c^2\] Now, we have the equation of the family of ellipses: \[\frac{x^2}{(2c)^2}+\frac{y^2}{3c^2}=1\]
02

Differentiate the equation

Next, we will differentiate the equation with respect to \(x\) using implicit differentiation. The derivative is denoted as \(\frac{dy}{dx}\). Differentiate both sides while treating \(c\) as a constant: \[\frac{2x}{(2c)^2} \cdot \frac{dx}{dx} + \frac{2y}{3c^2} \cdot \frac{dy}{dx} = 0\] Simplify the equation and solve for \(\frac{dy}{dx}\): \[\frac{dy}{dx} = -\frac{2x}{y}\]
03

Replace the derivative with its negative reciprocal

To find the orthogonal trajectories, we will replace \(\frac{dy}{dx}\) with its negative reciprocal, \(-\frac{y}{2x}\): \[-\frac{y}{2x} = -\frac{2x}{y}\]
04

Solve the new equation for \(y\)

Now, we will solve the new equation for \(y\): \[\frac{y^2}{2x} = \frac{4x^3}{y^2}\] Multiply both sides by \(y^2\): \[y^4 = 8x^4\]
05

Write the final equation

The final equation for the orthogonal trajectories is: \[y^4 = 8x^4\] This equation represents the orthogonal trajectories of the family of ellipses with center at the origin, a focus at the point \((c, 0)\), and semimajor axis of length \(2c\).

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Key Concepts

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

Orthogonal Trajectories
Orthogonal trajectories are an important concept in differential equations, especially when dealing with families of curves. These trajectories are curves that intersect a given family of curves at right angles (orthogonally). To find these trajectories, we need to determine a new family of curves from the original curves where the new curves intersect the original ones perpendicularly.

In the case of the family of ellipses given in this exercise, finding orthogonal trajectories involves taking the derivative of the ellipse equation and replacing the derivative with its negative reciprocal. This substitution results in a new differential equation representing the orthogonal trajectories. Solving this differential equation provides the equations of curves that cross each of the original ellipses at right angles, thereby forming a new family of curves orthogonal to the family of ellipses.
Implicit Differentiation
Implicit differentiation is a crucial tool used when differentiation is needed for equations where the dependent and independent variables are intermingled, and it is difficult to solve for one variable in terms of the other. In the exercise, the equation of the ellipse is not solved for one variable completely in terms of the other (like the usual y as a function of x), so implicit differentiation helps us find the derivative without doing so.

By keeping constant terms unaffected by the differentiation and applying the chain rule, we differentiate both x and y simultaneously. This requires careful application of calculus rules, as shown when differentiating the ellipse equation with respect to x, resulting in an equation that expresses the derivative \( \frac{dy}{dx} \) in terms of both x and y.
Family of Ellipses
A family of ellipses refers to a group of ellipses that share a common characteristic or a set of parameters, such as having the same center or a fixed ratio of axes lengths. In this problem, the ellipses are centered at the origin, have a focus at \((c, 0)\), and share a semimajor axis of length \(2c\).

This common property provides a parameter for the family — the distance from the center to the focus, \(c\), and establishes a relationship between the semi-major and semi-minor axes using the equation \(c^2 = a^2 - b^2\). Understanding this family structure is crucial because it allows us to express the general equation for any ellipse within the family and ultimately helps us in deriving the orthogonal trajectories.
Differentiation with Respect to x
Differentiation with respect to x means that x is treated as the variable of interest while any other variables, such as y in this case, are considered dependent on x. This is standard practice in calculus when dealing with functions where y depends on x, which is what's done with implicit differentiation in this exercise.

When differentiating the equation of the ellipses implicitly with respect to x, any derivatives of y are expressed through \( \frac{dy}{dx} \). This is necessary when the original equation contains y in terms of x without an explicit rule (function) separating them. By differentiating this way, we find relationships between the rates of change (slopes) of x and y for the given family of ellipses.

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Most popular questions from this chapter

The rate at which a certain substance dissolves in water is proportional at the product of the amount undissolved and the difference \(c_{1}-c_{2}\), where \(c_{1}\) is the concentration in the saturated solution and \(c_{2}\) is the concentration in the actual solution. If saturated, \(50 \mathrm{gm}\) of water would dissolve \(20 \mathrm{gm}\) of the substance. If \(10 \mathrm{gm}\) of the substance is placed in \(50 \mathrm{gm}\) of water and half of the substance is then dissolved in \(90 \mathrm{~min}\), how much will be dissolved in 3 hr?

Assume that the rate of change of the human population of the earth is proportional to the number of people on earth at any time, and suppose that this population is increasing at the rate of \(2 \%\) per year. The 1979 World Almanac gives the 1978 world population estimate as 4,219 million; assume this figure is in fact correct. (a) Using this data, express the human population of the earth as a function of time. (b) According to the formula of part (a), what was the population of the earth in \(1950 ?\) The 1979 World Almanac gives the 1950 world population estimate as 2,510 million. Assuming this estimate is very nearly correct, comment on the accuracy of the formula of part (a) in checking such past populations. (c) According to the formula of part (a), what will be the population of the earth in \(2000 ?\) Does this seem reasonable? (d) According to the formula of part (a), what was the population of the earth in \(1900 ?\) The 1979 World Almanac gives the 1900 world population estimate as 1,600 million. Assuming this estimate is very nearly correct, comment on the accuracy of the formula of part (a) in checking such past populations. (c) According to the formula of part (a), what will be the population of the earth in \(2100 ?\) Does this seem reasonable?

A tank initially contains 100 gal of pure water. Starting at \(t=0\), a brine containing \(4 \mathrm{lb}\) of salt per gallon flows into the tank at the rate of \(5 \mathrm{gal} / \mathrm{min}\). The mixture is kept uniform by stirring and the well-stirred mixture flows out at the slower rate of \(3 \mathrm{gal} / \mathrm{min}\). (a) How much salt is in the tank at the end of \(20 \mathrm{~min}\) ? (b) When is there \(50 \mathrm{lb}\) of salt in the tank?

A given family of curves is said to be self-orthogonal if its family of orthogonal trajectories is the same as the given family. Show that the family of parabolas \(y^{2}=2 c x+c^{2}\) is self-orthogonal.

A stone weighing \(4 \mathrm{lb}\) falls from rest toward the earth from a great height. As it falls it is acted upon by air resistance that is numercially equal to \(\frac{1}{2} v\) (in pounds), where \(v\) is the velocity (in feet per second). (a) Find the velocity and distance fallen at time \(t\) sec. (b) Find the velocity and distance fallen at the end of \(5 \mathrm{sec}\).

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