Ch. 9 - Differential Equations

Briggs - Calculus: Early Transcendentals 3rd Edition

Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook

Briggs 3rd Edition

Ch. 9 - Differential Equations

Problem 9.3.45

Chapter 9, Problem 9.3.45

Orthogonal trajectories Use the method in Exercise 44 to find the orthogonal trajectories for the family of circles x² + y² = a²

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Start with the given family of circles: \(x^{2} + y^{2} = a^{2}\), where \(a\) is a parameter representing the radius of each circle.

Differentiate both sides of the equation implicitly with respect to \(x\) to find the slope of the tangent to the circles. This gives: \(2x + 2y \frac{dy}{dx} = 0\).

Solve for \(\frac{dy}{dx}\) to find the slope of the tangent lines to the circles: \(\frac{dy}{dx} = -\frac{x}{y}\).

Since orthogonal trajectories intersect the original family at right angles, their slopes are negative reciprocals. Therefore, the slope of the orthogonal trajectories satisfies \(\frac{dy}{dx} = \frac{y}{x}\).

Set up the differential equation \(\frac{dy}{dx} = \frac{y}{x}\) and solve it by separating variables or using an appropriate method to find the family of orthogonal trajectories.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Orthogonal Trajectories

Orthogonal trajectories are curves that intersect a given family of curves at right angles (90 degrees). To find them, one typically derives a differential equation for the original family and then finds a new family whose slopes are negative reciprocals, ensuring perpendicularity.

Differential Equations of Families of Curves

A family of curves defined by an equation with a parameter can be expressed as a differential equation by differentiating with respect to x and eliminating the parameter. This differential equation characterizes the slope of the tangent lines to the curves in the family.

Slopes and Negative Reciprocal Relationship

For two curves to be orthogonal at their intersection, the product of their slopes must be -1. This means if the slope of the original family is dy/dx = m, the slope of the orthogonal trajectories is dy/dx = -1/m, which is used to find the differential equation of the orthogonal trajectories.

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