Chapter 8: Problem 32

Find the orthogonal trajectories of each of the following families of curves. In each case sketch several of the given curves and several of their orthogonal trajectories. Be careful to eliminate the constant from \(d y / d x\) for the original curves; this constant takes different values for different curves of the original family and you want an expression for \(d y / d x\) which is valid for all curves of the family crossed by the orthogonal trajectory you are trying to find. \(y=k x^{2}\)

Short Answer

Expert verified

The orthogonal trajectories are given by the family of curves: \( x^2 + 2y^2 = C \).

Step by step solution

- Differentiate the original curve

Given the family of curves: \[ y = kx^2 \] Differentiate both sides with respect to x to find the slope of the tangent to the curve: \[ \frac{d}{dx} y = \frac{d}{dx} (kx^2) \implies \frac{dy}{dx} = 2kx \]

- Eliminate the constant

Solve for the constant k from the original curve equation: \[ y = kx^2 \implies k = \frac{y}{x^2} \] Substitute this value of k back into the differentiated equation: \[ \frac{dy}{dx} = 2kx = 2 \cdot \frac{y}{x^2} \cdot x = \frac{2y}{x} \]

- Set up the orthogonal relationship

For orthogonal trajectories, the product of the slopes of the original and orthogonal trajectories must be -1: \[ \left( \frac{dy}{dx} \right)_{original} \cdot \left( \frac{dy}{dx} \right)_{orthogonal} = -1 \] So, if the slope of the original family is \( \frac{2y}{x} \), then: \[ \left( \frac{2y}{x} \right) \cdot \frac{dy}{dx}_{orth.} = -1 \implies \frac{dy}{dx}_{orth.} = -\frac{x}{2y} \]

- Separate variables and integrate

Separate the variables to integrate: \[ \frac{dy}{dx} = -\frac{x}{2y} \implies 2y \, dy = -x \, dx \] Integrate both sides: \[ \int 2y \, dy = \int -x \, dx \] This results in: \[ y^2 = -\frac{x^2}{2} + C \]

- Rearrange the equation

Rearrange the integrated result to express the family of orthogonal trajectories: \[ x^2 + 2y^2 = C \]

- Sketch the curves

The original family of curves \( y = kx^2 \) are parabolas opening upward or downward. The orthogonal trajectories \( x^2 + 2y^2 = C \) are ellipses centered at the origin. Sketch several of the given parabolas and several of their orthogonal ellipses to illustrate the orthogonality.

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Key Concepts

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

Differential Equations

A differential equation is a mathematical equation that involves functions and their derivatives. These equations are crucial in expressing relationships between the rate at which quantities change. In the context of our problem, we started with the family of curves given by: \[ y = kx^2 \]By differentiating, we found how y changes with x, resulting in the differential equation: \[ \frac{dy}{dx} = 2kx \]Understanding the differential equation allowed us to identify the slopes of the tangents to the curves, which is essential in finding orthogonal trajectories.

Family of Curves

A family of curves is a set of curves that can be described by a common equation with one or more parameters. In our problem, the family of curves is represented by:\[ y = kx^2 \]Here, the parameter k varies for different curves in the family. This parameter affects the shape and position of each curve but keeps the general parabolic shape consistent. When working with these families, we often need to eliminate the parameter to work with a general form that applies to all curves. For our orthogonal trajectories, we substituted k to form a more manageable differential equation:\[ \frac{dy}{dx} = \frac{2y}{x} \]

Integration

Integration is the process of finding the antiderivative of a function, which is the inverse of differentiation. It's a fundamental concept in calculus, used to find the area under curves or to reverse the process of differentiating functions. In this exercise, after setting up the differential equation for the orthogonal trajectories, we needed to separate the variables and integrate both sides to find the solution:\[ \frac{dy}{dx} = -\frac{x}{2y} \implies 2y \, dy = -x \, dx \]By integrating both sides, we found:\[ \int 2y \, dy = \int -x \, dx \rightarrow y^2 = -\frac{x^2}{2} + C \]This integration step gave us the family of orthogonal trajectories, which took the form of ellipses:\[ x^2 + 2y^2 = C \]