Problem 6 - Thomas Calculus

Question

In Exercises \(5-10,\) find the orthogonal trajectories of the family of curves. Sketch sketch skeveral members of each family. $$y=c x^{2}$$

Step-by-Step Solution

Answer

The orthogonal trajectories are ellipses defined by the equation \( y^2 + \frac{x^2}{2} = C \).

Step 1: Identify the Differential Equation

The given family of curves is represented as \( y = c x^{2} \). To find the differential equation associated with this family, we'll differentiate both sides with respect to \( x \). This gives \( \frac{dy}{dx} = 2cx \). Now, since \( y = cx^2 \), we can express \( c \) in terms of \( y \) and \( x \): \( c = \frac{y}{x^2} \). Substitute this into the derivative gives \( \frac{dy}{dx} = 2x \frac{y}{x^2} = \frac{2y}{x} \). So the differential equation is \( \frac{dy}{dx} = \frac{2y}{x} \).

Step 2: Derive the Orthogonal Trajectories

Orthogonal trajectories require that the product of slopes of the curves and their orthogonal trajectories equals \(-1\). Hence, if \( \frac{dy}{dx} = \frac{2y}{x} \), then for the orthogonal trajectories, the slope \( M_\perp \) must satisfy \( \frac{dy}{dx} \cdot M_\perp = -1 \). So, \( M_\perp = -\frac{x}{2y} \).

Step 3: Solve the Differential Equation of Orthogonal Trajectories

To solve \( \frac{dy}{dx} = -\frac{x}{2y} \), we can separate variables: \( 2y \frac{dy}{dx} = -x \). Rearranging gives \( 2y \, dy = -x \, dx \). Integrating both sides, \( \int 2y \, dy = \int -x \, dx \), which yields \( y^2 = -\frac{x^2}{2} + C \), where \( C \) is the constant of integration.

Step 4: Rearrange the Orthogonal Trajectories Equation

The equation for the orthogonal trajectories is \( y^2 + \frac{x^2}{2} = C \). This can be recognized as the general form of an ellipse centered at the origin.

Key Concepts

Differential Equations

Differential equations are mathematical tools used to describe relationships that involve rates of change. In the context of our exercise, the given family of curves is described by the equation \( y = c x^{2} \), where \( c \) is a constant. Our aim is to translate this equation into a differential equation which encapsulates the underlying behavior of this family of curves.

To start, consider the derivative \( \frac{dy}{dx} \), representing the slope of the tangent to the curve at any given point \( (x, y) \). By differentiating \( y = c x^2 \) with respect to \( x \), we obtain \( \frac{dy}{dx} = 2cx \). This allows us to replace \( c \) using \( c = \frac{y}{x^2} \), giving the differential equation \( \frac{dy}{dx} = \frac{2y}{x} \). This equation describes how the slope changes as you move along the curve.

To find orthogonal trajectories, which are curves intersecting the family of curves at right angles, you modify the differential equation. The slopes of orthogonal trajectories multiplied by the original slope must equal \(-1\), leading to a different differential equation for these new curves. In this exercise, the orthogonal trajectories’ differential equation is \( \frac{dy}{dx} = -\frac{x}{2y} \). Solving this, via separation of variables or other methods, provides the trajectories’ equation.

Family of Curves

A family of curves consists of multiple curves that share a common characteristic, typically represented by a variable parameter. In this exercise, the family of curves is given by \( y = c x^{2} \), where \( c \) is a constant that can take many values.

This family can be visualized by adjusting \( c \) to different values, producing different parabolas all with their vertex at the origin. Each parabola is similar in shape but may be stretched more or less widely depending on the specific value of \( c \). The parameter \( c \) determines the steepness and orientation of each curve, essentially acting as a dial to transform one curve into another within the same family.

Understanding this concept is vital for recognizing how small alterations in parameters lead to significant changes in the shape and behavior of the curves. Such understanding facilitates further exploration of complex curves and their interactions, like those seen with their orthogonal trajectories. Orthogonal trajectories consist of a different family of curves traversing our original family perpendicularly, illustrating the beautiful interplay between different mathematical structures.

Ellipses

Ellipses are geometric shapes that are defined by certain equations. In our exercise, orthogonal trajectories derived from the family of parabolas \( y = c x^{2} \) take the form of ellipses. This is shown in the derived equation \( y^2 + \frac{x^2}{2} = C \).

An ellipse is characterized by its two focal points. Any point on the ellipse has the property that the sum of its distances to these focal points is constant. In standard form, an ellipse centered at the origin with its major and minor axes aligned along the Cartesian plane axes is given by \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \).

In the trajectory equation \( y^2 + \frac{x^2}{2} = C \), the values \( a \) and \( b \) are implied by constants derived from the parameter \( C \) and scaling factors in the equation.

• If \( C \) is positive, it indicates possible real elliptical shapes.
• When sketching these ellipses, one might notice they are compressed or stretched versions depending on how \( C \) and related parameters adjust the axes.

Ellipses, through their connection to orthogonal trajectories, serve to demonstrate how modifying initial conditions and rational expressions can produce new mathematical relationships, emphasizing the interconnectedness of mathematical concepts.