Question

Orthogonal trajectories Two curves are orthogonal to each other if their tangent lines are perpendicular at each point of intersection. A family of curves forms orthogonal trajectories with another family of curves if each curve in one family is orthogonal to each curve in the other family. Use the following steps to find the orthogonal trajectories of the family of ellipses \(2 x^{2}+y^{2}=a^{2}\). a. Apply implicit differentiation to \(2 x^{2}+y^{2}=a^{2}\) to show that \(\frac{d y}{d x}=\frac{-2 x}{y}\) b. The family of trajectories orthogonal to \(2 x^{2}+y^{2}=a^{2}\) satisfies the differential equation \(\frac{d y}{d x}=\frac{y}{2 x} .\) Why? c. Solve the differential equation in part (b) to verify that \(y^{2}=e^{c}|x|\) and then explain why it follows that \(y^{2}=k x\) where \(k\) is an arbitrary constant. Therefore, the family of parabolas \(y^{2}=k x\) forms the orthogonal trajectories of the family of ellipses \(2 x^{2}+y^{2}=a^{2}\).

Short answer

#tag_title# Short Answer #tag_content# The family of parabolas \(y^2 = kx\) forms the orthogonal trajectories of the family of ellipses \(2x^2 + y^2 = a^2\). This is because their derivatives satisfy the condition that their product is equal to -1, as we found \(\frac{dy}{dx} = \frac{-2x}{y}\) for the ellipse and \(\frac{dy}{dx} = \frac{y}{2x}\) for the orthogonal trajectory.

Step-by-step solution

Apply implicit differentiation to \(2x^2 + y^2 = a^2\)

In this step, we will find the derivative with respect to \(x\): \(\frac{d}{dx} (2x^2 + y^2) = \frac{d}{dx} (a^2)\) Differentiate each term with respect to x: \(4x + 2yy' = 0\) Now, we can find \(\frac{dy}{dx}\): \(y' = \frac{-2x}{y}\) Step 2: Check if the product of their derivatives is equal to -1

Verify that the orthogonal trajectories satisfy the given differential equation

Given that \(\frac{dy}{dx} = \frac{-2x}{y}\), the family of trajectories orthogonal to \(2x^2 + y^2 = a^2\) will satisfy the differential equation: \(\frac{dy}{dx} = \frac{y}{2x}\), because the product of the original derivative, \(\frac{-2x}{y}\), and the orthogonal derivative, \(\frac{y}{2x}\), is equal to -1: \(\left(\frac{-2x}{y}\right)\left(\frac{y}{2x}\right) = -1\) Step 3: Solve the differential equation

Solve the differential equation \(\frac{dy}{dx} = \frac{y}{2x}\)

We can solve this first-order differential equation by using separation of variables: \(\frac{dy}{y} = \frac{dx}{2x}\) Now, integrate both sides: \(\int \frac{dy}{y} = \int \frac{dx}{2x}\) \(\ln |y| = \frac{1}{2}\ln |x| + C\) Exponentiate both sides: \(y^2 = e^C|x|\) Step 4: Determine the orthogonal trajectories

Explain why it follows that \(y^2 = kx\)

We are interested in the family of orthogonal trajectories, so we will substitute the constant \(k = e^C\): \(y^2 = kx\), where k is an arbitrary constant. Therefore, the family of parabolas \(y^2 = kx\) forms the orthogonal trajectories of the family of ellipses \(2x^2 + y^2 = a^2\).

Key concepts

Implicit Differentiation

Implicit differentiation is a technique used to find the derivative of a function when the relationship between the variables is given in an implicit form rather than an explicit equation. For example, in the original exercise, the equation of the ellipse is given implicitly as \(2x^2 + y^2 = a^2\).
Here are some helpful steps to apply implicit differentiation:

• Differentiating both sides of the equation with respect to \(x\): On the left side, use the power rule to differentiate the terms related to \(x\) and \(y\). Remember, when differentiating \(y\) terms, treat \(y\) as a function of \(x\) and multiply by \(\frac{dy}{dx}\).
• For example, differentiating \(y^2\) yields \(2y\frac{dy}{dx}\) because of the chain rule.
• Combine all terms and solve for \(\frac{dy}{dx}\) to find the derivative in terms of both \(x\) and \(y\).

This method allows us to find the rate of change of \(y\) with respect to \(x\) without having to solve for \(y\) explicitly. In our exercise, it was crucial for finding that the slope \(\frac{dy}{dx}\) of the ellipse was \(\frac{-2x}{y}\).

Differential Equations

Differential equations involve finding a function or a family of functions that satisfy a given mathematical relationship involving derivatives. These equations often describe physical phenomena, including motion, growth, and decay processes. In our context, differential equations help us find orthogonal trajectories.

The exercise asks us to establish the differential equation of the orthogonal trajectories of ellipses given by \(\frac{dy}{dx} = \frac{y}{2x}\). This result stems from the principle that the multiplication of slopes from each family of curves must equal -1 to be orthogonal. Orthogonality is key when dealing with curves intersecting at right angles.

• The differential equation \(\frac{dy}{dx} = \frac{y}{2x}\) was derived because it directly modifies the original derivative \(\frac{-2x}{y}\) to satisfy this orthogonality condition.
• Solving this equation requires rearranging and integrating, a classic method used in solving first-order differential equations by separation of variables.

Understanding and solving differential equations in this context is essential, as it reveals the inherent relationships between the curves in terms of direction and slope.

Family of Curves

The term "family of curves" refers to a set of related curves described by an equation with a parameter. By varying this parameter, different curves within the family are produced.
In the context of our exercise, we have two families of curves:

• The family of ellipses described by the equation \(2x^2 + y^2 = a^2\), where \(a\) is a parameter, representing the semi-axis length.
• Through the process of solving the differential equation, we determined that the family of parabolas described by \(y^2 = kx\) intersects orthogonally with the ellipses. Here, \(k\) is an arbitrary constant serving as a parameter, specifying different parabolas for varying values of \(k\).

Families of curves play a pivotal role in understanding complex relationships between different types of equations and their geometric interpretations. By analyzing how these curves interact, one can infer properties such as symmetry, intersection points, and dynamic relationships, aiding in both theoretical and practical problem-solving contexts.