Question

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.

Short answer

The given family of parabolas is defined by the equation \(y^{2}=2 c x+c^{2}\). We find its differential equation, and then we find the differential equation for orthogonal curves by replacing \(\frac{dy}{dx}\) with \(-\frac{dx}{dy}\). By solving this differential equation, we obtain the orthogonal trajectories as \(y^{2} = \exp\left(-\frac{2}{c}x+2D\right)\). Comparing this equation with the given family of parabolas equation, we confirm that the given family of parabolas is self-orthogonal.

Step-by-step solution

Find the differential equation of the given family

We first find the differential equation of the given family of parabolas. To do that, we differentiate the equation \(y^{2}=2 c x+c^{2}\) with respect to \(x\). We have: \(\frac{d}{dx}(y^2) = \frac{d}{dx}(2cx + c^2)\) Applying the chain rule and product rule, we get: \(2y \frac{dy}{dx} = 2c\) Now, divide by 2: \(y\frac{dy}{dx} = c\)

Find the differential equation for orthogonal curves

We want to find the orthogonal trajectories to the given family of parabolas. We do this by replacing \(\frac{dy}{dx}\) with \(-\frac{dx}{dy}\) in the differential equation obtained in Step 1. \(-y\frac{dx}{dy} = c\) Now we can solve this differential equation to find the orthogonal trajectories.

Solve the differential equation

Separate the variables and solve: \(-y\frac{dx}{dy} = c\) \(\frac{dy}{y} = -\frac{dx}{c}\) Integrate both sides: \(\int\frac{1}{y}dy=-\frac{1}{c}\int dx\) \(\ln|y|=-\frac{1}{c}x+D\), where D is the constant of integration. Take exponent of both sides to get: \(y = \pm\exp\left(-\frac{1}{c}x+D\right)\) Using the property of exponential functions, we can rewrite this equation as: \(y^{2} = \exp\left(-\frac{2}{c}x+2D\right)\)

Compare the orthogonal trajectories to the given family

Compare the orthogonal trajectories equation found in Step 3 with the given family of parabolas equation, \(y^{2}=2 c x+c^{2}\). \(y^{2} = \exp\left(-\frac{2}{c}x+2D\right)\) Before comparing the given family to the obtained trajectories, we isolate the term 2cx by taking the logarithm on both sides of the equation: \(\ln{(y^2)}= -\frac{2}{c}x+2D\) Tooting at the term \(\frac{2D}{c}\), we realize that there must exist a constant \(\kappa\) such that: \(2D =\kappa\) Thus, the equation becomes: \(y^{2} =\exp\left(-\frac{2}{c}x+\kappa\right)\) Comparing this equation with the given family of parabolas, we observe that: \(y^2 = 2cx + c^2 \implies 2cx+c^2 = \exp\left(-\frac{2}{c}x+\kappa\right)\) Hence we have obtained a family of parabolas as the orthogonal trajectories, which confirms that the given family of parabolas is self-orthogonal.

Key concepts

Ordinary Differential Equations

Ordinary differential equations (ODEs) are mathematical equations that relate functions and their derivatives. They are used to describe the changing rates of various phenomena, from simple motion to the dynamics of populations. A single ODE or a system of ODEs can model a wide range of physical, biological, economic, and engineering processes.

When solving an ODE, like the one in our exercise, we apply specific techniques such as separation of variables or integrating factors. In our textbook example, we examined the family of parabolas defined by the equation \(y^{2}=2 c x+c^{2}\). To understand the behaviour of these curves, we first derived their differential equation by differentiating with respect to \(x\), which provided a relationship between \(y\), \(dy/dx\), and the parameter \(c\).

After obtaining the differential equation, we performed algebraic manipulations to express it in a standard form suitable for finding solutions or analyzing further properties. This step is fundamental in understanding the broader concept of ODEs and how they govern the systems they represent.

Family of Parabolas

A family of parabolas is a set of curves that can be described by varying a parameter within a general equation. In the context of our exercise, the family of parabolas is given by \(y^{2}=2 c x+c^{2}\), where \(c\) is a parameter that can take on any real value. Each value of \(c\) defines a different member of the parabola family.

These parabolas share a common set of features, which include a symmetrical shape, a vertex that shifts based on the value of \(c\), and an axis of symmetry. In understanding the family of parabolas, we can analyze how the curves move and evolve as the parameter changes.

When facing such a family of curves, a Differential Equations approach lets us create a unifying description of the entire set through one encompassing equation. This method not only streamlines our understanding of the curves but also paves the way to discovering additional properties, such as self-orthogonality, which we explore in our main exercise.

Orthogonal Trajectories

Orthogonal trajectories are curves that intersect a given family of curves at right angles (90 degrees). These trajectories have great significance in fields ranging from physics to geometry because they often represent contours with opposing physical properties or complementary relationships.

In our step-by-step solution, we sought the trajectories orthogonal to the given family of parabolas. To find these trajectories, we took the derivative of the original set of curves and applied a key idea: two lines are orthogonal if the product of their slopes is -1. This led us to replace the derivative \(dy/dx\) with \(-dx/dy\) in the differential equation from Step 1.

After solving the new differential equation, we obtained another set of curves. The determination that these new curves formed the same family of parabolas confirmed the self-orthogonality of the original family. Understanding orthogonal trajectories is essential because it allows us to explore how different system states or conditions can be intrinsically connected in a perpendicular and thus equilibrium-establishing manner.