Question

Finding Orthogonal Trajectories In Exercises \(41-46,\) find the orthogonal trajectories of the family. Use a graphing utility to graph several members of each family. $$ x^{2}=C y $$

Short answer

The orthogonal trajectories of the given family of curves \(x^2 = Cy\) are \(y = -\frac{1}{2} x^2 \ln|x| + Kx\), for all real numbers \(K\).

Step-by-step solution

Find the Derivative

Start by differentiating the equation \(x^2 = Cy\) with respect to \(x\). This yields \(2x = C*y'\), where \(y'\) is the derivative of \(y\) with respect to \(x\). Solving for \(y'\) gives \(y' = \frac{2x}{C}\).

Calculate the Orthogonal Trajectory

The orthogonal trajectory to a given family of curves is found by taking the negative reciprocal of the derivative. This involves interchanging the coefficients of \(x\) and \(y\) in the derivative, and changing the sign. We compute: \( -\frac{C}{2x}\).

Solve the Differential Equatioin

Now, we need to solve the differential equation \(y' = -\frac{C}{2x}\). Integrating both sides with respect to \(x\) gives \(y = -\frac{C}{2} \ln|x| + K\), where \(K\) is an arbitrary constant (representing the constant of integration). This is the solution to the original differential equation, giving all possible orthogonal trajectories to the given family of curves.

Solve for Constant \(C\)

Substitute \(y = -\frac{C}{2} \ln|x| + K\) back into the original equation \(x^2 = Cy\) to find the constant \(C\). This yields \(x^2 = -2y \ln|x| + 2Ky\). Solving for \(K\) we find \(K = \frac{x^2 + 2y\ln|x|}{2y}\). Therefore, the orthogonal trajectories of the given family are given by \(y = -\frac{1}{2} x^2 \ln|x| + Kx\), for all real numbers \(K\).

Key concepts

Differential Equations

Differential equations are mathematical equations that describe the relationship between a function and its derivatives. They are crucial in modelling the behaviour of various physical systems, such as the motion of objects under the influence of forces, the flow of current in electrical circuits, and the growth and decay of populations in biology.

Differential equations can be ordinary (ODEs), involving derivatives with respect to only one variable, or partial (PDEs), involving derivatives with respect to multiple variables. The equation provided in the exercise, \( x^2 = Cy \), is an example of an implicit ODE, where the solution gives us a relationship between the two variables without explicitly solving for one in terms of the other.

To solve these equations, one typically starts with finding the derivative (rate of change) of the dependent variable with respect to the independent one, which represents the slope of the solution curve at any given point. The process often involves integration to find a function that satisfies the initial differential equation, leading to a general solution that includes an integration constant to account for initial conditions or particular solutions.

Derivative Calculation

The derivative calculation is a fundamental concept in calculus representing the rate at which a function changes at a particular point. It's defined as the limit of the average rate of change of the function over an interval as the interval becomes infinitesimally small. In simpler terms, when you differentiate a function, you're finding an equation for the slope of the tangent to the function's graph at any point.

In the provided problem, the derivative of the function \(x^2 = Cy\) with respect to \(x\) is found using explicit differentiation. Differentiation yields \(2x = C*y'\), where \(y'\) denotes the derivative of \(y\) with respect to \(x\). The solution to this step is critical to find orthogonal trajectories, which are curves that intersect the original family of curves at right angles.

Derivative calculation allows us to understand how a quantity changes in response to changes in another quantity; this is the core principle behind the study of rates of change in physical processes, economics, population studies, and other fields.

Integration Constants

When we integrate a function to solve a differential equation, we often add a symbol, typically \(C\), known as an integration constant. This represents an infinite family of solutions because when you differentiate a constant, it disappears, leaving behind an unspecified integral that could have been any number.

For example, when finding the orthogonal trajectories in the exercise, integration of \(y' = -\frac{C}{2x}\) yields \(y = -\frac{C}{2} \ln|x| + K\), where \(K\) is the integration constant. This constant is crucial for capturing all potential solutions that account for different initial conditions or peculiarities of the system being modeled.

Understanding how to properly incorporate integration constants is essential for crafting complete solutions to differential equations. This also lays the groundwork for grasping more advanced concepts in calculus, such as the initial value problem, where the constant is determined by a given condition at a specific point in the domain of the solution.