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. $$ x^{2}+y^{2}=C $$

Short answer

The orthogonal trajectories for the provided curve \(x^2+y^2=C\) are given by \(y=lnc|x|\).

Step-by-step solution

Write the Initial Differential Equation

Start from the given equation, \(x^{2}+y^{2}=C\). Take derivative with respect to \(x\), which gives \(2x+2yy'=0\). Solve for \(y'\) to obtain the differential equation for these curves. We get \(y'=-x/y\).

Find the Orthogonal Trajectories

The orthogonal trajectories are the solutions to the differential equation obtained by changing \(y'\) into \(-1/y'\). So replace \(y'\) in the equation obtained in step 1 with \(-1/y'\) to find the orthogonal trajectories. This yields \(y'=-y/x\).

Solve the New Differential Equation

The differential equation obtained is of separable type. Therefore, separate the variables and integrate. This gives \(\int dy = \int -y/x dx\). This can be integrated to \(y=lnc|x|\). Solve for \(y\) to obtain the equation of the orthogonal trajectories which is \(y=lnc|x|\).

Graph the Orthogonal Trajectories

Use a graphing utility to graph several members of this orthogonal trajectories. This helps to visually confirm whether these curves intersect the original family of curves at right angles.

Key concepts

Differential Equations

Differential equations are mathematical expressions that relate a function with its derivatives. These equations are fundamental in expressing physical laws and phenomena where the rate of change of a variable is relevant to its value. Imagine a scenario where you are tracking the growth of plants, and you need to find a relationship between the growth rate and time – it's a differential equation that comes to the rescue.

When we are presented with an equation such as \(x^{2}+y^{2}=C\), it describes a family of curves, circles in this case, where \(C\) represents the square of the radius. To study the rate of change of the variables involved, we take the derivative of this equation with respect to one of the variables - typically \(x\), holding \(C\) as a constant. This leads to a new equation involving \(y'\), which indicates the slope of the tangent line to the curve at any point. Understanding the slope is crucial in predicting the behavior of the curve and, in the case of orthogonal trajectories, in finding curves that intersect perpendicularly.

Separable Equations

A separable differential equation is one that can be separated into two parts: one that contains only the variable \(y\) (and its derivatives) and another that contains only the variable \(x\). The beauty of separable equations lies in their simplicity for solving, as once separated, we can integrate each side with respect to its own variable.

Consider the equation \(y'=-y/x\). This is a prime example of a separable equation. By rearranging the terms to isolate \(y\) terms on one side and \(x\) terms on the other, we have \(dy/y = -dx/x\), which sets us up perfectly for the process of integration. This property not only simplifies the process of finding solutions but also makes it an excellent tool for students to begin their journey into the intriguing world of differential equations.

Integration

Integration, one of the two main operations in calculus, is the process of finding the integral of a function, which represents the area under the curve of a graph of the function. It's like piecing together the whole from infinitely many infinitely small parts.

When we arrive at an expression like \(\int dy/y = -\int dx/x\), we're setting up to find the antiderivatives of the functions on each side of the equation. Executing this process results in \(ln|y| = -ln|x| + C\), where \(C\) is the constant of integration that represents the general solution to the differential equation. After integrating, we can manipulate the solution to find the exact relationship we need, such as expressing \(y\) in terms of \(x\), to describe the orthogonal trajectories of the original family of curves. Honing integration skills is vital for students tackling a variety of problems in calculus, physics, and beyond!