Find the orthogonal trajectories of each of the following families of curves. In each case, sketch or computer plot several of the given curves and several of their orthogonal trajectories. Be careful to eliminate the constant fromfor the original curves; this constant takes different values for different curves ${\mathit{y}}^{\mathbf{\text{'}}}$of the original family, and you want an expression for${\mathit{y}}^{\mathbf{\text{'}}}$which is valid for all curves of the family crossed by the orthogonal trajectory you are trying to find. See equations (2.10)to (2.12).

$\mathit{x}\mathit{y}\mathbf{=}\mathit{k}$

The orthogonal trajectory are the angles that are vertical to the family everyplace. In other words, the orthogonal circles are another family of angles within which each wind is vertical to the angles in original family.

The given family of curve is$xy=k$.

There need tofind the orthogonal trajectory of given family of curve and sketch the given curve and their orthogonal trajectories.

Firstly, draw the curve of the given family$xy=k$as shown in figure 1.

Find the slope of a line of the family (notice that,there need to substitute the value of the constant$xy=k$ ), therefore

$$\begin{gathered} y^{\text{'}}=-\frac{k}{x^2} \\ =-\frac{y}{x} \end{gathered}$$

but it knows that (based on the analytic geometry) the slopes of two perpendicular lines are negative reciprocals.

Therefore, the slope of the orthogonal trajectory is

$y^{\text{'}}=\frac{x}{y}$

Further, using the separation of variables method

$$\begin{gathered} \int ydy=\int xdx \\ \frac{y^2}{2}=\frac{x^2}{2}+C \\ y=\sqrt{x^2+C_1} \end{gathered}$$

Plot the orthogonal trajectory as shown in figure 2.

Now, combine the curves with their orthogonal trajectories in one graph as shown in Figure 3

Thus, the orthogonal trajectory is given by the equation$y=\sqrt{x^2+C_1}$.