Question

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 from $\mathit{y}\mathbf{\text{'}}$for the original curves; this constant takes different values for different curves 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 $\left(2.10\right)$to $\left(2.10\right)$

$\mathit{y}\mathbf{=}\mathit{k}{\mathit{x}}^{\mathbf{n}}$. (Assume that n is a given number; the different curves of the family have different values of k.)

Short answer

Answer

The orthogonal trajectory for $y=k{x}^n$is $y=\sqrt{-2x^n+C_1}$.

Step-by-step solution

Given information

The given equation is$y=k{x}^n$.

Definition of ordinary differential equation

An ordinary differential equation (ODE) is a differential equation containing one or even more functions of one independent variable and its derivatives.

Find the orthogonal trajectory of y=kxn

Draw the family of curves for $y=k{x}^n$with $n=2$.

Graph 1

Find the slope of a line in the family.

$$\begin{gathered} y\text{'}=nk{x}^{n-1} \\ y\text{'}=\frac{ny}{x} \end{gathered}$$

The slope of an orthogonal trajectory is $y\text{'}=\frac{-x}{ny}$.

Solve the above differential equation using the variable separable method.

$$\begin{gathered} n\int ydy=-\int x^{n-1}dx \\ \frac{n}{2}{y}^2=-\frac{x^n}{n}+C \\ y=\sqrt{-2x^n+C_1} \end{gathered}$$

Plot the above trajectory.

Graph 2

Combine the curves with their orthogonal trajectories in one graph.

Graph 3

Therefore, the orthogonal trajectory for $y=k{x}^n$is $y=\sqrt{2x^n+C_1}$.