Chapter 7.3, Problem 30E

To determine

To find : the orthogonal trajectories of the family of curves and draw members of family by using graphing device.

Answer to Problem 30E

Thus the orthogonal trajectories are the family of $y^2+\frac{2}{3}{x}^2=c$

Explanation of Solution

Given information :

The equation is $y^2=k{x}^3$ .

Calculation : Since the orthogonal trajectories to this family are those curves whose tangent lines are orthogonal to the tangent lines of the family, we should determine the slopes of the tangent lines to the given family of curves. To do so, differentiate implicitly:

  $\begin{array}{l}{y}^2=k{x}^3\\ 2y\frac{dy}{dx}=k3x^2\text{ [differentialte with respect to }x]\\ \frac{dy}{dx}=\frac{3k{x}^2}{2y}\text{ [divide by 2}y\text{]}\end{array}$

The differential equation depends on k but it is needed an equation that is valid for all values of k simultaneously. To eliminate k , substitute $k=\frac{y^2}{x^3}$ in the equation of slope,

  $\begin{array}{l}\frac{dy}{dx}=\frac{3k{x}^2}{2y}\\ \frac{dy}{dx}=\frac{3x^2}{2y}·\frac{y^2}{x^3}\text{ [substitute]}\\ \frac{dy}{dx}=\frac{3y}{2x}\text{ [divide out common factors]}\end{array}$

Since the orthogonal trajectories should have perpendicular tangent lines, their slopes will be negative reciprocals of the above. In other words, the orthogonal trajectories of the given family are the solutions of the differential equation

Therefore, the orthogonal trajectories must satisfy the differential equation,

  $\frac{dy}{dx}=-\frac{2x}{3y}$

The differential equation is separable, so to solve it

  $\begin{array}{l}\frac{dy}{dx}=-\frac{2x}{3y}\\ 3ydy=-2xdx\text{ [separate variables]}\\ {\displaystyle \int 3y=-{\displaystyle \int 2xdx}}\text{ [integrate both side]}\\ \text{3}·\frac{y^2}{2}=-2·\frac{x^2}{2}+c\text{ [simplify integrate]}\\ \frac{3y^2}{2}=-x^2+c\text{ [divide]}\\ y^2\text{=}\frac{-2}{3}{x}^2+c\text{ [multiply by 2 and divide by 3]}\\ y^2+\frac{2}{3}{x}^2=c\text{ [simplify]}\end{array}$

Where c is an arbitrary positive constant.

Thus the orthogonal trajectories are the family of $y^2+\frac{2}{3}{x}^2=c$

For members of family substitute c by different values ,

  $\begin{array}{l}{y}^2+\frac{2}{3}{x}^2=1\text{ [}c=1]\\ y^2+\frac{2}{3}{x}^2=2\text{ [}c=2]\\ y^2+\frac{2}{3}{x}^2=3\text{ [}c=3]\\ y^2+\frac{2}{3}{x}^2=4\text{ [}c=4]\end{array}$

Using implicit plotting capability of a CAS to graph the curve $y=\pm \sqrt{1-\frac{2}{3}{x}^2},y=\pm \sqrt{2-\frac{2}{3}{x}^2},y=\pm \sqrt{3-\frac{2}{3}{x}^2},y=\pm \sqrt{4-\frac{2}{3}{x}^2}$

The graph is obtained as :