Finding a Family of Orthogonal Trajectories

Find the family of orthogonal trajectories for the one-parameter family \(y^{2}=Cx^{3}\). See Figure 3 for several graphs in the family.

Solution

Step 1 We differentiate \(y^{2}=Cx^{3}\) with respect to \(x\). \[ 2yy^{\prime} =3Cx^{2} \]

Then we eliminate \(C\) using this equation and the equation \(y^{2}=Cx^{3}\) of the family. Since \(C=\dfrac{y^{2}}{x^{3}},\) we find \[ \begin{eqnarray*} 2yy^{\prime} & =&3\frac{y^{2}}{x^{3}}x^{2} \\ y^{\prime} & =&\frac{3y}{2x} \end{eqnarray*} \]

Step 2 Now we replace \(y^{\prime} \) with \(-\dfrac{1}{y^{\prime} },\) to obtain the differential equation for the family of orthogonal trajectories. \[ \begin{eqnarray*} -\dfrac{1}{y^{\prime} }& =&\frac{3y}{2x} \\ y^{\prime} & =&-\frac{2x}{3y} \end{eqnarray*} \]

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Step 3 The differential equation \(\dfrac{dy}{dx}=-\dfrac{2x}{3y}\) is separable and can be written as \[ \begin{equation*} 2x\,dx+3y\,dy=0 \end{equation*} \]

The general solution is \[ \begin{equation*} x^{2}+\frac{3}{2}y^{2}=K \end{equation*} \]

The orthogonal trajectories for the family \(y^{2}=Cx^{3}\) is a family of ellipses \(x^{2}+\dfrac{3}{2}y^{2}=K\), as shown in Figure 4.