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Find the orthogonal trajectories. First guess from the plot of the given curves. Plot or sketch some curves and trajectories. (Show the details of your work.) $y=c x^{3 / 2}$

Solution

VerifiedThe given equation of the curves can also be written as

$\begin{align*} yx^{-3/2}=c \end{align*}$

Differentiating both sides with respect to $x$ yields

$\begin{align*} & y'x^{-3/2}-\frac{3}{2}yx^{-3/2-1}=0\\ \implies & y'-\frac{3}{2}yx^{-1}=0\\ \implies & y'=\frac{3y}{2x} \end{align*}$

It follows that the differential equations of the orthogonal trajectories are

$\begin{align*} &y'=-\frac{2x}{3y}\\ \implies & 3ydy+2xdx=\frac{c^*}{2}\\ \implies & 3\int ydy+2\int xdx=\frac{c^*}{2}\\ \implies & \frac{3}{2}y^2+x^2=\frac{c^*}{2}\\ \implies & 3y^2+2x^2=c^* \end{align*}$

Therefore, the orthogonal trajectories of the given curves are

$3y^2+2x^2=c^*.$

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