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Question

Find a family of oblique trajectories that intersect the family of curves x+x+ y=cx2y=c x^{2} at angle aa such that tana=2\tan a=2.

Solution

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Answered 2 years ago
Answered 2 years ago
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Differentiate the equation of the family x+y=c x2x + y = c \ x^2 with respect to x. Then we have

1+dydx=2c xdydx=2c x1\begin{gather*} 1 + \dfrac{dy}{dx} = 2c \ x \\ \dfrac{dy}{dx} = 2c \ x - 1 \tag{1} \\ \end{gather*}

Finding the parameter c from the given equation. Then we obtain

x+y=c x2c=x+yx2\begin{gather*} x + y = c \ x^2\\ c = \dfrac{x + y}{x^2} \tag{2} \\ \end{gather*}

Substituting from equation (2) into equation (1), then we get the differential equation of the given family.

dydx=2 (x+yx2) x1dydx=2x+2yx1dydx=2x+2yxxdydx=x+2yx\begin{gather*} \dfrac{dy}{dx} = 2 \ \left(\dfrac{x + y}{x^2}\right) \ x - 1 \\ \dfrac{dy}{dx} = \dfrac{2x + 2y}{x} - 1 \\ \dfrac{dy}{dx} = \dfrac{2x + 2y - x}{x} \\\\ \dfrac{dy}{dx} = \dfrac{x + 2y}{x} \end{gather*}

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