The normal form of the equation of a straight | vedclass

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Question

(A) Step $1$: The equation of the line is $x \cos \alpha + y \sin \alpha = p$.Step $2$: Differentiate with respect to $x$: $\cos \alpha + \frac{dy}{dx

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$\frac{d^2y}{dx^2} = 0$

Vedclass · Answer

(A) Step $1$: The equation of the line is $x \cos \alpha + y \sin \alpha = p$.Step $2$: Differentiate with respect to $x$: $\cos \alpha + \frac{dy}{dx} \sin \alpha = 0$.Step $3$: This gives $\frac{dy}{dx} = - \frac{\cos \alpha}{\sin \alpha} = - \cot \alpha$.Step $4$: Differentiate again with respect to $x$: $\frac{d^2y}{dx^2} = \frac{d}{dx} (- \cot \alpha) = 0$ (since $\alpha$ is a constant).Step $5$: Thus, the differential equation is $\frac{d^2y}{dx^2} = 0$.

The normal form of the equation of a straight line is $x \cos \alpha + y \sin \alpha = p$. Find the differential equation of the family of all such lines where $p$ and $\alpha$ are arbitrary constants.

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