Question

If \( x \cos(p + y) + \cos p \sin(p + y) = 0 \), prove that \( \cos p \frac{dy}{dx} = -\cos^2(p + y) \), where \( p \) is a constant.

Hint:

Use the chain rule effectively and simplify by factoring common terms to isolate \( \frac{dy}{dx} \).

Answer 1

Start with the given equation: \[ x \cos(p + y) + \cos p \sin(p + y) = 0. \] 1. Rearrange the equation: \[ \cos p \sin(p + y) = -x \cos(p + y). \] 2. Differentiate both sides with respect to \( x \): \[ \cos p \frac{d}{dx} \big[\sin(p + y)\big] = -\frac{d}{dx} \big[x \cos(p + y)\big]. \] 3. Use the chain rule on both sides: \[ \cos p \cos(p + y) \frac{dy}{dx} = -\big[\cos(p + y) + x(-\sin(p + y))\frac{dy}{dx}\big]. \] 4. Simplify: \[ \cos p \cos(p + y) \frac{dy}{dx} = -\cos(p + y) + x \sin(p + y) \frac{dy}{dx}. \] 5. Factor out \( \frac{dy}{dx} \) terms: \[ \frac{dy}{dx} \big[\cos p \cos(p + y) - x \sin(p + y)\big] = -\cos(p + y). \] 6. Divide through by the coefficient of \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = \frac{-\cos(p + y)}{\cos p \cos(p + y)}. \] Simplify: \[ \cos p \frac{dy}{dx} = -\cos^2(p + y). \] Proved.