Find the equation of tangents to the curve y | vedclass

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(A) Given the curve $y = \cos(x + y)$. Differentiating with respect to $x$,we get $\frac{dy}{dx} = -\sin(x + y)(1 + \frac{dy}{dx})$.Solving for $\frac

Vedclass · Accepted Answer

$2x + 4y + 3\pi = 0$ and $2x + 4y - \pi = 0$

Vedclass · Answer

(A) Given the curve $y = \cos(x + y)$. Differentiating with respect to $x$,we get $\frac{dy}{dx} = -\sin(x + y)(1 + \frac{dy}{dx})$.Solving for $\frac{dy}{dx}$,we have $\frac{dy}{dx} = \frac{-\sin(x + y)}{1 + \sin(x + y)}$.The slope of the line $x + 2y = 0$ is $-\frac{1}{2}$.Since the tangents are parallel to the line,$\frac{-\sin(x + y)}{1 + \sin(x + y)} = -\frac{1}{2}$,which implies $2\sin(x + y) = 1 + \sin(x + y)$,so $\sin(x + y) = 1$.Thus,$x + y = 2n\pi + \frac{\pi}{2}$ for $n \in \mathbb{Z}$.Then $y = \cos(x + y) = \cos(2n\pi + \frac{\pi}{2}) = 0$.Substituting $y = 0$ into $x + y = 2n\pi + \frac{\pi}{2}$,we get $x = 2n\pi + \frac{\pi}{2}$.For $-2\pi \leq x \leq 2\pi$,the possible values for $x$ are $x = \frac{\pi}{2}$ (for $n=0$) and $x = -\frac{3\pi}{2}$ (for $n=-1$).The points of tangency are $(\frac{\pi}{2}, 0)$ and $(-\frac{3\pi}{2}, 0)$.The equation of the tangent at $(\frac{\pi}{2}, 0)$ is $y - 0 = -\frac{1}{2}(x - \frac{\pi}{2}) \Rightarrow 2x + 4y - \pi = 0$.The equation of the tangent at $(-\frac{3\pi}{2}, 0)$ is $y - 0 = -\frac{1}{2}(x + \frac{3\pi}{2}) \Rightarrow 2x + 4y + 3\pi = 0$.

Find the equation of tangents to the curve $y = \cos(x + y)$,$-2\pi \leq x \leq 2\pi$ that are parallel to the line $x + 2y = 0$.

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