## anonymous 5 years ago If r = -2cost, then dy/dx=

Hello watson, You can use the transformations from rectilinear to polar coordinates:$x=r \cos t, y = r \sin t$Given this, it follows$x^2+y^2=r^2\cos^ t + r^2 \sin^2 t= r^2 (\cos^2 t + \sin^2 t) = r^2$Your original equation can be rewritten as,$r=-2\frac{x}{r} \rightarrow r^2=-2x \rightarrow x^2+y^2=-2x \rightarrow y^2=-2x-x^2$Operating on the left and right sides with the derivative with respect to x,$\frac{d}{dx}y^2=\frac{d}{dx}(-2x-x^2)\rightarrow \frac{dy^2}{dy}\frac{dy}{dx}=-2-2x$$\rightarrow 2y \frac{dy}{dx}=-2(1+x) \rightarrow \frac{dy}{dx}=-\frac{1+x}{y}$