obtaining $y'$ from $r = 1 + \sin\theta$

550 Views Asked by At

Just want to check if this is the idea for this

$$x = r\cos\theta$$ $$y = r\sin\theta$$

so now we substitute

$$x = (1 + \sin\theta)\cos\theta$$ $$y = (1 + \sin\theta)\sin\theta$$

get the derivative of $x$ and $y$ in that form (after distributing) and make it $dy/dx$, thus

$$\frac{dy}{dx} = \frac{\cos\theta + 2\sin\theta \cos\theta}{-\sin\theta+\cos^2\theta-\sin^2\theta}$$

Dont know where to go from there. if there is any simplification or another step I'm not aware of. But is this the concept?

edit: Also as a follow up the $y''=d^2y/dx^2$ means we get the double prime of the top and bottom or just the top with the $dx$ being squared on the bottom?