Find dy/dx given x = phi (1- sin phi), y = phi cos phi

Find $\dfrac{dy}{dx}$ given $x = \varphi\bigl(1-\sin\varphi\bigr)$, $y = \varphi\cos\varphi$.

Step 1: Differentiate $x$

Product rule with the first factor $\varphi$ and the second $1-\sin\varphi$: \[x'_{\varphi} = \bigl(1-\sin\varphi\bigr) + \varphi\bigl(-\cos\varphi\bigr) = 1-\sin\varphi-\varphi\cos\varphi.\]

Step 2: Differentiate $y$

\[ y'_{\varphi} = \cos\varphi + \varphi\bigl(-\sin\varphi\bigr) = \cos\varphi - \varphi\sin\varphi. \]

Step 3: Form the ratio

\[ \frac{dy}{dx} = \frac{\cos\varphi-\varphi\sin\varphi}{1-\sin\varphi-\varphi\cos\varphi}. \]

No common factor appears, so the answer is left in this form.

Step 4: Verification

\[\varphi = 0:\quad (x,y) = (0,0),\quad \frac{dy}{dx} = \frac{1-0}{1-0-0} = 1 \;\;\checkmark\]

\[\varphi = \frac{\pi}{2}:\quad (x,y) = (0,0),\quad \frac{dy}{dx} = \frac{0-\frac{\pi}{2}}{1-1-0} \ \text{— undefined (vertical tangent)}\]

\[\varphi = \pi:\quad \frac{dy}{dx} = \frac{-1-0}{1-0+\pi} = \frac{-1}{1+\pi} \approx -0.24145 \;\;\checkmark\]

Step 5: A self-intersection

Both $\varphi = 0$ and $\varphi = \tfrac{\pi}{2}$ send the point to the origin, since $1-\sin\tfrac{\pi}{2} = 0$ and $\cos\tfrac{\pi}{2} = 0$. The curve therefore passes through the origin twice, with slope $1$ on the first visit and a vertical tangent on the second — a node, not a cusp.