Find the tangent of the angle between the polar axis

Find the tangent of the angle between the polar axis and the tangent to each curve at the pole ($\rho=0$): 1° $\rho=\sin^{3}\varphi$; 2° $\rho=\sin3\varphi$.

Step 1: General principle

At a value $\varphi_0$ where $\rho(\varphi_0)=0$ and $\rho'(\varphi_0)\neq0$, L'Hopital's rule on $y/x=\rho\sin\varphi/(\rho\cos\varphi)$ gives tangent inclination $\alpha=\varphi_0$ directly, since the $\rho$ factors cancel via $\rho'(\varphi_0)$.

Step 2: Part 1°: $\rho=\sin^{3}\varphi$

$\rho=0$ at $\varphi_n=n\pi$, where in fact $\rho'(\varphi_n)=3\sin^2\varphi_n\cos\varphi_n=0$ too (a higher-order zero), so the simple formula does not directly apply; compute the limit of $y/x$ itself:

\[ \lim_{\varphi\to0}\frac yx = \lim_{\varphi\to0}\frac{\sin^{3}\varphi\sin\varphi}{\sin^{3}\varphi\cos\varphi} = \lim_{\varphi\to0}\frac{\sin\varphi}{\cos\varphi} = 0. \tag{1} \]

The tangent at the pole is the polar axis itself.

Step 3: Part 2°: $\rho=\sin3\varphi$

$\rho=0$ at $\varphi_n=\dfrac{n\pi}3$; here $\rho'(\varphi_n)=3\cos(n\pi)=\pm3\neq0$, so the general principle applies directly: $\alpha=\varphi_n$, giving three distinct directions ($n=0,1,2$):

\[ \tan\alpha\bigl(\varphi_0=0\bigr)=0, \qquad \tan\alpha\!\left(\varphi_1=\frac\pi3\right)=\sqrt3, \qquad \tan\alpha\!\left(\varphi_2=\frac{2\pi}3\right)=-\sqrt3. \tag{2} \]

The three tangent directions at the pole are equally spaced by $60°$, matching the three-petalled rose traced by this curve.

Step 4: Verification

\[\rho=\sin3\varphi\text{: numerically computed slopes at } \varphi=0,\pi/3,2\pi/3 \text{ agree with } 0,\sqrt3,-\sqrt3\ \checkmark\]