Find the isogonal trajectories, at constant angle α , of the…

Find the isogonal trajectories, at constant angle $\alpha$, of the family of cardioids $r=a(1+\cos\varphi)$, and examine the orthogonal ($\alpha=90°$) limiting case.

Step 1: Re-derive $\tan\psi_1$ independently

\texttt{sympy.diff} gives $dr/d\varphi=-a\sin\varphi$ for $r=a(1+\cos\varphi)$. Then $\tan\psi_1=r/(dr/d\varphi)=-(1+\cos\varphi)/\sin\varphi$ ($a$ cancels). Numerically evaluating this against $-\cot(\varphi/2)$ at several $\varphi$ gives agreement to machine precision ($<10^{-15}$), confirming \[\tan\psi_1=-\cot\frac{\varphi}{2}\] independently of the source's half-angle-identity hand algebra.

Step 2: Angle addition and the isogonal ODE

With $k=\tan\alpha$, \texttt{sympy.together} on $(\tan\psi_1+k)/(1-k\tan\psi_1)$ and a numeric cross-check against the source's claimed form $\dfrac{r}{dr/d\varphi}=\dfrac{k\sin(\varphi/2)-\cos(\varphi/2)}{\sin(\varphi/2)+k\cos(\varphi/2)}$ agree exactly at multiple $(\varphi,k)$ sample pairs, confirming equation (5): \[\frac{dr}{d\varphi}=\frac{r\left(\sin\frac{\varphi}{2}+k\cos\frac{\varphi}{2}\right)}{k\sin\frac{\varphi}{2}-\cos\frac{\varphi}{2}}.\]

Step 3: Redo the separable integration from scratch

\texttt{sympy.integrate} on $\dfrac{\sin(\varphi/2)+k\cos(\varphi/2)}{k\sin(\varphi/2)-\cos(\varphi/2)}$ (computed independently, not by differentiating the source's guess) returns $2\ln\left|k\sin\frac{\varphi}{2}-\cos\frac{\varphi}{2}\right|$ up to sign convention, matching the source's claimed antiderivative. Exponentiating gives \[r=A\left(k\sin\frac{\varphi}{2}-\cos\frac{\varphi}{2}\right)^2.\] A direct \texttt{sympy} substitution of this $r(\varphi)$ back into the isogonal ODE from Step 2 gives residual $0$ identically — the solution is confirmed correct, not merely internally consistent with the source's own algebra.

Step 4: Rigorous $\alpha\to90°$ limit of the general formula (the key check)

As $k\to\infty$ with $A$ fixed, $A(k\sin\frac{\varphi}{2}-\cos\frac{\varphi}{2})^2\to\infty$ for generic $\varphi$ — a finite limit requires $A$ to be rescaled. Setting $A=B/k^2$ and expanding, \[\frac{B}{k^2}\left(k\sin\tfrac{\varphi}{2}-\cos\tfrac{\varphi}{2}\right)^2=B\sin^2\tfrac{\varphi}{2}-\frac{2B\sin\frac{\varphi}{2}\cos\frac{\varphi}{2}}{k}+\frac{B\cos^2\frac{\varphi}{2}}{k^2},\] and \texttt{sympy.limit} as $k\to\infty$ gives exactly $B\sin^2(\varphi/2)$. Using $\sin^2(\varphi/2)=\tfrac12(1-\cos\varphi)$, this is $\tfrac{B}{2}(1-\cos\varphi)$. Setting $B=2b$ reproduces $b(1-\cos\varphi)$ exactly — \texttt{sympy} confirms the symbolic difference is identically $0$. A numeric convergence check (fixed $b=3$, $A=2b/k^2$, $k=10,100,1000,10000$) shows the finite-$k$ curve converging to $b(1-\cos\varphi)$ pointwise with error $\to0$ like $1/k$.