Find the orthogonal trajectories of the family of curves…

Find the orthogonal trajectories of the family of curves $r^{2n}-2a^nr^n\cos n\varphi+a^{2n}=c^{2n}$, where $a$ and $n$ are fixed given constants and $c$ is the family's varying parameter.

Step 1: The source's original-family ODE is correct

\texttt{sympy} differentiates $r^{2n}-2a^nr^n\cos n\varphi+a^{2n}=c^{2n}$ implicitly in $\varphi$ and solves for $r'$, returning \[\frac{dr}{d\varphi}=\frac{-a^n r\sin n\varphi}{r^n-a^n\cos n\varphi},\] matching the source's section-3 boxed result exactly (residual $0$). This part of the source is right — the error begins only afterward.

Step 2: Where the source goes wrong — and how it self-detects it

The source guesses the orthogonal family $r^{2n}-2a^nr^n\sin n\varphi+a^{2n}=k^{2n}$ (replacing $\cos\to\sin$) and, in its own step 7, computes that family's slope as $\dfrac{a^nr\cos n\varphi}{r^n-a^n\sin n\varphi}$. Multiplying this by the true original-family slope from Step 1, \texttt{sympy.simplify} gives \[\frac{-a^nr\sin n\varphi}{r^n-a^n\cos n\varphi}\cdot\frac{a^nr\cos n\varphi}{r^n-a^n\sin n\varphi}=\frac{-a^{2n}r^2\sin n\varphi\cos n\varphi}{(a^n\sin n\varphi-r^n)(a^n\cos n\varphi-r^n)},\] which does not reduce to $-r^2$ (confirmed by a nonzero symbolic residual, and numerically: at $n=2,\,a=1.3,\,\varphi=0.4,\,r=1.7$ the product is $\approx-1.436$ while $-r^2=-2.89$) — exactly the failure the source itself flagged ("doesn't immediately simplify to $-r^2$") before boxing the guess regardless. The guess is confirmed wrong.

Step 3: The substitution $\rho=r^n,\ \theta=n\varphi$ is exact

Substituting $\rho=r^n,\ \theta=n\varphi$ into $\rho^2-2a^n\rho\cos\theta+a^{2n}=c^{2n}$ and simplifying with \texttt{sympy} reproduces the original family's left side exactly (residual $0$) — the substitution is algebraically lossless, not an approximation.

Step 4: Recognize concentric circles in the $(\rho,\theta)$ plane

With $X=\rho\cos\theta,\ Y=\rho\sin\theta$, \texttt{sympy} confirms \[(X-a^n)^2+Y^2-\left[\rho^2-2a^n\rho\cos\theta+a^{2n}\right]=0\] identically. So $\rho^2-2a^n\rho\cos\theta+a^{2n}=c^{2n}$ is exactly $(X-a^n)^2+Y^2=c^{2n}$: a family of circles all concentric about the fixed point $(a^n,0)$, with $c^n$ the varying radius.

Step 5: Orthogonal trajectories of concentric circles are the lines through the center (general fact, verified)

For the general family $(X-x_0)^2+Y^2=\text{const}$, implicit differentiation gives slope $-\dfrac{X-x_0}{Y}$. On a line $Y=m(X-x_0)$ through the center, \texttt{sympy} substitutes $X-x_0=Y/m$ and simplifies the circle's slope there to exactly $-1/m$; the product with the line's own slope $m$ is $-1$ identically. This is the elementary geometric fact that lines through the common center of a family of concentric circles cross every member perpendicularly, confirmed symbolically rather than assumed.

Step 6: Polar equation of a line through $(a^n,0)$, derived with \texttt{sympy}

A line through $(a^n,0)$ with inclination $\beta$ is $Y=\tan\beta\,(X-a^n)$. Substituting $X=\rho\cos\theta,\ Y=\rho\sin\theta$ and multiplying through by $\cos\beta$, \texttt{sympy.simplify}/\texttt{expand\_trig} reduces the relation exactly to \[\rho\sin(\beta-\theta)=a^n\sin\beta,\] with symbolic residual $0$ against this target form — derived from scratch, not hand-waved.

Step 7: Transform back to $(r,\varphi)$ and get the candidate orthogonal family

Substituting $\rho=r^n,\ \theta=n\varphi$: \[r^n\sin(\beta-n\varphi)=a^n\sin\beta,\qquad\beta\text{ an arbitrary constant.}\] Equivalently, writing $k=\tan\beta$ as the arbitrary constant (dividing through by $\cos\beta$): \[r^n(k\cos n\varphi-\sin n\varphi)=k\,a^n.\] (The earlier "replace $\cos\to\sin$" guess resembles the classical, genuinely-valid pattern for the simpler family $r^n=a^n\cos n\varphi\leftrightarrow r^n=b^n\sin n\varphi$; the extra $+a^{2n}$ term here breaks that analogy, which is presumably why the source's pattern-match failed.)

Step 8: Rigorous final check — exact residual $0$ against the source's own correct ODE

Implicitly differentiating $r^n\sin(\beta-n\varphi)=a^n\sin\beta$ and eliminating $\beta$ using $\tan\beta=\dfrac{r^n\sin n\varphi}{r^n\cos n\varphi-a^n}$ (from the family equation itself), \texttt{sympy} reduces the candidate's slope to \[\left(\frac{dr}{d\varphi}\right)_2=\frac{r\,(r^n-a^n\cos n\varphi)}{a^n\sin n\varphi}\] — which is identical (residual $0$) to the orthogonal ODE the source itself correctly derived in its own section 4, before abandoning it to guess instead. Multiplying by the Step 1 slope $\left(\frac{dr}{d\varphi}\right)_1=\dfrac{-a^nr\sin n\varphi}{r^n-a^n\cos n\varphi}$, \texttt{sympy.simplify} gives the product exactly $-r^2$, residual $0$. A numeric instance ($n=1,\ a=1.3,\ \varphi=0.7,\ k=0.9$, solving for $r\approx26.51$) reproduces the same exact match: slope product $\approx-702.59=-r^2$.