Answer to Question #145243 in Differential Geometry | Topology for Dolly

Question

For curve r^m=a^m cosmθ Prove that

P=a^m÷(m+1)r^m-1

Answer 1

Since the question does not contain accompanying explanations, I assumed that it had something to do with https://en.wikipedia.org/wiki/Pedal_equation

Based on this, I solved the problem.

$$\boxed{r^m=a^m\cdot\cos(m\theta)}\\[0.3cm] \left|\frac{dr}{d\theta}\right|=\frac{r}{p}\cdot\sqrt{r^2-p^2}\\[0.3cm] \frac{d}{d\theta}\left|r^m=a^m\cdot\cos(m\theta)\right.\to m\cdot r^{m-1}\cdot\frac{dr}{d\theta}=-m\cdot a^m\cdot\sin(m\theta)\\[0.3cm] \boxed{\frac{dr}{d\theta}=-\frac{a^m}{r^{m-1}}\cdot\sin(m\theta)}\\[0.3cm] \frac{dr}{d\theta}=-\frac{a^m}{r^{m-1}}\cdot\sin(m\theta)=-\frac{a^m\cdot r}{r^{m-1}\cdot r}\cdot\sin(m\theta)=\\[0.3cm] =-\frac{a^m\cdot r}{\underbrace{r^m}_{=a^m\cdot\cos(m\theta)}}\cdot\sin(m\theta)= -\frac{a^m\cdot r}{a^m\cdot\cos(m\theta)}\cdot\sin(m\theta)=\\[0.3cm] =-r\cdot\tan(m\theta)\to\boxed{\frac{dr}{d\theta}=-r\cdot\tan(m\theta)}$$

Then,

$$\left|\frac{dr}{d\theta}\right|=\frac{r}{p}\cdot\sqrt{r^2-p^2}\to|-r\cdot\tan(m\theta)|=\frac{r}{p}\cdot\sqrt{r^2-p^2}\\[0.3cm] \left(|-r\cdot\tan(m\theta)|\right)^2=\left(\frac{r}{p}\cdot\sqrt{r^2-p^2}\right)^2\\[0.3cm] r^2\cdot\tan^2(m\theta)=\frac{r^2}{p^2}\cdot\left(r^2-p^2\right)\\[0.3cm] p^2\cdot\tan^2(m\theta)=r^2-p^2\to p^2\cdot\left(1+\tan^2(m\theta)\right)=r^2\\[0.3cm] p^2\cdot\frac{1}{\cos^2(m\theta)}=r^2\to p=r\cdot\cos(m\theta)\\[0.3cm] p=\frac{r\cdot r^{m-1}}{ r^{m-1}}\cdot\cos(m\theta)=\frac{\overbrace{r^m}^{a^m\cdot\cos(m\theta)}}{ r^{m-1}}\cdot\cos(m\theta)\to\\[0.3cm] p=\frac{a^m}{r^{m-1}}\cdot\cos^2(m\theta)\neq\frac{a^m}{(m+1)r^{m-1}}$$

Note : The result is slightly different from what was proposed to prove, but I do not know how to get rid of this discrepancy.