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.
r m = a m ⋅ cos ( m θ ) ∣ d r d θ ∣ = r p ⋅ r 2 − p 2 d d θ ∣ r m = a m ⋅ cos ( m θ ) → m ⋅ r m − 1 ⋅ d r d θ = − m ⋅ a m ⋅ sin ( m θ ) d r d θ = − a m r m − 1 ⋅ sin ( m θ ) d r d θ = − a m r m − 1 ⋅ sin ( m θ ) = − a m ⋅ r r m − 1 ⋅ r ⋅ sin ( m θ ) = = − a m ⋅ r r m ⏟ = a m ⋅ cos ( m θ ) ⋅ sin ( m θ ) = − a m ⋅ r a m ⋅ cos ( m θ ) ⋅ sin ( m θ ) = = − r ⋅ tan ( m θ ) → d r d θ = − r ⋅ tan ( m θ ) \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)} r m = a m ⋅ cos ( m θ ) ∣ ∣ d θ d r ∣ ∣ = p r ⋅ r 2 − p 2 d θ d ∣ r m = a m ⋅ cos ( m θ ) → m ⋅ r m − 1 ⋅ d θ d r = − m ⋅ a m ⋅ sin ( m θ ) d θ d r = − r m − 1 a m ⋅ sin ( m θ ) d θ d r = − r m − 1 a m ⋅ sin ( m θ ) = − r m − 1 ⋅ r a m ⋅ r ⋅ sin ( m θ ) = = − = a m ⋅ c o s ( m θ ) r m a m ⋅ r ⋅ sin ( m θ ) = − a m ⋅ cos ( m θ ) a m ⋅ r ⋅ sin ( m θ ) = = − r ⋅ tan ( m θ ) → d θ d r = − r ⋅ tan ( m θ )
Then,
∣ d r d θ ∣ = r p ⋅ r 2 − p 2 → ∣ − r ⋅ tan ( m θ ) ∣ = r p ⋅ r 2 − p 2 ( ∣ − r ⋅ tan ( m θ ) ∣ ) 2 = ( r p ⋅ r 2 − p 2 ) 2 r 2 ⋅ tan 2 ( m θ ) = r 2 p 2 ⋅ ( r 2 − p 2 ) p 2 ⋅ tan 2 ( m θ ) = r 2 − p 2 → p 2 ⋅ ( 1 + tan 2 ( m θ ) ) = r 2 p 2 ⋅ 1 cos 2 ( m θ ) = r 2 → p = r ⋅ cos ( m θ ) p = r ⋅ r m − 1 r m − 1 ⋅ cos ( m θ ) = r m ⏞ a m ⋅ cos ( m θ ) r m − 1 ⋅ cos ( m θ ) → p = a m r m − 1 ⋅ cos 2 ( m θ ) ≠ a m ( m + 1 ) r m − 1 \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}} ∣ ∣ d θ d r ∣ ∣ = p r ⋅ r 2 − p 2 → ∣ − r ⋅ tan ( m θ ) ∣ = p r ⋅ r 2 − p 2 ( ∣ − r ⋅ tan ( m θ ) ∣ ) 2 = ( p r ⋅ r 2 − p 2 ) 2 r 2 ⋅ tan 2 ( m θ ) = p 2 r 2 ⋅ ( r 2 − p 2 ) p 2 ⋅ tan 2 ( m θ ) = r 2 − p 2 → p 2 ⋅ ( 1 + tan 2 ( m θ ) ) = r 2 p 2 ⋅ cos 2 ( m θ ) 1 = r 2 → p = r ⋅ cos ( m θ ) p = r m − 1 r ⋅ r m − 1 ⋅ cos ( m θ ) = r m − 1 r m a m ⋅ c o s ( m θ ) ⋅ cos ( m θ ) → p = r m − 1 a m ⋅ cos 2 ( m θ ) = ( m + 1 ) r m − 1 a m
Note : The result is slightly different from what was proposed to prove, but I do not know how to get rid of this discrepancy.
#339914 on Dec 2023
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