Question Number 155720 by ajfour last updated on 03/Oct/21
Commented by ajfour last updated on 03/Oct/21
$${If}\:{the}\:{polynomial}\:{curve}\:{is} \\ $$$$\:\:{y}={x}\left({x}^{\mathrm{2}} −{p}^{\mathrm{2}} \right)\:,\:{find}\:{r}\left({p}\right). \\ $$
Answered by TheSupreme last updated on 03/Oct/21
$${f}\left({x}\right)={x}\left({x}−{p}\right)\left({x}+{p}\right) \\ $$$${P}\:\left({x},{x}\left({x}^{\mathrm{2}} −{p}^{\mathrm{2}} \right)\right) \\ $$$${P}\:\left({r}\left(\mathrm{1}+{cos}\left({arctan}\left({f}'\left({P}\right)\right),{r}\left(\mathrm{1}−{sin}\left({arctan}\left({f}'\left({P}\right)\right)\right.\right.\right.\right.\right. \\ $$$${f}'\left({P}\right)=\left(\mathrm{3}{x}^{\mathrm{2}} −{p}^{\mathrm{2}} \right) \\ $$$$ \\ $$$$ \\ $$
Answered by mr W last updated on 03/Oct/21
![touching point A(a,b) with b=a(a^2 −p^2 ) tan θ=y′=3x^2 −p^2 =3a^2 −p^2 a=r(1+sin θ) b=r(1−cos θ) (b/a)=((1−cos θ)/(1+sin θ)) a^2 −p^2 =((1−cos θ)/(1+sin θ)) ⇒tan θ−((3(1−cos θ))/(1+sin θ))=2p^2 { ((p=(√((1/2)[tan θ−((3(1−cos θ))/(1+sin θ))])))),((r=(1/(1+sin θ))(√((1/2)[tan θ−((1−cos θ)/(1+sin θ))])))) :}](https://www.tinkutara.com/question/Q155725.png)
$${touching}\:{point}\:{A}\left({a},{b}\right)\:{with} \\ $$$${b}={a}\left({a}^{\mathrm{2}} −{p}^{\mathrm{2}} \right) \\ $$$$\mathrm{tan}\:\theta={y}'=\mathrm{3}{x}^{\mathrm{2}} −{p}^{\mathrm{2}} =\mathrm{3}{a}^{\mathrm{2}} −{p}^{\mathrm{2}} \\ $$$${a}={r}\left(\mathrm{1}+\mathrm{sin}\:\theta\right) \\ $$$${b}={r}\left(\mathrm{1}−\mathrm{cos}\:\theta\right) \\ $$$$\frac{{b}}{{a}}=\frac{\mathrm{1}−\mathrm{cos}\:\theta}{\mathrm{1}+\mathrm{sin}\:\theta} \\ $$$${a}^{\mathrm{2}} −{p}^{\mathrm{2}} =\frac{\mathrm{1}−\mathrm{cos}\:\theta}{\mathrm{1}+\mathrm{sin}\:\theta} \\ $$$$\Rightarrow\mathrm{tan}\:\theta−\frac{\mathrm{3}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)}{\mathrm{1}+\mathrm{sin}\:\theta}=\mathrm{2}{p}^{\mathrm{2}} \\ $$$$\begin{cases}{{p}=\sqrt{\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{tan}\:\theta−\frac{\mathrm{3}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)}{\mathrm{1}+\mathrm{sin}\:\theta}\right]}}\\{{r}=\frac{\mathrm{1}}{\mathrm{1}+\mathrm{sin}\:\theta}\sqrt{\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{tan}\:\theta−\frac{\mathrm{1}−\mathrm{cos}\:\theta}{\mathrm{1}+\mathrm{sin}\:\theta}\right]}}\end{cases} \\ $$
Commented by mr W last updated on 03/Oct/21
Commented by mr W last updated on 03/Oct/21
Commented by mr W last updated on 03/Oct/21
Answered by ajfour last updated on 01/Nov/21
![x=r+rcos θ y=c=r−rsin θ ; p=1 ⇒ c=r(1−2sin φcos φ) 2φ=90°−θ ⇒ r(1−cos 2φ)=x(x^2 −p^2 ) r(1+cos θ)=x 2rsin^2 ((π/4)−(θ/2))=2rcos^2 (θ/2)(4r^2 cos^4 (θ/2)−p^2 ) say (θ/2)=δ (cos δ−sin δ)^2 =cos^2 δ(4r^2 cos^4 δ−p^2 ) ⇒ 2(1−sin θ) =(1+cos θ)[((4c^2 )/((1−sin θ)^2 ))−1] 2(1−sin θ)^3 =(1+cos θ){4c^2 −(1−sin θ)^2 } ......](https://www.tinkutara.com/question/Q158257.png)
$$\:\:{x}={r}+{r}\mathrm{cos}\:\theta \\ $$$$\:\:{y}={c}={r}−{r}\mathrm{sin}\:\theta\:\:\:;\:\:{p}=\mathrm{1} \\ $$$$\:\:\Rightarrow\:\:{c}={r}\left(\mathrm{1}−\mathrm{2sin}\:\phi\mathrm{cos}\:\phi\right) \\ $$$$\mathrm{2}\phi=\mathrm{90}°−\theta \\ $$$$\Rightarrow\:{r}\left(\mathrm{1}−\mathrm{cos}\:\mathrm{2}\phi\right)={x}\left({x}^{\mathrm{2}} −{p}^{\mathrm{2}} \right) \\ $$$$\:\:\:\:\:{r}\left(\mathrm{1}+\mathrm{cos}\:\theta\right)={x} \\ $$$$\mathrm{2}{r}\mathrm{sin}\:^{\mathrm{2}} \left(\frac{\pi}{\mathrm{4}}−\frac{\theta}{\mathrm{2}}\right)=\mathrm{2}{r}\mathrm{cos}\:^{\mathrm{2}} \frac{\theta}{\mathrm{2}}\left(\mathrm{4}{r}^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{4}} \frac{\theta}{\mathrm{2}}−{p}^{\mathrm{2}} \right) \\ $$$${say}\:\:\frac{\theta}{\mathrm{2}}=\delta \\ $$$$\left(\mathrm{cos}\:\delta−\mathrm{sin}\:\delta\right)^{\mathrm{2}} =\mathrm{cos}\:^{\mathrm{2}} \delta\left(\mathrm{4}{r}^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{4}} \delta−{p}^{\mathrm{2}} \right) \\ $$$$\Rightarrow \\ $$$$\mathrm{2}\left(\mathrm{1}−\mathrm{sin}\:\theta\right) \\ $$$$\:\:\:\:=\left(\mathrm{1}+\mathrm{cos}\:\theta\right)\left[\frac{\mathrm{4}{c}^{\mathrm{2}} }{\left(\mathrm{1}−\mathrm{sin}\:\theta\right)^{\mathrm{2}} }−\mathrm{1}\right] \\ $$$$\mathrm{2}\left(\mathrm{1}−\mathrm{sin}\:\theta\right)^{\mathrm{3}} \\ $$$$\:=\left(\mathrm{1}+\mathrm{cos}\:\theta\right)\left\{\mathrm{4}{c}^{\mathrm{2}} −\left(\mathrm{1}−\mathrm{sin}\:\theta\right)^{\mathrm{2}} \right\} \\ $$$$…… \\ $$$$\:\:\:\:\: \\ $$