Question Number 41682 by ajfour last updated on 11/Aug/18
$${find}\:{radius}\:{of}\:{curvature}\:{to} \\ $$$${y}=\mathrm{sin}\:{x}\:\:{at}\:\:{x}=\pi/\mathrm{6}\:. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 11/Aug/18
$$ \\ $$$${ds}={rd}\theta \\ $$$${ds}=\sqrt{\left({dy}\right)^{\mathrm{2}} +\left({dx}\right)^{\mathrm{2}} }\: \\ $$$$\frac{{ds}}{{dx}}=\sqrt{\mathrm{1}+\left(\frac{{dy}}{{dx}}\right)^{\mathrm{2}} \:} \\ $$$${tan}\theta=\frac{{dy}}{{dx}} \\ $$$$\theta={tan}^{−\mathrm{1}} \left(\frac{{dy}}{{dx}}\right) \\ $$$$\frac{{d}\theta}{{dx}}=\frac{\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }}{\mathrm{1}+\left(\frac{{dy}}{{dx}}\right)^{\mathrm{2}} } \\ $$$${r}=\frac{{ds}}{{d}\theta}=\frac{\frac{{ds}}{{dx}}}{\frac{{d}\theta}{{dx}}}=\mid\frac{\left\{\mathrm{1}+\left(\frac{{dy}}{{dx}}\right)^{\mathrm{2}} \right\}^{\frac{\mathrm{3}}{\mathrm{2}}} }{\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }}\mid \\ $$$${y}={sinx} \\ $$$$\frac{{dy}}{{dx}}={cosx}\:\:\:{so}\:\left(\frac{{dy}}{{dx}}\right)_{{x}=\frac{\Pi}{\mathrm{6}}} \:=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=−{sinx}\:\:{so}\:\left(\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\right)_{{x}=\frac{\Pi}{\mathrm{6}}} =−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${r}=\mid\frac{\left\{\mathrm{1}+\frac{\mathrm{3}}{\mathrm{4}}\right\}^{\frac{\mathrm{3}}{\mathrm{2}}} }{−\frac{\mathrm{1}}{\mathrm{2}}}\mid.=\mathrm{2}.\left(\frac{\mathrm{7}}{\mathrm{4}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$$ \\ $$
Commented by ajfour last updated on 11/Aug/18
$${thank}\:{you}\:{Tanmay}\:{Sir}. \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 11/Aug/18
$${thank}\:{you}\:{also}\:{for}\:{posting}\:{variety}\:{of}\:{problem}\: \\ $$$${of}\:{different}\:{taste}… \\ $$
Answered by ajfour last updated on 11/Aug/18
![[x−((π/6)+rsin θ)]^2 +[y−((1/2)+rcos θ)]^2 =r^2 2[x−((π/6)+rsin θ)]+2[y−((1/2)+rcos θ)](dy/dx)=0 1+((dy/dx))^2 +[y−(1/2)−rcos θ](d^2 y/dx^2 )=0 ......(i) and ((d^2 (sin x))/dx^2 )∣_(x=(π/6)) = −(1/2) (dy/dx)∣_(x=(π/6)) = ((√3)/2) = tan θ using these in (i) ⇒ 1+(3/4)+((1/2)−(1/2)−((2r)/( (√7))))(−(1/2))=0 ⇒ ∣r∣ = ((7(√7))/4) .](https://www.tinkutara.com/question/Q41688.png)
$$\left[{x}−\left(\frac{\pi}{\mathrm{6}}+{r}\mathrm{sin}\:\theta\right)\right]^{\mathrm{2}} +\left[{y}−\left(\frac{\mathrm{1}}{\mathrm{2}}+{r}\mathrm{cos}\:\theta\right)\right]^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$$\mathrm{2}\left[{x}−\left(\frac{\pi}{\mathrm{6}}+{r}\mathrm{sin}\:\theta\right)\right]+\mathrm{2}\left[{y}−\left(\frac{\mathrm{1}}{\mathrm{2}}+{r}\mathrm{cos}\:\theta\right)\right]\frac{{dy}}{{dx}}=\mathrm{0} \\ $$$$\mathrm{1}+\left(\frac{{dy}}{{dx}}\right)^{\mathrm{2}} +\left[{y}−\frac{\mathrm{1}}{\mathrm{2}}−{r}\mathrm{cos}\:\theta\right]\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:……\left({i}\right) \\ $$$${and}\:\:\frac{{d}^{\mathrm{2}} \left(\mathrm{sin}\:{x}\right)}{{dx}^{\mathrm{2}} }\mid_{{x}=\frac{\pi}{\mathrm{6}}} \:=\:−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\:\:\frac{{dy}}{{dx}}\mid_{{x}=\frac{\pi}{\mathrm{6}}} =\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:=\:\mathrm{tan}\:\theta \\ $$$${using}\:{these}\:{in}\:\left({i}\right) \\ $$$$\Rightarrow\:\:\:\mathrm{1}+\frac{\mathrm{3}}{\mathrm{4}}+\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{2}{r}}{\:\sqrt{\mathrm{7}}}\right)\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{0} \\ $$$$\Rightarrow\:\:\mid{r}\mid\:=\:\frac{\mathrm{7}\sqrt{\mathrm{7}}}{\mathrm{4}}\:. \\ $$