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Question Number 41682 by ajfour last updated on 11/Aug/18

find radius of curvature to  y=sin x  at  x=π/6 .

$${find}\:{radius}\:{of}\:{curvature}\:{to} \\ $$$${y}=\mathrm{sin}\:{x}\:\:{at}\:\:{x}=\pi/\mathrm{6}\:. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 11/Aug/18

  ds=rdθ  ds=(√((dy)^2 +(dx)^2 ))   (ds/dx)=(√(1+((dy/dx))^2  ))  tanθ=(dy/dx)  θ=tan^(−1) ((dy/dx))  (dθ/dx)=((d^2 y/dx^2 )/(1+((dy/dx))^2 ))  r=(ds/dθ)=((ds/dx)/(dθ/dx))=∣(({1+((dy/dx))^2 }^(3/2) )/(d^2 y/dx^2 ))∣  y=sinx  (dy/dx)=cosx   so ((dy/dx))_(x=(Π/6))  =((√3)/2)  (d^2 y/dx^2 )=−sinx  so ((d^2 y/dx^2 ))_(x=(Π/6)) =−(1/2)  r=∣(({1+(3/4)}^(3/2) )/(−(1/2)))∣.=2.((7/4))^(3/2)

$$ \\ $$$${ds}={rd}\theta \\ $$$${ds}=\sqrt{\left({dy}\right)^{\mathrm{2}} +\left({dx}\right)^{\mathrm{2}} }\: \\ $$$$\frac{{ds}}{{dx}}=\sqrt{\mathrm{1}+\left(\frac{{dy}}{{dx}}\right)^{\mathrm{2}} \:} \\ $$$${tan}\theta=\frac{{dy}}{{dx}} \\ $$$$\theta={tan}^{−\mathrm{1}} \left(\frac{{dy}}{{dx}}\right) \\ $$$$\frac{{d}\theta}{{dx}}=\frac{\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }}{\mathrm{1}+\left(\frac{{dy}}{{dx}}\right)^{\mathrm{2}} } \\ $$$${r}=\frac{{ds}}{{d}\theta}=\frac{\frac{{ds}}{{dx}}}{\frac{{d}\theta}{{dx}}}=\mid\frac{\left\{\mathrm{1}+\left(\frac{{dy}}{{dx}}\right)^{\mathrm{2}} \right\}^{\frac{\mathrm{3}}{\mathrm{2}}} }{\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }}\mid \\ $$$${y}={sinx} \\ $$$$\frac{{dy}}{{dx}}={cosx}\:\:\:{so}\:\left(\frac{{dy}}{{dx}}\right)_{{x}=\frac{\Pi}{\mathrm{6}}} \:=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=−{sinx}\:\:{so}\:\left(\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\right)_{{x}=\frac{\Pi}{\mathrm{6}}} =−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${r}=\mid\frac{\left\{\mathrm{1}+\frac{\mathrm{3}}{\mathrm{4}}\right\}^{\frac{\mathrm{3}}{\mathrm{2}}} }{−\frac{\mathrm{1}}{\mathrm{2}}}\mid.=\mathrm{2}.\left(\frac{\mathrm{7}}{\mathrm{4}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$$ \\ $$

Commented by ajfour last updated on 11/Aug/18

thank you Tanmay Sir.

$${thank}\:{you}\:{Tanmay}\:{Sir}. \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 11/Aug/18

thank you also for posting variety of problem   of different taste...

$${thank}\:{you}\:{also}\:{for}\:{posting}\:{variety}\:{of}\:{problem}\: \\ $$$${of}\:{different}\:{taste}... \\ $$

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) .

$$\left[{x}−\left(\frac{\pi}{\mathrm{6}}+{r}\mathrm{sin}\:\theta\right)\right]^{\mathrm{2}} +\left[{y}−\left(\frac{\mathrm{1}}{\mathrm{2}}+{r}\mathrm{cos}\:\theta\right)\right]^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$$\mathrm{2}\left[{x}−\left(\frac{\pi}{\mathrm{6}}+{r}\mathrm{sin}\:\theta\right)\right]+\mathrm{2}\left[{y}−\left(\frac{\mathrm{1}}{\mathrm{2}}+{r}\mathrm{cos}\:\theta\right)\right]\frac{{dy}}{{dx}}=\mathrm{0} \\ $$$$\mathrm{1}+\left(\frac{{dy}}{{dx}}\right)^{\mathrm{2}} +\left[{y}−\frac{\mathrm{1}}{\mathrm{2}}−{r}\mathrm{cos}\:\theta\right]\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:......\left({i}\right) \\ $$$${and}\:\:\frac{{d}^{\mathrm{2}} \left(\mathrm{sin}\:{x}\right)}{{dx}^{\mathrm{2}} }\mid_{{x}=\frac{\pi}{\mathrm{6}}} \:=\:−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\:\:\frac{{dy}}{{dx}}\mid_{{x}=\frac{\pi}{\mathrm{6}}} =\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:=\:\mathrm{tan}\:\theta \\ $$$${using}\:{these}\:{in}\:\left({i}\right) \\ $$$$\Rightarrow\:\:\:\mathrm{1}+\frac{\mathrm{3}}{\mathrm{4}}+\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{2}{r}}{\sqrt{\mathrm{7}}}\right)\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{0} \\ $$$$\Rightarrow\:\:\mid{r}\mid\:=\:\frac{\mathrm{7}\sqrt{\mathrm{7}}}{\mathrm{4}}\:. \\ $$

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