Question

x=a(θ-sinθ) ,y=a(1-cosθ) at θ= π/3 radius of curvature

Answer 1

Topic :-

Radius Of Curvature

Given :-

$\sf{x=a(\theta-\sin\theta)}$

$\sf{y=a(1-\cos\theta)}$

To Find :-

$\sf{Radius\:of\:Curvature\:at\:\theta=\dfrac{\pi}{3}}$

Formula to be Used :-

$\sf{R_c=\left|\dfrac{[1+(y')^2]^{3/2}}{y''}\right|}$

where

$\sf{R_c=Radius \:of\:Curvature}$

$\sf {y'=\dfrac{dy}{dx}}$

$\sf {y''=\dfrac{d^2y}{dx^2}}$

Solution :-

Calculating y',

$\sf{y'=\dfrac{dy}{dx}=\dfrac{\dfrac{dy}{d\theta}}{\dfrac{dx}{d\theta}}=\dfrac{\dfrac{d[a(1-\cos\theta)]}{d\theta}}{\dfrac{d[a(\theta-\sin\theta)]}{d\theta}}}$

$\sf{\dfrac{a\cdot\dfrac{d[1-\cos\theta)]}{d\theta}}{a\cdot\dfrac{d[(\theta-\sin\theta)]}{d\theta}}=\dfrac{\dfrac{d[1-\cos\theta)]}{d\theta}}{\dfrac{d[(\theta-\sin\theta)]}{d\theta}}}$

$\sf{\left(\because \dfrac{d(k\cdot f(x))}{dx}=k\cdot\dfrac{d(f(x))}{dx}, when\:'k'\:is\:a\:constant.\right)}$

$\sf{\dfrac{\dfrac{d(1)}{d\theta}-\dfrac{d(\cos\theta)}{d\theta}}{\dfrac{d\theta}{d\theta}-\dfrac{d(\sin\theta)}{d\theta}}}$

$\sf{\left(\because \dfrac{d(f\pm g)}{dx}=\dfrac{d(f)}{dx}\pm \dfrac{d(g)}{dx}\right)}$

$\sf{\dfrac{0-\dfrac{d(\cos\theta)}{d\theta}}{\dfrac{d\theta}{d\theta}-\dfrac{d(\sin\theta)}{d\theta}}}$

$\sf{\left(\because \dfrac{d(k)}{dx}=0, when\:'k'\:is\:a\:constant.\right)}$

$\sf{\dfrac{0-(-\sin\theta)}{\dfrac{d\theta}{d\theta}-\dfrac{d(\sin\theta)}{d\theta}}}$

$\sf{\left(\because \dfrac{d(\cos\theta)}{d\theta}=-\sin\theta \right)}$

$\sf{\dfrac{\sin\theta}{1-\dfrac{d(\sin\theta)}{d\theta}}}$

$\sf{\left(\because \dfrac{d(\theta)}{d\theta}=1 \right)}$

$\sf{\dfrac{\sin\theta}{1-\cos\theta}}$

$\sf{\left(\because \dfrac{d(\sin\theta)}{d\theta}=\cos\theta \right)}$

$\sf{\dfrac{\sin\theta(1+\cos\theta)}{(1-\cos\theta)(1+\cos\theta)}=\dfrac{\sin\theta(1+\cos\theta)}{1-\cos^2\theta}=\dfrac{\sin\theta(1+\cos\theta)}{\sin^2\theta}}$

$\sf{\dfrac{1}{\sin\theta}+\dfrac{\cos\theta}{\sin\theta}=\csc\theta+\cot\theta}$

Calculating y'',

$\sf {y''=\dfrac{d^2y}{dx^2}=\dfrac{\dfrac{d}{d\theta}\left( \dfrac{dy}{dx} \right)}{\dfrac{dx}{d\theta}}=\dfrac{\dfrac{d(\csc\theta+\cot\theta)}{d\theta}}{a(1-\cos\theta)}}$

$\sf {\left(\because \dfrac{dx}{d\theta}=a(1-\cos\theta)\right)}$

$\sf {\dfrac{\dfrac{d(\csc\theta)}{d\theta}+\dfrac{d(\cot\theta)}{d\theta}}{a(1-\cos\theta)}}$

$\sf{\left(\because \dfrac{d(f\pm g)}{dx}=\dfrac{d(f)}{dx}\pm \dfrac{d(g)}{dx}\right)}$

$\sf {\dfrac{-\csc \theta\cdot\cot \theta+\dfrac{d(\cot\theta)}{d\theta}}{a(1-\cos\theta)}}$

$\sf{\left(\because \dfrac{d(\csc\theta)}{d\theta}=-\csc\theta\cdot\cot\theta \right)}$

$\sf {\dfrac{-\csc \theta\cdot\cot \theta-\csc^2\theta}{a(1-\cos\theta)}=\dfrac{\csc \theta\cdot\cot \theta+\csc^2\theta}{a(\cos\theta-1)}}$

Applying formula,

$\sf{R_c=\left|\dfrac{[1+(\csc\theta+\cot\theta)^2]^{3/2}}{\dfrac{\csc\theta\cdot\cot\theta+\csc^2\theta}{a(\cos\theta-1)}}\right|}$

$\sf{Substitute\:\theta=\dfrac{\pi}{3}=60^{\circ},}$

$\sf{R_c=\left|\dfrac{[1+(\csc60^{\circ}+\cot 60^{\circ})^2]^{3/2}}{\dfrac{\csc 60^{\circ}\cdot\cot 60^{\circ}+\csc^2 60^{\circ}}{a(\cos 60^{\circ}-1)}}\right|}$

$\sf{R_c=\left|\dfrac{\left[1+\left(\dfrac{2}{\sqrt{3}}+\dfrac{1}{\sqrt{3}}\right )^2\right]^{3/2}}{\dfrac{\dfrac{2}{\sqrt{3}}\cdot\dfrac{1}{\sqrt{3}}+\left(\dfrac{2}{\sqrt{3}}\right)^2}{a\left(\dfrac{1}{2}-1\right)}}\right|}$

$\left(\because \csc60^{\circ}=\dfrac{2}{\sqrt{3}}\right)$

$\left(\because \cot60^{\circ}=\dfrac{1}{\sqrt{3}}\right)$

$\sf{R_c=\left|\dfrac{\left[1+\left(\sqrt{3}\right )^2\right]^{3/2}}{\dfrac{\dfrac{2}{3}+\left(\dfrac{4}{3}\right)}{a\left(\dfrac{1}{2}-1\right)}}\right|=\left|\dfrac{\left[1+\left(\sqrt{3}\right )^2\right]^{3/2}}{\dfrac{\left(\dfrac{6}{3}\right)}{a\left(\dfrac{-1}{2}\right)}}\right|}$

$\sf{R_c=\left|\dfrac{a\cdot\left[1+3\right]^{3/2}}{-2\cdot 2}\right|=\left|\dfrac{a\cdot\left[1+3\right]^{3/2}}{-4}\right|}$

$\sf{R_c=\left|\dfrac{a\cdot\left[\left(2 \right)^2 \right]^{3/2}}{-4}\right|}$

$\sf{R_c=\left|\dfrac{a\cdot\left[2 \right]^{3}}{-4}\right|=\left|\dfrac{a\cdot8}{-4}\right|}$

$\sf{R_c=|-2a|=2a}$

Answer :-

$\sf{Radius\:of\:Curvature\:at\:\theta=\dfrac{\pi}{3}\:is\:\bold{2a}.}$