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


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

$f i n d r a d i u s o f c u r v a t u r e t o$ $y = \sin x a t x = π / 6 .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 11/Aug/18
	$ds=rdθ ds=(√((dy)^2 +(dx)^2 )) (ds/dx)=(√(1+((dy/dx))^2 )) tanθ=(dy/dx) θ=tan^(−1) ((dy/dx)) (dθ/dx)=((d^2 y/dx^2 )/(1+((dy/dx))^2 )) r=(ds/dθ)=((ds/dx)/(dθ/dx))=∣(({1+((dy/dx))^2 }^(3/2) )/(d^2 y/dx^2 ))∣ y=sinx (dy/dx)=cosx so ((dy/dx))_(x=(Π/6)) =((√3)/2) (d^2 y/dx^2 )=−sinx so ((d^2 y/dx^2 ))_(x=(Π/6)) =−(1/2) r=∣(({1+(3/4)}^(3/2) )/(−(1/2)))∣.=2.((7/4))^(3/2)$
	$d s = r d θ$ $d s = \sqrt{{(d y)}^{2} + {(d x)}^{2}}$ $\frac{d s}{d x} = \sqrt{1 + {(\frac{d y}{d x})}^{2}}$ $t a n θ = \frac{d y}{d x}$ $θ = {t a n}^{- 1} (\frac{d y}{d x})$ $\frac{d θ}{d x} = \frac{\frac{d^{2} y}{{d x}^{2}}}{1 + {(\frac{d y}{d x})}^{2}}$ $r = \frac{d s}{d θ} = \frac{\frac{d s}{d x}}{\frac{d θ}{d x}} =∣ \frac{{1 + {(\frac{d y}{d x})}^{2}}^{\frac{3}{2}}}{\frac{d^{2} y}{{d x}^{2}}} ∣$ $y = s i n x$ $\frac{d y}{d x} = c o s x s o {(\frac{d y}{d x})}_{x = \frac{Π}{6}} = \frac{\sqrt{3}}{2}$ $\frac{d^{2} y}{{d x}^{2}} = - s i n x s o {(\frac{d^{2} y}{{d x}^{2}})}_{x = \frac{Π}{6}} = - \frac{1}{2}$ $r =∣ \frac{{1 + \frac{3}{4}}^{\frac{3}{2}}}{- \frac{1}{2}} ∣ . = 2 . {(\frac{7}{4})}^{\frac{3}{2}}$
	Commented by ajfour last updated on 11/Aug/18

	$t h a n k y o u T a n m a y S i r .$
	Commented by tanmay.chaudhury50@gmail.com last updated on 11/Aug/18

	$t h a n k y o u a l s o f o r p o s t i n g v a r i e t y o f p r o b l e m$ $o f d i f f e r e n t t a s t e . . .$
	Answered by ajfour last updated on 11/Aug/18

	${[x - (\frac{π}{6} + r \sin θ)]}^{2} + {[y - (\frac{1}{2} + r \cos θ)]}^{2} = r^{2}$ $2 [x - (\frac{π}{6} + r \sin θ)] + 2 [y - (\frac{1}{2} + r \cos θ)] \frac{d y}{d x} = 0$ $1 + {(\frac{d y}{d x})}^{2} + [y - \frac{1}{2} - r \cos θ] \frac{d^{2} y}{{d x}^{2}} = 0$ $. . . . . . (i)$ $a n d \frac{d^{2} (\sin x)}{{d x}^{2}} ∣_{x = \frac{π}{6}} = - \frac{1}{2}$ $\frac{d y}{d x} ∣_{x = \frac{π}{6}} = \frac{\sqrt{3}}{2} = \tan θ$ $u s i n g t h e s e i n (i)$ $\Rightarrow 1 + \frac{3}{4} + (\frac{1}{2} - \frac{1}{2} - \frac{2 r}{\sqrt{7}}) (- \frac{1}{2}) = 0$ $\Rightarrow ∣ r ∣ = \frac{7 \sqrt{7}}{4} .$
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