find-radius-of-curvature-to-y-sin-x-at-x-pi-6-

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

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 \dots

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

\dots \dots (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} .

Facebook Tweet Pin