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Question Number 200035 by ajfour last updated on 12/Nov/23

Commented by ajfour last updated on 12/Nov/23

$F i n d e q u a t i o n o f p a r a b o l a h a v i n g$ $s a m e c u r v a t u r e a s \sin x,$ $a t s h o w n p o i n t$
Commented by Matica last updated on 13/Nov/23

$U s e a p p r o x i t a t i o n m e t h o d s :$ $- M a c l u r i n e s e r i e$ $- P a d e a p p r o x i m a t i o n$
	Answered by mr W last updated on 12/Nov/23

	$y = \sin x a t x = \frac{π}{2} :$ $y^{'} = \cos x ∣_{x = \frac{π}{2}} = 0$ $y^{″} = - \sin x ∣_{x = \frac{π}{2}} = - 1$ $R =∣ \frac{{(1 + 0)}^{\frac{3}{2}}}{- 1} ∣= 1$ $y = - a {(x - \frac{π}{2})}^{2} + 1 a t x = \frac{π}{2} :$ $y^{'} = - 2 a (x - \frac{π}{2}) ∣_{x = \frac{π}{2}} = 0$ $y^{″} = - 2 a$ $R =∣ \frac{{(1 + 0)}^{\frac{3}{2}}}{- 2 a} ∣= \frac{1}{2 a} = 1 \Rightarrow a = \frac{1}{2}$ $\Rightarrow t h e p a r a b o l a i s y = - \frac{1}{2} {(x - \frac{π}{2})}^{2} + 1$
	Commented by ajfour last updated on 13/Nov/23

	$y e s s i r, t h i s i s p e r f e c t l y c o r r e c t .$ $T h a n k s .$
	Commented by mr W last updated on 12/Nov/23

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