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Question Number 203115 by Mathstar last updated on 10/Jan/24

Commented by Mathstar last updated on 10/Jan/24

$Send an explanatory solution, thank you$
	Answered by mr W last updated on 10/Jan/24

	$y = x^{2} e^{x}$ $y^{'} = e^{x} (x^{2} + 2 x)$ $c e n t e r o f c i r c l e (0, r)$ $s a y t h e c i r c l e t a n g e n t s t h e c u r v e$ $a d d i t i o n a l l y a t P (p, q) .$ $q = p^{2} e^{p}$ $\tan θ = y^{'} = e^{p} (p^{2} + 2 p)$ $p = r \sin θ = 2 r \sin \frac{θ}{2} \cos \frac{θ}{2}$ $q = r (1 - \cos θ) = 2 r \sin^{2} \frac{θ}{2}$ $\tan \frac{θ}{2} = \frac{q}{p} = \frac{p^{2} e^{p}}{p} = {p e}^{p}$ $\tan θ = \frac{2 \tan \frac{θ}{2}}{1 - \tan^{2} \frac{θ}{2}}$ $e^{p} (p^{2} + 2 p) = \frac{2 {p e}^{p}}{1 - p^{2} e^{2 p}}$ $\Rightarrow (p + 2) (1 - p^{2} e^{2 p}) = 2$ $\Rightarrow p = 0 \lor p \approx 0.2619$ $r_{m a x} = \frac{p}{\sin θ} = \sqrt{p^{2} + \frac{p^{2}}{\tan^{2} θ}}$ $= \sqrt{p^{2} + \frac{1}{e^{2 p} {(p + 2)}^{2}}} \approx 0.4294$
	Commented by Mathstar last updated on 10/Jan/24
	Thank you. I will look out for a video that can explain better.
	Commented by mr W last updated on 10/Jan/24

	Commented by mr W last updated on 10/Jan/24

	$A l g e b r a$ $G e o m e t r y$ $T r i g o n o m e t r y$ $D e r i v a t i v e$ $E q u a t i o n s$
	Commented by Mathstar last updated on 10/Jan/24
	Thank you so much Mr W. Please how can I understand something like this?
	Commented by Mathstar last updated on 10/Jan/24
	Could you please enumerate the scope in this type of question?
	Commented by mr W last updated on 11/Jan/24

	Commented by mr W last updated on 11/Jan/24

	$s u c h t h a t a c i r c l e o n l y t o u c h e s a$ $c u r v e a t a p o i n t, t h e r a d i u s o f t h e$ $c i r c l e m a y n o t b e l a r g e r t h a n t h e$ $r a d i u s R o f t h e o s c u l a t i n g c i r c l e$ $w i t h R =∣ \frac{{(1 + y^{' 2})}^{\frac{3}{2}}}{y^{″}} ∣ .$ $i n o u r c a s e R = 0.5 .$ $b u t t h i s c i r c l e m a y i n t e r s e c t t h e$ $c u r v e a t o t h e r p o i n t s . s u c h t h a t$ $t h i s {d o e s n}^{'} t h a p p e n, w e m u s t$ $e n s u r e t h a t t h e c i r c l e w h i c h t o u c h e s$ $t h e c u r c e a t p o i n t (0, 0) m a x i m a l l y$ $t o u c h e s t h e c u r v e a t a n o t h e r p o i n t .$ $t h i s c i r c l e i s t h e n t h e l a r g e s t c i r c l e$ $w h i c h f u l f i l l s t h e r e q u i r e m e n t o f$ $t h e q u e s t i o n .$ $t h i s i s t h e ‘ ‘ {t h e o r y}^{″} b e h i n d m y$ $s o l u t i o n .$
	Commented by mr W last updated on 11/Jan/24

	Commented by Mathstar last updated on 11/Jan/24
	Thank you so much, Mr W. You've been of great support to me ever since I joined this platform ��‍♂️
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