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Question Number 214199 by mr W last updated on 01/Dec/24

Commented by mr W last updated on 01/Dec/24

$[s e e Q 214176]$ $f i n d t h e r a d i u s o f t h e m a x i m a l$ $c i r c l e i n s c r i b e d b e t w e e n t h e c u r v e s$ $f (x) a n d g (x) a s s h o w n .$
	Answered by issac last updated on 01/Dec/24

	$Osculate circle radious = \frac{1}{κ}$ $κ = curvature of f (x) at x = α$ $κ = \frac{∣ f^{(2)} (α) ∣}{\sqrt{{(1 + {(f^{(1)} (α))}^{2})}^{3}}}$
	Answered by mr W last updated on 01/Dec/24

	$d u e t o s y m m e t r y t h e c e n t e r o f c i r c l e$ $l i e s o n t h e y - a x i s, s a y a t (0, h) .$ $t h e e q u a t i o n o f c i r c l e i s t h e n$ $x^{2} + {(y - h)}^{2} = R^{2}$ $w e o n l y n e e d t o c o n s i d e r x > 0 .$ $s a y t h e c i r c l e t o u c h e s g (x) a t (s, t)$ $a n d t o u c h e s f (x) a t (p, q) .$ $a t t o u c h i n g p o i n t c i r c l e a n d c u r v e$ $h a v e s a m e t a n g e n t l i n e .$ $x^{2} + {(y - h)}^{2} = R^{2}$ $2 x + 2 (y - h) \frac{d y}{d x} = 0$ $y = \ln x$ $\frac{d y}{d x} = \frac{1}{x}$ $s + (t - h) \frac{1}{s} = 0$ $\Rightarrow t - h = - s^{2}$ $s^{2} + s^{4} = R^{2}$ $\Rightarrow s^{2} = \frac{\sqrt{1 + 4 R^{2}} - 1}{2}$ $t = h - s^{2} = h - \frac{\sqrt{1 + 4 R^{2}} - 1}{2}$ $t = \ln s = \frac{1}{2} \ln (\frac{\sqrt{1 + 4 R^{2}} - 1}{2})$ $\Rightarrow h - \frac{\sqrt{1 + 4 R^{2}} - 1}{2} = \frac{1}{2} \ln (\frac{\sqrt{1 + 4 R^{2}} - 1}{2})$ $\Rightarrow h = \frac{1}{2} [\sqrt{1 + 4 R^{2}} - 1 + \ln (\frac{\sqrt{1 + 4 R^{2}} - 1}{2})]$ $y = \frac{3}{x^{3}} + \frac{1}{15 x}$ $\frac{d y}{d x} = - \frac{9}{x^{4}} - \frac{1}{15 x^{2}}$ $2 p - 2 (q - h) (\frac{9}{p^{4}} + \frac{1}{15 p^{2}}) = 0$ $\Rightarrow q - h = \frac{15 p^{5}}{p^{2} + 135}$ $p^{2} + {(\frac{15 p^{5}}{p^{2} + 135})}^{2} = R^{2}$ $\Rightarrow R = \sqrt{p^{2} + {(\frac{15 p^{5}}{p^{2} + 135})}^{2}} . . . (i)$ $q = h + \frac{15 p^{5}}{p^{2} + 135}$ $q = \frac{3}{p^{3}} + \frac{1}{15 p}$ $\Rightarrow h + \frac{15 p^{5}}{p^{2} + 135} = \frac{3}{p^{3}} + \frac{1}{15 p}$ $\Rightarrow h = \frac{3}{p^{3}} + \frac{1}{15 p} - \frac{15 p^{5}}{p^{2} + 135}$ $\frac{1}{2} [\sqrt{1 + 4 R^{2}} - 1 + \ln (\frac{\sqrt{1 + 4 R^{2}} - 1}{2})]$ $= \frac{3}{p^{3}} + \frac{1}{15 p} - \frac{15 p^{5}}{p^{2} + 135} . . . (i i)$ $w e c a n s o l v e (i) a n d (i i) a n d g e t$ $p \approx 1.312274765822$ $R \approx 1.379982232083$
	Commented by mr W last updated on 01/Dec/24

	Commented by a.lgnaoui last updated on 01/Dec/24

	$thank you very much$
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