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Question Number 56011 by mr W last updated on 08/Mar/19

Commented by mr W last updated on 08/Mar/19

$P o i n t A i s c h o o s e n a t r a n d o m i n a$ $s q u a r e w i t h s i d e l e n g t h a, s e e f i g u r e .$ $W h a t i s i t s a v e r a g e d i s t a n c e t o t h e$ $o r i g i n ?$ $T a k e P (h, k) = (0, 0) a t f i r s t .$
	Answered by ajfour last updated on 08/Mar/19

	$F o r h > k (I f h < k w e w i l l s w a p h w i t h k)$ $a^{2} {\bar{r}}_{a v g} = \int_{\sqrt{h^{2} + k^{2}}}^{\sqrt{h^{2} + {(k + a)}^{2}}} [\cos^{- 1} (\frac{h}{r}) - \sin^{- 1} (\frac{k}{r})] r^{2} d r$ $+ \int_{\sqrt{h^{2} + {(k + a)}^{2}}}^{\sqrt{{(h + a)}^{2} + k^{2}}} [\sin^{- 1} (\frac{k + a}{r}) - \sin^{- 1} (\frac{k}{r})] r^{2} d r$ $+ \int_{\sqrt{{(h + a)}^{2} + k^{2}}}^{\sqrt{{(h + a)}^{2} + {(k + a)}^{2}}} [\sin^{- 1} (\frac{k + a}{r}) - \cos^{- 1} (\frac{h + a}{r})] r^{2} d r .$ $\underline{I f (h, k) \equiv (0, 0)}$ $a^{2} r_{a v g} = \frac{π}{2} \int_{0}^{a} r^{2} d r + \int_{a}^{a \sqrt{2}} r^{2} [\sin^{- 1} (\frac{a}{r}) - \cos^{- 1} (\frac{a}{r})] d r$ $= \frac{π a^{3}}{6} + \int_{a}^{a \sqrt{2}} r^{2} [\frac{π}{2} - {2 \cos}^{- 1} (\frac{a}{r})] d r$ $= \frac{π a^{3}}{6} + \frac{π a^{3}}{6} (2 \sqrt{2} - 1) - 2 I$ $= \frac{2 \sqrt{2} π a^{3}}{6} - 2 I$ $I = \int_{a}^{a \sqrt{2}} r^{2} \cos^{- 1} (\frac{a}{r}) d r$ $l e t \cos θ = \frac{a}{r} \Rightarrow r = a \sec θ$ $a n d d r = a \sec θ \tan θ d θ$ $\Rightarrow I = \int_{0}^{π / 4} a^{3} θ \sec^{3} θ \tan θ d θ$ $= a^{3} [θ \int_{0}^{π / 4} \sec^{2} θ (\sec θ \tan θ d θ)$ $- \frac{1}{3} \int_{0}^{π / 4} \sec^{3} θ d θ]$ $= a^{3} [\frac{π}{4} \times \frac{2 \sqrt{2}}{3} - \frac{1}{3} \int_{0}^{π / 4} \sec^{3} θ d θ]$ $\int \sec^{3} θ d θ = \sec θ \tan θ - \int \sec θ \tan^{2} θ d θ$ $2 \int \sec^{3} θ d θ = \sec θ \tan θ + \int \sec θ d θ$ $2 \int \sec^{3} θ d θ = \sec θ \tan θ + \ln ∣ \sec θ + \tan θ ∣ + c$ $\Rightarrow 2 \int_{0}^{π / 4} \sec^{3} θ = \sqrt{2} + \ln (\sqrt{2} + 1)$ $a^{2} r_{a v g} = \frac{2 \sqrt{2} π a^{3}}{6} - 2 a^{3} [\frac{π}{4} \times \frac{2 \sqrt{2}}{3} - \frac{1}{3} \int_{0}^{π / 4} \sec^{3} θ d θ]$ $a^{2} r_{a v g} = \frac{2 \sqrt{2} π a^{3}}{6} - 2 a^{3} [\frac{π}{4} \times \frac{2 \sqrt{2}}{3} - \frac{1}{6} (\sqrt{2} + \ln (\sqrt{2} + 1))]$ $\Rightarrow r_{a v g} = \frac{a}{3} [\sqrt{2} + \ln (\sqrt{2} + 1)]$ $\Rightarrow r_{a v g} \approx 0.765 a ◼$
	Commented by mr W last updated on 08/Mar/19

	$t h a n k s a l o t s i r!$
	Answered by mr W last updated on 09/Mar/19

	$f o r P (h, k) = O :$ $A (x, y) = (r, θ)$ $d = \sqrt{x^{2} + y^{2}} = r$ $d_{A v} = \frac{2 \int_{0}^{\frac{π}{4}} \int_{0}^{\frac{a}{\cos θ}} r^{2} d r d θ}{a^{2}}$ $d_{A v} = \frac{2 a}{3} \int_{0}^{\frac{π}{4}} \frac{1}{\cos^{3} θ} d θ$ $d_{A v} = \frac{2 a}{3} \int_{0}^{\frac{π}{4}} \frac{1}{{(1 - \sin^{2} θ)}^{2}} d \sin θ$ $d_{A v} = \frac{a}{6} {[\ln \frac{1 + \sin θ}{1 - \sin θ} + \frac{2 \tan θ}{\cos θ}]}_{0}^{\frac{π}{4}}$ $d_{A v} = \frac{a}{6} (\ln \frac{2 + \sqrt{2}}{2 - \sqrt{2}} + 2 \sqrt{2})$ $d_{A v} = \frac{a}{3} [\ln (1 + \sqrt{2}) + \sqrt{2}] \approx 0.765 a$
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