From Förberedande kurs i matematik 1

Revision as of 09:39, 9 September 2008 (edit) Tekbot (Talk | contribs) m (Lösning 4.1:9 moved to Solution 4.1:9: Robot: moved page) ← Previous diff		Current revision (11:50, 8 October 2008) (edit) (undo) Tek (Talk | contribs) m
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-	{{NAVCONTENT_START}}	+	10 seconds corresponds to 1/6 minutes, so that during that time period, the second hand sweeps over 1/6 of a turn, i.e. the sector of a circle with angle
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		+	{{Displayed math||<math>\alpha = \frac{1}{6}\cdot 2\pi\ \text{radians} = \frac{\pi}{3}\ \text{radians.}</math>}}
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	<center> [[Image:4_1_9_.gif]] </center>		<center> [[Image:4_1_9_.gif]] </center>
-	<center> [[Image:4_1_9.gif]] </center>	+	The area of the sector is
-	{{NAVCONTENT_STOP}}	+
		+	{{Displayed math||<math>\text{Area} = \frac{1}{2}\alpha r^{2} = \frac{1}{2}\cdot \frac{\pi}{3}\cdot (8\ \text{cm})^2 = \frac{32\pi}{3}\ \text{cm}^{2} \approx 33\textrm{.}5\ \text{cm}^{2}\,\textrm{.}</math>}}

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Current revision

10 seconds corresponds to 1/6 minutes, so that during that time period, the second hand sweeps over 1/6 of a turn, i.e. the sector of a circle with angle

\displaystyle \alpha = \frac{1}{6}\cdot 2\pi\ \text{radians} = \frac{\pi}{3}\ \text{radians.}

The area of the sector is

\displaystyle \text{Area} = \frac{1}{2}\alpha r^{2} = \frac{1}{2}\cdot \frac{\pi}{3}\cdot (8\ \text{cm})^2 = \frac{32\pi}{3}\ \text{cm}^{2} \approx 33\textrm{.}5\ \text{cm}^{2}\,\textrm{.}