Solution 4.3:3d - Revision history

Tek at 13:44, 9 October 2008

2008-10-09T13:44:01Z

←Older revision		Revision as of 13:44, 9 October 2008
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-	The expression for the angle	+	The expression for the angle <math>\pi/2 - v</math> differs from <math>\pi/2</math> by as much as <math>-v</math> differs from <math>0</math>. This means that <math>\pi/2-v</math> makes the same angle with the positive ''y''-axis as <math>-v</math> makes with the positive ''x''-axis.
-	<math>{\pi }/{2~~}\;~~-v</math>	+
-	differs from	+
-	<math>{\pi }/{2~~}\;~~</math>	+
-	by as much as	+
-	<math>-v~~\text{ }~~</math>	+
-	differs from	+
-	<math>0</math>. This means that	+
-	<math>{\pi }/{2~~}\;~~</math>	+
-	makes the same angle with the positive	+
-	~~<math>~~y~~</math>~~	+
-	-axis as	+
-	<math>-v~~\text{ }~~</math>	+
-	makes with the positive	+
-	~~<math>~~x~~</math>~~	+
-	-axis.	+

		+	{\| align="center"
		+	\| align="center" \|[[Image:4_3_3_d-1.gif]]
		+	\| width="20px"\|
		+	\| align="center" \|[[Image:4_3_3_d-2.gif]]
		+	\|-
		+	\| align="center" \|<small>Angle ''v''</small>
		+	\|\|
		+	\| align="center" \|<small>Angle π/2 - ''v''</small>
		+	\|}

-	~~[[Image:4_3_3_d~~.~~gif\|center]]~~	+	Therefore, the angle <math>\pi/2 - v</math> has a ''y''-coordinate which is equal to the ''x''-coordinate for the angle ''v'', i.e.

-	~~Angle~~	+	{{Displayed math\|\|<math>\sin\Bigl(\frac{\pi}{2} - v\Bigr) = \cos v</math>}}
-	<math~~>v</math>~~	+
-	~~angle~~	+
-	<math>\pi -v</math>	+
-		+

-	~~Therefore~~, ~~the angle~~	+	and from exercise c, we know that <math>\cos v = \sqrt{1-a^2}\,</math>,
-	<math>{\~~pi }/~~{2}\~~;-v</math>~~	+
-	~~has a~~	+
-	~~<math>y</math>~~	+
-	~~-coordinate which is equal to the~~	+
-	~~<math>x</math>~~	+
-	~~-coordinate for the angle~~	+
-	~~<math>v~~</math>, ~~i.e.~~	+

-		+	{{Displayed math\|\|<math>\sin\Bigl(\frac{\pi}{2}-v\Bigr) = \sqrt{1-a^2}\,\textrm{.}</math>}}
-	~~<math>\sin \left(~~ {~~\pi }/~~{~~2}\;-v \right)=\cos v</~~math>	+
-		+
-		+
-	~~and from exercise c, we know that~~	+
-	~~<math>\cos v=\sqrt{1-a^{2}}</math>~~	+
-		+
-		+
-		+
-	<math>\sin \~~left~~( \frac{\pi }{2}-v \~~right~~)=\sqrt{1-a^{2}}</math>	+

Ian at 11:11, 29 September 2008

2008-09-29T11:11:58Z

←Older revision		Revision as of 11:11, 29 September 2008
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-	{{~~NAVCONTENT_START~~}}	+	The expression for the angle
-	<~~center~~> ~~[[Image:4_3_3d.gif]]~~ </~~center~~>	+	<math>{\pi }/{2}\;-v</math>
-	{{~~NAVCONTENT_STOP~~}}	+	differs from
		+	<math>{\pi }/{2}\;</math>
		+	by as much as
		+	<math>-v\text{ }</math>
		+	differs from
		+	<math>0</math>. This means that
		+	<math>{\pi }/{2}\;</math>
		+	makes the same angle with the positive
		+	<math>y</math>
		+	-axis as
		+	<math>-v\text{ }</math>
		+	makes with the positive
		+	<math>x</math>
		+	-axis.
		+
		+
	[[Image:4_3_3_d.gif\|center]]		[[Image:4_3_3_d.gif\|center]]
		+
		+	Angle
		+	<math>v</math>
		+	angle
		+	<math>\pi -v</math>
		+
		+
		+	Therefore, the angle
		+	<math>{\pi }/{2}\;-v</math>
		+	has a
		+	<math>y</math>
		+	-coordinate which is equal to the
		+	<math>x</math>
		+	-coordinate for the angle
		+	<math>v</math>, i.e.
		+
		+
		+	<math>\sin \left( {\pi }/{2}\;-v \right)=\cos v</math>
		+
		+
		+	and from exercise c, we know that
		+	<math>\cos v=\sqrt{1-a^{2}}</math>
		+
		+
		+
		+	<math>\sin \left( \frac{\pi }{2}-v \right)=\sqrt{1-a^{2}}</math>

Tekbot: Lösning 4.3:3d moved to Solution 4.3:3d: Robot: moved page

2008-09-09T09:53:03Z

Lösning 4.3:3d moved to Solution 4.3:3d: Robot: moved page

←Older revision

Revision as of 09:53, 9 September 2008

Tekbot: Robot: Automated text replacement (-[[Bild: +[[Image:)

2008-08-21T07:13:39Z

Robot: Automated text replacement (-[[Bild: +[[Image:)

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Madde at 14:15, 18 August 2008

2008-08-18T14:15:58Z

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2008-04-03T09:45:12Z

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