{"id":1245,"date":"2015-08-22T20:17:41","date_gmt":"2015-08-22T11:17:41","guid":{"rendered":"http:\/\/www.roundown.net\/nyushi\/?p=1245"},"modified":"2021-09-24T17:27:02","modified_gmt":"2021-09-24T08:27:02","slug":"tok201504","status":"publish","type":"post","link":"https:\/\/www.roundown.net\/nyushi\/tok201504\/","title":{"rendered":"\u6771\u5de5\u59272015\uff1a\u7b2c4\u554f"},"content":{"rendered":"<hr \/>\n<p>\\(xy\\) \u5e73\u9762\u4e0a\u3092\u904b\u52d5\u3059\u308b\u70b9 P \u306e\u6642\u523b \\(t \\ ( t \\gt 0 )\\) \u306b\u304a\u3051\u308b\u5ea7\u6a19 \\(( x , y )\\) \u304c\r\n\\[\r\nx = t^2 \\cos t , \\ y = t^2 \\sin t\r\n\\]\r\n\u3067\u8868\u3055\u308c\u3066\u3044\u308b. \u539f\u70b9\u3092 O \u3068\u3057, \u6642\u523b \\(t\\) \u306b\u304a\u3051\u308b P \u306e\u901f\u5ea6\u30d9\u30af\u30c8\u30eb\u3092 \\(\\overrightarrow{v}\\) \u3068\u3059\u308b.<\/p>\r\n<ol>\r\n<li><p><strong>(1)<\/strong>\u3000\\(\\overrightarrow{\\text{OP}}\\) \u3068 \\(\\overrightarrow{v}\\) \u306e\u306a\u3059\u89d2\u3092 \\(\\theta (t)\\) \u3068\u3059\u308b\u3068\u304d, \u6975\u9650\u5024 \\(\\displaystyle\\lim _ {t \\rightarrow \\infty} \\theta (t)\\) \u3092\u6c42\u3081\u3088.<\/p><\/li>\r\n<li><p><strong>(2)<\/strong>\u3000\\(\\overrightarrow{v}\\) \u304c \\(y\\) \u8ef8\u306b\u5e73\u884c\u306b\u306a\u308b\u3088\u3046\u306a \\(t \\ ( t \\gt 0 )\\) \u306e\u3046\u3061, \u6700\u3082\u5c0f\u3055\u3044\u3082\u306e\u3092 \\(t _ 1\\) , \u6b21\u306b\u5c0f\u3055\u3044\u3082\u306e\u3092 \\(t _ 2\\) \u3068\u3059\u308b. \u3053\u306e\u3068\u304d, \u4e0d\u7b49\u5f0f \\(t _ 2 -t _ 1 \\lt \\pi\\) \u3092\u793a\u305b.<\/p><\/li>\r\n<\/ol>\r\n<hr>\r\n<!--more-->\r\n<h4>\u3010 \u89e3 \u7b54 \u3011<\/h4>\r\n<p><strong>(1)<\/strong><\/p>\r\n<p>\\[\r\n\\overrightarrow{\\text{OP}} = \\left( \\begin{array}{c} x \\\\ y \\end{array} \\right) = t^2 \\left( \\begin{array}{c} \\cos t \\\\ \\sin t \\end{array} \\right)\r\n\\]\r\n\u307e\u305f\r\n\\[\\begin{align}\r\n\\dfrac{dx}{dt} & = 2t \\cos t -t^2 \\sin t , \\\\\r\n\\dfrac{dy}{dt} & = 2t \\sin t +t^2 \\cos t\r\n\\end{align}\\]\r\n\u306a\u306e\u3067\r\n\\[\\begin{align}\r\n\\overrightarrow{v} & = \\left( \\begin{array}{c} \\dfrac{dx}{dt} \\\\ \\dfrac{dy}{dt} \\end{array} \\right) = t \\left( \\begin{array}{c} 2 \\cos t -t \\sin t \\\\ 2 \\sin t +t \\cos t \\end{array} \\right)\r\n\\end{align}\\]\r\n\u3053\u3053\u3067, \u8907\u7d20\u5e73\u9762\u3092\u8003\u3048\u3066\r\n\\[\\begin{align}\r\n\\alpha (t) & = \\cos t +i \\sin t , \\\\\r\n\\beta (t) & = ( 2 \\cos t -t \\sin t ) +i ( 2 \\sin t +t \\cos t )\r\n\\end{align}\\]\r\n\u3068\u304a\u3051\u3070\r\n\\[\r\n\\theta (t) = \\left| \\arg \\beta (t) -\\arg \\alpha (t) \\right| = \\left| \\arg \\dfrac{\\beta (t)}{\\alpha (t)} \\right| \\quad ... [1]\r\n\\]\r\n\u3068\u8868\u305b\u308b.\r\n\\[\\begin{align}\r\n\\dfrac{\\beta (t)}{\\alpha (t)} & = \\left\\{ ( 2 \\cos t -t \\sin t ) +i ( 2 \\sin t +t \\cos t ) \\right\\} ( \\cos t -i \\sin t ) \\\\\r\n& = 2 +i t \\\\\r\n& = \\sqrt{t^2 +4} \\left( \\dfrac{2}{\\sqrt{t^2 +4}} +\\dfrac{i t}{\\sqrt{t^2 +4}} \\right)\r\n\\end{align}\\]\r\n\u306a\u306e\u3067, [1] \u3088\u308a\r\n\\[\r\n\\cos \\theta (t) = \\dfrac{2}{\\sqrt{t^2 +4}} , \\ \\sin \\theta (t) = \\dfrac{t}{\\sqrt{t^2 +4}} \\quad ... [2]\r\n\\]\r\n\u3053\u3053\u3067, \\(t \\rightarrow \\infty\\) \u306e\u3068\u304d\u3092\u8003\u3048\u308b\u3068\r\n\\[\r\n\\cos \\theta (t) \\rightarrow 0 , \\ \\sin \\theta (t) \\rightarrow 1\r\n\\]\r\n\u306a\u306e\u3067, \u6c42\u3081\u308b\u6975\u9650\u5024\u306f\r\n\\[\r\n\\displaystyle\\lim _ {t \\rightarrow \\infty} \\theta (t) = \\underline{\\dfrac{\\pi}{2}}\r\n\\]\r\n<p><strong>(2)<\/strong><\/p>\r\n[2] \u3088\u308a \\(t \\gt 0\\) \u306b\u304a\u3044\u3066, \\(\\cos \\theta (t) , \\sin \\theta (t)\\) \u306f\u305d\u308c\u305e\u308c\u5358\u8abf\u6e1b\u5c11, \u5358\u8abf\u5897\u52a0\u3059\u308b\u306e\u3067, \\(\\theta (t)\\) \u306f\u5358\u8abf\u5897\u52a0\u3059\u308b.<br \/>\r\n\u305d\u308c\u305e\u308c\u306e\u5897\u6e1b\u306f\u4e0b\u8868\u306e\u3088\u3046\u306b\u306a\u308b.\r\n\\[\r\n\\begin{array}{c|ccc} t & (0) & \\cdots & ( \\infty ) \\\\ \\hline \\cos \\theta (t) & (1) & \\nearrow & (0) \\\\ \\hline \\sin \\theta (t) & (0) & \\searrow & (1) \\\\ \\hline \\theta (t) & (0) & \\nearrow & \\left( \\dfrac{\\pi}{2} \\right) \\end{array}\r\n\\]\r\n\\(\\overrightarrow{v}\\) \u304c \\(x\\) \u8ef8\u6b63\u65b9\u5411\u3068\u306a\u3059\u89d2\u306f \\(t +\\theta (t)\\) \u306a\u306e\u3067, \\(t _ 1 , t _ 2\\) \u306b\u3064\u3044\u3066\r\n\\[\\begin{align}\r\nt _ 1 +\\theta ( t _ 1 ) & = \\dfrac{\\pi}{2} , \\\\\r\nt _ 2 +\\theta ( t _ 2 ) & = \\dfrac{3 \\pi}{2}\r\n\\end{align}\\]\r\n\\(0 \\lt \\theta ( t _ 1 ) \\lt \\theta ( t _ 2 ) \\lt \\dfrac{\\pi}{2}\\) \u3088\u308a\r\n\\[\r\n0 \\lt \\theta ( t _ 2 ) -\\theta ( t _ 1 ) \\lt \\dfrac{\\pi}{2}\r\n\\]\r\n\u3053\u308c\u3092\u7528\u3044\u308c\u3070\r\n\\[\\begin{align}\r\nt _ 2 -t _ 1 & = \\left( \\dfrac{3 \\pi}{2} -\\theta ( t _ 2 ) \\right) -\\left( \\dfrac{\\pi}{2} -\\theta ( t _ 1 ) \\right) \\\\\r\n& = \\pi -\\left( \\theta ( t _ 2 ) -\\theta ( t _ 1 ) \\right) \\lt \\pi\r\n\\end{align}\\]\r\n<h2>\u3010 \u5225 \u89e3 \u3011\uff08\u884c\u5217\u3092\u7528\u3044\u308b\uff09<\/h2>\r\n<p>\\[\\begin{align}\r\n\\overrightarrow{v} & = \\left( \\begin{array}{c} \\dfrac{dx}{dt} \\\\ \\dfrac{dy}{dt} \\end{array} \\right) = t \\left( \\begin{array}{c} 2 \\cos t -t \\sin t \\\\ 2 \\sin t +t \\cos t \\end{array} \\right) \\\\\r\n& = t \\underline{\\left( \\begin{array}{cc} 2 & -t \\\\ t & 2 \\end{array} \\right)} _ {[1]} \\left( \\begin{array}{c} \\cos t \\\\ \\sin t \\end{array} \\right)\r\n\\end{align}\\]\r\n[1] \u3092\u884c\u5217 \\(A\\) \u3068\u304a\u3051\u3070, \\(\\theta (t)\\) \u306f, \\(A\\) \u304c\u3042\u3089\u308f\u3059\u539f\u70b9\u3092\u4e2d\u5fc3\u3068\u3059\u308b\u56de\u8ee2\u79fb\u52d5\u306e\u56de\u8ee2\u89d2\u306b\u3042\u305f\u308b.<br \/>\r\n\u3057\u305f\u304c\u3063\u3066\r\n\\[\\begin{align}\r\nA & = \\dfrac{1}{\\sqrt{t^2 +4}} \\left( \\begin{array}{cc} \\cos \\theta (t) & -\\sin \\theta (t) \\\\ \\sin \\theta (t) & \\cos \\theta (t) \\end{array} \\right) \\\\\r\n\\text{\u2234} \\quad & \\cos \\theta (t) = \\dfrac{2}{\\sqrt{t^2 +4}} , \\ \\sin \\theta (t) = \\dfrac{t}{\\sqrt{t^2 +4}} \\quad ... [2]\r\n\\end{align}\\]\r\n","protected":false},"excerpt":{"rendered":"\\(xy\\) \u5e73\u9762\u4e0a\u3092\u904b\u52d5\u3059\u308b\u70b9 P \u306e\u6642\u523b \\(t \\ ( t \\gt 0 )\\) \u306b\u304a\u3051\u308b\u5ea7\u6a19 \\(( x , y )\\) \u304c \\[ x = t^2 \\cos t , \\ y = t^2 \\sin t \\] \u3067\u8868\u3055\u308c &hellip; <a href=\"https:\/\/www.roundown.net\/nyushi\/tok201504\/\">\u7d9a\u304d\u3092\u8aad\u3080 <span class=\"meta-nav\">&rarr;<\/span><\/a>","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"inline_featured_image":false,"footnotes":""},"categories":[128],"tags":[141,137],"class_list":["post-1245","post","type-post","status-publish","format-standard","hentry","category-toko_2015","tag-toko","tag-137"],"_links":{"self":[{"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/posts\/1245","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/comments?post=1245"}],"version-history":[{"count":0,"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/posts\/1245\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/media?parent=1245"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/categories?post=1245"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/tags?post=1245"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}