{"id":77,"date":"2011-11-26T22:50:56","date_gmt":"2011-11-26T13:50:56","guid":{"rendered":"http:\/\/roundown.main.jp\/nyushi\/?p=77"},"modified":"2021-09-30T09:58:18","modified_gmt":"2021-09-30T00:58:18","slug":"kbr201103","status":"publish","type":"post","link":"https:\/\/www.roundown.net\/nyushi\/kbr201103\/","title":{"rendered":"\u7b51\u6ce2\u5927\u7406\u7cfb2011\uff1a\u7b2c3\u554f"},"content":{"rendered":"
\n

\\(\\alpha\\) \u3092 \\(0 \\lt \\alpha \\lt \\dfrac{\\pi}{2}\\) \u3092\u6e80\u305f\u3059\u5b9a\u6570\u3068\u3059\u308b.\r\n\u5186 \\(C : \\ x^2+( y+ \\sin \\alpha )^2 = 1\\) \u304a\u3088\u3073, \u305d\u306e\u4e2d\u5fc3\u3092\u901a\u308b\u76f4\u7dda \\(l : \\ y = ( \\tan \\alpha )x -\\sin \\alpha\\) \u3092\u8003\u3048\u308b. \u3053\u306e\u3068\u304d, \u4ee5\u4e0b\u306e\u554f\u3044\u306b\u7b54\u3048\u3088.<\/p>\r\n

    \r\n

  1. (1)<\/strong>\u3000\u76f4\u7dda \\(l\\) \u3068\u5186 \\(C\\) \u306e \\(2\\) \u3064\u306e\u4ea4\u70b9\u306e\u5ea7\u6a19\u3092 \\(\\alpha\\) \u3092\u7528\u3044\u3066\u8868\u305b.<\/p><\/li>\r\n

  2. (2)<\/strong>\u3000\u7b49\u5f0f\r\n\\[\r\n2 \\displaystyle\\int _ {\\cos \\alpha}^1 \\sqrt{1-x^2} \\, dx +\\displaystyle\\int _ {-\\cos \\alpha}^{\\cos \\alpha} \\sqrt{1-x^2} \\, dx = \\dfrac{\\pi}{2}\r\n\\]\r\n\u304c\u6210\u308a\u7acb\u3064\u3053\u3068\u3092\u793a\u305b.<\/p><\/li>\r\n

  3. (3)<\/strong>\u3000\u9023\u7acb\u4e0d\u7b49\u5f0f\r\n\\[\r\n\\left\\{ \\begin{array}{l} y \\leqq ( \\tan \\alpha )x -\\sin \\alpha \\\\ x^2+( y+ \\sin \\alpha )^2 \\leqq 1 \\end{array} \\right.\r\n\\]\r\n\u306e\u8868\u3059 \\(xy\\) \u5e73\u9762\u4e0a\u306e\u56f3\u5f62\u3092 \\(D\\) \u3068\u3059\u308b. \u56f3\u5f62 \\(D\\) \u3092 \\(x\\) \u8ef8\u306e\u307e\u308f\u308a\u306b \\(1\\) \u56de\u8ee2\u3055\u305b\u3066\u3067\u304d\u308b\u7acb\u4f53\u306e\u4f53\u7a4d\u3092\u6c42\u3081\u3088.<\/p><\/li>\r\n<\/ol>\r\n

    \r\n\r\n

    \u3010 \u89e3 \u7b54 \u3011<\/h2>\r\n

    (1)<\/strong><\/p>\r\n

    \u76f4\u7dda \\(l\\) \u306f \\(C\\) \u306e\u4e2d\u5fc3\u3092\u901a\u308a, \u50be\u304d\u304c \\(\\tan \\alpha = \\dfrac{\\sin \\alpha}{\\cos \\alpha}\\) \u3067\u3042\u308b.<\/p>\r\n\"\"\r\n

    \\(C\\) \u306e\u534a\u5f84\u306f \\(1 \\ \\left( = \\sin^2 \\alpha +\\cos^2 \\alpha \\right)\\) \u3067\u3042\u308b\u3053\u3068\u304b\u3089, \u4ea4\u70b9\u306f, \u4e2d\u5fc3 \\(( 0 , \\sin \\alpha )\\) \u304b\u3089 \\(\\left( \\begin{array}{c} \\pm \\cos \\alpha \\\\ \\pm \\sin \\alpha \\end{array} \\right) \\ ( \\text{\u8907\u53f7\u540c\u9806} )\\) \u79fb\u52d5\u3057\u305f\u70b9\u3067\u3042\u308b.
    \r\n\u3086\u3048\u306b\u6c42\u3081\u308b\u4ea4\u70b9\u306f\r\n\\[\r\n\\underline{( -\\cos \\alpha , -2 \\sin \\alpha ) , ( \\cos \\alpha , 0 )}\r\n\\]\r\n

    (2)<\/strong><\/p>\r\n

    \u4e0e\u5f0f\u3092 \\(I\\) \u3068\u304a\u304f.
    \r\n\u95a2\u6570 \\(f(x) = \\sqrt{1-x^2}\\) \uff08 \\(-1 \\leqq x \\leqq 1\\) \uff09\u306f, \\(f(-x) = f(x)\\) \u304c\u6210\u7acb\u3059\u308b\u5947\u95a2\u6570\u306a\u306e\u3067\r\n\\[\r\n\\displaystyle\\int _ {\\cos \\alpha}^1 \\sqrt{1-x^2} \\, dx = \\displaystyle\\int _ {-1}^{-\\cos \\alpha} \\sqrt{1-x^2} \\, dx\r\n\\]\r\n\u3057\u305f\u304c\u3063\u3066,\r\n\\[\\begin{align}\r\nI & = \\displaystyle\\int _ {-1}^{-\\cos \\alpha} \\sqrt{1-x^2} \\, dx +\\displaystyle\\int _ {-\\cos \\alpha}^{\\cos \\alpha} \\sqrt{1-x^2} \\, dx +\\displaystyle\\int _ {\\cos \\alpha}^1 \\sqrt{1-x^2} \\, dx \\\\\r\n& = \\displaystyle\\int _ {-1}^1 \\sqrt{1-x^2} \\, dx\r\n\\end{align}\\]\r\n\u3053\u308c\u306f, \u534a\u5f84 \\(1\\) \u306e\u534a\u5186\u306e\u9762\u7a4d\u306b\u7b49\u3057\u3044\u306e\u3067\r\n\\[\r\nI = \\dfrac{1}{2} \\cdot 1^2 \\cdot \\dfrac{\\pi}{2} = \\dfrac{\\pi}{2}\r\n\\]\r\n

    (3)<\/strong><\/p>\r\n\"\"\r\n

    \u6c42\u3081\u308b\u4f53\u7a4d \\(V\\) \u306f, \u4e0a\u56f3\u306e\u56de\u8ee2\u4f53\u306e\u4f53\u7a4d \\(V _ 1 , V _ 2 , V _ 3\\) \u3092\u7528\u3044\u3066\r\n\\[\r\nV = V _ 1 -V _ 2 -V _ 3\r\n\\]\r\n\u3068\u8868\u305b\u308b.\r\n\\[\\begin{align}\r\nV _ 1 & = \\pi \\displaystyle\\int _ {-\\cos \\alpha}^1 \\left( -\\sin \\alpha -\\sqrt{1-x^2} \\right)^2 \\, dx \\\\\r\n& = \\pi \\displaystyle\\int _ {-\\cos \\alpha}^1 ( 1-x^2) \\, dx \\\\\r\n& \\qquad +2 \\pi \\sin \\alpha \\displaystyle\\int _ {-\\cos \\alpha}^1 \\sqrt{1-x^2} \\, dx +\\pi \\sin^2 \\alpha ( 1 +\\cos \\alpha ) , \\\\\r\nV _ 2 & = \\dfrac{1}{3} \\cdot \\pi ( 2 \\sin \\alpha )^2 \\cdot 2 \\cos \\alpha = \\dfrac{8 \\pi}{3} \\sin^2 \\alpha \\cos \\alpha , \\\\\r\nV _ 3 & = \\pi \\displaystyle\\int _ {\\cos \\alpha}^1 \\left( -\\sin \\alpha +\\sqrt{1-x^2} \\right)^2 \\, dx \\\\\r\n& = \\pi \\displaystyle\\int _ {\\cos \\alpha}^1 ( 1-x^2) \\, dx \\\\\r\n& \\qquad +2 \\pi \\sin \\alpha \\displaystyle\\int _ {\\cos \\alpha}^1 \\sqrt{1-x^2} \\, dx +\\pi \\sin^2 \\alpha ( 1 -\\cos \\alpha )\r\n\\end{align}\\]\r\n\u4ee5\u4e0a\u3092\u7528\u3044\u308c\u3070\r\n\\[\\begin{align}\r\nV & = \\pi \\displaystyle\\int _ {-\\cos \\alpha}^{\\cos \\alpha} ( 1-x^2 ) \\, dx \\\\\r\n& \\quad +2 \\pi \\sin \\alpha \\left( \\displaystyle\\int _ {-\\cos \\alpha}^1 \\sqrt{1-x^2} \\, dx\r\n+\\displaystyle\\int _ {\\cos \\alpha}^1 \\sqrt{1-x^2} \\, dx \\right) \\\\\r\n& \\qquad -\\dfrac{2 \\pi}{3} \\sin^2 \\alpha \\cos \\alpha \\\\\r\n& = 2 \\pi \\displaystyle\\int _ {0}^{\\cos \\alpha} ( 1-x^2) \\, dx \\\\\r\n& \\quad +2 \\pi \\sin \\alpha \\left( 2 \\displaystyle\\int _ {\\cos \\alpha}^1 \\sqrt{1-x^2} \\, dx +\\displaystyle\\int _ {-\\cos \\alpha}^{\\cos \\alpha} \\sqrt{1-x^2} \\, dx \\right) \\\\\r\n& \\qquad -\\dfrac{2 \\pi}{3} \\sin^2 \\alpha \\cos \\alpha \\\\\r\n& = 2 \\pi \\left[ x -\\dfrac{x^3}{3} \\right] _ 0^{\\cos \\alpha} +2\\pi \\sin \\alpha \\cdot \\dfrac{\\pi}{2} \\\\\r\n& \\qquad -\\dfrac{2 \\pi}{3}( 1 -\\cos^2 \\alpha ) \\cos \\alpha \\quad ( \\ \\text{\u2235} \\ \\text{(2)\u306e\u7d50\u679c} ) \\\\\r\n& = 2\\pi \\left( \\cos \\alpha -\\dfrac{\\cos^3 \\alpha}{3} \\right) +\\pi^2 \\sin \\alpha -\\dfrac{2 \\pi}{3} ( 1 -\\cos^2 \\alpha ) \\cos \\alpha \\\\\r\n& = \\underline{\\dfrac{4\\pi}{3} \\cos \\alpha +\\pi^2 \\sin \\alpha}\r\n\\end{align}\\]\r\n","protected":false},"excerpt":{"rendered":"\\(\\alpha\\) \u3092 \\(0 \\lt \\alpha \\lt \\dfrac{\\pi}{2}\\) \u3092\u6e80\u305f\u3059\u5b9a\u6570\u3068\u3059\u308b. \u5186 \\(C : \\ x^2+( y+ \\sin \\alpha )^2 = 1\\) \u304a\u3088\u3073, \u305d\u306e\u4e2d\u5fc3 … \u7d9a\u304d\u3092\u8aad\u3080 →<\/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":[41],"tags":[144,13],"class_list":["post-77","post","type-post","status-publish","format-standard","hentry","category-tsukuba_r_2011","tag-tsukuba_r","tag-13"],"_links":{"self":[{"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/posts\/77","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=77"}],"version-history":[{"count":0,"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/posts\/77\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/media?parent=77"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/categories?post=77"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/tags?post=77"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}