{"id":6550,"date":"2011-03-01T15:29:50","date_gmt":"2011-03-01T15:29:50","guid":{"rendered":"https:\/\/www.acasinhadamatematica.pt\/?p=6550"},"modified":"2021-12-26T17:46:17","modified_gmt":"2021-12-26T17:46:17","slug":"o-perimetro-de-uma-circunferencia","status":"publish","type":"post","link":"https:\/\/www.acasinhadamatematica.pt\/?p=6550","title":{"rendered":"O comprimento de uma circunfer\u00eancia"},"content":{"rendered":"<p><ul id='GTTabs_ul_6550' class='GTTabs' style='display:none'>\n<li id='GTTabs_li_0_6550' class='GTTabs_curr'><a  id=\"6550_0\" onMouseOver=\"GTTabsShowLinks('Enunciado'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Enunciado<\/a><\/li>\n<li id='GTTabs_li_1_6550' ><a  id=\"6550_1\" onMouseOver=\"GTTabsShowLinks('Resolu\u00e7\u00e3o'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Resolu\u00e7\u00e3o<\/a><\/li>\n<\/ul>\n\n<div class='GTTabs_divs GTTabs_curr_div' id='GTTabs_0_6550'>\n<span class='GTTabs_titles'><b>Enunciado<\/b><\/span><\/p>\n<p>O per\u00edmetro $P$ de um c\u00edrculo de raio $r$ \u00e9 dado pela express\u00e3o $P=2\\pi r$.<\/p>\n<ol>\n<li>Calcule a taxa m\u00e9dia de varia\u00e7\u00e3o de $P$ em cada um dos intervalos: $\\left[ 2,9 \\right]$, $\\left[ 2;2,5 \\right]$, $\\left[ 2;2,1 \\right]$, $\\left[ 2;2,001 \\right]$ e $\\left[ 2,2+h \\right]$.<\/li>\n<li>Qual \u00e9 o valor da taxa de varia\u00e7\u00e3o do per\u00edmetro para $r=2$?<\/li>\n<\/ol>\n<p><div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_next'><a href='#GTTabs_ul_6550' onClick='GTTabs_show(1,6550)'>Resolu\u00e7\u00e3o &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_1_6550'>\n<span class='GTTabs_titles'><b>Resolu\u00e7\u00e3o<\/b><\/span><!--more--><\/p>\n<ol>\n<li>Ora,<br \/>\n\\[tm{{v}_{\\left[ 2,9 \\right]}}=\\frac{2\\pi \\times 9-2\\pi \\times 2}{9-2}=\\frac{7\\times 2\\pi }{7}=2\\pi \\]\\[tm{{v}_{\\left[ 2;2,25 \\right]}}=\\frac{2\\pi \\times 2,25-2\\pi \\times 2}{2,25-2}=\\frac{0,25\\times 2\\pi }{0,25}=2\\pi \\]<\/p>\n<p>\\[tm{{v}_{\\left[ 2;2,1 \\right]}}=\\frac{2\\pi \\times 2,1-2\\pi \\times 2}{2,1-2}=\\frac{0,1\\times 2\\pi }{0,1}=2\\pi \\]<\/p>\n<p>\\[tm{{v}_{\\left[ 2;2,001 \\right]}}=\\frac{2\\pi \\times 2,001-2\\pi \\times 2}{2,001-2}=\\frac{0,001\\times 2\\pi }{0,001}=2\\pi \\]<\/p>\n<p>\\[tm{{v}_{\\left[ 2;2+h \\right]}}=\\frac{2\\pi \\times (2+h)-2\\pi \\times 2}{2+h-2}=\\frac{h\\times 2\\pi }{h}=2\\pi \\]<\/p>\n<\/li>\n<li>O valor da taxa de varia\u00e7\u00e3o do per\u00edmetro\u00a0para $r=2$ \u00e9 $P'(2)=2\\pi $.<\/li>\n<\/ol>\n<div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_prev'><a href='#GTTabs_ul_6550' onClick='GTTabs_show(0,6550)'>&lt;&lt; Enunciado<\/a><\/span><\/div><\/div>\n\n","protected":false},"excerpt":{"rendered":"<p>Enunciado Resolu\u00e7\u00e3o Enunciado O per\u00edmetro $P$ de um c\u00edrculo de raio $r$ \u00e9 dado pela express\u00e3o $P=2\\pi r$. Calcule a taxa m\u00e9dia de varia\u00e7\u00e3o de $P$ em cada um dos intervalos: $\\left[ 2,9 \\right]$,&#46;&#46;&#46;<\/p>\n","protected":false},"author":1,"featured_media":19442,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[98,97,134],"tags":[135],"series":[],"class_list":["post-6550","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-11--ano","category-aplicando","category-derivadas","tag-taxa-media-de-variacao"],"views":1581,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/03\/Circulo_preto.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/6550","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=6550"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/6550\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/media\/19442"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=6550"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=6550"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=6550"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=6550"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}