{"id":1089,"date":"2015-02-10T18:11:25","date_gmt":"2015-02-10T17:11:25","guid":{"rendered":"http:\/\/www.unimath.fr\/?p=1089"},"modified":"2015-02-10T18:11:25","modified_gmt":"2015-02-10T17:11:25","slug":"dm-guide-derivee-et-variations-mathx-1s-n49-p-119","status":"publish","type":"post","link":"http:\/\/www.unimath.fr\/?p=1089","title":{"rendered":"DM Guid\u00e9 D\u00e9riv\u00e9e et variations – Math’x 1S – n\u00b049 p 119"},"content":{"rendered":"

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    \n
  1. Montrer que\u00a0r^2=36-h^2
    \n[peekaboo_link name=\u00a0\u00bbquestion1″]Aide 1[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion1″]On applique le th\u00e9or\u00e8me de Pythagore dans un triangle rectangle. Rappel : la sph\u00e8re a pour rayon 6[\/peekaboo_content]
    \n[peekaboo_link name=\u00a0\u00bbquestion2″]Aide 2[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion2″]On prend comme triangle rectangle un carr\u00e9 bleu coup\u00e9 en diagonale[\/peekaboo_content]<\/li>\n
  2. Intervalle pour h
    \n[peekaboo_link name=\u00a0\u00bbquestion3b\u00a0\u00bb]R\u00e9ponse[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion3b\u00a0\u00bb][0 ; 6] car la hauteur ne peut pas \u00eatre sup\u00e9rieure au rayon de la sph\u00e8re[\/peekaboo_content]<\/li>\n
  3. a.\u00a0Volume du cylindre
    \n[peekaboo_link name=\u00a0\u00bbquestion4″]Rappel : formule[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion4″]Aire de la base \\times\u00a0hauteur [\/peekaboo_content]
    \nb.\u00a0Dimensions du cylindre de volume maximal
    \n[peekaboo_link name=\u00a0\u00bbquestion4b\u00a0\u00bb]Aide 1[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion4b\u00a0\u00bb]Il faut donc d\u00e9terminer les variations de la fonction V [\/peekaboo_content]
    \n[peekaboo_link name=\u00a0\u00bbquestion4c\u00a0\u00bb]Aide 2[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion4c\u00a0\u00bb]Pour calculer la d\u00e9riv\u00e9e, on se rappelle que le nombre\u00a0\\pi est une constante[\/peekaboo_content]
    \n[peekaboo_link name=\u00a0\u00bbquestion5″]R\u00e9ponse[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion5″]La d\u00e9riv\u00e9e de V donne -3\\pi h^2+36\\pi[\/peekaboo_content]
    \n[peekaboo_link name=\u00a0\u00bbquestion6″]Signe de la d\u00e9riv\u00e9e[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion6″]On reconna\u00eet un trin\u00f4me du second degr\u00e9 en h (on peut mettre\u00a03\\pi en facteur pour simplifier) (trin\u00f4me du second degr\u00e9 \u00ab\u00a0incomplet\u00a0\u00bb, inutile de calculer le discriminant \\Delta pour trouver les racines) [\/peekaboo_content]
    \n[peekaboo_link name=\u00a0\u00bbquestion9″]Variations de V[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion9″]Les racines du trin\u00f4me sont 2\\sqrt{3} et \u00a0-2\\sqrt{3}. La fonction V est croissante de 0 \u00e0\u00a02\\sqrt{3}\u00a0puis d\u00e9croissante ensuite donc l’aire est maximale pour \u00a0\u00a0h=2\\sqrt{3} cm[\/peekaboo_content]
    \n[peekaboo_link name=\u00a0\u00bbquestion10″]Volume maximal[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion10″]Calculer V(2\\sqrt{3}).[\/peekaboo_content]
    \n[peekaboo_link name=\u00a0\u00bbquestion11″]R\u00e9ponse[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion11″]Le volume maximal vaut 48 \\pi \\sqrt{3} ~~cm^3[\/peekaboo_content]
    \n[peekaboo_link name=\u00a0\u00bbquestion12″]Calcul du rayon du cylindre[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion12″]On calcule le rayon r d’apr\u00e8s\u00a0r^2=36-h^2[\/peekaboo_content]
    \n[peekaboo_link name=\u00a0\u00bbquestion13″]R\u00e9ponse[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion13″] r=2\\sqrt{6} cm donc les dimensions du cylindre qui donne le volume maximal sont : hauteur 2\\sqrt{3} cm et rayon r=2\\sqrt{6} cm\u00a0[\/peekaboo_content]<\/li>\n<\/ol>\n","protected":false},"excerpt":{"rendered":"

      Montrer que\u00a0 [peekaboo_link name=\u00a0\u00bbquestion1″]Aide 1[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion1″]On applique le th\u00e9or\u00e8me de Pythagore dans un triangle rectangle. Rappel : la sph\u00e8re a pour rayon 6[\/peekaboo_content] [peekaboo_link name=\u00a0\u00bbquestion2″]Aide 2[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion2″]On prend comme triangle rectangle un carr\u00e9 bleu coup\u00e9 en diagonale[\/peekaboo_content] Intervalle pour h [peekaboo_link name=\u00a0\u00bbquestion3b\u00a0\u00bb]R\u00e9ponse[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion3b\u00a0\u00bb][0 ; 6] car la hauteur ne peut pas \u00eatre sup\u00e9rieure au […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":[],"categories":[1],"tags":[],"_links":{"self":[{"href":"http:\/\/www.unimath.fr\/index.php?rest_route=\/wp\/v2\/posts\/1089"}],"collection":[{"href":"http:\/\/www.unimath.fr\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"http:\/\/www.unimath.fr\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"http:\/\/www.unimath.fr\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"http:\/\/www.unimath.fr\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=1089"}],"version-history":[{"count":6,"href":"http:\/\/www.unimath.fr\/index.php?rest_route=\/wp\/v2\/posts\/1089\/revisions"}],"predecessor-version":[{"id":1095,"href":"http:\/\/www.unimath.fr\/index.php?rest_route=\/wp\/v2\/posts\/1089\/revisions\/1095"}],"wp:attachment":[{"href":"http:\/\/www.unimath.fr\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=1089"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"http:\/\/www.unimath.fr\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=1089"},{"taxonomy":"post_tag","embeddable":true,"href":"http:\/\/www.unimath.fr\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=1089"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}