{"id":1074,"date":"2015-02-10T12:13:38","date_gmt":"2015-02-10T11:13:38","guid":{"rendered":"http:\/\/www.unimath.fr\/?p=1074"},"modified":"2015-02-10T12:13:38","modified_gmt":"2015-02-10T11:13:38","slug":"dm-guide-derivation-et-variations-mathx-1s-n46-p-119","status":"publish","type":"post","link":"http:\/\/www.unimath.fr\/?p=1074","title":{"rendered":"DM Guid\u00e9 D\u00e9rivation et variations – Math’x 1S – n\u00b046 p 119"},"content":{"rendered":"

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  1. Coordonn\u00e9es de A et B, points d’intersection de la parabole et de la droite\u00a0u_2
    \n[peekaboo_link name=\u00a0\u00bbquestion1″]Aide[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion1″] On r\u00e9sout x^2=k[\/peekaboo_content]
    \n[peekaboo_link name=\u00a0\u00bbquestion2″]R\u00e9ponse[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion2″] A (- \\sqrt{k} ; k) et B\u00a0( \\sqrt{k} ; k)[\/peekaboo_content]<\/li>\n
  2. a. Calcul de A_{k}(x)
    \n[peekaboo_link name=\u00a0\u00bbquestion3b\u00a0\u00bb]Longueur CC'[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion3b\u00a0\u00bb]L’abscisse de C est x >0 et l’abscisse de C’ est -x donc CC’ = 2x [\/peekaboo_content]
    \nLongueur CM
    \n[peekaboo_link name=\u00a0\u00bbquestion4″]Coordonn\u00e9es de M[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion4″]M appartient \u00e0 la parabole et l’abscisse de M est x donc M (x ; x ^2)[\/peekaboo_content]
    \n[peekaboo_link name=\u00a0\u00bbquestion4b\u00a0\u00bb]R\u00e9ponse longueur CM[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion4b\u00a0\u00bb]CM = k - x^2[\/peekaboo_content]
    \n[peekaboo_link name=\u00a0\u00bbquestion4c\u00a0\u00bb]Expression de\u00a0 A_{k}(x)[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion4c\u00a0\u00bb]A_{k}(x)=-2x^3+2kx avec x \\in [0 ; \\sqrt{k}] [\/peekaboo_content]
    \nb. Variation de\u00a0A_{k}\u00a0sur\u00a0[0 ; \\sqrt{k}]
    \n[peekaboo_link name=\u00a0\u00bbquestion5″]D\u00e9riv\u00e9e[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion5″]-6x^2+2k[\/peekaboo_content]
    \n[peekaboo_link name=\u00a0\u00bbquestion6″]Signe de la d\u00e9riv\u00e9e[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion6″]On cherche le signe de -6x^2+2k (trin\u00f4me du second degr\u00e9 \u00ab\u00a0incomplet\u00a0\u00bb, on factorise par x pour trouver les racines) [\/peekaboo_content]
    \n[peekaboo_link name=\u00a0\u00bbquestion8″]variations[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion8″]La fonction est croissante de 0 \u00e0\u00a0\u00a0\\sqrt{\\frac{k}{3}} puis d\u00e9croissante ensuite donc l’aire est maximale pour \u00a0x=\\sqrt{\\frac{k}{3}}\u00a0[\/peekaboo_content]
    \nc.\u00a0[peekaboo_link name=\u00a0\u00bbquestion9″]Point C associ\u00e9 \u00e0 l’aire maximale[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion9″] C a pour coordonn\u00e9es (\u00a0\\sqrt{\\frac{k}{3}} ; \u00a0k) [\/peekaboo_content]
    \n[peekaboo_link name=\u00a0\u00bbquestion10″]Aide[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion10″]Calculer 3x_{C}^2. Que constatez-vous ?[\/peekaboo_content]
    \n[peekaboo_link name=\u00a0\u00bbquestion11″]R\u00e9ponse[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion11″] Les coordonn\u00e9es de C v\u00e9rifient l’\u00e9quation 3x^2=y\u00a0donc les points C appartiennent \u00e0 la parabole d’\u00e9quation\u00a0y=3x^2\u00a0[\/peekaboo_content]<\/li>\n<\/ol>\n","protected":false},"excerpt":{"rendered":"

      Coordonn\u00e9es de A et B, points d’intersection de la parabole et de la droite\u00a0 [peekaboo_link name=\u00a0\u00bbquestion1″]Aide[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion1″] On r\u00e9sout [\/peekaboo_content] [peekaboo_link name=\u00a0\u00bbquestion2″]R\u00e9ponse[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion2″] A et B\u00a0[\/peekaboo_content] a. Calcul de [peekaboo_link name=\u00a0\u00bbquestion3b\u00a0\u00bb]Longueur CC'[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion3b\u00a0\u00bb]L’abscisse de C est et l’abscisse de C’ est donc CC’ = [\/peekaboo_content] Longueur CM [peekaboo_link name=\u00a0\u00bbquestion4″]Coordonn\u00e9es de M[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion4″]M appartient \u00e0 […]<\/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\/1074"}],"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=1074"}],"version-history":[{"count":10,"href":"http:\/\/www.unimath.fr\/index.php?rest_route=\/wp\/v2\/posts\/1074\/revisions"}],"predecessor-version":[{"id":1087,"href":"http:\/\/www.unimath.fr\/index.php?rest_route=\/wp\/v2\/posts\/1074\/revisions\/1087"}],"wp:attachment":[{"href":"http:\/\/www.unimath.fr\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=1074"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"http:\/\/www.unimath.fr\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=1074"},{"taxonomy":"post_tag","embeddable":true,"href":"http:\/\/www.unimath.fr\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=1074"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}