From d7ab3078c542c9c2d98bc00db13e7ea7d4699166 Mon Sep 17 00:00:00 2001 From: Bertrand Benjamin Date: Tue, 7 Oct 2025 17:51:00 +0200 Subject: [PATCH] =?UTF-8?q?feat(1G=5Fmath):=20s=C3=A9quence=20sur=20le=20n?= =?UTF-8?q?ombre=20d=C3=A9riv=C3=A9?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- .../1B_taux_accroissement.tex | 71 ++ .../1B_taux_de_variations.pdf | Bin 0 -> 21116 bytes .../1B_taux_de_variations.tex | 79 ++ .../2B_tangente.pdf | Bin 0 -> 20150 bytes .../2B_tangente.tex | 54 ++ .../3B_nombre_derive.pdf | Bin 0 -> 21419 bytes .../3B_nombre_derive.tex | 59 ++ .../4B_equation_tangente.pdf | Bin 0 -> 17811 bytes .../4B_equation_tangente.tex | 32 + .../exercises.tex | 743 ++++++++++++++++++ .../index.rst | 88 +++ .../plan_de_travail.pdf | Bin 0 -> 49660 bytes .../plan_de_travail.tex | 61 ++ .../solutions.pdf | Bin 0 -> 38962 bytes .../solutions.tex | 28 + 15 files changed, 1215 insertions(+) create mode 100644 1G_math/04_Derivation_point_de_vue_local/1B_taux_accroissement.tex create mode 100644 1G_math/04_Derivation_point_de_vue_local/1B_taux_de_variations.pdf create mode 100644 1G_math/04_Derivation_point_de_vue_local/1B_taux_de_variations.tex create mode 100644 1G_math/04_Derivation_point_de_vue_local/2B_tangente.pdf create mode 100644 1G_math/04_Derivation_point_de_vue_local/2B_tangente.tex create mode 100644 1G_math/04_Derivation_point_de_vue_local/3B_nombre_derive.pdf create mode 100644 1G_math/04_Derivation_point_de_vue_local/3B_nombre_derive.tex create mode 100644 1G_math/04_Derivation_point_de_vue_local/4B_equation_tangente.pdf create mode 100644 1G_math/04_Derivation_point_de_vue_local/4B_equation_tangente.tex create mode 100644 1G_math/04_Derivation_point_de_vue_local/exercises.tex create mode 100644 1G_math/04_Derivation_point_de_vue_local/index.rst create mode 100644 1G_math/04_Derivation_point_de_vue_local/plan_de_travail.pdf create mode 100644 1G_math/04_Derivation_point_de_vue_local/plan_de_travail.tex create mode 100644 1G_math/04_Derivation_point_de_vue_local/solutions.pdf create mode 100644 1G_math/04_Derivation_point_de_vue_local/solutions.tex diff --git a/1G_math/04_Derivation_point_de_vue_local/1B_taux_accroissement.tex b/1G_math/04_Derivation_point_de_vue_local/1B_taux_accroissement.tex new file mode 100644 index 0000000..aa5d6b6 --- /dev/null +++ b/1G_math/04_Derivation_point_de_vue_local/1B_taux_accroissement.tex @@ -0,0 +1,71 @@ +\documentclass[a4paper,10pt]{article} +\usepackage{myXsim} +\usepackage{tikz} +\usepackage{pgfplots} + +\author{Benjamin Bertrand} +\title{Nombre dérivé et tangente - Cours} +\date{novembre 2022} + +\pagestyle{empty} + +\begin{document} + +\maketitle + +\section{Taux d'accroissement} + +\begin{definition}[Taux d'accroissement] + \begin{minipage}{0.5\linewidth} + Soit $f$ une fonction, $a$ et $b$ deux nombres. + + \textbf{Le taux d'accroissement} de la fonction $f$ entre $a$ et $b$ se calcule par + \[ + \frac{f(b) - f(a)}{b-a} + \] + + \bigskip + + On interprète ce nombre comme la pente de la droite qui relie les points de la droite d'abscisse $a$ et $b$. Cette droite est appelé \textbf{corde}. + \end{minipage} + \hfill + \begin{minipage}{0.45\linewidth} + \begin{tikzpicture} + \begin{axis}[ + axis lines = center, + grid= both, + xlabel = {$x$}, + xtick distance=1, + ylabel = {$f(x)$}, + ytick distance=1, + ] + \addplot[domain=0:5,samples=20, color=red, very thick]{0.1*x^3 - 1.5*x + 1}; + \end{axis} + \end{tikzpicture} + \end{minipage} +\end{definition} + +\paragraph{Exemples} +\begin{itemize} + \item Calcul du taux d'accroissement entre $x = 1$ et $x = 4$ sur le graphique ci-dessus. + + \vspace{2cm} + + \item Soit $f(t) = 3t^2 + 2$ le taux d'accroissement entre $t=3$ et $t = 10$ est calculé: + \vspace{2cm} +\end{itemize} + +\afaire{Traiter les exemples} + + +\paragraph{Remarques} +\begin{itemize} + \item Le taux d'accroissement est parfois nommé \textbf{taux de variations}. + \item En économie, quand la fonction $f$ représente les coûts, le taux d'accroissement est appelé \textbf{coût marginal}. Il permet de savoir quel sera le coût si l'on décide d'ajouter une unité. + \item En physique, quand la fonction $f$ représente la position, le taux d'accroissement est appelé \textbf{vitesse moyenne}. + \[ + v_{moyenne} = \frac{\Delta p}{\Delta t} = \frac{p(t_2) - p(t_1)}{t_2 - t_1} + \] +\end{itemize} + +\end{document} diff --git a/1G_math/04_Derivation_point_de_vue_local/1B_taux_de_variations.pdf b/1G_math/04_Derivation_point_de_vue_local/1B_taux_de_variations.pdf new file mode 100644 index 0000000000000000000000000000000000000000..ce80d2d22437da022d28f6206d156c4487c2b778 GIT binary patch literal 21116 zcmce;1#l!wk|x}0F|*WSW@ct)DzVgJW@ct)W@ct)sl`%@nHgHl@aygVck|A6Y;Wz% zX3DBE(;_0%!!x`*!oP5L5_w@!8U|W6D3XbVzWJf$nZaQwW_)^l8+~&qZf<-!fSaum zKAoDJzPXWsBR-vylfL7Bl!)5cTjA6Fb^MPO4ES_{f;O)BnqRll<1^6z=T94-PDans z-t4b(hX1VjkKb~*%q$&^?7z;IdX7fIMg}&9Uv#C7tW6wE z@fld~d3o{w*9XeM(cVbU3d(hL4t`3Wd}{mqH}H3$D<~st!~ehL>*}xmvwc1J-^U00 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+\documentclass[a4paper,10pt]{article} +\usepackage{myXsim} +\usepackage{tikz} +\usepackage{pgfplots} + +\author{Benjamin Bertrand} +\title{Dérivation point de vue local- Cours} +\date{octobre 2025} + +\pagestyle{empty} + +\begin{document} + +\maketitle + +\section{Taux de variation} + +\begin{definition}[Taux de variation] + \begin{minipage}{0.5\linewidth} + Soit $f$ une fonction, $a$ et $b$ deux nombres. + + \textbf{Le taux de variation} de la fonction $f$ entre $a$ et $b$ se calcule par + \[ + \frac{f(b) - f(a)}{b-a} + \] + + \bigskip + + On interprète ce nombre comme la pente de la droite qui relie les points de la droite d'abscisse $a$ et $b$. Cette droite est appelé \textbf{corde}. + + Pour exprimer le taux de variation d'une quantité $y$ par rapport à une quantité $x$, on peut utiliser la notation + \[ + \frac{\Delta y}{\Delta x} + \] + \end{minipage} + \hfill + \begin{minipage}{0.45\linewidth} + \begin{tikzpicture} + \begin{axis}[ + axis lines = center, + grid= both, + xlabel = {$x$}, + xtick distance=1, + ylabel = {$f(x)$}, + ytick distance=1, + ] + \addplot[domain=0:5,samples=20, color=red, very thick]{0.1*x^3 - 1.5*x + 1}; + \end{axis} + \end{tikzpicture} + \end{minipage} +\end{definition} + +\paragraph{Exemples}~ +\begin{multicols}{2} + \begin{itemize} + \item Calcul du taux de variation entre $x = 1$ et $x = 4$ sur le graphique ci-dessus. + + \vspace{2cm} + + \item Soit $f(t) = 3t^2 + 2$ le taux de variation entre $t=3$ et $t = 10$ est calculé: + \vspace{2cm} +\end{itemize} +\end{multicols} + + +\afaire{Traiter les exemples} + + +\paragraph{Remarques} +\begin{itemize} + \item En géométrie, quand la fonction $f$ représente une courbe, le taux de variation représente la \textbf{pente} de la corde. + \item En économie, quand la fonction $f$ représente les coûts, le taux de variation est appelé \textbf{coût marginal}. Il permet de savoir quel sera le coût si l'on décide d'ajouter une unité. + \item En physique, quand la fonction $f$ représente la position, le taux de variation est appelé \textbf{vitesse moyenne}. + \[ + v_{moyenne} = \frac{\Delta p}{\Delta t} = \frac{p(t_2) - p(t_1)}{t_2 - t_1} + \] +\end{itemize} + +\end{document} diff --git a/1G_math/04_Derivation_point_de_vue_local/2B_tangente.pdf b/1G_math/04_Derivation_point_de_vue_local/2B_tangente.pdf new file mode 100644 index 0000000000000000000000000000000000000000..1ba77aa4b96cc73e02a6d4e7de84f2248e898ac6 GIT binary patch literal 20150 zcmce;1#o1$k|u0sW@cu_HZwCbGjp4v?Y7-!X1mSI%*@Qp%*^)s&iwP=yD@j)zL?ne zDyq(@kW`YglqZ$nCn-tgg+*x@>DZx2$L71|`j@8q2B2677zpePEup@BBcNCIus0!~ zSC=!iG%<1}pjUD+bpDSNQ9DN)0{TC$|51XGfL>70&YeK(>!%C^j12$z(IKFhF>rRY z_#>V1KXd-$`jxI|;$-LIXk_9_*IpKwX=!i*WKE{*+kgH$jq>@{^&pZSIz%EIXM1h za&Z1z6D#ANM*dF|f377FrQHnM@mT5ZyEVLm-esPh ze!&*4HI6YvI8=;C4({o9P3p^sUoT}>?D&WS+xuSfL=l!pL5CDh|EckG^Xxw+e+rCnB}X&EoPWO4tUu7YKl-?z zh0mE3d@HrJMlo>@Di>1dzaRP3Ga8-jkt_b z-sM|!Tf&RNQ50n00HTm6-^c0bCU*peS$#;CDx}FNbBXaMs`Jg;{`RB>qoiWNrj>BXZ z6GCD4lGIVI08^P~$q{)80ht^idBU4nxRNj@Xqlc1O#fMT(C1o2i2J4hxM}4Bg+6$P zA5|SIv+*wFpmH>KaRR-_-o&(VGci7&+J287p+i&oP#ZL#S`c0lzNaBCsG6kXbert> zq)F0_6fdhc(F-){P5`7{03WU4kl#Pm@ji|)hSDj?5){c~FbIj<(b5xBi%c+-dw)S0 zATF6Yv_c+uAh@`Lw$0oHQyS&lO$`;92!v&0$3LM6K+GSp%Ny;oO2l`W^w{w*IL#Lf zS+mPB0pvmX!vJP{;{pyt1JlR6!@6I!QgkkTlS<;N9Ic$jI1M)UB(fzu2-pp1n50MT z)adVYOFd^<(O)ig!5_U%wfXkUHLm>&b7GL6sHXDw=lkK}<>(#=JdjULD^-6G%%~;G z(G`E+WRFBogDNpC0&4|dKSu{tsT4)$`4nHMLb-htY|{8dv0mLRs= znIZQhKJaaN%s58HvIamlYbW-bkcOn%dlS$5ozWe`LBm8IGfkKOBo{NbRdg#EIgM5E zZnx;mAQN&>spIj&{ofMC}8_MB5SDh^h zKaN;(t$4`TdUc7{299nmSujzgtr?i#6-Xbl_(nKy*%@X$ZXjqKN<$W7s)0&vH9m4_ zY|J?Cp;n!Rsr>EhlK-*93{ z$s(}9*Bu>ZK%0#CDqmso$=gj5C*Rh$QNvQc3YB;g3Ddc3jH4E^-tEizL~$vfgi!)@bp0&g zNrJoa!x2TX6Ph~tMpF?N#FrwT8MWBwGA4*^%%PM@6&S`m9-EUGOB3|Yyp!JB#JQbR z)Gayg$l`K@QAU}vhgCcrF-MRYEoiYvoY_#td_XYYP+tB8vGgnL*Yxi_oERvEU69c> zK^p8Bs=6z>t-A8Tny4`~`|db@MhlEjO&1R9fNn}T->?rV8AH#F)c zohqiB!sY&ew09;d-@d%xOI6nZ-=Nyi8=c77H9)CB5h`?o2_2Wi6{|NQw@VhW){V$n zIz4Hg6#$BvhJJ7m&q5G*SdmHN)Wbeq#C_fZtczFReHML|cmD0JtikIf)e)duHL3%d zwV#Ga{>*e}W#lc#n^R5O`?_}<%Q<)dIFdAN_S`4h0qGD+@8l!Z0f5YgCwEZho#y9^ zS0XJAL^Rk$s4qajB*M=L%|Sc?WV*08@R}J^%75jgZr+cgX15lgNM!X(tdqjZq^Ta1 z`Fr<-T4>R@>zK1BUjEpa+QuY_Fv#^K+7y9Q!;LN=vn%;lWL8SF8>lw*?JJTX3`yyg zgtOzFC?by&g@l^WtLVCJZ;)Qx4Ixm#b(tU8B?{d(9@iA2jb7EjZHh7AD|6c-9jy-au;+P=KK}PfOb5M= zf#(#jfZqO=D_ua=c*+aiwqNVt2_r$Qjom=b)cF$RiPk+j9W?vL+HXYGh%$KNE27up z${D?>u$cfS&OJ?>=F)aA6X=Z4pT}~%BF-{{BJPzSBg9l*Y+qr>$duhAldHpn9^`vO zUFCWPUR>zR*ajJ2#UOkAdmF1C1AF<6t^JYwCy1}_G(-0rCkHQmqFoJtoBAj%0vczfYc1+U|xn|Gls%PgB1=ZSXTsOI)9d%2YWB z(vl5G28Gt&VOJ#m=ZKws#0~b@YP6-5K#CNBvLquc3Q`W0E=W*vtGJOz&wSX*r{K5v zE6F@yZu=ts1V7PRuS7v^E`$QHog(0Y~T znf8XFI;*A#2V;D;ptHkeP%7m5bJs9W-XHzuyF9x*O z#Zbe@GfhCf4eUi7^NN(5awBXlGU|jZRaYwh36GMN#0Doc7CLmgYRH|6twDg`Ajh!7xQj~t_e56I^ zn14a#)zxJwB*o&1#5vy7W=A_b>&Xc7imj1qaW^gQ*s%a<)m1uXvGBw;>LeAbeqp89 zWu23wqxCd=LQp?F@$R*vp>p7x7x_+Cm=0tVZ=ENh`|NWTb|PaAHwY5m+}vO}@3Yr* zbF+1XTXA#u5v@)K_j*S{!pKwAOkQ!l(tOclmO@K*4g4#t!kZ3PXCx)=v^Ml~gbL1* zrmcf+E8bz;Mcc0kzwoyjG6zW>D2)q?ET`V&6D_iqw99vBl{3Lp)0X=k?)sSZix=wh zXK!#xn+3agLr=}kA=l+xCYJtre89eNZ|+ZZ0ydDkXQ{#!rNy8eFx}vTrX?>$EY{$} zW(tvG7o}TuOO3UJ%WR!$i^6vR!BeqU2h*c$EjzPG3@MAv-!V6o#2+Vc9>R`MeH^A? zSdg(|J+yGyS4W@^fju8M_B>8-(!{|4~ z`q9mm+j4)7xdg_x0EL-uPFH3NJfdo1ek`PHOp>uiLm_rC9`@-zw+QDo!3w4lyi=H2(o*!C+ z6BYlh!KTrUGUI5U)sr|2<%KuN# z2}uZx+Sxk)!ObxU{bO>C^{an>vTJ{7)mY!aK;PIH1^U(j%GGWDf$mC%W_Su#Qigt9 zf(B^V{X_mKXK9@HQ7Cvjgk~3>vjG2QjE_X!N9u-Q^3dQJo&GjXAlNepW)NLf0}O0H z)dzu7KtA{$I|?a!cyBTn22U@7 z(P5e)wJCAH+q)V5Z#!e6571M1KmCA|0eYOIhvg*ir6tCxhiQt9_4W1NM_^#E2=op0 z4UO+s@|W5l2h-Yr@hh6pJF$;OMP>o|axFc4JX5T^8_l@2ZpLZ&Qor#MysTfm8be(6 zN3s9j)N}p>b5I@dCvEwZHr$7%=e_G~KJz>w95{yW@b&2W_U_qsWEWw@wz=Gy=9Bbh zTP0lK1y`EYnvrXTE=HbK(Yt>+ab}5yHxA>k`{K{_TS6;nfH!!z}wVsOOl?_6CC6p=#i*X3^lKmuy_F_SC zvm0kVpN`+*)F~#D!r&%?zj}#@y!7hP#O6SIewoE!0|Y;bdx1Mj{`e@D-q#BaK;Cpq z`8VqL2j})*bkaXkc4k(Vzg0W$SUsyA29*BITgnSO5bvV$uqp{e)6!h)kM6LudNMWG zC6c^jq0aZ%GOAxSg3yhJ>3d1D54x@%R}*`Xfx`>|JgPMdD1pX06f>v!u?wV7wF1UN z7nZDn8Hp@n!0#kd%-!Q%smua)ti7$Jt94z_=yKt!;4k2BaFKf9Z$_A2B`yXP-FWo( zKCOyIV*3$IzQWndXtPo$AW)O7tYqP-MhXW{w!p)~8uR7ee82pgf3+6?Q0J;Q`? z(-;_q1sQK~i47#^6KNy*DWSC_4lp_EDQcqZneO~O<)%f=P5hm4Q(QNi#n@a}KZm^F zU?yD+R>un{5Cz??erY766}0-B?MzVwRcDEHRw*-;DGMl|?*x!EfC@YPRC`E8MmdYx zD3CS#PFv?9r>1l&t4BV|;-U2-T-sb#x&=R`43l9~S8k*a?s=%*;*`^hGoy z(vuVlQ{oHC3``6Ri>yoxbyEuV6BHvfbi?#R;wtvCW73mS(zHOu(^A0>I5T02O7}Bi zEG|tW%u-A!0gB;D;UkJlC~G&jk|=LTPD-fOZO~_TO;hCzbs(As#wEF1vB^m(skvK; 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+\setcounter{section}{2} +\section{Nombre dérivé} + +\begin{definition}[Fonction dérivable et nombre dérivé] + Soit $f$ une fonction définie sur l'intervalle $I$, et $a$ un réel appartenant à $I$. + + \bigskip + $f$ est \textbf{dérivable} si $\dfrac{f(x) - f(a)}{x-a}$ le taux de variation de $f$ entre $a$ et $x$ se rapproche d'un certain nombre quand $x$ se rapproche de $a$ sans y être égal. + + \bigskip + Ce nombre est appelé \textbf{nombre dérivé de $f$ en $a$} et on le note $f'(a)$. +\end{definition} + +\begin{definition}[autre formulation] + On reprend les hypothèses de la définition précédente et on note $h = x - a$ l'écart entre $a$ et $x$. + + \bigskip + $f$ est \textbf{dérivable} si $\dfrac{f(a+h) - f(a)}{h}$ le taux de variation de $f$ entre $a$ et $a+h$ se rapproche d'un certain nombre quand $h$ tend vers 0. + + \bigskip + Ce nombre est appelé \textbf{nombre dérivé de $f$ en $a$} et on le note $f'(a)$. + +\end{definition} + +\paragraph{Exemple:} +\begin{itemize} + \item Calcul du nombre dérivé de $f(x) = 3x^2$ en $a=3$ + \afaire{} + \item Calcul du nombre dérivé de $f(x) = \frac{1}{x}$ en $a=1$ + \afaire{} +\end{itemize} + +\paragraph{Remarque}: le concept de dérivé a été construit en meme temps par deux mathématiciens au XVII siècle: Isaac Newton et Gottfried Wilhelm Leibniz. Newton utilisait une notation proche de celle des définition précédente ($\dot{f}(a)$). Tandis que Leibniz utilisait une autre notation encore largement utilisée en physique pour désigner le nombre dérivé de $f$ en $a$: +\[ + \frac{df}{dx}(a) +\] + +\paragraph{Remarque}: +\begin{itemize} + \item En géométrie, quand la fonction $f$ représente une courbe $\mathcal{C}_f$, le nombre dérivé en $a$ est le \textbf{coefficient directeur} de la tangente à la courbe au point $A(a, f(a))$. + \item En physique, quand la fonction $f$ représente la position, le nombre dérivé en $a$ est la vitesse instantanée au moment $a$. +\end{itemize} + +\end{document} diff --git a/1G_math/04_Derivation_point_de_vue_local/4B_equation_tangente.pdf b/1G_math/04_Derivation_point_de_vue_local/4B_equation_tangente.pdf new file mode 100644 index 0000000000000000000000000000000000000000..dcbfc947779a73de33beb63af65673673d091a49 GIT binary patch literal 17811 zcmch<1#lcqvMwrSmMkWVnVFfH(PCy8FXgVzi8O>@cJg^SwU?7iR{BU|0wl2<;3lVR(25=~dkA zO$g~Vvs%e=8BQbF?9(|GocP4MswGAt5_ALhX-R83-8}{{GV?q?a{t 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@@ -0,0 +1,32 @@ +\documentclass[a4paper,10pt]{article} +\usepackage{myXsim} +\usepackage{tikz} +\usepackage{pgfplots} + +\author{Benjamin Bertrand} +\title{Dérivation point de vue local- Cours} +\date{octobre 2025} + +\pagestyle{empty} + +\begin{document} + +\maketitle + +\setcounter{section}{3} +\section{Equation de la tangente} + +\begin{propriete}[Equation de la tangente] + Soit $f$ une fonction dérivable en $a$, alors une équation de la tangente à la courbe représentative de $f$ au point d'abscisse $A$ est + \[ + y = f'(a)(x-a) + f(a) + \] +\end{propriete} + +\paragraph{Exemple}: +Soit $f(x) = 3x^3$, dans un exemple précédent on avait déterminé que $f'(3) = 18$. + +Déterminer l'équation de la tangente à $\mathcal{C}_f$ au point d'abscisse $3$. +\afaire{} + +\end{document} diff --git a/1G_math/04_Derivation_point_de_vue_local/exercises.tex b/1G_math/04_Derivation_point_de_vue_local/exercises.tex new file mode 100644 index 0000000..6fe7c64 --- /dev/null +++ b/1G_math/04_Derivation_point_de_vue_local/exercises.tex @@ -0,0 +1,743 @@ +\begin{exercise}[subtitle={Vitesse moyenne d'une balle}, step={1}, origin={ma tête}, topics={ Nombre dérivé et tangente }, tags={ Dérivation }, mode={\searchMode}] + \noindent + \begin{minipage}{0.6\linewidth} + On lance une balle et on décrit la hauteur ($h$ en m) en fonction du temps ($t$ en secondes) dans le graphique ci-contre + \begin{enumerate} + \item Quelle est la hauteur de la balle après 5 s ? + \item Calculer la vitesse moyenne verticale entre $t=0$ et $t=4$. + \item Calculer la vitesse moyenne verticale entre $t=2$ et $t=10$. + \item Calculer la vitesse moyenne verticale entre $t=10$ et $t=16$. + \item Comment peut on déterminer graphiquement et sans calculs le signe de la vitesse moyenne? + \end{enumerate} + \end{minipage} + \hfill + \begin{minipage}{0.4\linewidth} + \begin{tikzpicture} + \begin{axis}[ + axis lines = center, + grid = both, + xlabel = {$t$ en s}, + xtick distance=2, + ylabel = {$h$ en m}, + ytick distance=1, + ] + \addplot[domain=0:20,samples=20, color=red, very thick]{-0.1*x^2+2*x}; + \end{axis} + \end{tikzpicture} + \end{minipage} +\end{exercise} + +\begin{solution} + \begin{enumerate} + \item La hauteur de la balle après 5 s est $h(5) = -0,1 \times 5^2 + 2 \times 5 = -2,5 + 10 = 7,5$ m. + \item La vitesse moyenne verticale entre $t=0$ et $t=4$ est : + \[ + \frac{h(4) - h(0)}{4 - 0} = \frac{(-0,1 \times 16 + 8) - 0}{4} = \frac{6,4}{4} = 1,6 \text{ m/s} + \] + \item La vitesse moyenne verticale entre $t=2$ et $t=10$ est : + \[ + \frac{h(10) - h(2)}{10 - 2} = \frac{(-0,1 \times 100 + 20) - (-0,1 \times 4 + 4)}{8} = \frac{10 - 3,6}{8} = \frac{6,4}{8} = 0,8 \text{ m/s} + \] + \item La vitesse moyenne verticale entre $t=10$ et $t=16$ est : + \[ + \frac{h(16) - h(10)}{16 - 10} = \frac{(-0,1 \times 256 + 32) - 10}{6} = \frac{6,4 - 10}{6} = \frac{-3,6}{6} = -0,6 \text{ m/s} + \] + \item Graphiquement, le signe de la vitesse moyenne correspond au signe du coefficient directeur de la droite qui relie les deux points considérés : + \begin{itemize} + \item Si la droite monte (de gauche à droite), la vitesse moyenne est positive. + \item Si la droite descend (de gauche à droite), la vitesse moyenne est négative. + \item Si la droite est horizontale, la vitesse moyenne est nulle. + \end{itemize} + \end{enumerate} +\end{solution} + +\begin{exercise}[subtitle={Résultats d'une entreprise}, step={1}, origin={ma tête}, topics={ Nombre dérivé et tangente }, tags={ Dérivation }, mode={\searchMode}] + On souhaite évaluer la situation financière d'une entreprise. Pour cela, nous avons les chiffres d'affaires de quelques années + + \begin{center} + \begin{tabular}{|p{7cm}|*{5}{c|}} + \hline + Année & 1980 & 1995 & 2000 & 2008 & 2020 \\ + \hline + Chiffre d'affaires (en milliers d'euros) & 10 & 18 & 29 & 45 & 50 \\ + \hline + \end{tabular} + \end{center} + + \begin{enumerate} + \item Tracer un repère et y placer les points pour représenter graphiquement le tableau. + \item Sur quel période, la progression du chiffre d'affaires a été le plus rapide ? Proposez une réponse grace à la lecture graphique. + \item Traduire votre méthode graphique en calcul pour proposer un classement des périodes en fonction de la "vitesse de progression" rigoureux. + \end{enumerate} +\end{exercise} + +\begin{annexe} + \begin{tabular}{|p{4cm}|*{4}{p{3cm}|}} + \hline + Période & 1980-1995 & 1995-2000 & 2000-2008 & 2008-2020 \\ + \hline + Écart horizontal & & & & \\ + \hline + Écart vertical & & & & \\ + \hline + Rapport (vertical sur horizontal) & & & & \\ + \hline + \end{tabular} +\end{annexe} + +\begin{solution} + \begin{enumerate} + \item Graphique : placer les points $(1980, 10)$, $(1995, 18)$, $(2000, 29)$, $(2008, 45)$ et $(2020, 50)$ dans un repère. + \item Graphiquement, on observe que la progression semble la plus rapide entre 1995 et 2000 (la pente de la droite reliant ces deux points est la plus importante). + \item Calcul des vitesses de progression (taux de variation) : + \begin{itemize} + \item 1980-1995 : $\frac{18 - 10}{1995 - 1980} = \frac{8}{15} \approx 0,53$ milliers d'euros/an + \item 1995-2000 : $\frac{29 - 18}{2000 - 1995} = \frac{11}{5} = 2,2$ milliers d'euros/an + \item 2000-2008 : $\frac{45 - 29}{2008 - 2000} = \frac{16}{8} = 2$ milliers d'euros/an + \item 2008-2020 : $\frac{50 - 45}{2020 - 2008} = \frac{5}{12} \approx 0,42$ milliers d'euros/an + \end{itemize} + + Classement des périodes par vitesse de progression décroissante : + \begin{enumerate} + \item 1995-2000 : $2,2$ milliers d'euros/an + \item 2000-2008 : $2$ milliers d'euros/an + \item 1980-1995 : $\approx 0,53$ milliers d'euros/an + \item 2008-2020 : $\approx 0,42$ milliers d'euros/an + \end{enumerate} + \end{enumerate} + + Tableau complété : + \begin{center} + \begin{tabular}{|p{4cm}|*{4}{p{3cm}|}} + \hline + Période & 1980-1995 & 1995-2000 & 2000-2008 & 2008-2020 \\ + \hline + Écart horizontal & 15 & 5 & 8 & 12 \\ + \hline + Écart vertical & 8 & 11 & 16 & 5 \\ + \hline + Rapport (vertical sur horizontal) & $\frac{8}{15} \approx 0,53$ & $\frac{11}{5} = 2,2$ & $\frac{16}{8} = 2$ & $\frac{5}{12} \approx 0,42$ \\ + \hline + \end{tabular} + \end{center} +\end{solution} + +\begin{exercise}[subtitle={Taux de variations}, step={1}, origin={ma tête}, topics={ Nombre dérivé et tangente }, tags={ Dérivation }, mode={\trainMode}] + \begin{enumerate} + \item Calculer le taux de variation de la fonction $f(x) = 3x + 1$ entre $x = 1$ et $x = 5$. + \item Calculer le taux de variation de la fonction $g(x) = x^2 + x + 1$ entre $x = 5$ et $x = 10$. + \item Calculer le taux de variation de la fonction $h(x) = \dfrac{1}{x}$ entre $x = -1$ et $x = -3$. + \item Calculer le taux de variation de la fonction $s(x) = \sqrt{x}$ entre $x = 1$ et $x = 0.5$. + \end{enumerate} +\end{exercise} + +\begin{solution} + \begin{enumerate} + \item $\frac{f(5) - f(1)}{5 - 1} = \frac{(3 \times 5 + 1) - (3 \times 1 + 1)}{4} = \frac{16 - 4}{4} = \frac{12}{4} = 3$ + \item $\frac{g(10) - g(5)}{10 - 5} = \frac{(10^2 + 10 + 1) - (5^2 + 5 + 1)}{5} = \frac{111 - 31}{5} = \frac{80}{5} = 16$ + \item $\frac{h(-3) - h(-1)}{-3 - (-1)} = \frac{\frac{1}{-3} - \frac{1}{-1}}{-2} = \frac{-\frac{1}{3} + 1}{-2} = \frac{\frac{2}{3}}{-2} = -\frac{1}{3}$ + \item $\frac{s(0,5) - s(1)}{0,5 - 1} = \frac{\sqrt{0,5} - \sqrt{1}}{-0,5} = \frac{\frac{\sqrt{2}}{2} - 1}{-0,5} = \frac{\frac{\sqrt{2} - 2}{2}}{-0,5} = \frac{\sqrt{2} - 2}{-1} = 2 - \sqrt{2} \approx 0,59$ + \end{enumerate} +\end{solution} + +\begin{exercise}[subtitle={Tangente}, step={2}, origin={ma tête}, topics={ Nombre dérivé et tangente }, tags={ Dérivation }, mode={\searchMode}] + Dans cet exercice, nous allons étudier comment se comporte le taux d'accroissement et la corde quand on fixe un point et que l'on fait se rapprocher l'autre point. L'étude de ce comportement mènera au concept de tangente. + + \begin{enumerate} + \item Pour la fonction $f(x) = (x-3)^2 + 1$ + + \begin{minipage}{0.45\linewidth} + \begin{tikzpicture} + \begin{axis}[ + axis lines = center, + grid= both, + xlabel = {$x$}, + xtick distance=1, + ylabel = {$f(x)$}, + ytick distance=1, + ymin = 0, + ymax = 11, + ] + \addplot[domain=0:5,samples=20, color=red, very thick]{(x-3)^2 + 1}; + \end{axis} + \end{tikzpicture} + \end{minipage} + \begin{minipage}{0.5\linewidth} + \begin{enumerate} + \item On fixe le point $A$ qui est sur la courbe à l'abscisse 1. Repérer ce point sur le graphique. Quelle est la valeur de $f(1)$? + \item Représenter la corde entre $A$ et le point d'abscisse 5. Calculer le taux de variations entre 1 et 5. + \item Représenter la corde entre $A$ et le point d'abscisse 4. Calculer le taux de variations entre 1 et 4. + \item Faire la même chose pour l'abscisse 3, 2 puis 1,5. + \end{enumerate} + \end{minipage} + + \item Pour la fonction $g(x) = \sqrt{x}$ + + \begin{minipage}{0.45\linewidth} + \begin{tikzpicture} + \begin{axis}[ + axis lines = center, + grid= both, + xlabel = {$x$}, + xtick distance=1, + ylabel = {$f(x)$}, + ytick distance=1, + ymin = 0, + ymax = 2.5, + ] + \addplot[domain=0:5,samples=20, color=red, very thick]{sqrt(x)}; + \end{axis} + \end{tikzpicture} + \end{minipage} + \begin{minipage}{0.5\linewidth} + \begin{enumerate} + \item On fixe le point $A$ qui est sur la courbe à l'abscisse 1. Repérer ce point sur le graphique. Quelle est la valeur exacte de $f(1)$? $f(2)$? + \item Représenter la corde entre $A$ et le point d'abscisse 5. Calculer le taux de variations entre 1 et 5. + \item Représenter la corde entre $A$ et le point d'abscisse 4. Calculer le taux de variations entre 1 et 4. + \item En gardant le point $A$ comme de départ départ, tracer les cordes avec des points qui d'approche le plus en plus du point $A$. Déterminer le coefficient directeur de la droit ainsi obtenu. + + \end{enumerate} + \end{minipage} + \end{enumerate} +\end{exercise} + +\begin{solution} + \begin{enumerate} + \item Pour la fonction $f(x) = (x-3)^2 + 1$ + \begin{enumerate} + \item $f(1) = (1-3)^2 + 1 = 4 + 1 = 5$ + \item Taux de variation entre 1 et 5 : $\frac{f(5) - f(1)}{5 - 1} = \frac{((5-3)^2 + 1) - 5}{4} = \frac{5 - 5}{4} = 0$ + \item Taux de variation entre 1 et 4 : $\frac{f(4) - f(1)}{4 - 1} = \frac{((4-3)^2 + 1) - 5}{3} = \frac{2 - 5}{3} = -1$ + \item Taux de variation entre 1 et 3 : $\frac{f(3) - f(1)}{3 - 1} = \frac{((3-3)^2 + 1) - 5}{2} = \frac{1 - 5}{2} = -2$ + + Taux de variation entre 1 et 2 : $\frac{f(2) - f(1)}{2 - 1} = \frac{((2-3)^2 + 1) - 5}{1} = \frac{2 - 5}{1} = -3$ + + Taux de variation entre 1 et 1,5 : $\frac{f(1,5) - f(1)}{1,5 - 1} = \frac{((1,5-3)^2 + 1) - 5}{0,5} = \frac{3,25 - 5}{0,5} = -3,5$ + + On observe que plus le point se rapproche de $A$, plus le taux de variation diminue (en valeur absolue, il augmente). La limite semble être $-4$ (coefficient directeur de la tangente en $x=1$). + \end{enumerate} + + \item Pour la fonction $g(x) = \sqrt{x}$ + \begin{enumerate} + \item $g(1) = \sqrt{1} = 1$ et $g(2) = \sqrt{2}$ + \item Taux de variation entre 1 et 5 : $\frac{g(5) - g(1)}{5 - 1} = \frac{\sqrt{5} - 1}{4} \approx 0,31$ + \item Taux de variation entre 1 et 4 : $\frac{g(4) - g(1)}{4 - 1} = \frac{2 - 1}{3} = \frac{1}{3} \approx 0,33$ + \item En faisant se rapprocher le point de $A$, on trace des cordes dont le coefficient directeur tend vers $\frac{1}{2} = 0,5$ qui est le coefficient directeur de la tangente à la courbe en $x=1$. + \end{enumerate} + \end{enumerate} +\end{solution} + +\begin{exercise}[subtitle={Tracer des tangentes}, step={3}, origin={ma tête}, topics={ Nombre dérivé et tangente }, tags={ Dérivation }, mode={\trainMode}] + Tracer les tangentes aux points marqués sur les graphiques + + \pgfkeys{tikz/.cd} + \tikzset{tangent/.style={black,thick}, + tangent at/.style={postaction={decorate,decoration={markings, + mark=at position #1 with {\fill[tangent] (axis direction cs:0,0) circle (2pt);} + }}}, + } + + \begin{minipage}{0.5\linewidth} + \begin{tikzpicture} + \begin{axis}[ + axis lines = center, + grid= both, + xlabel = {$x$}, + xtick distance=1, + ylabel = {$f(x)$}, + ytick distance=1, + ymin = 0, + ymax = 11, + ] + \addplot[tangent at/.list={0.29,0.645,0.795},domain=0:5,samples=20, color=red, very thick]{(x-3)^2 + 1}; + \end{axis} + \end{tikzpicture} + \end{minipage} + \begin{minipage}{0.5\linewidth} + \begin{tikzpicture} + \begin{axis}[ + axis lines = center, + grid= both, + xlabel = {$x$}, + xtick distance=1, + ylabel = {$f(x)$}, + ytick distance=1, + ] + \addplot[domain=-2:2,samples=50, color=red, very thick,tangent at/.list={0.25,0.5,0.865}]{sin(deg(x*pi/2))*5}; + \end{axis} + \end{tikzpicture} + \end{minipage} +\end{exercise} + +\begin{solution} + \pgfkeys{tikz/.cd, + tangent length/.store in=\TangentLength, + tangent length=30mm + } + \tikzset{tangent/.style={black,thick}, + tangent at/.style={postaction={decorate,decoration={markings, + mark=at position #1 with {\draw[tangent] (axis direction cs:-\TangentLength,0) -- (axis direction cs:\TangentLength,0); + \fill[tangent] (axis direction cs:0,0) circle (2pt);}}}}, + } + \begin{minipage}{0.5\linewidth} + \begin{tikzpicture} + \begin{axis}[ + axis lines = center, + grid= both, + xlabel = {$x$}, + xtick distance=1, + ylabel = {$f(x)$}, + ytick distance=1, + ymin = 0, + ymax = 11, + ] + \addplot[tangent at/.list={0.29,0.64,0.795},domain=0:5,samples=20, color=red, very thick]{(x-3)^2 + 1}; + \end{axis} + \end{tikzpicture} + \end{minipage} + \begin{minipage}{0.5\linewidth} + \begin{tikzpicture} + \begin{axis}[ + axis lines = center, + grid= both, + xlabel = {$x$}, + xtick distance=1, + ylabel = {$f(x)$}, + ytick distance=1, + ymin=-6, ymax=6, + ] + \addplot[domain=-2:2,samples=50, color=red, very thick,tangent at/.list={0.24715,0.5,0.865}]{sin(deg(x*pi/2))*5}; + \end{axis} + \end{tikzpicture} + \end{minipage} +\end{solution} + +\begin{exercise}[subtitle={Tracer une courbe}, step={3}, origin={ma tête}, topics={ Nombre dérivé et tangente }, tags={ Dérivation }, mode={\trainMode}] + \begin{multicols}{2} + \begin{enumerate} + \item Tracer une courbe passant par les points. + + \begin{tikzpicture}[yscale=1.2] + \begin{axis}[ + axis lines = center, + grid= both, + xlabel = {$x$}, + xtick distance=1, + ylabel = {$f(x)$}, + ytick distance=1, + ymin = -6, + ymax = 6, + ] + \addplot [black, mark=*, very thick, only marks] coordinates {(-2,-3) (-1,-5) (0,0) (1.5,5) (2,2)}; + \end{axis} + \end{tikzpicture} + \columnbreak + \item Tracer une courbe passant par les points en respectant les tangentes. + + \begin{tikzpicture}[yscale=1.2] + \begin{axis}[ + axis lines = center, + grid= both, + xlabel = {$x$}, + xtick distance=1, + ylabel = {$f(x)$}, + ytick distance=1, + ymin = -6, + ymax = 6, + ] + \addplot [black, mark=*, very thick, only marks] coordinates {(-2,-3) (-1,-5) (0,0) (1.5,5) (2,2)}; + \addplot [mark=, very thick] coordinates {(-2,-3) (-1.8, -3.5)}; + \addplot [mark=, very thick] coordinates {(-1.2,-5) (-0.8, -5)}; + \addplot [mark=, very thick] coordinates {(-0.2,0) (0.2, 0)}; + \addplot [mark=, very thick] coordinates {(1.3, 4.8) (1.7, 5.2)}; + \addplot [mark=, very thick] coordinates {(1.8, 2) (2, 2)}; + \end{axis} + \end{tikzpicture} + \end{enumerate} + \end{multicols} +\end{exercise} + +\begin{solution} + \begin{multicols}{2} + \begin{enumerate} + \item + + \begin{tikzpicture} + \begin{axis}[ + axis lines = center, + grid= both, + xlabel = {$x$}, + xtick distance=1, + ylabel = {$f(x)$}, + ytick distance=1, + ymin = -6, + ymax = 6, + ] + \addplot [black, mark=*, very thick] coordinates {(-2,-3) (-1,-5) (0,0) (1.5,5) (2,2)}; + \end{axis} + \end{tikzpicture} + + \item + + \begin{tikzpicture} + \begin{axis}[ + axis lines = center, + grid= both, + xlabel = {$x$}, + xtick distance=1, + ylabel = {$f(x)$}, + ytick distance=1, + ymin = -6, + ymax = 6, + ] + \addplot [black, mark=*, very thick, only marks] coordinates {(-2,-3) (-1,-5) (0,0) (1.5,5) (2,2)}; + \addplot [mark=, very thick] coordinates {(-2,-3) (-1.8, -3.5)}; + \addplot [mark=, very thick] coordinates {(-1.2,-5) (-0.8, -5)}; + \addplot [mark=, very thick] coordinates {(-0.2,0) (0.2, 0)}; + \addplot [mark=, very thick] coordinates {(1.3, 4.8) (1.7, 5.2)}; + \addplot [mark=, very thick] coordinates {(1.8, 2) (2, 2)}; + \end{axis} + \end{tikzpicture} + \end{enumerate} + \end{multicols} + \begin{enumerate} + \setcounter{enumi}{2} + \item + Tracer une courbe qui respecte les points et les tangentes représentées dans les graphiques suivants. + + \pgfkeys{tikz/.cd, + tangent length/.store in=\TangentLength, + tangent length=7mm + } + \tikzset{tangent/.style={black,thick}, + tangent at/.style={postaction={decorate,decoration={markings, + mark=at position #1 with {\draw[tangent] (-\TangentLength,0) -- (\TangentLength,0); + \fill[tangent] (0,0) circle (2pt);}}}}, + } + + \begin{tikzpicture}[scale=1] + % Axes + \draw [-latex] (-0.5,0) -- (8,0) node [above] {$x$}; + \draw [-latex] (0,-0.5) -- (0,4) node [right] {$y$}; + % Origin + \node at (0,0) [below left] {$0$}; + % Points + \coordinate (start) at (0,-0.8); + \coordinate (c1) at (3,3); + \coordinate (c2) at (5,1.5); + \coordinate (c3) at (6,4); + \coordinate (end) at (8,2); + % show the points + % \foreach \n in {start,c1,c2,c3,end} \fill [black] (\n) + % circle (2pt) node [below] {}; + % join the coordinates + \draw [tangent at/.list={0.15,0.3,...,1}] (start) to[out=70,in=180] (c1) to[out=0,in=180] + (c2) to[out=0,in=180] (c3) to[out=0,in=150] (end); + \end{tikzpicture} + \hfill + \begin{tikzpicture}[scale=1] + % Axes + \draw [-latex] (-4,0) -- (4,0) node [above] {$x$}; + \draw [-latex] (0,-3) -- (0,3) node [right] {$y$}; + % Origin + \node at (0,0) [below left] {$0$}; + % Points + \coordinate (start) at (-4,-1); + \coordinate (c1) at (-2,3); + \coordinate (c2) at (0,1); + \coordinate (c3) at (2,-2); + \coordinate (end) at (4,0); + % show the points + % \foreach \n in {start,c1,c2,c3,end} \fill [black] (\n) + % circle (2pt) node [below] {}; + % join the coordinates + \draw [tangent at/.list={0.2,0.4,...,1}] (start) to[out=70,in=180] (c1) to[out=0,in=180] + (c2) to[out=0,in=180] (c3) to[out=0,in=150] (end); + \end{tikzpicture} + \end{enumerate} +\end{solution} + +\begin{exercise}[subtitle={Calculer une vitesse}, step={4}, origin={ma tête}, topics={ Nombre dérivé et tangente }, tags={ Dérivation }, mode={\trainMode}] +On lance un caillou du haut d'un point. La distance parcourue par le caillou au bout de $t$ secondes avant de toucher le sol est $d(t) = 4,9t^2$ + \begin{enumerate} + \item Exprimer le taux de variations de la fonction $d$ entre $2$ et $2+h$ où $h\neq0$ et $h>-2$. + \item Déterminer la vitesse instantanée du caillou au bout de 2 secondes. + \item En reprenant les deux questions précédentes, déterminer la vitesse instantanée du caillou au bout de 10 secondes. + \end{enumerate} +\end{exercise} + +\begin{solution} + \begin{enumerate} + \item Le taux de variation de $d$ entre 2 et $2+h$ est : + \begin{align*} + \frac{d(2+h) - d(2)}{(2+h) - 2} &= \frac{4,9(2+h)^2 - 4,9 \times 2^2}{h} \\ + &= \frac{4,9(4 + 4h + h^2) - 19,6}{h} \\ + &= \frac{19,6 + 19,6h + 4,9h^2 - 19,6}{h} \\ + &= \frac{19,6h + 4,9h^2}{h} \\ + &= 19,6 + 4,9h + \end{align*} + \item La vitesse instantanée au bout de 2 secondes est obtenue quand $h \to 0$ : + \[ + 19,6 + 4,9h \to 19,6 \text{ m/s} + \] + \item Le taux de variation de $d$ entre 10 et $10+h$ est : + \begin{align*} + \frac{d(10+h) - d(10)}{h} &= \frac{4,9(10+h)^2 - 4,9 \times 10^2}{h} \\ + &= \frac{4,9(100 + 20h + h^2) - 490}{h} \\ + &= \frac{490 + 98h + 4,9h^2 - 490}{h} \\ + &= 98 + 4,9h + \end{align*} + La vitesse instantanée au bout de 10 secondes est obtenue quand $h \to 0$ : + \[ + 98 + 4,9h \to 98 \text{ m/s} + \] + \end{enumerate} +\end{solution} + +\begin{exercise}[subtitle={Calculer un nombre dérivé}, step={4}, origin={ma tête}, topics={ Nombre dérivé et tangente }, tags={ Dérivation }, mode={\trainMode}] + \begin{enumerate} + \item Soit $f(x) = x^2$ + \begin{enumerate} + \item Exprimer le taux de variations de la fonction $f$ entre $1$ et $1+h$ où $h\neq0$ + \item Déterminer le nombre dérivé de $f$ en $1$. + \end{enumerate} + \item Soit $f(x) = 2x^2+x$ + \begin{enumerate} + \item Exprimer le taux de variations de la fonction $f$ entre $1$ et $1+h$ où $h\neq0$ + \item Déterminer le nombre dérivé de $f$ en $1$. + \end{enumerate} + \item (*) Soit $f(x) = \dfrac{1}{x}$ + \begin{enumerate} + \item Exprimer le taux de variations de la fonction $f$ entre $2$ et $2+h$ où $h\neq0$ + \item Déterminer le nombre dérivé de $f$ en $2$. + \end{enumerate} + \item (*) Soit la fonction $f:x\mapsto 2x - 1$ définie sur $\R$. + \begin{enumerate} + \item Démontrer que pour tout réel $a$ et pour tout $h\neq0$, le taux de variation de $f$ entre $a$ et $a+h$ est égal à 2. + \item En déduire la valeur du nombre dérivé $f'(a)$. + \item Représenter graphiquement la fonction $f$ ainsi que la tangente à la courbe représentative de $f$ au point 1. Que penser du résultat de la question précédente? + \end{enumerate} + \end{enumerate} +\end{exercise} + +\begin{solution} + \begin{enumerate} + \item Soit $f(x) = x^2$ + \begin{enumerate} + \item Le taux de variation de $f$ entre 1 et $1+h$ est : + \begin{align*} + \frac{f(1+h) - f(1)}{h} &= \frac{(1+h)^2 - 1^2}{h} \\ + &= \frac{1 + 2h + h^2 - 1}{h} \\ + &= \frac{2h + h^2}{h} \\ + &= 2 + h + \end{align*} + \item Le nombre dérivé de $f$ en 1 est : quand $h \to 0$, $2 + h \to 2$ donc $f'(1) = 2$ + \end{enumerate} + \item Soit $f(x) = 2x^2+x$ + \begin{enumerate} + \item Le taux de variation de $f$ entre 1 et $1+h$ est : + \begin{align*} + \frac{f(1+h) - f(1)}{h} &= \frac{(2(1+h)^2 + (1+h)) - (2 \times 1^2 + 1)}{h} \\ + &= \frac{(2(1 + 2h + h^2) + 1 + h) - 3}{h} \\ + &= \frac{2 + 4h + 2h^2 + 1 + h - 3}{h} \\ + &= \frac{5h + 2h^2}{h} \\ + &= 5 + 2h + \end{align*} + \item Le nombre dérivé de $f$ en 1 est : quand $h \to 0$, $5 + 2h \to 5$ donc $f'(1) = 5$ + \end{enumerate} + \item (*) Soit $f(x) = \dfrac{1}{x}$ + \begin{enumerate} + \item Le taux de variation de $f$ entre 2 et $2+h$ est : + \begin{align*} + \frac{f(2+h) - f(2)}{h} &= \frac{\frac{1}{2+h} - \frac{1}{2}}{h} \\ + &= \frac{\frac{2 - (2+h)}{2(2+h)}}{h} \\ + &= \frac{\frac{-h}{2(2+h)}}{h} \\ + &= \frac{-1}{2(2+h)} + \end{align*} + \item Le nombre dérivé de $f$ en 2 est : quand $h \to 0$, $\frac{-1}{2(2+h)} \to \frac{-1}{4}$ donc $f'(2) = \frac{-1}{4}$ + \end{enumerate} + \item (*) Soit la fonction $f:x\mapsto 2x - 1$ définie sur $\R$. + \begin{enumerate} + \item Pour tout réel $a$ et pour tout $h\neq0$ : + \begin{align*} + \frac{f(a+h) - f(a)}{h} &= \frac{(2(a+h) - 1) - (2a - 1)}{h} \\ + &= \frac{2a + 2h - 1 - 2a + 1}{h} \\ + &= \frac{2h}{h} \\ + &= 2 + \end{align*} + \item Le nombre dérivé : quand $h \to 0$, $2 \to 2$ donc $f'(a) = 2$ pour tout réel $a$. + \item La fonction $f$ est une fonction affine de coefficient directeur 2. La tangente à la courbe en tout point est confondue avec la droite elle-même, donc elle a le même coefficient directeur 2. Cela confirme que $f'(1) = 2$. + \end{enumerate} + \end{enumerate} +\end{solution} + +\begin{exercise}[subtitle={Nombre dérivé graphique et équation tangente}, step={5}, origin={ma tête}, topics={ Nombre dérivé et tangente }, tags={ Dérivation }, mode={\trainMode}] + \begin{center} + \begin{tikzpicture}[x=2.4cm, y=1.2cm] + % Clip pour ne pas dépasser du repère + \clip (-2.5,-6) rectangle (4.5,2); + + % Sub-grille (tous les 0.5) + %\draw[gray!30, very thin] (-2.5,-6) grid[step=0.5] (4.5,2); + + % Grille principale + \draw[gray!60, thin] (-2.5,-6) grid[step=0.5] (4.5,2); + + % Axes + \draw[->, thick] (-2.5,0) -- (4.5,0) node[right] {$x$}; + \draw[->, thick] (0,-6) -- (0,2) node[above] {$y$}; + + % Origine + \node[below right] at (0,0) {$O$}; + \node[below left] at (0,2) {$y$}; + \node[below left] at (4.5,0) {$x$}; + \node[left] at (0,1) {$1$}; + \node[below] at (1,0) {$1$}; + + % Courbe f(x) = -x² + 2x + \draw[very thick, domain=-2:4, samples=100] plot (\x, {-\x*\x + 2*\x}); + + % Point C (sommet) + \filldraw (1,1) circle (2pt) node[above left] {$C$}; + + % Point A + \filldraw (-1,-3) circle (2pt) node[left] {$A$}; + + % Point D + \filldraw (2,0) circle (2pt) node[below left] {$D(2;0)$}; + + \filldraw (3,-3) circle (2pt) node[below left] {$E$}; + % Point F + + % Tangente T₁ au point C(1;1) - horizontale car c'est le sommet + \draw[dashed, domain=-2.5:4.5] plot (\x, 1); + \node[left] at (-1.3,1.3) {$T_1$}; + + % Tangente T₋₁ au point A(-1;-3) + \draw[dashed, domain=-2.5:0.5] plot (\x, {4*\x + 1}); + \node[left] at (-1.5,-5) {$T_{-1}$}; + + % Tangente T₂ au point D(2;0) + \draw[dotted, thick, domain=0:5] plot (\x, {-2*\x + 4}); + \node[right] at (4,-4) {$T_2$}; + + % Tangente T₃ au point F(3;-3) + \draw[dashed, domain=1:4.5] plot (\x, {-4*\x + 9}); + \node[right] at (1.9,1.5) {$T_3 : y = -4x + 9$}; + \filldraw (3,-2) circle (2pt) node[right] {$F(3;-2)$}; + + \end{tikzpicture} + \end{center} + + Sur le graphique ci-dessus, on a représenté la courbe $\mathscr{C}_f$ de la fonction $f$ définie sur $[-2;4]$ par $f(x) = -x^2 + 2x$. + + On admet que $f$ est dérivable en $-1$, $0$, $1$, $2$ et $3$ et on a tracé les tangentes à $\mathscr{C}_f$ : + + \begin{minipage}{0.5\textwidth} + \begin{itemize} + \item $T_1$ au point $C(1;1)$ ; + \item $T_{-1}$ au point $A(-1;-3)$ ; + \end{itemize} + \end{minipage}% + \begin{minipage}{0.5\textwidth} + \begin{itemize} + \item $T_2$ au point $D(2;0)$ ; + \item $T_3$ au point $F(3;-3)$ ; + \end{itemize} + \end{minipage} + + \begin{enumerate} + \item Avec les éléments présents sur le graphique, déterminer les nombres dérivés $f'(1)$, $f'(-1)$, $f'(2)$ et $f'(3)$ puis les équations réduites des tangentes $T_1$, $T_{-1}$, $T_2$ et $T_3$. + \item Soit $h$ un réel non nul, vérifier que le taux de variation de $f$ entre $0$ et $0+h$ pour tout $h\neq0$ est égal à $-h+2$. + \item Faire tendre $h$ vers $0$ et en déduire le nombre dérivé de $f$ en $0$. + \item Représenter graphiquement la tangente $T_0$ à $\mathscr{C}_f$ au point $O(0;0)$ et déterminer son équation réduite. + \end{enumerate} +\end{exercise} + +\begin{solution} + \begin{enumerate} + \item Détermination des nombres dérivés et équations des tangentes : + \begin{itemize} + \item $f'(1) = 0$ (la tangente $T_1$ est horizontale). Équation de $T_1$ : $y = 1$ + \item $f'(-1) = 4$ (coefficient directeur de $T_{-1}$). Équation de $T_{-1}$ : $y = 4(x - (-1)) + (-3) = 4x + 1$ + \item $f'(2) = -2$ (on peut le déterminer à partir de la tangente $T_2$ qui passe par $(2;0)$ et $(0;4)$). Équation de $T_2$ : $y = -2(x - 2) + 0 = -2x + 4$ + \item $f'(3) = -4$ (coefficient directeur donné dans l'équation $T_3 : y = -4x + 9$). Équation de $T_3$ : $y = -4x + 9$ + \end{itemize} + \item Calcul du taux de variation : + \begin{align*} + \frac{f(0+h) - f(0)}{h} &= \frac{f(h) - f(0)}{h} \\ + &= \frac{(-h^2 + 2h) - 0}{h} \\ + &= \frac{-h^2 + 2h}{h} \\ + &= -h + 2 + \end{align*} + \item Le nombre dérivé de $f$ en $0$ est : quand $h \to 0$, $-h + 2 \to 2$ donc $f'(0) = 2$ + \item L'équation de la tangente $T_0$ en $(0;0)$ est : $y = f'(0)(x - 0) + f(0) = 2x + 0 = 2x$ + + La tangente $T_0$ passe par l'origine et a un coefficient directeur de 2. + \end{enumerate} +\end{solution} + +\begin{exercise}[subtitle={Calculer équation tangente}, step={5}, origin={ma tête}, topics={ Nombre dérivé et tangente }, tags={ Dérivation }, mode={\trainMode}] + \begin{enumerate} + \item Soit $f(x) = x^2$ + \begin{enumerate} + \item Calculer $f(2)$ + \item Exprimer le taux de variations de la fonction $f$ entre $2$ et $2+h$ où $h\neq0$ + \item Déterminer $f'(2)$ + \item Déterminer l'équation de la tangente à $f$ en $x=2$. + \end{enumerate} + \item Soit $f(x) = 2x^2+4$ + \begin{enumerate} + \item Exprimer le taux de variations de la fonction $f$ entre $0$ et $0+h$ où $h\neq0$ + \item Déterminer le nombre dérivé de $f$ en $0$. + \item Déterminer l'équation de la tangente à $f$ en $x=0$. + \end{enumerate} + \item (*) Soit $f(x) = \dfrac{1}{x}$ + \begin{enumerate} + \item Exprimer le taux de variations de la fonction $f$ entre $1$ et $1+h$ où $h\neq0$ + \item Déterminer la valeur de $f'(1)$ + \item En déduire l'équation de la tangente à $f$ en $x=1$. + \end{enumerate} + \end{enumerate} +\end{exercise} + +\begin{solution} + \begin{enumerate} + \item Soit $f(x) = x^2$ + \begin{enumerate} + \item $f(2) = 2^2 = 4$ + \item Le taux de variation de $f$ entre 2 et $2+h$ est : + \begin{align*} + \frac{f(2+h) - f(2)}{h} &= \frac{(2+h)^2 - 2^2}{h} \\ + &= \frac{4 + 4h + h^2 - 4}{h} \\ + &= \frac{4h + h^2}{h} \\ + &= 4 + h + \end{align*} + \item Quand $h \to 0$, $4 + h \to 4$ donc $f'(2) = 4$ + \item L'équation de la tangente à $f$ en $x=2$ est : $y = f'(2)(x - 2) + f(2) = 4(x - 2) + 4 = 4x - 4$ + \end{enumerate} + \item Soit $f(x) = 2x^2+4$ + \begin{enumerate} + \item Le taux de variation de $f$ entre 0 et $0+h$ est : + \begin{align*} + \frac{f(0+h) - f(0)}{h} &= \frac{(2h^2 + 4) - 4}{h} \\ + &= \frac{2h^2}{h} \\ + &= 2h + \end{align*} + \item Quand $h \to 0$, $2h \to 0$ donc $f'(0) = 0$ + \item L'équation de la tangente à $f$ en $x=0$ est : $y = f'(0)(x - 0) + f(0) = 0 \cdot x + 4 = 4$ + \end{enumerate} + \item (*) Soit $f(x) = \dfrac{1}{x}$ + \begin{enumerate} + \item Le taux de variation de $f$ entre 1 et $1+h$ est : + \begin{align*} + \frac{f(1+h) - f(1)}{h} &= \frac{\frac{1}{1+h} - 1}{h} \\ + &= \frac{\frac{1 - (1+h)}{1+h}}{h} \\ + &= \frac{\frac{-h}{1+h}}{h} \\ + &= \frac{-1}{1+h} + \end{align*} + \item Quand $h \to 0$, $\frac{-1}{1+h} \to -1$ donc $f'(1) = -1$ + \item L'équation de la tangente à $f$ en $x=1$ est : $y = f'(1)(x - 1) + f(1) = -1(x - 1) + 1 = -x + 2$ + \end{enumerate} + \end{enumerate} +\end{solution} diff --git a/1G_math/04_Derivation_point_de_vue_local/index.rst b/1G_math/04_Derivation_point_de_vue_local/index.rst new file mode 100644 index 0000000..41f279f --- /dev/null +++ b/1G_math/04_Derivation_point_de_vue_local/index.rst @@ -0,0 +1,88 @@ +Dérivation point de vue local +############################# + +:date: 2025-10-13 +:modified: 2025-10-13 +:authors: Benjamin Bertrand +:tags: dérivation, fonction, représentation graphique +:category: 1G_math +:summary: Construction de la notion de dérivée d'un point de vue local: taux de variation, nombre dérivé et tangente + + +Éléments du programme +===================== + +Construction de la notion de nombre dérivé comme limite du taux de variations et représentation à travers la notion de tangente. + +Contenus +-------- + +- Taux de variation. Sécantes à la courbe représentative d’une fonction en un point donné. +- Nombre dérivé d’une fonction en un point, comme limite du taux de variation. Notation ƒ’(a). +- Tangente à la courbe représentative d’une fonction en un point, comme « limite des sécantes ». Pente. Équation : la tangente à la courbe représentative de ƒ au point d’abscisse a est la droite d’équation y = ƒ(a) + ƒ’(a)(x - a). + +Capacités attendues +------------------- + +- Calculer un taux de variation, la pente d’une sécante. +- Interpréter le nombre dérivé en contexte : pente d’une tangente, vitesse instantanée, coût marginal… +- Déterminer graphiquement un nombre dérivé par la pente de la tangente. Construire la tangente en un point à une courbe représentative connaissant le nombre dérivé. +- Déterminer l’équation de la tangente en un point à la courbe représentative d’une fonction. +- À partir de la définition, calculer le nombre dérivé en un point ou la fonction dérivée de la fonction carré, de la fonction inverse. + +Commentaires +------------ + +Progression +=========== + +Plan de travail + +.. image:: ./plan_de_travail.pdf + :height: 200px + :alt: Plan de travail + +Solutions (vérifiées globalement -- à prendre avec esprit critique) + +.. image:: ./solutions.pdf + :height: 200px + :alt: Solutions + +Étape 1: Taux de variations +--------------------------- + +Bilan: + +.. image:: ./1B_taux_de_variations.pdf + :height: 200px + :alt: Taux de variations + +Étape 2: Limite du taux +----------------------- + +Étape 3: Tangente +----------------- + +Bilan: + +.. image:: ./2B_tangente.pdf + :height: 200px + :alt: tangente + +Étape 4: Nombre dérivé +---------------------- + +Bilan: + +.. image:: ./3B_nombre_derive.pdf + :height: 200px + :alt: Nombre dérivé + +Étape 5: Equation de la tangente +-------------------------------- + +Bilan: + +.. image:: ./4B_equation_tangente.pdf + :height: 200px + :alt: équation de la tangente diff --git a/1G_math/04_Derivation_point_de_vue_local/plan_de_travail.pdf b/1G_math/04_Derivation_point_de_vue_local/plan_de_travail.pdf new file mode 100644 index 0000000000000000000000000000000000000000..50f6aa88b166b807ba64c6294cbed19932deda52 GIT binary patch literal 49660 zcmce8W00jmvUb~;Y1{5;+tapf+qP}nwr$(Cr)|6Y>)Cs6+>Q7)?!Vm=@2Pq!vy@R; zS*NnmBLR56rEKtM~i@ghjD>DN_P>guAc-DGmP@J52GzzXZhIlk8(t2iw`VM$B z@{W2A|DXt2+gak#{0;w+gAR{|kI&i#PyM?nEgl{1e|`edC>HdrJkMNr= zXJ~KjXs2&zkH_>)lKZa8)WX5g?mM>7bubh#)VDVHZdbz4%GkjKkB$kCn;Y+cK2Y`! zc80o^P%dlp@KZ8mQ#(Ka;2(h3P=;0p|IPVEfBVn!UGslV4%UC09BltlIq3e5;J;@; z=>96fe{MwR=>L6^{AYEG?%$_e^nZu@KQ|`ye^ub$qh0h2|5-`?n#q5YB+UO=N&YJH zzef_Lzf2U1}shmNSiPGZyj-MI+EJ@g0CcU;P={Yqy7Ztdxegd&LiOC( 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