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Warning: this is an htmlized version!
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% (defun o () (interactive) (find-LATEX "2022-2-C2-VR.tex"))
% (defun u () (interactive) (find-latex-upload-links "2022-2-C2-VS"))
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% (code-eec-LATEX "2022-2-C2-VS")
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% file:///home/edrx/LATEX/2022-2-C2-VS.pdf
% file:///tmp/2022-2-C2-VS.pdf
% file:///tmp/pen/2022-2-C2-VS.pdf
% http://angg.twu.net/LATEX/2022-2-C2-VS.pdf
% (find-LATEX "2019.mk")
% (find-sh0 "cd ~/LUA/; cp -v Pict2e1.lua Pict2e1-1.lua Piecewise1.lua ~/LATEX/")
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% (find-sh0 "cd ~/LUA/; cp -v C2Subst1.lua C2Formulas1.lua ~/LATEX/")
% (find-CN-aula-links "2022-2-C2-VS" "2" "c2m222vs" "c2vs")
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% Video (not yet):
% (find-ssr-links "c2m222vs" "2022-2-C2-VS")
% (code-eevvideo "c2m222vs" "2022-2-C2-VS")
% (code-eevlinksvideo "c2m222vs" "2022-2-C2-VS")
% (find-c2m222vsvideo "0:00")
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%%L dofile "Piecewise1.lua" -- (find-LATEX "Piecewise1.lua")
%%L dofile "QVis1.lua" -- (find-LATEX "QVis1.lua")
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% «defs-T-and-B» (to ".defs-T-and-B")
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\def\T(Total: #1 pts){{\bf(Total: #1)}}
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%
% «title» (to ".title")
% (c2m222vsp 1 "title")
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\thispagestyle{empty}
\begin{center}
\vspace*{1.2cm}
{\bf \Large Cálculo 2 - 2022.2}
\bsk
Prova suplementar (VS)
\bsk
Eduardo Ochs - RCN/PURO/UFF
\url{http://angg.twu.net/2022.2-C2.html}
\end{center}
\newpage
% «links» (to ".links")
% (c2m222p1p 5 "questao-1-gab")
% (c2m222p1a "questao-1-gab")
% (c2m221atisp 50 "exemplo")
% (c2m221atisa "exemplo")
\newpage
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%
% «questao-1» (to ".questao-1")
% 2fT135: (c2m222vsp 2 "questao-1")
% (c2m222vsa "questao-1")
%
%L namedang("EDOVSintro", "", [[
%L \begin{array}{rcl}
%L \ga{[M]} &=& <EDOVSG> \\ \\[-5pt]
%L \ga{[F]} &=& <EDOVSP> \\
%L \end{array}
%L ]])
%L namedang("metodo", "", [[ <EDOVSG> ]])
%L metodo:sa("FOO"):output()
\pu
\scalebox{0.6}{\def\colwidth{9.5cm}\firstcol{
{\bf \large Questão 1}
\T(Total: 6.0 pts)
\ssk
A primeira EDO com variáveis separáveis (``EDOVS'') que nós vimos
no curso foi $\frac{dy}{dx}=-\frac{x}{y}$. As soluções
particulares dela eram pedaços das curvas de nível de $x^2+y^2=C$
--- e esses pedaços eram ou semicírculos acima do eixo $x$ ou
semicírculos abaixo do eixo $y$.
\msk
Na P2 eu pus uma questão sobre uma EDOVS um pouco mais complicada
do que essa dos semicírculos, e muitas pessoas fizeram erros de
conta horríveis {\sl que eu acho que foram causados por
desorganização na hora de fazer as contas}... por exemplo,
várias pessoas escreveram ``$H(x)=\sqrt{x}$'' num lugar das contas
e ``$H(x)=\sqrt{x+3}$'' em outro --- o que OBVIAMENTE é um erro
conceitual GRAVÍSSIMO, né, {\sl a menos que você explique em
português direitinho que $H(x)$ vai ser $\sqrt{x}$ em um
contexto e $\sqrt{x+3}$ em outro contexto separado...}
\msk
}\anothercol{
\vspace*{0.85cm}
Seja $(*)$ esta EDOVS:
%
$$\frac{dy}{dx} = \frac{4x^3}{4(y+3)^3} \qquad (*)
$$
Encontre as duas soluções gerais da EDO $(*)$ --- uma ``positiva'' e
outra ``negativa'' ---, encontre as soluções particulares que passam
pelos pontos $(1,-1)$ e $(1,-5)$, e teste tudo.
\msk
{\sl Nesta questão eu vou avaliar principalmente se você sabe usar
direito os truques do anexo da página 4.} Ou seja, não vai bastar
você usar o ``método'' para resolver EDOVSs, que é esse aqui:
%
$$\ga{[M]} \;=\; \scalebox{0.85}{$\ga{FOO}$}$$
}}
% «questao-1-maxima» (to ".questao-1-maxima")
% (c2m222vsp 2 "questao-1-maxima")
% (c2m222vsa "questao-1-maxima")
% (setq eepitch-preprocess-regexp "^")
% (setq eepitch-preprocess-regexp "^%T ")
%
%T * (eepitch-maxima)
%T * (eepitch-kill)
%T * (eepitch-maxima)
%T define(G(x), x^4);
%T define(H(y), (y+3)^4);
%T
%T define(g(x), diff(G(x),x));
%T define(h(y), diff(H(y),y));
%T edo : dydx = g(x) / h(y);
%T
%T eq_a : x = H(y);
%T solve(eq_a, y);
%T
%T getsol(n,C31,x1,y1) := (
%T eq_b : solve(eq_a, y)[n],
%T define(Hinv(x), rhs(eq_b)),
%T eq_c : y = Hinv(G(x) + C3),
%T eq_d : y = Hinv(G(x) + C31),
%T define(f (x), rhs(eq_c)),
%T define(f1 (x), rhs(eq_d))
%T );
%T
%T getsol(4,15, 1,-1);
%T f(x);
%T f1(x);
%T f1(1);
%T
%T getsol(3,15, 1,-5);
%T f(x);
%T f1(x);
%T f1(1);
\newpage
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%
% «questao-2» (to ".questao-2")
% (c2m222vsp 3 "questao-2")
% (c2m222vsa "questao-2")
\scalebox{0.7}{\def\colwidth{9cm}\firstcol{
{\bf \large Questão 2}
\T(Total: 4.0 pts)
\ssk
Calcule a integral abaixo usando pelo menos três mudanças de
variáveis.
%
$$\intx{ \frac{8e^{4x}\ln(e^{4x}+2)}{e^{4x}+2} }
$$
Obs: no curso nós vimos que qualquer integral que pode ser resolvida
por uma sequência de mudanças de variáveis também pode ser resolvida
por uma mudança de variável só, mas aqui é pra usar pelo menos três!
% «questao-2-maxima» (to ".questao-2-maxima")
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%
%T * (eepitch-maxima)
%T * (eepitch-kill)
%T * (eepitch-maxima)
%T b(x) := exp(x);
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%T F : d(c(b(x)));
%T f : diff(F, x);
%T integrate(f, x);
}\anothercol{
}}
\newpage
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% (c2m222p2p 5 "questao-1-gab")
% (c2m222p2a "questao-1-gab")
% «anexo-L» (to ".anexo-L")
\def\anexoL{
A substituição é:
%
$$\ga{[S]} \;=\;
\bmat{
G(x) := x^4 + 5 \\
H(y) := y^2 + 3 \\
g(x) := 4x^3 \\
h(y) := 2y \\
H^{-1}(x) := \sqrt{x-3} \\
}
$$
a) Seja:
%
$$\frac{dy}{dx} = \frac{4x^3}{2y} \qquad (*)$$
b)
%
$\begin{array}[t]{lrcl}
\text{Seja:} & H^{-1}(x) &=& \sqrt{x-3}. \\
\text{Temos:} & H^{-1}(H(y)) &=& \sqrt{H(y)-3} \\
& &=& \sqrt{(y^2+3)-3} \\
& &=& y. \\
\end{array}
$
\msk
c) $\begin{array}[t]{lrcl}
& y &=& H^{-1}(G(x)+C_3) \\
&&=& \sqrt{(G(x)+C_3)-3} \\
&&=& \sqrt{((x^4+5)+C_3)-3} \\
&&=& \sqrt{x^4+2+C_3} \\
\text{Seja:} &
f(x) &=& \sqrt{x^4+2+C_3}. \\
\end{array}
$
}
% «anexo-R» (to ".anexo-R")
\def\anexoR{
d) $\begin{array}[t]{l}
\text{Será que $f(x)$ obedece $(*)$?} \\
\text{Temos }
f'(x) = \frac{2x^3}{\sqrt{x^4 + 2 + C_3}},
\text{ e com isso:}
\\
\\[-5pt]
\left(
f'(x) = \frac{4x^3}{2f(x)}
\right)
\bmat{
f(x) = \sqrt{x^4+2+C_3} \\
f'(x) = \frac{2x^3}{\sqrt{x^4 + 2 + C_3}} \\
}
\\
= \;\;
\left(
\frac{2x^3}{\sqrt{x^4 + 2 + C_3}}
= \frac{4x^3}{2\sqrt{x^4+2+C_3}}
\right)
\qquad \smile \\
\end{array}
$
\bsk
e) $\begin{array}[t]{lrcl}
\text{Se} & f(x_1) &=& y_1, \\
\text{i.e.,} & f(1) &=& 2, \\
\text{então} & f(1) &=& \sqrt{1^4+2+C_3} \\
&&=& \sqrt{3+C_3} \\
&&=& 2 \\
& 2^2 &=& \sqrt{3+C_3}^2 \\
& 4 &=& 3+C_3 \\
& C_3 &=& 1 \\
& f(x) &=& \sqrt{x^4+2+C_3} \\
& &=& \sqrt{x^4+3} \\
\text{Seja:} & f_1(x) &=& \sqrt{x^4+3}. \\
\end{array}
$
\bsk
f) $\begin{array}[t]{lrcl}
\text{Será que} & f_1(x_1) &=& y_1, \\
\text{i.e.,} & f_1(1) &=& 2? \\
& \sqrt{1^4+3} &=& \sqrt{4} \\
&&=& 2 \qquad \smile \\
\end{array}
$
}
% «anexo» (to ".anexo")
\scalebox{0.6}{\def\colwidth{9cm}\firstcol{
\vspace*{-0.5cm}
{\bf Anexo: gabarito de}
{\bf uma questão da P2}
\ssk
\anexoL
}\anothercol{
\anexoR
}}
\GenericWarning{Success:}{Success!!!} % Used by `M-x cv'
\end{document}
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%
% «djvuize» (to ".djvuize")
% (find-LATEXgrep "grep --color -nH --null -e djvuize 2020-1*.tex")
* (eepitch-shell)
* (eepitch-kill)
* (eepitch-shell)
# (find-fline "~/2022.2-C2/")
# (find-fline "~/LATEX/2022-2-C2/")
# (find-fline "~/bin/djvuize")
cd /tmp/
for i in *.jpg; do echo f $(basename $i .jpg); done
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f () { rm -fv $1.png $1.pdf; djvuize WHITEBOARDOPTS="-m 0.25" $1.pdf; xpdf $1.pdf }
f () { cp -fv $1.png $1.pdf ~/2022.2-C2/
cp -fv $1.pdf ~/LATEX/2022-2-C2/
cat <<%%%
% (find-latexscan-links "C2" "$1")
%%%
}
f 20201213_area_em_funcao_de_theta
f 20201213_area_em_funcao_de_x
f 20201213_area_fatias_pizza
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%
% <make>
* (eepitch-shell)
* (eepitch-kill)
* (eepitch-shell)
# (find-LATEXfile "2019planar-has-1.mk")
make -f 2019.mk STEM=2022-2-C2-VS veryclean
make -f 2019.mk STEM=2022-2-C2-VS pdf
% Local Variables:
% coding: utf-8-unix
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% ee-tla: "c2m222vs"
% End: