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Warning: this is an htmlized version!
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% (find-es "tex" "geometry")
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%L dofile "Piecewise1.lua" -- (find-LATEX "Piecewise1.lua")
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\sa{[M]}{\CFname{M}{}}
\sa{[F]}{\CFname{F}{}}
\sa{[S]}{\CFname{S}{}}
% (find-LATEXgrep "grep --color=auto -nH --null -e mname 202{1,2}*.tex")
\def\sumiN#1{\sum_{i=1}^N #1 (b_i-a_i)}
\def\mname#1{\text{[#1]}}
\def\Smile{\emoji{slightly-smiling-face}}
% «defs-T-and-B» (to ".defs-T-and-B")
\long\def\ColorOrange#1{{\color{orange!90!black}#1}}
\def\T(Total: #1 pts){{\bf(Total: #1)}}
\def\T(Total: #1 pts){{\bf(Total: #1 pts)}}
\def\T(Total: #1 pts){\ColorRed{\bf(Total: #1 pts)}}
\def\B (#1 pts){\ColorOrange{\bf(#1 pts)}}
% _____ _ _ _
% |_ _(_) |_| | ___ _ __ __ _ __ _ ___
% | | | | __| |/ _ \ | '_ \ / _` |/ _` |/ _ \
% | | | | |_| | __/ | |_) | (_| | (_| | __/
% |_| |_|\__|_|\___| | .__/ \__,_|\__, |\___|
% |_| |___/
%
% «title» (to ".title")
% (c2m222p2p 1 "title")
% (c2m222p2a "title")
\thispagestyle{empty}
\begin{center}
\vspace*{1.2cm}
{\bf \Large Cálculo 2 - 2022.2}
\bsk
P2 (Segunda prova)
\bsk
Eduardo Ochs - RCN/PURO/UFF
\url{http://angg.twu.net/2022.2-C2.html}
\end{center}
\newpage
% «links» (to ".links")
% (c2m222p2p 2 "links")
% (c2m222p2a "links")
% (c2m222dp2p 2 "links")
% (c2m222dp2a "links")
% «links-edovs» (to ".links-edovs")
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% (c2m222p2a "links-edovs")
% (c2m222edovsp 2 "links")
% (c2m222edovsa "links")
% (c2m221vsbp 5 "questao-4")
% (c2m221vsba "questao-4")
% (find-es "maxima" "separable-2")
% «links-edolcc» (to ".links-edolcc")
% (c2m222edolsp 2 "links")
% (c2m222edolsa "links")
% (c2m222dp2p 3 "somas-de-riemann")
% (c2m222dp2a "somas-de-riemann")
\newpage
% _ _____ ____ _____ ______
% / | | ____| _ \ / _ \ \ / / ___|
% | | | _| | | | | | | \ \ / /\___ \
% | |_ | |___| |_| | |_| |\ V / ___) |
% |_(_) |_____|____/ \___/ \_/ |____/
%
% «questao-1» (to ".questao-1")
% (c2m222p2p 2 "questao-1")
% (c2m222p2a "questao-1")
% «edovs» (to ".edovs")
% (c2m222p2p 2 "edovs")
% (c2m222p2a "edovs")
% (find-es "maxima" "separable-2")
% (find-es "maxima" "2022-2-C2-P2-edovs")
{\bf Questão 1}
%L namedang("EDOVSintro", "", [[
%L \begin{array}{rcl}
%L \ga{[M]} &=& <EDOVSG> \\ \\[-5pt]
%L \ga{[F]} &=& <EDOVSP> \\
%L \end{array}
%L ]])
%L EDOVSintro:sa("FOO"):output()
\pu
\scalebox{0.55}{\def\colwidth{10cm}\firstcol{
\vspace*{-0.4cm}
\T(Total: 6.0 pts)
Lembre que nós vimos que o ``método'' para resolver EDOs com variáveis
separáveis --- ``EDOVSs'' --- pode ser escrito como a demonstração
$\ga{[M]}$ abaixo, e a ``fórmula'' para resolver EDOVSs pode ser
escrita como $\ga{[F]}$:
\bsk
$\ga{FOO}$
}\anothercol{
Quando a gente quer criar exercícios de EDOVSs que sejam fácil de
resolver a gente começa escolhendo $G(x)$ e $H(y)$, não $g(x)$ e
$h(y)$.
Digamos que $G(x)=x^4+5$ e $H(y)=y^2+3$.
\msk
a) \B (0.5 pts) Diga qual é a EDO da forma
$\frac{dy}{dx} = \frac{g(x)}{h(y)}$ associada a esta escolha de
$G(c)$ e $H(y)$. Chame-a de $(*)$. Não esqueça do ``Seja''!
\ssk
b) \B (0.5 pts) Escolha uma função $H^{-1}$ adequada. Defina ela com
um ``Seja'' e verifique que ela obedece o que esperamos dela.
\ssk
c) \B (1.0 pts) Encontre a solução geral da EDO $(*)$. Chame-a de
$f(x)$ e defina ela com um ``Seja''.
\ssk
d) \B (1.5 pts) Verifique que essa função $f(x)$ obedece $(*)$.
\ssk
e) \B (1.0 pts) Encontre uma solução $f_1(x)$ que passe pelo ponto
$(x_1,y_1)=(1,2)$. Defina-a com um ``Seja''.
\ssk
f) \B (1.5 pts) Teste a sua solução $f_1(x)$.
% (find-es "maxima" "2022-2-C2-P2")
}}
\newpage
% ____ _____ ____ ___ _ ____ ____
% |___ \ | ____| _ \ / _ \| | / ___/ ___|___
% __) | | _| | | | | | | | | | | | | / __|
% / __/ _ | |___| |_| | |_| | |__| |__| |___\__ \
% |_____(_) |_____|____/ \___/|_____\____\____|___/
%
% «questao-2» (to ".questao-2")
% (c2m222p2p 3 "questao-2")
% (c2m222p2a "questao-2")
% «edolccs» (to ".edolccs")
% (c2m222p2p 3 "edolccs")
% (c2m222p2a "edolccs")
% (find-es "maxima" "2022-2-C2-P2-edolccs")
{\bf Questão 2}
\scalebox{0.6}{\def\colwidth{9cm}\firstcol{
\vspace*{-0.4cm}
\T(Total: 3.0 pts)
No curso nós vimos um modo de resolver EDOs lineares com coeficientes
constantes --- ``EDOLCCs'' --- no qual a gente traduzia a EDO ``pra
Álgebra Linear'', fatorava uma ``matriz'', e aí encontrava as soluções
básicas dessa EDO e tratava elas como ``vetores''... por exemplo,
%
$$\begin{array}{rcl}
y'' + 5y' + 6y &=& 0 \\
(D^2 + 5D + 6)f &=& 0 \\
(D+2)(D+3)f &=& 0 \\
M &=& (D+2)(D+3) \\
M e^{-2x} &=& 0 \\
M e^{-3x} &=& 0 \\
M e^{-2x} &=& 0 \\
M(42e^{-2x} + 99e^{-3x}) &=& 0 \\
\end{array}
$$
Seja $(*)$ esta EDO:
%
$$y'' + y' - 20y \;=\; 0 \qquad (*)
$$
}\anothercol{
a) \B (0.2 pts) Traduza a EDO $(*)$ para ``Álgebra Linear'' e
fatore-a. Chame essa versão fatorada de $(**)$, e defina-a com um
``Seja''.
\msk
b) \B (0.3 pts) Encontre as duas soluções básicas para a EDO $(*)$.
Chame elas de $f_1$ e $f_2$. Não esqueça o ``Sejam''!
\msk
c) \B (0.5 pts) Encontre a solução geral para a EDO $(*)$ e chame-a
de $f$. Não esqueça o ``Seja''!
\msk
d) \B (2.0 pts) Encontre uma solução $g$ para a EDO $(*)$ que
obedeça $g(0)=7$ e $g'(0)=1$. Defina esta $g$ com um ``seja'' e
verifique que ela realmente obedece $g(0)=7$ e $g'(0)=1$.
% (find-es "maxima" "2022-2-C2-P2")
}}
\newpage
% ___ _ _____
% / _ \ _ _ ___ ___| |_ __ _ ___ |___ /
% | | | | | | |/ _ \/ __| __/ _` |/ _ \ |_ \
% | |_| | |_| | __/\__ \ || (_| | (_) | ___) |
% \__\_\\__,_|\___||___/\__\__,_|\___/ |____/
%
% «questao-3» (to ".questao-3")
% (c2m222p2p 4 "questao-3")
% (c2m222p2a "questao-3")
%L Pict2e.bounds = PictBounds.new(v(0,0), v(7,6))
%L spec = "(0,1)--(1,1)--(2,4)--(3,5)--(4,4)o (4,3)c (4,1)o--(6,3)--(7,3)"
%L pws = PwSpec.from(spec)
%L pws:topict():prethickness("1pt"):pgat("pgatc"):sa("F(x)"):output()
\pu
\unitlength=10pt
{\bf Questão 3}
\scalebox{0.55}{\def\colwidth{10.5cm}\firstcol{
\vspace*{-0.25cm}
\T(Total: 1.5 pts)
Lembre que nós vimos estes tipos de Somas de Riemann,
%
$$\scalebox{0.95}{$
\begin{array}{ccl}
\mname{L} &=& \sumiN {f(a_i)} \\[2pt]
\mname{R} &=& \sumiN {f(b_i)} \\[2pt]
\mname{Trap} &=& \sumiN {\frac{f(a_i) + f(b_i)}{2}} \\[2pt]
\mname{M} &=& \sumiN {f(\frac{a_i+b_i}{2})} \\[2pt]
\mname{min} &=& \sumiN {\min(f(a_i), f(b_i))} \\[2pt]
\mname{max} &=& \sumiN {\max(f(a_i), f(b_i))} \\[2pt]
\mname{inf} &=& \sumiN {\inf(f([a_i,b_i]))} \\[2pt]
\mname{sup} &=& \sumiN {\sup(f([a_i,b_i]))} \\
\end{array}
$}
$$
e vimos que o $\mname{Trap}$ pode ser interpretado tanto como uma soma
de trapézios como como uma soma de retângulos.
\msk
Seja $f(x)$ a função dos gráficos à direita.
Represente graficamente:
\msk
a) $\mname{inf}_{\{1,2,3,4\}}$
b) $\mname{sup}_{\{1,2,3,4\}}$
c) $\mname{M}_{\{1,3,5\}}$
d) $\mname{Trap}_{\{1,3,5\}}$ usando retângulos
e) $\mname{Trap}_{\{1,3,5\}}$ usando trapézios
\msk
Indique claramente qual desenho é a resposta final de cada item e
quais desenhos são rascunhos.
}\anothercol{
\vspace*{0cm}
\def\Fx{\scalebox{1.2}{$\ga{F(x)}$}}
$\begin{matrix}
\Fx & \Fx & \Fx \\ \\[-5pt]
\Fx & \Fx & \Fx \\ \\[-5pt]
\Fx & \Fx & \Fx \\ \\[-5pt]
\Fx & \Fx & \Fx \\
\end{matrix}
$
}}
\newpage
% «questao-1-gab» (to ".questao-1-gab")
% (c2m222p2p 5 "questao-1-gab")
% (c2m222p2a "questao-1-gab")
\scalebox{0.6}{\def\colwidth{9cm}\firstcol{
{\bf Questão 1: gabarito}
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}
$
}\anothercol{
\vspace*{0cm}
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}
$
}}
\newpage
% «questao-2-gab» (to ".questao-2-gab")
% (c2m222p2p 6 "questao-2-gab")
% (c2m222p2a "questao-2-gab")
\scalebox{0.6}{\def\colwidth{9cm}\firstcol{
{\bf Questão 2: gabarito}
\msk
a) Temos: $D^2 + D - 20 = (D+5)(D-4)$.
\phantom{a)}
Seja $(**)$ esta EDO:
%
$$(D+5)(D-4)f \; = \; 0. \qquad (**)
$$
\msk
b) Sejam $f_1(x) = e^{4x}$,
$f_2(x) = e^{-5x}$,
\msk
c) Seja
%
$$\begin{array}{rcl}
f(x) &=& af_1(x) + bf_2(x) \\
&=& ae^{4x} + be^{-5x}. \\
\end{array}
$$
d)
%
$\begin{array}[t]{lrcl}
\text{Digamos que} &
g(x) &=& af_1(x) + bf_2(x) \\
&&=& ae^{4x} + be^{-5x}, \\
& g(0) &=& 7, \\
& g'(0) &=& 1. \\
\text{Então:}
& g(0) &=& ae^0 + be^0, \\
&&=& a + b, \\
& g'(0) &=& a·4e^0 + b·(-5)e^0, \\
&&=& 4a -5b, \\
& a &=& 4, \\
& b &=& 3, \\
& g(x) &=& 4e^{4x} +3e^{-5x}, \\
& g(0) &=& 4 + 3 \;\;=\;\; 7, \qquad \smile \\
& g'(0) &=& 16 - 15 \;\;=\;\; 1, \quad\, \smile. \\
\end{array}
$
}\anothercol{
% «questao-3-gab» (to ".questao-3-gab")
% (c2m222p2p 6 "questao-3-gab")
% (c2m222p2a "questao-3-gab")
{\bf Questão 3: gabarito (sem desenhos)}
\bsk
\def\Item#1{\text{#1) }}
$\begin{array}{lcl}
\Item{a} \mname{inf}_{\{1,2,3,4\}}
&=& 1(2-1) + 4(3-2) + 3(4-3) \\
\Item{b} \mname{sup}_{\{1,2,3,4\}}
&=& 4(2-1) + 5(3-2) + 5(4-3) \\
\Item{c} \mname{M} _{\{1,3,5\}}
&=& 4(3-1) + 3(5-3) \\
\Item{d} \mname{Trap}_{\{1,3,5\}}
&=& 3(3-1) + 3.5(5-3) \\
\Item{e} \mname{Trap}_{\{1,3,5\}}
&=& \frac{1+5}{2}(3-1) + \frac{5+2}{2}(5-3) \\
\end{array}
$
\bsk
\bsk
$$\unitlength=20pt
\ga{F(x)}
$$
}}
%L Pict2e.bounds = PictBounds.new(v(0,0), v(7,6))
%L spec = "(0,1)--(1,1)--(2,4)--(3,5)--(4,4)o (4,3)c (4,1)o--(6,3)--(7,3)"
%L pws = PwSpec.from(spec)
%L pws:topict():prethickness("1pt"):pgat("pgatc"):sa("F(x)"):output()
\pu
%\printbibliography
\GenericWarning{Success:}{Success!!!} % Used by `M-x cv'
\end{document}
% ____ _ _
% | _ \(_)_ ___ _(_)_______
% | | | | \ \ / / | | | |_ / _ \
% | |_| | |\ V /| |_| | |/ / __/
% |____// | \_/ \__,_|_/___\___|
% |__/
%
% «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
f () { rm -v $1.pdf; textcleaner -f 50 -o 5 $1.jpg $1.png; djvuize $1.pdf; xpdf $1.pdf }
f () { rm -v $1.pdf; textcleaner -f 50 -o 10 $1.jpg $1.png; djvuize $1.pdf; xpdf $1.pdf }
f () { rm -v $1.pdf; textcleaner -f 50 -o 20 $1.jpg $1.png; djvuize $1.pdf; xpdf $1.pdf }
f () { rm -fv $1.png $1.pdf; djvuize $1.pdf }
f () { rm -fv $1.png $1.pdf; djvuize WHITEBOARDOPTS="-m 1.0 -f 15" $1.pdf; xpdf $1.pdf }
f () { rm -fv $1.png $1.pdf; djvuize WHITEBOARDOPTS="-m 1.0 -f 30" $1.pdf; xpdf $1.pdf }
f () { rm -fv $1.png $1.pdf; djvuize WHITEBOARDOPTS="-m 1.0 -f 45" $1.pdf; xpdf $1.pdf }
f () { rm -fv $1.png $1.pdf; djvuize WHITEBOARDOPTS="-m 0.5" $1.pdf; xpdf $1.pdf }
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
% __ __ _
% | \/ | __ _| | _____
% | |\/| |/ _` | |/ / _ \
% | | | | (_| | < __/
% |_| |_|\__,_|_|\_\___|
%
% <make>
* (eepitch-shell)
* (eepitch-kill)
* (eepitch-shell)
# (find-LATEXfile "2019planar-has-1.mk")
make -f 2019.mk STEM=2022-2-C2-P2 veryclean
make -f 2019.mk STEM=2022-2-C2-P2 pdf
% Local Variables:
% coding: utf-8-unix
% ee-tla: "c2p2"
% ee-tla: "c2m222p2"
% End: