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Warning: this is an htmlized version!
The original is here, and the conversion rules are here. |
% (find-angg "LATEX/2017-2-GA-P2.tex")
% (defun c () (interactive) (find-LATEXsh "lualatex -record 2017-2-GA-P2.tex"))
% (defun d () (interactive) (find-xpdfpage "~/LATEX/2017-2-GA-P2.pdf"))
% (defun b () (interactive) (find-zsh "bibtex 2017-2-GA-P2; makeindex 2017-2-GA-P2"))
% (defun e () (interactive) (find-LATEX "2017-2-GA-P2.tex"))
% (defun u () (interactive) (find-latex-upload-links "2017-2-GA-P2"))
% (find-xpdfpage "~/LATEX/2017-2-GA-P2.pdf")
% (find-sh0 "cp -v ~/LATEX/2017-2-GA-P2.pdf /tmp/")
% (find-sh0 "cp -v ~/LATEX/2017-2-GA-P2.pdf /tmp/pen/")
% file:///home/edrx/LATEX/2017-2-GA-P2.pdf
% file:///tmp/2017-2-GA-P2.pdf
% file:///tmp/pen/2017-2-GA-P2.pdf
% http://angg.twu.net/LATEX/2017-2-GA-P2.pdf
% «.gab-1» (to "gab-1")
% «.gab-2a-2j» (to "gab-2a-2j")
% «.gab-2k-2s» (to "gab-2k-2s")
% «.gab-3» (to "gab-3")
% «.gab-4» (to "gab-4")
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%
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%
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%
\begin{document}
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\def\pu{\directlua{pu()}}
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%L V.__tostring = function (v) return format("(%.3f,%.3f)", v[1], v[2]) end
\pu
\def\V(#1){\VEC{#1}}
\def\und#1#2{\underbrace{#1}_{#2}}
% ____ _ _ _
% / ___|__ _| |__ ___ ___ __ _| | |__ ___
% | | / _` | '_ \ / _ \/ __/ _` | | '_ \ / _ \
% | |__| (_| | |_) | __/ (_| (_| | | | | | (_) |
% \____\__,_|_.__/ \___|\___\__,_|_|_| |_|\___/
%
{\setlength{\parindent}{0em}
\footnotesize
\par Geometria Analítica
\par PURO-UFF - 2017.2
\par P2 - 11/dez/2017 - Eduardo Ochs
\par Respostas sem justificativas não serão aceitas.
\par Diagramas muito ambíguos {\sl serão} interpretados errado.
\par Proibido usar quaisquer aparelhos eletrônicos.
}
\bsk
\bsk
\setlength{\parindent}{0em}
\def\T(Total: #1 pts){{\bf(Total: #1 pts)}}
\def\T(Total: #1 pts){{\bf(Total: #1)}}
\def\B (#1 pts){{\bf(#1 pts)}}
% Usage:
% 1) \T(Total: 2.34 pts) Foo
% a) \B(0.45 pts) Bar
Lembre que uma equação de cônica é uma equação da forma $ax^2 + bxy +
cy^2 + dx + ey + f = 0$; $4+(x+y)(x-y)=5y$ não é uma equação de cônica
mas é equivalente a uma: $x^2-y^2-5y+4=0$. E o truque pra gente se
livrar das duas raízes quadradas em $√A + √B = C$ ou $√A - √B = C$ é:
%
$$\begin{array}{rcl}
√A + √B = C &⇒& C^2(C^2 - 2(A+B)) + (A-B)^2 = 0 \\
√A - √B = C &⇒& C^2(C^2 - 2(A+B)) + (A-B)^2 = 0 \\
\end{array}
$$
% Nas questões 3 e 4 vamos usar a abreviação $[\text{equação}] =
% \setofxyzst{\text{equação}}$.
\bsk
1) \T(Total: 1.5 pts) Use as fórmulas acima para transformar
%
$$d((x,y),(-2,0))-d((x,y),(2,0))=2$$
%
numa equação de cônica e verifique se os seguintes pontos obedecem a
equação original e a equação da cônica: $(1,0)$, $(2,3)$, $(-2,3)$,
$(-1,0)$.
\bsk
% \def\myu {\frac{x-4}{2}}
% \def\myv { y-\frac{x}{2} }
\def\myu {y-x/2-3}
\def\myv {y+x/2-3}
\def\myuu{\und{\textstyle\myu}{u}}
\def\myvv{\und{\textstyle\myv}{v}}
2) \T(Total: 4.5 pts) Faça esboços das cônicas com as equações abaixo.
Algumas delas são degeneradas. Em todos os itens abaixo considere que
$u=2y$ e $v=x+y$ --- ou que $u$ e $v$ são {\sl abreviações} para $2y$
e $x+y$.
%
$$
\begin{tabular}[t]{rl}
a) (0.1 pts) & $x^2+y^2=1$ \\
b) (0.1 pts) & $x^2+y^2=4$ \\
c) (0.1 pts) & $x^2+y^2=0$ \\
d) (0.1 pts) & $xy=1$ \\
e) (0.1 pts) & $xy=4$ \\
f) (0.1 pts) & $xy=0$ \\
g) (0.1 pts) & $xy=-1$ \\
h) (0.1 pts) & $x^2=y$ \\
i) (0.1 pts) & $x=y^2$ \\
j) (0.1 pts) & $x^2=y^2$ \\
\end{tabular}
\quad
\begin{tabular}[t]{rl}
k) (0.5 pts) & $u(u-1)=0$ \\
l) (0.5 pts) & $v(v-1)=0$ \\[4pt]
m) (0.2 pts) & $u^2+v^2=0$ \\
n) (0.3 pts) & $u^2+v^2=1$ \\[4pt]
o) (0.2 pts) & $uv=0$ \\
p) (0.4 pts) & $uv=1$ \\
q) (0.4 pts) & $uv=-1$ \\[4pt]
r) (0.5 pts) & $u^2=v$ \\
s) (0.5 pts) & $u=v^2$ \\
\end{tabular}
$$
\bsk
3) \T(Total: 1.5 pts) Dê uma parametrização para a reta que pertence
aos planos $x+y+z=3$ e $x+2z=3$.
\bsk
4) \T(Total: 2.5 pts) Sejam:
%
$$\begin{array}{rcl}
r &=& \setofst{(t,t,0)}{t∈\R} \\
s &=& \setofst{(2t,-2t,4)}{t∈\R} \\
r' &=& \setofst{(t,t,0)}{t∈\R} \\
s' &=& \setofst{(t,0,3-t}{t∈\R} \\
\end{array}
$$
a) \B(0.1 pts) Calcule $d(r,s)$ no olhômetro.
a) \B(0.9 pts) Calcule $d(r,s)$ pela fórmula.
c) \B(1.5 pts) Calcule $d(r',s')$ pela fórmula.
\newpage
% ____ _ _ _
% / ___| __ _| |__ __ _ _ __(_) |_ ___
% | | _ / _` | '_ \ / _` | '__| | __/ _ \
% | |_| | (_| | |_) | (_| | | | | || (_) |
% \____|\__,_|_.__/ \__,_|_| |_|\__\___/
%
{\bf Gabarito} (versão não revisada)
\msk
% _
% / |
% | |
% | |
% |_|
%
% «gab-1» (to ".gab-1")
1) $d((x,y),(-2,0))-d((x,y),(2,0))=2$
$⇒ \sqrt{(x+2)^2+y^2} - \sqrt{(x-2)^2+y^2} = 2$. Sejam $C=2$,
$A = (x+2)^2+y^2 = x^2+4x+4+y^2$,
$B = (x-2)^2+y^2 = x^2-4x+4+y^2$; então $A-B=8x$ e
$A+B=2(x^2+4+y^2)$.
$√A - √B = C ⇒ C^2(C^2 - 2(A+B)) + (A-B)^2 = 0$
$⇒ 4(4 - 4(x^2+4+y^2)) + (8x)^2 = 0$
$⇒ (-16x^2 - 16y^2 - 64 + 16) + 64x^2 = 0$
$⇒ 48x^2 - 16y^2 - 48 = 0$
\msk
$\begin{array}{ccc}
(x,y) & d((x,y),(-2,0))-d((x,y),(2,0))=0 & 48x^2 - 16y^2 - 48=0 \\\hline
(1,0) & 3-1=2 & 48-0-48 = 0 \\
(2,3) & 5-3=2 & 192-144-48=0 \\
(-2,3) & 5-3=2 & 192-144-48=0 \\
(-1,0) & 1-2≠-2 & 48-0-48=0 \\
\end{array}
$
% ____
% |___ \
% __) |
% / __/
% |_____|
%
\unitlength=5pt
\def\closeddot{\circle*{0.6}}
\def\pictpoint#1{\put(#1){\closeddot}}
\def\pictline#1{{\linethickness{1.0pt}\expr{Line.new(#1):pict()}}}
\def\pictlinethin#1{{\linethickness{0.2pt}\expr{Line.new(#1):pict()}}}
\def\pictLine(#1)(#2)#3{%
\vcenter{\hbox{%
\beginpicture(#1)(#2)%
\pictaxes%
\pictline{#3}
\end{picture}%
}}%
}
\def\pictellipse#1{{\linethickness{1.0pt}\expr{Ellipse.new(#1):pict()}}}
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\vcenter{\hbox{%
\beginpicture(#1)(#2)%
\pictaxes%
\pictellipse{#3}
\end{picture}%
}}%
}
\def\pictEllipseF(#1)(#2)#3(#4)(#5){%
\vcenter{\hbox{%
\beginpicture(#1)(#2)%
\pictaxes%
\pictellipse{#3}
\put(#4){\closeddot}
\put(#5){\closeddot}
\end{picture}%
}}%
}
\def\picthyperbole#1#2{{\linethickness{1.0pt}\expr{Hyperbole.new(#1):pict(#2)}}}
\def\pictparabola #1#2{{\linethickness{1.0pt}\expr{Parabola .new(#1):pict(#2)}}}
% (find-LATEX "edrxtikz.lua" "Line")
\def\mygrid{
\pictlinethin{v(0, 0), v(1, 0), -3, 3} % 2y = 0
\pictlinethin{v(0, .5), v(1, 0), -3, 3} % 2y = 1
\pictlinethin{v(0,-.5), v(1, 0), -3, 3} % 2y = -1
\pictlinethin{v(0, 0), v(1, -1), -2, 2} % x+y = 0
\pictlinethin{v(0, 1), v(1, -1), -1, 3} % x+y = 1
\pictlinethin{v(0,-1), v(1, -1), -3, 1} % x+y = -1
}
% ____ ____ _
% |___ \ __ _ |___ \ (_)
% __) / _` |_____ __) || |
% / __/ (_| |_____/ __/ | |
% |_____\__,_| |_____|/ |
% |__/
%
% «gab-2a-2j» (to ".gab-2a-2j")
\unitlength=7.5pt
\def\closeddot{\circle*{0.4}}
2a) % a) x^2+y^2=1
$\vcenter{\hbox{%
\beginpicture(-2,-2)(2,2)%
\pictaxes%
\pictellipse{v(0,0), v(1,0), v(0,1)}
% \pictline{v(0,0), v(1,0), -2, 2}
% \pictline{v(0,0), v(0,1), -2, 2}
\end{picture}%
}}$
\;
2b) % b) x^2+y^2=4
$\vcenter{\hbox{%
\beginpicture(-2,-2)(2,2)%
\pictaxes%
\pictellipse{v(0,0), v(2,0), v(0,2)}
%\pictline{v(0,0), v(1,1), -2, 2}
%\pictline{v(0,0), v(1,-1), -2, 2}
\end{picture}%
}}$
\;
2c) % c) x^2+y^2=0
$\vcenter{\hbox{%
\beginpicture(-2,-2)(2,2)%
\pictaxes%
\put(0,0){\closeddot}
%\pictline{v(0,1), v(1,0), -2, 2}
%\pictline{v(0,2), v(1,0), -2, 2}
\end{picture}%
}}$
%
\quad
%
2d) % d) xy=1
$\vcenter{\hbox{%
\beginpicture(-4,-4)(4,4)%
\pictaxes%
\picthyperbole{v(0,0), v(1,0), v(0,1), 1}{10, -4, -1/4, 1/4, 4}
\put(1,1){\closeddot}
\put(-1,-1){\closeddot}
%\pictline{v(0,0), v(-1,1), -2, 2}
\end{picture}%
}}$
\;
2e) % e) xy=4
$\vcenter{\hbox{%
\beginpicture(-4,-4)(4,4)%
\pictaxes%
\picthyperbole{v(0,0), v(2,0), v(0,2), 1}{10, -2, -1/2, 1/2, 2}
\put(2,2){\closeddot}
\put(-2,-2){\closeddot}
%\pictline{v(0,0), v(-1,1), -2, 2}
\end{picture}%
}}$
\;
2f) % f) xy=0
$\vcenter{\hbox{%
\beginpicture(-3,-3)(3,3)%
\pictaxes%
\pictline{v(0,0), v(1,0), -3, 3}
\pictline{v(0,0), v(0,1), -3, 3}
\end{picture}%
}}$
\;
2g) % g) xy=-1
$\vcenter{\hbox{%
\beginpicture(-4,-4)(4,4)%
\pictaxes%
\picthyperbole{v(0,0), v(1,0), v(0,-1), 1}{10, -4, -1/4, 1/4, 4}
\put(1,-1){\closeddot}
\put(-1,1){\closeddot}
\end{picture}%
}}$
\;
2h) % h) x^2=y
$\vcenter{\hbox{%
\beginpicture(-2,-2)(2,2)%
\pictaxes%
\pictparabola{v(0,0), v(1,0), v(0,1), 2}{10, -1.4, 1.4}
\end{picture}%
}}$
\quad
2i) % i) x=y^2
$\vcenter{\hbox{%
\beginpicture(-2,-2)(2,2)%
\pictaxes%
\pictparabola{v(0,0), v(0,1), v(1,0), 2}{10, -1.4, 1.4}
\end{picture}%
}}$
\quad
2j) % j) x^2=y^2
$\vcenter{\hbox{%
\beginpicture(-2,-2)(2,2)%
\pictaxes%
\pictline{v(0,0), v(1,1), -2, 2}
\pictline{v(0,0), v(1,-1), -2, 2}
\end{picture}%
}}$
% ____ _ ____
% |___ \| | __ |___ \ ___
% __) | |/ /____ __) / __|
% / __/| <_____/ __/\__ \
% |_____|_|\_\ |_____|___/
%
% «gab-2k-2s» (to ".gab-2k-2s")
\unitlength=7.5pt
\def\closeddot{\circle*{0.4}}
2k) % k) u(u-1)=0
$\vcenter{\hbox{%
\beginpicture(-3,-2)(3,2)%
\pictaxes%
\mygrid
\pictline{v(0, 0), v(.5,-.5), -3, 3}
\pictline{v(0, 1), v(.5,-.5), -1, 4}
\end{picture}%
}}$
\;
2l) % l) v(v-1)=0
$\vcenter{\hbox{%
\beginpicture(-3,-2)(3,2)%
\pictaxes%
\mygrid
\pictline{v(0, 0), v(1,0), -2, 2}
\pictline{v(0,.5), v(1,0), -2.5, 1.5}
\end{picture}%
}}$
\;
2m) % m) u^2+v^2=0
$\vcenter{\hbox{%
\beginpicture(-3,-2)(3,2)%
\pictaxes%
\mygrid
\put(0,0){\closeddot}
\end{picture}%
}}$
\;
2n) % n) u^2+v^2=1
$\vcenter{\hbox{%
\beginpicture(-3,-2)(3,2)%
\pictaxes%
\mygrid
\pictellipse{v(0,0), v(1,0), v(.5,-.5)}
% \picthyperbole{v(2,-2), v(1,0), v(-1,1), 1}{10, -4, -1/2, 1/4, 4}
% \put(2,-2){\closeddot}
% \pictline{v(0,-2), v(1,0), -1, 5} % y-2 = 0
% \pictline{v(0, 0), v(1,-1), -2, 4} % x+y = 0
\end{picture}%
}}$
\;
2o) % o) uv=0
$\vcenter{\hbox{%
\beginpicture(-3,-2)(3,2)%
\pictaxes%
\mygrid
\pictline{v(0,0), v(1,0), -3, 3}
\pictline{v(0,0), v(.5,-.5), -3, 3}
\end{picture}%
}}$
\quad
2p) % p) uv=1
$\vcenter{\hbox{%
\beginpicture(-3,-2)(3,2)%
\pictaxes%
\mygrid
\picthyperbole{v(0,0), v(1,0), v(-.5,.5), 1}{10, -3, -1/3, 1/3, 3}
\end{picture}%
}}$
\;
2q) % q) uv=-1
$\vcenter{\hbox{%
\beginpicture(-4,-2)(4,2)%
\pictaxes%
\mygrid
\picthyperbole{v(0,0), v(1,0), v(.5,-.5), 1}{10, -3, -1/3, 1/3, 3}
\end{picture}%
}}$
\;
2r) % r) u^2=v
$\vcenter{\hbox{%
\beginpicture(-4,-2)(4,2)%
\pictaxes%
\mygrid
\pictparabola{v(0,0), v(1,0), v(-.5,.5), 2}{10, -1.4, 1.4}
\end{picture}%
}}$
\;
2s)
$\vcenter{\hbox{%
\beginpicture(-4,-2)(4,2)%
\pictaxes%
\mygrid
\pictparabola{v(0,0), v(-.5,.5), v(1,0), 2}{10, -1.4, 1.4}
% \pictellipse{v(4,6), v(2,0), v(0,3)}
\end{picture}%
}}$
\bsk
% _____
% |___ /
% |_ \
% ___) |
% |____/
%
% «gab-3» (to ".gab-3")
3) $z=\frac{3-x}{2}$, $y = 3-x-2 = 3 - x - \frac{3-x}{2} =
\frac{3-x}{2}$, $r=\setofst{(x, \frac{3-x}{2} \frac{3-x}{2})}{x∈\R}$
\bsk
% _ _
% | || |
% | || |_
% |__ _|
% |_|
%
% «gab-4» (to ".gab-4")
4) Sejam:
$\begin{array}{lll}
r = \setofst{(0,0,0)+\VEC{1,1,0}}{t∈\R}, & A=(0,0,0), & \uu=\VEC{1,1,0}, \\
s = \setofst{(0,0,4)+\VEC{2,-2,0}}{t∈\R}, & B=(0,0,4), & \vv=\VEC{2,-2,0}, \\
r' = \setofst{(0,0,0)+\VEC{1,1,0}}{t∈\R}, & A'=(0,0,0), & \uu'=\VEC{1,1,0}, \\
s' = \setofst{(0,0,3)+\VEC{1,0,-1}}{t∈\R}, & B'=(0,0,3), & \vv'=\VEC{1,0,-1}, \\
\end{array}
$
4a) 4
4b) $d(r,s) = |([\uu,\vv,\Vec{AB}]/\area(\uu,\vv))| = |(16/||\VEC{0,0,4}||)| = 4$
4c) $d(r',s') = |([\uu',\vv',\Vec{A'B'}]/\area(\uu',\vv'))| = |(\vsm{1
& 1 & 0 \\ 1 & 0 & -1 \\ 0 & 0 & 3 } / ||\VEC{1,1,-1}||)| =
|3/\sqrt3| = \sqrt3$
\end{document}
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