|
Warning: this is an htmlized version!
The original is here, and the conversion rules are here. |
% (find-angg "LATEX/2015-2-GA-P2.tex")
% (find-angg "LATEX/2015-2-GA-P2.lua")
% (defun c () (interactive) (find-LATEXsh "lualatex -record 2015-2-GA-P2.tex"))
% (defun d () (interactive) (find-xpdfpage "~/LATEX/2015-2-GA-P2.pdf"))
% (defun e () (interactive) (find-LATEX "2015-2-GA-P2.tex"))
% (defun u () (interactive) (find-latex-upload-links "2015-2-GA-P2"))
% (defun z () (interactive) (find-zsh "flsfiles-tgz 2015-2-GA-P2.fls 2015-2-GA-P2.tgz")
% (find-xpdfpage "~/LATEX/2015-2-GA-P2.pdf")
% (find-sh0 "cp -v ~/LATEX/2015-2-GA-P2.pdf /tmp/")
% (find-sh0 "cp -v ~/LATEX/2015-2-GA-P2.pdf /tmp/pen/")
% file:///home/edrx/LATEX/2015-2-GA-P2.pdf
% file:///tmp/2015-2-GA-P2.pdf
% file:///tmp/pen/2015-2-GA-P2.pdf
% http://angg.twu.net/LATEX/2015-2-GA-P2.pdf
\documentclass[oneside]{book}
\usepackage[colorlinks]{hyperref} % (find-es "tex" "hyperref")
%\usepackage[latin1]{inputenc}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{tikz}
%
\usepackage{edrx15} % (find-angg "LATEX/edrx15.sty")
\input edrxaccents.tex % (find-angg "LATEX/edrxaccents.tex")
\input edrxheadfoot.tex % (find-dn4ex "edrxheadfoot.tex")
\input istanbuldefs % (find-LATEX "istanbuldefs.tex")
%
\begin{document}
\catcode`\^^J=10
\directlua{dednat6dir = "dednat6/"}
\directlua{dofile(dednat6dir.."dednat6.lua")}
\directlua{texfile(tex.jobname)}
\directlua{verbose()}
%\directlua{output(preamble1)}
\def\expr#1{\directlua{output(tostring(#1))}}
\def\eval#1{\directlua{#1}}
\def\pu{\directlua{pu()}}
\directlua{dofile "edrxtikz.lua"} % (find-LATEX "edrxtikz.lua")
%L V.__tostring = function (v) return format("(%.3f,%.3f)", v[1], v[2]) end
% ____ _ _ _
% / ___|__ _| |__ ___ ___ __ _| | |__ ___
% | | / _` | '_ \ / _ \/ __/ _` | | '_ \ / _ \
% | |__| (_| | |_) | __/ (_| (_| | | | | | (_) |
% \____\__,_|_.__/ \___|\___\__,_|_|_| |_|\___/
%
{\setlength{\parindent}{0em}
\footnotesize
\par Geometria Analítica
\par PURO-UFF - 2015.2
\par P2 - 21/mar/2016 - Eduardo Ochs
% \par Versão: 14/mar/2016
% \par Links importantes:
% \par \url{http://angg.twu.net/2015.2-C2.html} (página do curso)
% \par \url{http://angg.twu.net/2015.2-C2/2015.2-C2.pdf} (quadros)
% \par \url{http://angg.twu.net/LATEX/2015-2-C2-material.pdf}
% \par {\tt eduardoochs@gmail.com} (meu e-mail)
}
\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)}}
1) \T(Total: 3.0 pts) Sejam $F=(9,0)$ e $r: x=25$.
a) \B(0.5 pts) Encontre 4 pontos de $\R^2$ tais que $d(P,F) = \frac35
d(P,r)$.
b) \B(2.0 pts) Seja $E$ o conjunto dos pontos que obedecem $d(P,F) =
\frac35 d(P,r)$. Dê a equação reduzida da elipse $E$, as coordenadas
dos seus focos, os comprimentos dos seus semi-eixos maior e menor e a
sua excentricidade.
c) \B(0.5 pts) Represente graficamente tudo o que você obteve nos
itens anteriores.
\bsk
2) \T(Total: 1.0 pts) Represente graficamente a hipérbole $(2(x-3) +
(y-4)) (2(x-3) - (y-4)) = 4$ e suas assíntotas.
\bsk
3) \T(Total: 3.0 pts) Seja $π_1$ o plano com equação $2x+3y+4z=6$.
a) \B(0.5 pts) Dê uma parametrização para a reta $r$ que passa pelo
ponto $A=(1,1,0)$ e é ortogonal a $π_1$.
b) \B(0.5 pts) Dê as coordenadas do ponto $B∈π_1∩r$.
c) \B(1.0 pts) Dê as coordenadas de um ponto $C∈r$ tal que
$d(C,π_1)=1$.
d) \B(1.0 pts) Dê a equação de um plano $π_2 \parallel π_1$ tal que
$d(π_1,π_2)=2$.
\bsk
4) \T(Total: 3.0 pts) Sejam $A=(2,0,0)$, $B=(0,2,0)$, $C=(0,0,5)$,
$D=(0,5,0)$. Sejam $r$ uma reta que passa por $A$ e $B$, $r'$ uma reta
que passa por $C$ e $D$, e $s$ uma reta ortogonal a $r$ e $r'$ que
corta ambas.
a) \B(0.5 pts) Calcule a distância entre $r$ e $r'$.
b) \B(1.5 pts) Dê uma parametrização para $s$.
c) \B(1.0 pts) Dê a equação de um plano $π$ paralelo a $r$ e $r'$ e
equidistante de ambas.
\bsk
\newpage
% _ _ _
% | |_(_) | __ ____
% | __| | |/ /|_ /
% | |_| | < / /
% \__|_|_|\_\/___|
%
% Dots, labels, vectors
%
\def\uu{\vec u}
\def\vv{\vec v}
\def\ww{\vec w}
\def\nn{\vec n}
\def\VEC#1{{\overrightarrow{(#1)}}}
\def\nm#1{\|#1\|}
\def\Reg#1{(#1)}
\def\setofxyst #1{\setofst{(x,y) ∈\R^2}{#1}}
\def\setofxyzst#1{\setofst{(x,y,z)∈\R^3}{#1}}
\def\setofet #1{\setofst{#1}{t∈\R}}
\def\setofeu #1{\setofst{#1}{u∈\R}}
\def\setofpt #1 #2 #3 #4 {\setofet{(#1,#2)+t\VEC{#3,#4}}}
\def\setofpu #1 #2 #3 #4 {\setofeu{(#1,#2)+u\VEC{#3,#4}}}
\def\setofeaa #1{\setofst{#1}{α∈\R}}
\def\setofebb #1{\setofst{#1}{β∈\R}}
% \mygrid and \myaxes
% (find-es "tikz" "mygrid")
\tikzset{mycurve/.style=very thick}
\tikzset{axis/.style=semithick}
\tikzset{tick/.style=semithick}
\tikzset{grid/.style=gray!20,very thin}
\tikzset{anydot/.style={circle,inner sep=0pt,minimum size=1.2mm}}
\tikzset{opdot/.style={anydot, draw=black,fill=white}}
\tikzset{cldot/.style={anydot, draw=black,fill=black}}
%
\def\mygrid(#1,#2) (#3,#4){
\clip (#1-0.4, #2-0.4) rectangle (#3+0.4, #4+0.4);
\draw[step=1,grid] (#1-0.2, #2-0.2) grid (#3+0.2, #4+0.2);
\draw[axis] (-10,0) -- (10,0);
\draw[axis] (0,-10) -- (0,10);
\foreach \x in {-10,...,10} \draw[tick] (\x,-0.2) -- (\x,0.2);
\foreach \y in {-10,...,10} \draw[tick] (-0.2,\y) -- (0.2,\y);
}
\def\myaxes(#1,#2) (#3,#4){
\clip (#1-0.4, #2-0.4) rectangle (#3+0.4, #4+0.4);
%\draw[step=1,grid] (#1-0.2, #2-0.2) grid (#3+0.2, #4+0.2);
\draw[axis] (-30,0) -- (30,0);
\draw[axis] (0,-30) -- (0,30);
\foreach \x in {-30,...,30} \draw[tick] (\x,-0.2) -- (\x,0.2);
\foreach \y in {-30,...,30} \draw[tick] (-0.2,\y) -- (0.2,\y);
}
% Grid color
\tikzset{grid/.style=gray!50,very thin}
\def\tikzp#1{\mat{\begin{tikzpicture}#1\end{tikzpicture}}}
\def\mydraw #1;{\draw [mycurve] \expr{#1};}
\def\mydot #1;{\node [cldot] at \expr{#1} [] {};}
\def\myldot #1 #2 #3;{\node [cldot] at \expr{#1} [label=#2:${#3}$] {};}
\def\myseg #1 #2;{\draw [mycurve] \expr{#1} -- \expr{#2};}
\def\mylabel #1 #2 #3;{\node [] at \expr{#1} [label=#2:${#3}$] {};}
\def\myseggrid #1 #2;{\draw [grid] \expr{#1} -- \expr{#2};}
% (find-dn6 "picture.lua" "V")
%L V.__tostring = function (v) return format("(%.3f,%.3f)", v[1], v[2]) end
%L V.__div = function (v, k) return v*(1/k) end
%L V.__index.tow = function (A, B, t) return A+(B-A)*t end -- towards
%L V.__index.mid = function (A, B) return A+(B-A)*0.5 end -- midpoint
% _ _ _
% __ _ __ _| |__ __ _ _ __(_) |_ ___
% / _` |/ _` | '_ \ / _` | '__| | __/ _ \
% | (_| | (_| | |_) | (_| | | | | || (_) |
% \__, |\__,_|_.__/ \__,_|_| |_|\__\___/
% |___/
Gabarito:
(Primeira versão, não revisada)
\bsk
%L e = Ellipse.new(v(0,0), v(15,0), v(0,12))
\pu
1) \quad
$\tikzp{[scale=0.2,auto]
\myaxes(-23,-15) (32,15)
\draw [mycurve] (-9,0) -- (0,12) -- (9,0);
\draw [mycurve] (0,12) -- (25,12);
\draw [mycurve] (25,-14) -- (25,14);
\myldot v(0,12) 45 (0,12);
\myldot v(0,-12) 315 (0,-12);
\myldot v(9,0) 270 (9,0);
\myldot v(-9,0) 270 (-9,0);
\myldot v(-15,0) 225 (-15,0);
\myldot v(15,0) 315 (15,0);
\myldot v(25,0) 315 (25,0);
\myldot v(25,12) 0 (25,12);
\mydraw e:draw();
}
$
\bsk
\bsk
%L h = Hyperbole.new(v(3,4), v(1,-2)/2, v(1,2)/2, 8)
\pu
2) Sejam $u = 2(x-3)$, $v=y-4$. Então
$\begin{array}[t]{rcl}
H &=& \setofxyst{(2(x-3) + (y-4)) (2(x-3) - (y-4)) = 4=4} \\
&=& \setofxyst{(u+v)(u-v)=4} \\
\end{array}
$.
$H$ tem assíntotas em $u+v=0$ e em $u-v=0$, e
$H$ passa pelo ponto que tem $(u+v)=(u-v)=2$, ou seja, $u=2$ e $v=0$.
\msk
$\def\s#1{{\tiny#1}}
\def\s#1{{\scriptscriptstyle#1}}
\def\s#1{{\scriptstyle#1}}
%
\tikzp{[scale=0.4,auto]
\myaxes(-2,-4) (14,12)
\draw [grid] (2,-10) -- (2,11); \mylabel v(2,11) 90 \s{u{=}-2};
\draw [grid] (3,-10) -- (3,10.5); \mylabel v(3,10.5) 90 \s{u{=}0};
\draw [grid] (4,-10) -- (4,10); \mylabel v(4,10) 90 \s{u{=}2};
\draw [grid] (-10,2) -- (10,2); \mylabel v(10,2) 0 \s{v{=}-2};
\draw [grid] (-10,4) -- (10,4); \mylabel v(10,4) 0 \s{v{=}0};
\draw [grid] (-10,6) -- (10,6); \mylabel v(10,6) 0 \s{v{=}2};
\myaxes(-2,-4) (14,12)
\mydraw h:draw(10);
\mydraw h:drawau(-8,8);
\mydraw h:drawav(-8,8);
}
$
\newpage
3) $\begin{array}[t]{rcl}
π_1 &=& \setofxyzst{2x + 3y + 4z = 6} \\
\nn &=& \VEC{2,3,4} \\
A &=& (1,1,0) \\
\end{array}
$
3a) $\begin{array}[t]{rcl}
r &=& \setofet{A+t\nn} \\
&=& \setofet{(1+2t, 1+3t, 4t)} \\
\end{array}
$
3b) Se $B∈π_1∩r$ então
$\begin{array}[t]{rcl}
6 &=& 2(1+2t) + 3(1+3t) + 4(4t) \\
&=& 2+4t+3+9t+16t \\
&=& 29t+5 \\
% 29t &=& 6 - 5 = 1 \\
t &=& \frac{1}{29} \\
B &=& (1+\frac{2}{29}, 1+\frac{3}{29}, \frac{4}{29})
= (\frac{31}{29}, \frac{32}{29}, \frac{4}{29})
\end{array}
$
3c) Seja $\nn' = \frac{\nn}{||\nn||} = \VEC{2,3,4}/\sqrt{4+9+16} = \VEC{2,3,4}/\sqrt{29}$.
Seja $C = B+\nn'$.
3d) Seja $D = B+2\nn'$.
Seja $π_2 = \setofxyzst{2x + 3y + 4z = a}$, onde $a$ é tal que $D∈π_2$.
Então $a = 2D_1 + 3D_2 + 4D_3$.
\bsk
\bsk
4) Sejam
$\begin{array}[t]{rcl}
\uu &=& \Vec{AB} = \VEC{-2,2,0}, \\
\vv &=& \Vec{CD} = \VEC{0,5,-5}, \\
r &=& \setofeaa{A+α\uu} \\
&=& \setofeaa{(2,0,0)+α\VEC{-2,2,0}} \\
&=& \setofeaa{(2-2α,2α,0)}, \\
r' &=& \setofebb{C+β\vv} \\
&=& \setofebb{(0,0,5)+β\VEC{0,5,-5}} \\
&=& \setofebb{(0,5β,5-5β)}, \\
P_α &=& (2-2α,2α,0), \\
Q_β &=& (0,5β,5-5β), \\
\end{array}
$
Então $\uu×\vv = \VEC{-10,-10,-10}$.
4a) $\begin{array}[t]{rcl}
\uu×\vv &=& \VEC{-10,-10,-10} \\
\Vec{AC} &=& \VEC{-2,0,5} \\
\end{array}
$
$ \frac{[\uu,\vv,\Vec{AC}]}{||\uu×\vv||}
= \frac{(\uu×\vv)·\Vec{AC}]}{||\uu×\vv||}
= \frac{\VEC{-10,-10,-10}·\VEC{-2,0,5}}{10\sqrt{3}}
= \frac{20-50}{10\sqrt{3}}
= - \sqrt{3}
$
$d(r,r') = \left| \frac{[\uu,\vv,\Vec{AC}]}{||\uu×\vv||} \right | = \sqrt{3}$
\newpage
\def\PaQb{\Vec{P_αQ_β}}
4b) Queremos que $s$ passe por $P_α$ e $Q_β$ e que $\PaQb⊥\uu$ e $\PaQb⊥\vv$.
$\begin{array}[t]{rcl}
\PaQb &=& \VEC{2α-2, 5β-2α, 5-5β} \\
\PaQb·\uu &=& (4-4α)+(10β-4α) = 4 - 8α + 10β = 0 \\
\PaQb·\vv &=& (25β-10α)+(25β-25) = 25 - 10α + 50β = 0\\
10α &=& 50β-25 \\
α &=& 5β-\frac52 \\
4 - 8α + 10β &=& 4 - 8(5β-\frac52) + 10β \\
&=& 4 - 40β + 20 + 10β \\
&=& 24 - 30β = 0 \\
β &=& \frac{24}{30} = \frac45 \\
α &=& 5\frac45 - \frac52 = \frac82 - \frac52 = \frac32 \\
P_α &=& (2-2\frac32, 2\frac32, 0) = (-1, 3, 0) \\
Q_β &=& (0, 5\frac45, 5 - 5\frac45, 0) = (0, 4, 1) \\
\PaQb &=& \VEC{1,1,1} \\
s &=& \setofet{P_α + t \PaQb} \\
&=& \setofet{(-1,3,0) + t\VEC{1,1,1}} \\
\end{array}
$
\msk
4c) Seja $M = \frac{P_α+Q_β}{2} = (-\frac12, \frac72, \frac12).$
Seja $π = \setofxyzst{x+y+z=a}$, onde $a$ é tal que $M∈π$.
Temos $-\frac12 + \frac72 + \frac12 = a = \frac72$,
$π = \setofxyzst{x+y+z=\frac72}$.
$\begin{array}[t]{l}
\end{array}
$
\end{document}
\end{document}
% Local Variables:
% coding: utf-8-unix
% End: