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[SPAM] Re: [obm-l] tangentes ortogonais
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--0-26793332-1212885378=:65083
Content-Type: text/plain; charset=utf-8
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Parece que =C3=A9 y =3D 3x + 55/8....
Quanto =C3=A0s tangentes ortogonais, t=C3=ADtulo do assunto, =C3=A9 u=
ma pergunta um pouco estranha:
em pares de pontos de abcissas x1 e x2 tais que x1*x2 =3D - 1/16 as tangent=
es ser=C3=A3o=20
ortogonais (exemplos : x1=3D -x2 =3D 1/4; x1=3D 1/8 e x2=
=3D -1/2 ,etc.)=20
--- Em sex, 6/6/08, Rafael Ando <rafael.ando@xxxxxxxxx> escreveu:
De: Rafael Ando <rafael.ando@xxxxxxxxx>
Assunto: Re: [obm-l] tangentes ortogonais
Para: obm-l@xxxxxxxxxxxxxx
Data: Sexta-feira, 6 de Junho de 2008, 8:49
A reta dada eh y=3D3x+1, logo as paralelas tem formato y=3D3x+a, com a real=
.
A inclinacao da reta tangente eh portanto y'=3D3. A derivada da funcao de s=
egundo grau sendo y'=3D4x, temos 4x=3D3 o q implica x=3D3/4.
O ponto de tangencia tem coordenada y =3D 2x=C2=B2 + 8 =3D 57/8, logo a =3D=
y - 3x =3D 39/8.
A equacao da reta tangente eh entao y=3D3x+ 39 / 8
On 6/5/08, Vivi H. <xjxjbo@xxxxxxxxx> wrote:
Pessoal, preciso desesperadamente desta quest=C3=A3o. Se algu=C3=A9m puder =
dar alguma luz, agrade=C3=A7o.
"Ache a equa=C3=A7=C3=A3o da reta tangente a curva y =3D 2x=C2=B2 + 8 que =
=C3=A9 paralela a reta 3x - y + 1 =3D0.
Verifique se a fun=C3=A7=C3=A3o possui tangentes ortogonais."
--=20
Rafael=20
=0A=0A=0A Abra sua conta no Yahoo! Mail, o =C3=BAnico sem limite de es=
pa=C3=A7o para armazenamento!=0Ahttp://br.mail.yahoo.com/
--0-26793332-1212885378=:65083
Content-Type: text/html; charset=utf-8
Content-Transfer-Encoding: quoted-printable
<table cellspacing=3D'0' cellpadding=3D'0' border=3D'0' background=3D'none'=
style=3D'font-family:arial;font-size:10pt;color:rgb(51, 51, 51);background=
-color:rgb(255, 255, 255);width:100%;'><tr><td valign=3D'top' style=3D'font=
: inherit;'>Parece que =C3=A9 y =3D 3x + 55/8....<br><br>Quanto =C3=
=A0s tangentes ortogonais, t=C3=ADtulo do assunto, =C3=A9 uma pergunt=
a um pouco estranha:<br><br>em pares de pontos de abcissas x1 e x2 tais qu=
e x1*x2 =3D - 1/16 as tangentes ser=C3=A3o <br><br> ortogonais (=
exemplos : x1=3D -x2 =3D 1/4; x1=3D 1/8 e x2=3D -1/2 ,etc.) <br><br>-=
-- Em <b>sex, 6/6/08, Rafael Ando <i><rafael.ando@xxxxxxxxx></i></b> =
escreveu:<br><blockquote style=3D"border-left: 2px solid rgb(16, 16, 255); =
margin-left: 5px; padding-left: 5px;">De: Rafael Ando <rafael.ando@gmail=
.com><br>Assunto: Re: [obm-l] tangentes ortogonais<br>Para: obm-l@xxxxxx=
c-rio.br<br>Data: Sexta-feira, 6 de Junho de 2008, 8:49<br><br><div id=3D"y=
iv1364920009"><div>A reta dada eh y=3D3x+1,
logo as paralelas tem formato y=3D3x+a, com a real.</div>
<div>A inclinacao da reta tangente eh portanto y'=3D3. A derivada da funcao=
de segundo grau sendo y'=3D4x, temos 4x=3D3 o q implica x=3D3/4.</div>
<div> </div>
<div>O ponto de tangencia tem coordenada y =3D 2x=C2=B2 + 8 =3D 57/8, logo =
a =3D y - 3x =3D 39/8.</div>
<div> </div>
<div>A equacao da reta tangente eh entao y=3D3x+ 39 / 8</div>
<div> </div>
<div><span class=3D"gmail_quote">On 6/5/08, <b class=3D"gmail_sendername">V=
ivi H.</b> <<a rel=3D"nofollow" target=3D"_blank" href=3D"mailto:xjxjbo@=
gmail.com">xjxjbo@xxxxxxxxx</a>> wrote:</span>
<blockquote class=3D"gmail_quote" style=3D"border-left: 1px solid rgb(204, =
204, 204); margin: 0px 0px 0px 0.8ex; padding-left: 1ex;">
<div>Pessoal, preciso desesperadamente desta quest=C3=A3o. Se algu=C3=A9m p=
uder dar alguma luz, agrade=C3=A7o.</div>
<div> </div>
<div>"Ache a equa=C3=A7=C3=A3o da reta tangente a curva y =3D 2x=C2=B2 + 8 =
que =C3=A9 paralela a reta 3x - y + 1 =3D0.</div>
<div>Verifique se a fun=C3=A7=C3=A3o possui tangentes ortogonais." </d=
iv></blockquote></div><br><br clear=3D"all"><br>-- <br>Rafael=20
</div></blockquote></td></tr></table><br>=0A=0A=0A <hr size=3D1>Abra s=
ua conta no <a href=3D"http://br.rd.yahoo.com/mail/taglines/mail/*http://br=
.mail.yahoo.com/">Yahoo! Mail</a>, o =C3=BAnico sem limite de espa=C3=A7o p=
ara armazenamento! =0A
--0-26793332-1212885378=:65083--
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