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	<title>Campo de Arroz</title>
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	<description>Blog com assuntos relacionados às Ciências Farmacêuticas</description>
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		<title>Campo de Arroz</title>
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		<title>Ensino Básico no Japão</title>
		<link>http://campodearroz.wordpress.com/2009/11/04/ensino-basico-no-japao/</link>
		<comments>http://campodearroz.wordpress.com/2009/11/04/ensino-basico-no-japao/#comments</comments>
		<pubDate>Wed, 04 Nov 2009 20:54:33 +0000</pubDate>
		<dc:creator>Augusto Maeda</dc:creator>
				<category><![CDATA[Japonês]]></category>

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		<description><![CDATA[Crianças entram na escola após o primeiro mês de abril após seu 6º aniversário. Primeiramente, entram no shougakkou (小学校) que corresponde aos 6 primeiros anos. Depois entram no chuugakkou (中学校) que corresponde a mais três anos. Shougakkou e Chuugakkou correspondem a educação obrigatória.
Mais sobre a edução no Japão:
http://www.city.otaru.hokkaido.jp/soumu/e_r/school_education.pdf
http://en.wikipedia.org/wiki/Education_in_Japan
Em
http://www.geocities.jp/mutasanjp/print/01nensei/index_kokugo.html
pode-se encontrar materiais de kokugo (こくご &#8211; 国語 [...]<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=campodearroz.wordpress.com&blog=2694588&post=380&subd=campodearroz&ref=&feed=1" />]]></description>
			<content:encoded><![CDATA[<div class='snap_preview'><br /><p>Crianças entram na escola após o primeiro mês de abril após seu 6º aniversário. Primeiramente, entram no shougakkou (小学校) que corresponde aos 6 primeiros anos. Depois entram no chuugakkou (中学校) que corresponde a mais três anos. Shougakkou e Chuugakkou correspondem a educação obrigatória.</p>
<p>Mais sobre a edução no Japão:</p>
<p>http://www.city.otaru.hokkaido.jp/soumu/e_r/school_education.pdf</p>
<p>http://en.wikipedia.org/wiki/Education_in_Japan</p>
<p>Em</p>
<p><a href="http://www.geocities.jp/mutasanjp/print/01nensei/index_kokugo.html">http://www.geocities.jp/mutasanjp/print/01nensei/index_kokugo.html</a></p>
<p>pode-se encontrar materiais de kokugo (こくご &#8211; 国語 &#8211; Aulas de Japonês) e sansuu (さんすう &#8211; 算数 &#8211; Aritmética) do Shougakkou.</p>
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			<media:title type="html">campodearroz</media:title>
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		<title>Língua Japonesa na Internet (Inglês)</title>
		<link>http://campodearroz.wordpress.com/2009/11/04/lingua-japonesa-na-internet-ingles/</link>
		<comments>http://campodearroz.wordpress.com/2009/11/04/lingua-japonesa-na-internet-ingles/#comments</comments>
		<pubDate>Wed, 04 Nov 2009 20:52:51 +0000</pubDate>
		<dc:creator>Augusto Maeda</dc:creator>
				<category><![CDATA[Japonês]]></category>

		<guid isPermaLink="false">http://campodearroz.wordpress.com/?p=378</guid>
		<description><![CDATA[Jim Breen&#8217;s WWWJDIC
Word Search: Dicionário de Japonês. Entrada em Kanji, hiragana ou katakana, romaji ou outras línguas
Translate Words: Permite a tradução de palavras contidas em um texto.
Kanji Lookup: pode-se colocar o kanji ou sua leitura em kana ou romaji e verificar detalhes sobre o kanji, como número de traços, forma de escrever, leitura, significado.
Multi-Radical kanji: [...]<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=campodearroz.wordpress.com&blog=2694588&post=378&subd=campodearroz&ref=&feed=1" />]]></description>
			<content:encoded><![CDATA[<div class='snap_preview'><br /><p><strong>Jim Breen&#8217;s WWWJDIC</strong></p>
<p>Word Search: Dicionário de Japonês. Entrada em Kanji, hiragana ou katakana, romaji ou outras línguas</p>
<p>Translate Words: Permite a tradução de palavras contidas em um texto.</p>
<p>Kanji Lookup: pode-se colocar o kanji ou sua leitura em kana ou romaji e verificar detalhes sobre o kanji, como número de traços, forma de escrever, leitura, significado.</p>
<p>Multi-Radical kanji: Procura de Kanji por seus radicais.</p>
<p><a href="http://www.csse.monash.edu.au/~jwb/cgi-bin/wwwjdic.cgi?1C">http://www.csse.monash.edu.au/~jwb/cgi-bin/wwwjdic.cgi?1C</a></p>
<p><strong>Popjisyo</strong></p>
<p><a href="http://www.popjisyo.com/WebHint/Portal_e.aspx">http://www.popjisyo.com/WebHint/Portal_e.aspx</a></p>
<p>Ao colocar o endereço de uma página escrita em Japonês, pode-se ver os significados das palavras apenas apontado o cursos do mouse sobre elas.</p>
<p><strong>Hiragana Megane</strong></p>
<p>Ao colocar o endereço de uma página escrita em Japonês, pode-se ver a leitura do kanji em hiragana. Se desejar, pode-se colocar a página gerada no Popjisyo nesta outra ferramenta.</p>
<p><a href="http://www.hiragana.jp/">http://www.hiragana.jp/</a></p>
<p><strong>Japanese Audiobooks</strong></p>
<p>Post de um fórum com vários links para textos escritos em japonês, juntamente com o audio e sua leitura.</p>
<p><a href="http://how-to-learn-any-language.com/forum/forum_posts.asp?TID=6241&amp;PN=1&amp;TPN=1">http://how-to-learn-any-language.com/forum/forum_posts.asp?TID=6241&amp;PN=1&amp;TPN=1</a></p>
<p><strong>Japanese Verbs</strong></p>
<p>Guia de verbos na língua japonesa.</p>
<p><a href="http://www.timwerx.net/language/jpverbs/index.htm#contents">http://www.timwerx.net/language/jpverbs/index.htm#contents</a></p>
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		<title>Difusão numa matriz monolítica &#8211; Droga dissolvida em carreador polimérico</title>
		<link>http://campodearroz.wordpress.com/2009/07/30/difusao-numa-matriz-monolitica-droga-dissolvida-em-carreador-polimerico/</link>
		<comments>http://campodearroz.wordpress.com/2009/07/30/difusao-numa-matriz-monolitica-droga-dissolvida-em-carreador-polimerico/#comments</comments>
		<pubDate>Thu, 30 Jul 2009 21:31:44 +0000</pubDate>
		<dc:creator>Augusto Maeda</dc:creator>
				<category><![CDATA[Equações diferenciais]]></category>
		<category><![CDATA[Farmacocinética]]></category>
		<category><![CDATA[Físico-Farmácia]]></category>
		<category><![CDATA[Tecnologia Farmacêutica]]></category>

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		<description><![CDATA[Uma matriz monolítica é a forma mais simples e mais barata de controlar a liberação de fármacos. Para fabricá-la, um polímero ou outro material é homogeneamente distribuído com a droga por mistura da droga com o material polimérico. Os interstícios controlam a liberação da droga. O grau do controle da difusão da droga pela matriz é [...]<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=campodearroz.wordpress.com&blog=2694588&post=258&subd=campodearroz&ref=&feed=1" />]]></description>
			<content:encoded><![CDATA[<div class='snap_preview'><br /><p style="text-align:justify;line-height:14.25pt;background:white;"><span style="font-size:10pt;font-family:&quot;color:black;">Uma matriz monolítica é a forma mais simples e mais barata de controlar a liberação de fármacos. Para fabricá-la, um polímero ou outro material é homogeneamente distribuído com a droga por mistura da droga com o material polimérico. Os interstícios controlam a liberação da droga. O grau do controle da difusão da droga pela matriz é determinado pelas propriedades do polímero e do fármaco.</span></p>
<p style="text-align:justify;line-height:14.25pt;background:white;"><span style="font-size:10pt;font-family:&quot;color:black;">Ou a droga está totalmente dissolvida no polímero ou dispersa como partículas sólidas na matriz. A última condição prevalece quando a concentração da droga é muito maior que sua solubilidade no polímero. A cinética de liberação nestes dois estados é diferente.</span></p>
<p style="text-align:justify;line-height:14.25pt;background:white;"><span style="font-size:10pt;font-family:&quot;color:black;">Aqui será descrita a liberação para o caso da droga dissolvida na matriz. Quando a matriz entra em contato com o solvente, a droga começa a se difundir para fora dos interstícios da estrutura polimérica. A expressão matemática que rege a difusão pela matriz é uma equação diferencial parcial, a segunda lei de Fick:</span></p>
<p style="text-align:justify;line-height:14.25pt;background:white;"><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle%5Cfrac%7B%5Cpartial+C%7D%7B%5Cpartial+t%7D%3DD%5Cnabla%5E%7B2%7DC&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle\frac{\partial C}{\partial t}=D\nabla^{2}C' title='\displaystyle\frac{\partial C}{\partial t}=D\nabla^{2}C' class='latex' /></p>
<p style="text-align:justify;line-height:14.25pt;background:white;">Iremos analisar uma situação em que a matriz polimérica seja esférica. Em coordenadas esféricas, esta expressão é modificada para:</p>
<p style="text-align:justify;line-height:14.25pt;background:white;"><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle%5Cfrac%7B%5Cpartial+C%7D%7B%5Cpartial+t%7D%3D%5Cfrac%7B1%7D%7Br%5E%7B2%7D%7D%5Cleft%5C%7B%5Cfrac%7B%5Cpartial%7D%7B%5Cpartial+r%7D%5Cleft%28Dr%5E%7B2%7D%5Cfrac%7B%5Cpartial+C%7D%7B%5Cpartial+r%7D%5Cright%29%2B%5Cfrac%7B1%7D%7Bsen%5Ctheta%7D%5Cfrac%7B%5Cpartial%7D%7B%5Cpartial+%5Ctheta%7D%5Cleft%28Dsen%5Ctheta%5Cfrac%7B%5Cpartial+C%7D%7B%5Cpartial+%5Ctheta%7D%5Cright%29%2B%5Cfrac%7BD%7D%7Bsen%5E%7B2%7D%5Ctheta%7D%5Cfrac%7B%5Cpartial%5E%7B2%7DC%7D%7B%5Cpartial%5Cphi%5E%7B2%7D%7D%5Cright%5C%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle\frac{\partial C}{\partial t}=\frac{1}{r^{2}}\left\{\frac{\partial}{\partial r}\left(Dr^{2}\frac{\partial C}{\partial r}\right)+\frac{1}{sen\theta}\frac{\partial}{\partial \theta}\left(Dsen\theta\frac{\partial C}{\partial \theta}\right)+\frac{D}{sen^{2}\theta}\frac{\partial^{2}C}{\partial\phi^{2}}\right\}' title='\displaystyle\frac{\partial C}{\partial t}=\frac{1}{r^{2}}\left\{\frac{\partial}{\partial r}\left(Dr^{2}\frac{\partial C}{\partial r}\right)+\frac{1}{sen\theta}\frac{\partial}{\partial \theta}\left(Dsen\theta\frac{\partial C}{\partial \theta}\right)+\frac{D}{sen^{2}\theta}\frac{\partial^{2}C}{\partial\phi^{2}}\right\}' class='latex' /></p>
<p style="text-align:justify;line-height:14.25pt;background:white;"><span style="font-size:10pt;font-family:&quot;color:black;">Considerado a difusão como radial, a equação se reduz a:</span></p>
<p style="text-align:justify;line-height:14.25pt;background:white;"><span style="font-size:10pt;font-family:&quot;color:black;"><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle%5Cfrac%7B%5Cpartial+C%7D%7B%5Cpartial+t%7D%3D%5Cfrac%7BD%7D%7Br%5E%7B2%7D%7D%5Cleft%5C%7B%5Cfrac%7B%5Cpartial%7D%7B%5Cpartial+r%7D%5Cleft%28r%5E%7B2%7D%5Cfrac%7B%5Cpartial+C%7D%7B%5Cpartial+r%7D%5Cright%29%5Cright%5C%7D%3DD%5Cleft%5C%7B%5Cfrac%7B%5Cpartial%5E%7B2%7D+C%7D%7B%5Cpartial+r%5E%7B2%7D%7D%2B%5Cfrac%7B2%7D%7Br%7D%5Cfrac%7B%5Cpartial+C%7D%7B%5Cpartial+r%7D%5Cright%5C%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle\frac{\partial C}{\partial t}=\frac{D}{r^{2}}\left\{\frac{\partial}{\partial r}\left(r^{2}\frac{\partial C}{\partial r}\right)\right\}=D\left\{\frac{\partial^{2} C}{\partial r^{2}}+\frac{2}{r}\frac{\partial C}{\partial r}\right\}' title='\displaystyle\frac{\partial C}{\partial t}=\frac{D}{r^{2}}\left\{\frac{\partial}{\partial r}\left(r^{2}\frac{\partial C}{\partial r}\right)\right\}=D\left\{\frac{\partial^{2} C}{\partial r^{2}}+\frac{2}{r}\frac{\partial C}{\partial r}\right\}' class='latex' /> (1)</span></p>
<p style="text-align:justify;line-height:14.25pt;background:white;"><span style="font-size:10pt;font-family:&quot;color:black;">onde D é o coeficiente de difusão, C é a concentração da droga e r é a distância do centro da esfera. </span></p>
<p style="text-align:justify;line-height:14.25pt;background:white;"><span style="font-size:10pt;font-family:&quot;color:black;">Podemos então determinar as condições iniciais e de fronteira. Ao colocar a forma farmacêutica em contato com a água, a concentração da droga na superfície da esfera é zero, enquanto que a concentração dentro da esfera é única para todos os pontos. Se a esfera tem raio a, segue que:</span></p>
<p style="text-align:justify;line-height:14.25pt;background:white;"><img src='http://l.wordpress.com/latex.php?latex=C%28a%2Ct%29%3D0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='C(a,t)=0' title='C(a,t)=0' class='latex' />, <img src='http://l.wordpress.com/latex.php?latex=t%3E0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='t&gt;0' title='t&gt;0' class='latex' /> (2)</p>
<p style="text-align:justify;line-height:14.25pt;background:white;"><img src='http://l.wordpress.com/latex.php?latex=C%28r%2C0%29%3DC_%7B0%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='C(r,0)=C_{0}' title='C(r,0)=C_{0}' class='latex' />, <img src='http://l.wordpress.com/latex.php?latex=0%5Cle+r+%3C+a&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='0\le r &lt; a' title='0\le r &lt; a' class='latex' /> (3)</p>
<p style="text-align:justify;line-height:14.25pt;background:white;">Podemos resolver este problema por separação de variáveis. Podemos primeiro fazer uma substituição que facilitará a resolução:</p>
<p style="text-align:justify;line-height:14.25pt;background:white;"><img src='http://l.wordpress.com/latex.php?latex=u%3DCr&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='u=Cr' title='u=Cr' class='latex' /> (4)</p>
<p style="text-align:justify;line-height:14.25pt;background:white;">Perceba que:</p>
<p style="text-align:justify;line-height:14.25pt;background:white;"><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle%5Cfrac%7B%5Cpartial+u%7D%7B%5Cpartial+t%7D+%3D+r%5Cfrac%7B%5Cpartial+C%7D%7B%5Cpartial+t%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle\frac{\partial u}{\partial t} = r\frac{\partial C}{\partial t}' title='\displaystyle\frac{\partial u}{\partial t} = r\frac{\partial C}{\partial t}' class='latex' /></p>
<p style="text-align:justify;line-height:14.25pt;background:white;">Isolando <img src='http://l.wordpress.com/latex.php?latex=%5Cpartial+C%2F%5Cpartial+t&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\partial C/\partial t' title='\partial C/\partial t' class='latex' />,</p>
<p style="text-align:justify;line-height:14.25pt;background:white;"><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle%5Cfrac%7B%5Cpartial+C%7D%7B%5Cpartial+t%7D%3D%5Cfrac%7B1%7D%7Br%7D%5Cfrac%7B%5Cpartial+u%7D%7B%5Cpartial+t%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle\frac{\partial C}{\partial t}=\frac{1}{r}\frac{\partial u}{\partial t}' title='\displaystyle\frac{\partial C}{\partial t}=\frac{1}{r}\frac{\partial u}{\partial t}' class='latex' /> (5)</p>
<p style="text-align:justify;line-height:14.25pt;background:white;">Perceba também que:</p>
<p style="text-align:justify;line-height:14.25pt;background:white;"><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle%5Cfrac%7B%5Cpartial%5E%7B2%7D+u%7D%7B%5Cpartial+r%5E%7B2%7D%7D+%3D+r%5Cfrac%7B%5Cpartial%5E%7B2%7D+C%7D%7B%5Cpartial+r%5E%7B2%7D%7D%2B2%5Cfrac%7B%5Cpartial+C%7D%7B%5Cpartial+r%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle\frac{\partial^{2} u}{\partial r^{2}} = r\frac{\partial^{2} C}{\partial r^{2}}+2\frac{\partial C}{\partial r}' title='\displaystyle\frac{\partial^{2} u}{\partial r^{2}} = r\frac{\partial^{2} C}{\partial r^{2}}+2\frac{\partial C}{\partial r}' class='latex' /></p>
<p style="text-align:justify;line-height:14.25pt;background:white;">Assim:</p>
<p style="text-align:justify;line-height:14.25pt;background-image:initial;background-repeat:initial;background-attachment:initial;background-color:white;background-position:initial initial;"><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle%5Cfrac%7B%5Cpartial%5E%7B2%7D+C%7D%7B%5Cpartial+r%5E%7B2%7D%7D%2B%5Cfrac%7B2%7D%7Br%7D%5Cfrac%7B%5Cpartial+C%7D%7B%5Cpartial+r%7D%3D%5Cfrac%7B1%7D%7Br%7D%5Cfrac%7B%5Cpartial%5E%7B2%7D+u%7D%7B%5Cpartial+r%5E%7B2%7D%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle\frac{\partial^{2} C}{\partial r^{2}}+\frac{2}{r}\frac{\partial C}{\partial r}=\frac{1}{r}\frac{\partial^{2} u}{\partial r^{2}}' title='\displaystyle\frac{\partial^{2} C}{\partial r^{2}}+\frac{2}{r}\frac{\partial C}{\partial r}=\frac{1}{r}\frac{\partial^{2} u}{\partial r^{2}}' class='latex' /> (6)</p>
<p style="text-align:justify;line-height:14.25pt;background:white;"><span style="font-size:10pt;font-family:&quot;color:black;">Substituindo (5) em (6) em (1), assim como (2) e (3) em (4), chegamos ao PVIF (cuja EDP é análoga a conhecida equação do calor em uma dimensão):</span></p>
<p style="text-align:justify;line-height:14.25pt;background:white;"><span style="font-size:10pt;font-family:&quot;color:black;"><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle%5Cfrac%7B%5Cpartial+u%7D%7B%5Cpartial+t%7D%3DD%5Cfrac%7B%5Cpartial%5E%7B2%7D+u%7D%7B%5Cpartial+r%5E%7B2%7D%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle\frac{\partial u}{\partial t}=D\frac{\partial^{2} u}{\partial r^{2}}' title='\displaystyle\frac{\partial u}{\partial t}=D\frac{\partial^{2} u}{\partial r^{2}}' class='latex' /> (7)</span></p>
<p style="text-align:justify;line-height:14.25pt;background:white;"><img src='http://l.wordpress.com/latex.php?latex=u%280%2Ct%29%3Du%28a%2Ct%29%3D0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='u(0,t)=u(a,t)=0' title='u(0,t)=u(a,t)=0' class='latex' /> , <img src='http://l.wordpress.com/latex.php?latex=t%3E0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='t&gt;0' title='t&gt;0' class='latex' /> ( 8 )</p>
<p style="text-align:justify;line-height:14.25pt;background:white;"><img src='http://l.wordpress.com/latex.php?latex=u%28r%2C0%29%3DrC_%7B0%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='u(r,0)=rC_{0}' title='u(r,0)=rC_{0}' class='latex' />, <img src='http://l.wordpress.com/latex.php?latex=0%3Cr%3Ca&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='0&lt;r&lt;a' title='0&lt;r&lt;a' class='latex' /> (9)</p>
<p style="text-align:justify;line-height:14.25pt;background:white;">Podemos resolver por separação de variáveis:</p>
<p style="text-align:justify;line-height:14.25pt;background:white;"><img src='http://l.wordpress.com/latex.php?latex=u%28r%2Ct%29%3DA%28r%29B%28t%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='u(r,t)=A(r)B(t)' title='u(r,t)=A(r)B(t)' class='latex' /> (10)</p>
<p style="text-align:justify;line-height:14.25pt;background:white;">Substituindo (10) em (7):</p>
<p style="text-align:justify;line-height:14.25pt;background:white;"><img src='http://l.wordpress.com/latex.php?latex=A%28r%29B%27%28t%29%3DDA%27%27%28r%29B%28t%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='A(r)B&#039;(t)=DA&#039;&#039;(r)B(t)' title='A(r)B&#039;(t)=DA&#039;&#039;(r)B(t)' class='latex' /></p>
<p style="text-align:justify;line-height:14.25pt;background:white;"><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle%5Cfrac%7BB%27%28t%29%7D%7BDB%28t%29%7D%3D%5Cfrac%7BA%27%27%28r%29%7D%7BA%28r%29%7D%3D%5Csigma&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle\frac{B&#039;(t)}{DB(t)}=\frac{A&#039;&#039;(r)}{A(r)}=\sigma' title='\displaystyle\frac{B&#039;(t)}{DB(t)}=\frac{A&#039;&#039;(r)}{A(r)}=\sigma' class='latex' /></p>
<p style="text-align:justify;line-height:14.25pt;background-image:initial;background-repeat:initial;background-attachment:initial;background-color:white;background-position:initial initial;">Chegamos, então, a duas EDO&#8217;s com as condições determinadas por (8) e (9)</p>
<p style="text-align:justify;line-height:14.25pt;background-image:initial;background-repeat:initial;background-attachment:initial;background-color:white;background-position:initial initial;"><img src='http://l.wordpress.com/latex.php?latex=A%27%27%28r%29-%5Csigma+A%28r%29%3D0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='A&#039;&#039;(r)-\sigma A(r)=0' title='A&#039;&#039;(r)-\sigma A(r)=0' class='latex' /> (11)</p>
<p style="text-align:justify;line-height:14.25pt;background-image:initial;background-repeat:initial;background-attachment:initial;background-color:white;background-position:initial initial;"><img src='http://l.wordpress.com/latex.php?latex=A%280%29%3DA%28a%29%3D0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='A(0)=A(a)=0' title='A(0)=A(a)=0' class='latex' /> (12)</p>
<p style="text-align:justify;line-height:14.25pt;background-image:initial;background-repeat:initial;background-attachment:initial;background-color:white;background-position:initial initial;"><img src='http://l.wordpress.com/latex.php?latex=B%27%28t%29-%5Csigma+DB%28t%29%3D0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='B&#039;(t)-\sigma DB(t)=0' title='B&#039;(t)-\sigma DB(t)=0' class='latex' /> (13)</p>
<p style="text-align:justify;line-height:14.25pt;background-image:initial;background-repeat:initial;background-attachment:initial;background-color:white;background-position:initial initial;">Se multiplicarmos a equação (11) por A(r) e fizermos uma integração de 0 a a:</p>
<p style="text-align:justify;line-height:14.25pt;background-image:initial;background-repeat:initial;background-attachment:initial;background-color:white;background-position:initial initial;"><img src='http://l.wordpress.com/latex.php?latex=%5Cint_%7B0%7D%5E%7Ba%7D+A%27%27%28r%29A%28r%29dr-%5Csigma%5Cint_%7B0%7D%5E%7Ba%7D+A%5E%7B2%7D%28r%29dr%3D0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\int_{0}^{a} A&#039;&#039;(r)A(r)dr-\sigma\int_{0}^{a} A^{2}(r)dr=0' title='\int_{0}^{a} A&#039;&#039;(r)A(r)dr-\sigma\int_{0}^{a} A^{2}(r)dr=0' class='latex' /></p>
<p style="text-align:justify;line-height:14.25pt;background-image:initial;background-repeat:initial;background-attachment:initial;background-color:white;background-position:initial initial;">Usando integração por partes na primeira integral:</p>
<p style="text-align:justify;line-height:14.25pt;background-image:initial;background-repeat:initial;background-attachment:initial;background-color:white;background-position:initial initial;"><img src='http://l.wordpress.com/latex.php?latex=%28A%27%28r%29A%28r%29%29%7C_%7Br%3D0%7D%5E%7Br%3Da%7D-%5Cint_%7B0%7D%5E%7Ba%7D+A%27%28r%29%5E%7B2%7Ddr-%5Csigma%5Cint_%7B0%7D%5E%7Ba%7D+A%5E%7B2%7D%28r%29dr%3D0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='(A&#039;(r)A(r))|_{r=0}^{r=a}-\int_{0}^{a} A&#039;(r)^{2}dr-\sigma\int_{0}^{a} A^{2}(r)dr=0' title='(A&#039;(r)A(r))|_{r=0}^{r=a}-\int_{0}^{a} A&#039;(r)^{2}dr-\sigma\int_{0}^{a} A^{2}(r)dr=0' class='latex' /></p>
<p style="text-align:justify;line-height:14.25pt;background-image:initial;background-repeat:initial;background-attachment:initial;background-color:white;background-position:initial initial;">Chegamos a conclusão que só temos autofunções para autovalores <img src='http://l.wordpress.com/latex.php?latex=%5Csigma%5Cle+0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\sigma\le 0' title='\sigma\le 0' class='latex' />:</p>
<p style="text-align:justify;line-height:14.25pt;background-image:initial;background-repeat:initial;background-attachment:initial;background-color:white;background-position:initial initial;"><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle%5Csigma%3D-%5Cfrac%7B%5Cint_%7B0%7D%5E%7Ba%7D+A%27%28r%29%5E%7B2%7Ddr%7D%7B%5Cint_%7B0%7D%5E%7Ba%7D+A%5E%7B2%7D%28r%29dr%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle\sigma=-\frac{\int_{0}^{a} A&#039;(r)^{2}dr}{\int_{0}^{a} A^{2}(r)dr}' title='\displaystyle\sigma=-\frac{\int_{0}^{a} A&#039;(r)^{2}dr}{\int_{0}^{a} A^{2}(r)dr}' class='latex' /></p>
<p style="text-align:justify;line-height:14.25pt;background-image:initial;background-repeat:initial;background-attachment:initial;background-color:white;background-position:initial initial;">Se <img src='http://l.wordpress.com/latex.php?latex=%5Csigma%3D0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\sigma=0' title='\sigma=0' class='latex' />, <img src='http://l.wordpress.com/latex.php?latex=A%27%27%28r%29%3D0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='A&#039;&#039;(r)=0' title='A&#039;&#039;(r)=0' class='latex' /> e, pelas condições em (11), não há autofunções.</p>
<p style="text-align:justify;line-height:14.25pt;background-image:initial;background-repeat:initial;background-attachment:initial;background-color:white;background-position:initial initial;">Se <img src='http://l.wordpress.com/latex.php?latex=%5Csigma+%3C0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\sigma &lt;0' title='\sigma &lt;0' class='latex' />,</p>
<p style="text-align:justify;line-height:14.25pt;background-image:initial;background-repeat:initial;background-attachment:initial;background-color:white;background-position:initial initial;"><img src='http://l.wordpress.com/latex.php?latex=A%28r%29%3DK_%7B1%7D%5Ccos%28%5Csqrt%7B-%5Csigma%7Dr%29%2BK_%7B2%7Dsen%28%5Csqrt%7B-%5Csigma%7Dr%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='A(r)=K_{1}\cos(\sqrt{-\sigma}r)+K_{2}sen(\sqrt{-\sigma}r)' title='A(r)=K_{1}\cos(\sqrt{-\sigma}r)+K_{2}sen(\sqrt{-\sigma}r)' class='latex' /></p>
<p style="text-align:justify;line-height:14.25pt;background-image:initial;background-repeat:initial;background-attachment:initial;background-color:white;background-position:initial initial;">Pela condição <img src='http://l.wordpress.com/latex.php?latex=A%280%29%3D0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='A(0)=0' title='A(0)=0' class='latex' />, <img src='http://l.wordpress.com/latex.php?latex=K_%7B1%7D%3D0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='K_{1}=0' title='K_{1}=0' class='latex' /></p>
<p style="text-align:justify;line-height:14.25pt;background-image:initial;background-repeat:initial;background-attachment:initial;background-color:white;background-position:initial initial;">Pela condição <img src='http://l.wordpress.com/latex.php?latex=A%28a%29%3D0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='A(a)=0' title='A(a)=0' class='latex' />, <img src='http://l.wordpress.com/latex.php?latex=K_%7B2%7Dsen%28a%5Csqrt%7B-%5Csigma%7D%29%3D0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='K_{2}sen(a\sqrt{-\sigma})=0' title='K_{2}sen(a\sqrt{-\sigma})=0' class='latex' />, que só produz autofunções para:</p>
<p style="text-align:justify;line-height:14.25pt;background-image:initial;background-repeat:initial;background-attachment:initial;background-color:white;background-position:initial initial;"><img src='http://l.wordpress.com/latex.php?latex=a%5Csqrt%7B-%5Csigma%7D%3Dn%5Cpi&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='a\sqrt{-\sigma}=n\pi' title='a\sqrt{-\sigma}=n\pi' class='latex' /></p>
<p style="text-align:justify;line-height:14.25pt;background-image:initial;background-repeat:initial;background-attachment:initial;background-color:white;background-position:initial initial;">e, portanto:</p>
<p style="text-align:justify;line-height:14.25pt;background-image:initial;background-repeat:initial;background-attachment:initial;background-color:white;background-position:initial initial;"><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle%5Csigma%3D-%5Cleft%28%5Cfrac%7Bn%5Cpi%7D%7Ba%7D%5Cright%29%5E%7B2%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle\sigma=-\left(\frac{n\pi}{a}\right)^{2}' title='\displaystyle\sigma=-\left(\frac{n\pi}{a}\right)^{2}' class='latex' />, <img src='http://l.wordpress.com/latex.php?latex=n%3D1%2C2%2C3%2C...&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='n=1,2,3,...' title='n=1,2,3,...' class='latex' /> (14)</p>
<p style="text-align:justify;line-height:14.25pt;background-image:initial;background-repeat:initial;background-attachment:initial;background-color:white;background-position:initial initial;"><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle+A_%7Bn%7D%28r%29%3Dsen%5Cleft%28%5Cfrac%7Bn%5Cpi+r%7D%7Ba%7D%5Cright%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle A_{n}(r)=sen\left(\frac{n\pi r}{a}\right)' title='\displaystyle A_{n}(r)=sen\left(\frac{n\pi r}{a}\right)' class='latex' /> (15)</p>
<p style="text-align:justify;line-height:14.25pt;background-image:initial;background-repeat:initial;background-attachment:initial;background-color:white;background-position:initial initial;">Substituindo (14) em (13):</p>
<p style="text-align:justify;line-height:14.25pt;background-image:initial;background-repeat:initial;background-attachment:initial;background-color:white;background-position:initial initial;"><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle+B%27%28t%29%2B%5Cleft%28%5Cfrac%7Bn%5Cpi%7D%7Ba%7D%5Cright%29%5E%7B2%7DDB%28t%29%3D0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle B&#039;(t)+\left(\frac{n\pi}{a}\right)^{2}DB(t)=0' title='\displaystyle B&#039;(t)+\left(\frac{n\pi}{a}\right)^{2}DB(t)=0' class='latex' /></p>
<p style="text-align:justify;line-height:14.25pt;background-image:initial;background-repeat:initial;background-attachment:initial;background-color:white;background-position:initial initial;">Logo, <img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle+B_%7Bn%7D%28t%29%3D%5Calpha_%7Bn%7D+e%5E%7B%5Cdisplaystyle+-%5Cfrac%7Bn%5E%7B2%7D%5Cpi%5E%7B2%7D+D%7D%7Ba%5E%7B2%7D%7D+t%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle B_{n}(t)=\alpha_{n} e^{\displaystyle -\frac{n^{2}\pi^{2} D}{a^{2}} t}' title='\displaystyle B_{n}(t)=\alpha_{n} e^{\displaystyle -\frac{n^{2}\pi^{2} D}{a^{2}} t}' class='latex' /> (16)</p>
<p style="text-align:justify;line-height:14.25pt;background-image:initial;background-repeat:initial;background-attachment:initial;background-color:white;background-position:initial initial;">Através de (10), (15) e (16), e pelo princípio da superposição, chegamos à solução geral:</p>
<p style="text-align:justify;line-height:14.25pt;background-image:initial;background-repeat:initial;background-attachment:initial;background-color:white;background-position:initial initial;"><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle+u%28r%2Ct%29%3D%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%5Calpha_%7Bn%7D+e%5E%7B%5Cdisplaystyle-%5Cfrac%7Bn%5E%7B2%7D%5Cpi%5E%7B2%7D+D%7D%7Ba%5E%7B2%7D%7D+t%7Dsen%5Cleft%28%5Cfrac%7Bn%5Cpi+r%7D%7Ba%7D%5Cright%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle u(r,t)=\sum_{n=1}^{\infty}\alpha_{n} e^{\displaystyle-\frac{n^{2}\pi^{2} D}{a^{2}} t}sen\left(\frac{n\pi r}{a}\right)' title='\displaystyle u(r,t)=\sum_{n=1}^{\infty}\alpha_{n} e^{\displaystyle-\frac{n^{2}\pi^{2} D}{a^{2}} t}sen\left(\frac{n\pi r}{a}\right)' class='latex' /> (17)</p>
<p style="text-align:justify;line-height:14.25pt;background-image:initial;background-repeat:initial;background-attachment:initial;background-color:white;background-position:initial initial;">Falta agora, determinar os coeficientes <img src='http://l.wordpress.com/latex.php?latex=%5Calpha_%7Bn%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\alpha_{n}' title='\alpha_{n}' class='latex' />. Usaremos a condição (9)</p>
<p style="text-align:justify;line-height:14.25pt;background-image:initial;background-repeat:initial;background-attachment:initial;background-color:white;background-position:initial initial;"><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle+u%28r%2C0%29%3D%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%5Calpha_%7Bn%7D+sen%5Cleft%28%5Cfrac%7Bn%5Cpi+r%7D%7Ba%7D%5Cright%29+%3D+rC_%7B0%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle u(r,0)=\sum_{n=1}^{\infty}\alpha_{n} sen\left(\frac{n\pi r}{a}\right) = rC_{0}' title='\displaystyle u(r,0)=\sum_{n=1}^{\infty}\alpha_{n} sen\left(\frac{n\pi r}{a}\right) = rC_{0}' class='latex' /></p>
<p style="text-align:justify;line-height:14.25pt;background-image:initial;background-repeat:initial;background-attachment:initial;background-color:white;background-position:initial initial;">Assim,</p>
<p style="text-align:justify;line-height:14.25pt;background-image:initial;background-repeat:initial;background-attachment:initial;background-color:white;background-position:initial initial;"><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle%5Calpha_%7Bn%7D%3D%5Cfrac%7B2C_%7B0%7D%7D%7Ba%7D%5Cint_%7B0%7D%5E%7Ba%7Dr+sen%5Cleft%28%5Cfrac%7Bn%5Cpi+r%7D%7Ba%7D%5Cright%29dr%3D-%5Cfrac%7B2C_%7B0%7Da%5E%7B2%7D%5Ccos%28n%5Cpi%29%7D%7Bna%5Cpi%7D%3D-%5Cfrac%7B2C_%7B0%7Da%28-1%29%5E%7Bn%7D%7D%7Bn%5Cpi%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle\alpha_{n}=\frac{2C_{0}}{a}\int_{0}^{a}r sen\left(\frac{n\pi r}{a}\right)dr=-\frac{2C_{0}a^{2}\cos(n\pi)}{na\pi}=-\frac{2C_{0}a(-1)^{n}}{n\pi}' title='\displaystyle\alpha_{n}=\frac{2C_{0}}{a}\int_{0}^{a}r sen\left(\frac{n\pi r}{a}\right)dr=-\frac{2C_{0}a^{2}\cos(n\pi)}{na\pi}=-\frac{2C_{0}a(-1)^{n}}{n\pi}' class='latex' /> (18)</p>
<p style="text-align:justify;line-height:14.25pt;background-image:initial;background-repeat:initial;background-attachment:initial;background-color:white;background-position:initial initial;">E, finalmente, usando (4), (17) e (18):</p>
<p style="text-align:justify;line-height:14.25pt;background-image:initial;background-repeat:initial;background-attachment:initial;background-color:white;background-position:initial initial;"><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle+C%28r%2Ct%29%3D-%5Cfrac%7B2C_%7B0%7Da%7D%7Br%5Cpi%7D%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%5Cfrac%7B%28-1%29%5E%7Bn%7D%7D%7Bn%7D+e%5E%7B%5Cdisplaystyle-%5Cfrac%7Bn%5E%7B2%7D%5Cpi%5E%7B2%7D+D%7D%7Ba%5E%7B2%7D%7D+t%7Dsen%5Cleft%28%5Cfrac%7Bn%5Cpi+r%7D%7Ba%7D%5Cright%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle C(r,t)=-\frac{2C_{0}a}{r\pi}\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n} e^{\displaystyle-\frac{n^{2}\pi^{2} D}{a^{2}} t}sen\left(\frac{n\pi r}{a}\right)' title='\displaystyle C(r,t)=-\frac{2C_{0}a}{r\pi}\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n} e^{\displaystyle-\frac{n^{2}\pi^{2} D}{a^{2}} t}sen\left(\frac{n\pi r}{a}\right)' class='latex' /></p>
<p style="text-align:justify;line-height:14.25pt;background-image:initial;background-repeat:initial;background-attachment:initial;background-color:white;background-position:initial initial;">Referências:</p>
<p>C.-J Kim. Controlled release dosage form design. Technomic, Lancaster PA, 2000, 301 pp.</p>
<p>J. Crank. The mathematics of diffusion, Second Edition, Claremdon Press, Oxford, 1975.</p>
<p style="text-align:justify;line-height:14.25pt;background-image:initial;background-repeat:initial;background-attachment:initial;background-color:white;background-position:initial initial;">
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		<title>Fármacos usados no tratamento da AIDS</title>
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		<pubDate>Sat, 11 Jul 2009 02:37:49 +0000</pubDate>
		<dc:creator>Augusto Maeda</dc:creator>
				<category><![CDATA[Farmacodinâmica]]></category>
		<category><![CDATA[Farmacologia]]></category>

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		<description><![CDATA[Há cinco categorias de fármacos usados no combate do HIV:

Inibidores nucleosídicos/nucleotídicos da transcriptase reversa
Inibidores não-nucleosídicos da transcriptase reversa
Inibidores da protease
Inibidores da fusão do HIV com as células do hospedeiro
Inibidores da Integrase

Seguem alguns vídeos do youtube ilustrando a ação destes fármacos (não repare na propaganda que alguns vídeos fazem)
Inibidores Nucleosídeos/Nucleotídicos da Transcriptase Reversa (NRTIs - Nucleoside Reverse [...]<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=campodearroz.wordpress.com&blog=2694588&post=245&subd=campodearroz&ref=&feed=1" />]]></description>
			<content:encoded><![CDATA[<div class='snap_preview'><br /><p>Há cinco categorias de fármacos usados no combate do HIV:</p>
<ol>
<li>Inibidores nucleosídicos/nucleotídicos da transcriptase reversa</li>
<li>Inibidores não-nucleosídicos da transcriptase reversa</li>
<li>Inibidores da protease</li>
<li>Inibidores da fusão do HIV com as células do hospedeiro</li>
<li>Inibidores da Integrase</li>
</ol>
<p>Seguem alguns vídeos do youtube ilustrando a ação destes fármacos (não repare na propaganda que alguns vídeos fazem)</p>
<p>Inibidores Nucleosídeos/Nucleotídicos da Transcriptase Reversa (NRTIs - Nucleoside Reverse Transcriptase Inhibitors; NTRTIs &#8211; Nucleotide Reverse Transcriptase Inhibitors)</p>
<p style="text-align:center;"><span style="text-align:center; display: block;"><a href="http://campodearroz.wordpress.com/2009/07/10/farmacos-usados-no-tratamento-da-aids/"><img src="http://img.youtube.com/vi/qYUnDzDO-Ic/2.jpg" alt="" /></a></span></p>
<h5 style="text-align:center;"><span style="font-weight:normal;">NRTIs and NTRTIs are antiretroviral drugs which following absorption into the body must be metabolized in the affected cell in order to become active. The NRTI metabolism involves three steps, while NTRTIs require only two steps. NRTIS are analogs of DNA building blocks. NRTIs work by imitating natural DNA building blocks. When building a new viral DNA chain, reverse transcriptase enzyme binds to NRTIs instead of binding to the natural occurring DNA building blocks. Because the structure of the NRTIs, there is no allowed attachment to the next DNA building block. DNA chain growth is terminated. HIV has developed two drug resistance mechanisms to NRTIs. Mechanism number one is decreased binding of the enzyme reverse transcriptase to NRTIs. ?(The remaining mutations that have this sort of impact on the binding affinity). The second mechanism to NRTI resistance is the increased removal of the NRTI from the elongating DNA chain.</span></h5>
<p><span style="font-weight:normal;"><br />
</span></p>
<p class="MsoNormal">
<p class="MsoNormal" style="text-align:center;">AZT</p>
<p class="MsoNormal" style="text-align:center;"><span style="text-align:center; display: block;"><a href="http://campodearroz.wordpress.com/2009/07/10/farmacos-usados-no-tratamento-da-aids/"><img src="http://img.youtube.com/vi/I0Y5R7r2P6g/2.jpg" alt="" /></a></span></p>
<p class="MsoNormal" style="text-align:center;">
<p class="MsoNormal" style="text-align:center;">
<p class="MsoNormal">
<p class="MsoNormal">Inibidores não-nucleosídicos da transcriptase reversa (NNRTIs &#8211; Non-nucleoside reverse transcriptase inhibitors)</p>
<p class="MsoNormal" style="text-align:center;"><span style="text-align:center; display: block;"><a href="http://campodearroz.wordpress.com/2009/07/10/farmacos-usados-no-tratamento-da-aids/"><img src="http://img.youtube.com/vi/3n_MYZEfnxU/2.jpg" alt="" /></a></span></p>
<p style="text-align:center;">Instead of competing with naturally occurring DNA building blocks as do the NRTIs, NNRTIs bind tightly to the enzyme reverse transcriptase thereby preventing viral RNA from being converted to DNA. In contrast to the NRTIs mutations, drug resistance to NNRTIs is normally associated with mutations that are proximal to the drug binding site on reverse transcriptase. And that distortion of this binding pocket is the mechanism of resistance.</p>
<p style="text-align:center;">
<p style="text-align:center;">
<p class="MsoNormal" style="text-align:center;">
<p class="MsoNormal">
<p class="MsoNormal" style="text-align:center;"><span style="text-align:center; display: block;"><a href="http://campodearroz.wordpress.com/2009/07/10/farmacos-usados-no-tratamento-da-aids/"><img src="http://img.youtube.com/vi/RUUyd5bE9vQ/2.jpg" alt="" /></a></span></p>
<p class="MsoNormal" style="text-align:center;">HIV1 is a virus with a high replication rate. Within infected CD4 cells the viral nucleocapsid breaks open releasing two RNA strands and essential replication enzymes such as HIV1 reverse transcriptase. It is a heterodimer with a p51 subunit and a p66 subunit. The p66 subunit contains a finger upon and a thumb region resembling a (____ hand?). Reverse transcriptase has two catalytic domains, the ribonuclease H active site and the polymerase active site. Here single stranded viral RNA is transcribed into a RNA-DNA double helix. Ribonuclease H breaks down the RNA. The polymerase then completes the remaining DNA strand to form a DNA double helix. This proviral DNA contains the genetic material of HIV1. Therapeutical suppression of viral replication slows down the decline of CD4 cells and the disease progression. Nucleoside reverse transcriptase inhibitors (NRTIs) inhibit the polymerase active site. After metabolism to non function nucleotides their incorporation causes chain termination. The non-nucleoside reverse transcriptase inhibitors (NNRTIs) form another class of powerful antiretroviral agents. They inhibit reverse transcriptase by reducing its conformational flexibility. The thumb region of reverse transcriptase is flexible. It opens and closes like a hand. Only the closed position allows transcription of RNA. The base of the thumb has a hydrophobic pocket-like binding site. This is the target of NNRTIs. Nevirapine is an important representative of this class. It does not need to be metabolized. In its native form, nevirapine binds in the pocket. This locks the thumb in the open position and prevents the transcription of RNA. Thus nevirapine stops viral replication. Due to this mechanism, nevirapine is a potent partner in the combination therapy of HIV1 infection.</p>
<p class="MsoNormal" style="text-align:center;">
<p class="MsoNormal" style="text-align:center;">
<p class="MsoNormal">
<p class="MsoNormal"><span lang="EN-US">Inibidores da Protease (PIs &#8211; Protease Inhibitors)<br />
</span></p>
<p class="MsoNormal" style="text-align:center;"><span style="text-align:center; display: block;"><a href="http://campodearroz.wordpress.com/2009/07/10/farmacos-usados-no-tratamento-da-aids/"><img src="http://img.youtube.com/vi/yYZgFndtfzc/2.jpg" alt="" /></a></span></p>
<p style="text-align:center;">Protease inhibitors slow down HIV replication of the viruses integrated into the host cells DNA in acutely and chronically infected cells. During the maturation process of new virions, the HIV protease enzyme cuts newly produced elements of virus (____)? polyproteins into essential functional protein products. This critical process occurs as each new virion grows from the membrane of an HIV infected cell and continues after the immature virus is released from the cell. The host cell is eventually destroyed in the process. If the polyproteins are not cut the new virus fails to mature and is incapable of infecting a new cell. Protease inhibitors are able to interfere with the functioning of the native protease enzyme. They disable the enzyme before it can cleave the new viral polyprotein into its essential protein products thereby (_____)? the new virion immature and non infectious. The mechanism of resistance to protease inhibitors may be more complex than originally thought and is not as well understood as NRTI and NNRTI resistance. When PI resistance occurs, PI interference with mutant protease is no longer adequate.</p>
<h5 style="text-align:center;"><span style="font-weight:normal;"><br />
</span></h5>
<p class="MsoNormal">
<p class="MsoNormal" style="text-align:center;"><span style="text-align:center; display: block;"><a href="http://campodearroz.wordpress.com/2009/07/10/farmacos-usados-no-tratamento-da-aids/"><img src="http://img.youtube.com/vi/yLDpTM0eSgM/2.jpg" alt="" /></a></span></p>
<p class="MsoNormal" style="text-align:center;">The ability of HIV to infect cells is essential to its replication. Infection of a suitable host cell such as a CD4+ T lymphocyte leads to integration of the proviral DNA into the host cell genome. It now contains the genetic information for the building blocks of new HIV virus including two viral RNA strands and three viral enzymes, one of which is the HIV protease. The protease plays a key role in the formation of infectious virus. Activation of the cell induces transcription of a proviral DNA into a messenger RNA. The viral messenger RNA migrates into the cytoplasm where components of a new virus are synthesized. Some of these components have to be processed by the virus protease which cleaves longer proteins into smaller core proteins. This step is crucial to create an infectious virus. Two viral RNA strands, the viral enzymes and core proteins are assembled. This immature viral particle leaves the cell, acquiring a new (envelope?) host and viral proteins. The virus matures and becomes ready to infect other cells. Inhibition of HIV protease can stop this replication cycle. The introduction of peptidic protease inhibitors represented a milestone in treatment of HIV infection. These types of protease inhibitors bind to protease via an extensive network of hydrogen bonds resulting in the drug being attached to the active site. This blocks the action of the protease preventing HIV replication. Unfortunately mutations in HIV protease occur frequently and may limit the use of current protease inhibitors. A change in even a few amino acids can prevent these drugs binding to protease leading to broad cross resistance. Therefore there is a need for novel protease inhibitors with high potency and improved ability to overcome the diversity of mutations in the protease. Tipranavir is a novel non-peptidic protease inhibitor that displays the essential features of such an improved inhibitor. Unbound tipranavir displays the bioactive conformation more often than current protease inhibitors. Therefore less energy is required for tipranavir to adapt its overall conformation to the requirements of the binding site. Tipranavir establishes a strong network ()? energy bond interactions with conserved elements of the protease active site that cannot be mutated without the enzyme losing its function. In addition, tipranavir makes a direct hydrogen bond with the backbone atoms of the isoleucine amino acids at position 50 on each subunit of the protease. All current protease inhibitors make this interaction indirectly through a water molecule expending energy to immobilize it. The release of this water molecule by tipranavir is an energetic favorable event. The direct bond to this conserved regions results in an improved binding of tipranavir to the protease. Finally, for current protease inhibitor mutations in the protease generally weaken the bonding interactions. Tipranavir compensates for the impact of mutations in a thermodynamically unique manner by conserving and even enhancing these important contacts. As a result of these important features tipranavir retains entire ()? even against HIV strains with extensive resistance to current protease inhibitors. In summary, tipranavir has an improved ability to overcome the diversity of mutations in the protease by adopting the bioactive conformation more frequently; by estabilishing a strong network hydrogen bonds with conserved elements of the protease;  By binding directly to isoleucine 50; and by compensating for the impact of mutations by enhancing important contacts. These characteristics contribute to tipranavir’s unique resistance profile and make tipranavir an innovative option for the treatment of HIV disease.</p>
<p class="MsoNormal" style="text-align:center;">
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<p class="MsoNormal">Inibidores da fusão do HIV com as células do hospedeiro</p>
<p class="MsoNormal" style="text-align:center;"><span style="text-align:center; display: block;"><a href="http://campodearroz.wordpress.com/2009/07/10/farmacos-usados-no-tratamento-da-aids/"><img src="http://img.youtube.com/vi/jM4TJg5hd0s/2.jpg" alt="" /></a></span></p>
<p class="MsoNormal" style="text-align:center;">Occuring after attachment and co-receptor binding the third step in HIV cell entry, fusion, also represents a target for anti-viral drug development. One model of fusion requires gp41 to undergo extensive structural reorganization in order to destabilize the viral and cell membranes. Compounds that bind gp41 and interfere with this process have the potential to prevent HIV cell entry. Drug candidates that block gp41, known as fusion Inhibitors, are currently in clinical development.</p>
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<p class="MsoNormal" style="text-align:center;">
<p class="MsoNormal">
<p class="MsoNormal">
<p class="MsoNormal">Inibidores da Integrase</p>
<p class="MsoNormal" style="text-align:center;"><span lang="EN-US"><span style="text-align:center; display: block;"><a href="http://campodearroz.wordpress.com/2009/07/10/farmacos-usados-no-tratamento-da-aids/"><img src="http://img.youtube.com/vi/wNgrjePOD_I/2.jpg" alt="" /></a></span></span></p>
<p class="MsoNormal" style="text-align:center;">
<p style="text-align:center;">Integrase is an essential enzyme that allows HIV to integrate its proviral DNA into the host cell cromossomes. Integrase Inhibitors are on the development as a new class of anti-HIV drugs.</p>
<p class="MsoNormal" style="text-align:center;">
<p class="MsoNormal" style="text-align:left;">
<p class="MsoNormal" style="text-align:left;"><span lang="EN-US">Nota: Existem inibidores da fusão e da integrase que são comercializados.<br />
</span></p>
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		<title>Infusão Intravenosa &#8211; Um Compartimento</title>
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		<pubDate>Tue, 06 Jan 2009 19:50:46 +0000</pubDate>
		<dc:creator>Augusto Maeda</dc:creator>
				<category><![CDATA[Equações diferenciais]]></category>
		<category><![CDATA[Farmacocinética]]></category>

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		<description><![CDATA[Entre as vias de administração de fármacos, temos as parenterais, que incluem: intravenosa, subcutânea e intramuscular. Soluções injetadas intravenosamente podem ser dadas em bolus (injetadas completamente de uma só vez) ou infudidas lentamente a uma taxa constante no plasma.
Usando o modelo de um compartimento, podemos representar a situação de infusão intravenosa pelo esquema abaixo:

Seguindo o [...]<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=campodearroz.wordpress.com&blog=2694588&post=181&subd=campodearroz&ref=&feed=1" />]]></description>
			<content:encoded><![CDATA[<div class='snap_preview'><br /><p class="contentBody">Entre as vias de administração de fármacos, temos as parenterais, que incluem: intravenosa, subcutânea e intramuscular. Soluções injetadas intravenosamente podem ser dadas em bolus (injetadas completamente de uma só vez) ou infudidas lentamente a uma taxa constante no plasma.</p>
<p class="contentBody">Usando o modelo de um compartimento, podemos representar a situação de infusão intravenosa pelo esquema abaixo:</p>
<p class="contentBody"><img class="alignnone size-full wp-image-191" title="Infusão - Um compartimento" src="http://campodearroz.files.wordpress.com/2009/01/infusao.jpg?w=100&#038;h=158" alt="Infusão - Um compartimento" width="100" height="158" /></p>
<p class="contentBody">Seguindo o princípio</p>
<ul>
<li>Taxa de variação da concentração da droga no compartimento = Taxa de entrada da droga &#8211; Taxa de saída da droga</li>
</ul>
<p>E considerando que,</p>
<ol>
<li>A droga é infudida no sangue a uma taxa constante (Taxa de ordem zero)</li>
<li>A eliminação da droga é diretamente proporcional a concentração desta no compartimento (Taxa de eliminação de primeira ordem)</li>
</ol>
<p>Temos que:</p>
<ul>
<li><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle%5Cfrac%7BdC_%7Bp%7D%7D%7Bdt%7D%3DI-kC_%7Bp%7D+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle\frac{dC_{p}}{dt}=I-kC_{p} ' title='\displaystyle\frac{dC_{p}}{dt}=I-kC_{p} ' class='latex' /></li>
</ul>
<p>Esta equação diferencial, por ser separável, pode ser resolvida por integração direta. Observe que a concentração inicial da droga no compartimento (tempo zero) é zero.</p>
<ul>
<li><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle%5Cfrac%7BdC_%7Bp%7D%7D%7BI-kC_%7Bp%7D%7D%3Ddt&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle\frac{dC_{p}}{I-kC_{p}}=dt' title='\displaystyle\frac{dC_{p}}{I-kC_{p}}=dt' class='latex' /></li>
<li><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle%5Cint_%7B0%7D%5E%7BC_%7Bp%7D%7D%5Cfrac%7BdC_%7Bp%7D%7D%7BI-kC_%7Bp%7D%7D+%3D+%5Cint_%7B0%7D%5E%7Bt%7Ddt&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle\int_{0}^{C_{p}}\frac{dC_{p}}{I-kC_{p}} = \int_{0}^{t}dt' title='\displaystyle\int_{0}^{C_{p}}\frac{dC_{p}}{I-kC_{p}} = \int_{0}^{t}dt' class='latex' /> (1)</li>
</ul>
<p>A integral do segundo membro é trivial. Já a do primeiro membro pode ser resolvida por uma substituição simples:</p>
<ul>
<li><img src='http://l.wordpress.com/latex.php?latex=u%3DI-kC_%7Bp%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='u=I-kC_{p}' title='u=I-kC_{p}' class='latex' /> (2)</li>
<li><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle%5Cfrac%7Bd%7D%7BdC_%7Bp%7D%7D%28u%29%3D%5Cfrac%7Bd%7D%7BdC_%7Bp%7D%7D%28I%29-%5Cfrac%7Bd%7D%7BdC_%7Bp%7D%7D%28kC_%7Bp%7D%29+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle\frac{d}{dC_{p}}(u)=\frac{d}{dC_{p}}(I)-\frac{d}{dC_{p}}(kC_{p}) ' title='\displaystyle\frac{d}{dC_{p}}(u)=\frac{d}{dC_{p}}(I)-\frac{d}{dC_{p}}(kC_{p}) ' class='latex' /></li>
<li><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle%5Cfrac%7Bdu%7D%7BdC_%7Bp%7D%7D%3D0-k+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle\frac{du}{dC_{p}}=0-k ' title='\displaystyle\frac{du}{dC_{p}}=0-k ' class='latex' /></li>
<li><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle+dC_%7Bp%7D%3D-%5Cfrac%7Bdu%7D%7Bk%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle dC_{p}=-\frac{du}{k}' title='\displaystyle dC_{p}=-\frac{du}{k}' class='latex' /> (3)</li>
</ul>
<p>A substituição exige a mudança do intervalo de integração:</p>
<ul>
<li><img src='http://l.wordpress.com/latex.php?latex=C_%7Bp%7D%3D0+%5CRightarrow+u%3DR+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='C_{p}=0 \Rightarrow u=R ' title='C_{p}=0 \Rightarrow u=R ' class='latex' /> (4)</li>
<li><img src='http://l.wordpress.com/latex.php?latex=C_%7Bp%7D%3DC_%7Bp%7D+%5CRightarrow+u%3DI-kC_%7Bp%7D+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='C_{p}=C_{p} \Rightarrow u=I-kC_{p} ' title='C_{p}=C_{p} \Rightarrow u=I-kC_{p} ' class='latex' /> (5)</li>
</ul>
<p>Substituindo (2), (3), (4) e (5) em (1):</p>
<ul>
<li><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle%5Cint_%7BI%7D%5E%7BI-kC_%7Bp%7D%7D%5Cfrac%7B-%5Cfrac%7Bdu%7D%7Bk%7D%7D%7Bu%7D%3Dt-0+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle\int_{I}^{I-kC_{p}}\frac{-\frac{du}{k}}{u}=t-0 ' title='\displaystyle\int_{I}^{I-kC_{p}}\frac{-\frac{du}{k}}{u}=t-0 ' class='latex' /></li>
<li><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle-%5Cfrac%7B1%7D%7Bk%7D%5Cint_%7BI%7D%5E%7BI-kC_%7Bp%7D%7D%5Cfrac%7Bdu%7D%7Bu%7D%3Dt+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle-\frac{1}{k}\int_{I}^{I-kC_{p}}\frac{du}{u}=t ' title='\displaystyle-\frac{1}{k}\int_{I}^{I-kC_{p}}\frac{du}{u}=t ' class='latex' /></li>
<li><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle-%5Cfrac%7B1%7D%7Bk%7Dln%28%5Cfrac%7BI-kC_%7Bp%7D%7D%7BI%7D%29%3Dt+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle-\frac{1}{k}ln(\frac{I-kC_{p}}{I})=t ' title='\displaystyle-\frac{1}{k}ln(\frac{I-kC_{p}}{I})=t ' class='latex' /></li>
<li><img src='http://l.wordpress.com/latex.php?latex=ln%28%5Cfrac%7BI-kC_%7Bp%7D%7D%7BI%7D%29%3D-kt+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='ln(\frac{I-kC_{p}}{I})=-kt ' title='ln(\frac{I-kC_{p}}{I})=-kt ' class='latex' /></li>
<li><img src='http://l.wordpress.com/latex.php?latex=%5Cfrac%7BI-kC_%7Bp%7D%7D%7BI%7D%3De%5E%7B-kt%7D+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\frac{I-kC_{p}}{I}=e^{-kt} ' title='\frac{I-kC_{p}}{I}=e^{-kt} ' class='latex' /></li>
<li><img src='http://l.wordpress.com/latex.php?latex=I-kC_%7Bp%7D%3DIe%5E%7B-kt%7D+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='I-kC_{p}=Ie^{-kt} ' title='I-kC_{p}=Ie^{-kt} ' class='latex' /></li>
<li><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle+C_%7Bp%7D%3D%5Cfrac%7BI%7D%7Bk%7D%281-e%5E%7B-kt%7D%29+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle C_{p}=\frac{I}{k}(1-e^{-kt}) ' title='\displaystyle C_{p}=\frac{I}{k}(1-e^{-kt}) ' class='latex' />, que representa a equação concentração plasmática x tempo para este modelo</li>
</ul>
<p>No começo <img src='http://l.wordpress.com/latex.php?latex=C_%7Bp%7D%3D0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='C_{p}=0' title='C_{p}=0' class='latex' /> e, portanto:</p>
<ul>
<li><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle%5Cfrac%7BdC_%7Bp%7D%7D%7Bdt%7D%3DI+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle\frac{dC_{p}}{dt}=I ' title='\displaystyle\frac{dC_{p}}{dt}=I ' class='latex' /></li>
</ul>
<p>o que indica que a função <img src='http://l.wordpress.com/latex.php?latex=C_%7Bp%7D%28t%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='C_{p}(t)' title='C_{p}(t)' class='latex' /> é crescente no tempo t=0. A medida que o tempo prossegue, <img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle%5Cfrac%7BdC_%7Bp%7D%7D%7Bdt%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle\frac{dC_{p}}{dt}' title='\displaystyle\frac{dC_{p}}{dt}' class='latex' /> vai dimuindo, mas continua positiva, mostrando que a função <img src='http://l.wordpress.com/latex.php?latex=C_%7Bp%7D%28t%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='C_{p}(t)' title='C_{p}(t)' class='latex' />   é crescente em todo o seu domínio. Quando t tende ao infinito, temos que a taxa de infusão se iguala a taxa de eliminação do fármaco, situação que é chamada de Estado de Equilíbrio estável (Steady State). A concentração plasmática tende a um valor denominado <img src='http://l.wordpress.com/latex.php?latex=C_%7BSS%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='C_{SS}' title='C_{SS}' class='latex' />. Ele é calculado por:</p>
<p><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle+C_%7BSS%7D%3D%5Clim_%7Bt%5Cto%5Cinfty%7DC_%7Bp%7D%28t%29%3D%5Clim_%7Bt%5Cto%5Cinfty%7D%5Cfrac%7BI%7D%7Bk%7D%281-e%5E%7B-kt%7D%29%3D%5Cfrac%7BI%7D%7Bk%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle C_{SS}=\lim_{t\to\infty}C_{p}(t)=\lim_{t\to\infty}\frac{I}{k}(1-e^{-kt})=\frac{I}{k}' title='\displaystyle C_{SS}=\lim_{t\to\infty}C_{p}(t)=\lim_{t\to\infty}\frac{I}{k}(1-e^{-kt})=\frac{I}{k}' class='latex' /></p>
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			<media:title type="html">Infusão - Um compartimento</media:title>
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		<title>Administração Intravenosa em bolus &#8211; Modelo de 2 compartimentos</title>
		<link>http://campodearroz.wordpress.com/2008/12/26/administracao-intravenosa-em-bolus-modelo-de-2-compartimentos/</link>
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		<pubDate>Sat, 27 Dec 2008 00:56:11 +0000</pubDate>
		<dc:creator>Augusto Maeda</dc:creator>
				<category><![CDATA[Equações diferenciais]]></category>
		<category><![CDATA[Farmacocinética]]></category>
		<category><![CDATA[Farmacologia]]></category>

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		<description><![CDATA[Considerações para o uso deste modelo
Modelo de 2 compartimentos:

A droga no sangue não se equilibra rapidamente com tecidos extravasculares;
Há dois compartimentos: um central e um periférico (tecidual). O central representa o sangue e tecidos com alta perfusão. O periférico representa os tecidos em que a droga se equilibra mais devagar.

Administração em bolus:

Droga é administrada de [...]<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=campodearroz.wordpress.com&blog=2694588&post=91&subd=campodearroz&ref=&feed=1" />]]></description>
			<content:encoded><![CDATA[<div class='snap_preview'><br /><p><strong>Considerações para o uso deste modelo</strong></p>
<p>Modelo de 2 compartimentos:</p>
<ul>
<li>A droga no sangue não se equilibra rapidamente com tecidos extravasculares;</li>
<li>Há dois compartimentos: um central e um periférico (tecidual). O central representa o sangue e tecidos com alta perfusão. O periférico representa os tecidos em que a droga se equilibra mais devagar.</li>
</ul>
<p>Administração em bolus:</p>
<ul>
<li>Droga é administrada de uma vez no compartimento central (injeção intravenosa rápida);</li>
</ul>
<p>Eliminação e transferência entre compartimentos:</p>
<ul>
<li>Tanto a eliminação quanto a transferência entre compartimentos são processos de primeira ordem</li>
</ul>
<p><strong>Equação Concentração x Tempo</strong></p>
<div>
<div id="attachment_93" class="wp-caption alignnone" style="width: 357px"><img class="size-full wp-image-93" title="bolus2" src="http://campodearroz.files.wordpress.com/2008/12/bolus2.jpg?w=347&#038;h=169" alt="Administração IV em bolus - 2 compartimentos" width="347" height="169" /><p class="wp-caption-text">Administração IV em bolus - 2 compartimentos</p></div>
<div>A modelagem segue o seguinte princípio:</div>
<div>
<ul>
<li>Taxa de variação da concentração da droga no compartimento = Taxa de entrada da droga &#8211; Taxa de saída da droga</li>
</ul>
</div>
</div>
<div>Para o compartimento 1, assume-se que a taxa de entrada é diretamente proporcional a concentração da droga no compartimento 2, sendo <img src='http://l.wordpress.com/latex.php?latex=k_%7B21%7D+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='k_{21} ' title='k_{21} ' class='latex' /> a constante de proporcionalidade.</div>
<div>
<ul>
<li>Taxa de entrada da droga (1) = <img src='http://l.wordpress.com/latex.php?latex=k_%7B21%7DC_%7Bt%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='k_{21}C_{t}' title='k_{21}C_{t}' class='latex' /></li>
</ul>
</div>
<div>
<div>Ainda para o compartimento 1, assume-se que a taxa de saída é diretamente proporcional a concentração da droga no compartimento 1. Entretanto, temos duas formas de saída:</div>
<div>
<ul>
<li>Taxa de saída da droga (1) = <img src='http://l.wordpress.com/latex.php?latex=k_%7B12%7DC_%7Bp%7D%2Bk+C_%7Bp%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='k_{12}C_{p}+k C_{p}' title='k_{12}C_{p}+k C_{p}' class='latex' /></li>
</ul>
<p>Para o compartimento 2, fazemos de forma análoga ao que fizemos para o compartimento 1:</p></div>
</div>
<div>
<div>
<ul>
<li>Taxa de entrada da droga (2) = <img src='http://l.wordpress.com/latex.php?latex=k_%7B12%7DC_%7Bp%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='k_{12}C_{p}' title='k_{12}C_{p}' class='latex' /></li>
<li>Taxa de saída da droga (2) = <img src='http://l.wordpress.com/latex.php?latex=k_%7B21%7DC_%7Bt%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='k_{21}C_{t}' title='k_{21}C_{t}' class='latex' /></li>
</ul>
<div>Observe que a taxa de variação da concentração da droga em relação ao tempo para um compartimento é dada por <img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle%5Cfrac%7BdC%7D%7Bdt%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle\frac{dC}{dt}' title='\displaystyle\frac{dC}{dt}' class='latex' />. A modelagem para o problema em questão gera, então, um sistema de duas equações diferenciais:</div>
</div>
<div>
<ul>
<li><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle%5Cfrac%7BdC_%7Bp%7D%7D%7Bdt%7D%3Dk_%7B21%7DC_%7Bt%7D-%28k_%7B12%7DC_%7Bp%7D%2Bk+C_%7Bp%7D%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle\frac{dC_{p}}{dt}=k_{21}C_{t}-(k_{12}C_{p}+k C_{p})' title='\displaystyle\frac{dC_{p}}{dt}=k_{21}C_{t}-(k_{12}C_{p}+k C_{p})' class='latex' /> (1)</li>
<li><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle%5Cfrac%7BdC_%7Bt%7D%7D%7Bdt%7D%3Dk_%7B12%7DC_%7Bp%7D-k_%7B21%7DC_%7Bt%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle\frac{dC_{t}}{dt}=k_{12}C_{p}-k_{21}C_{t}' title='\displaystyle\frac{dC_{t}}{dt}=k_{12}C_{p}-k_{21}C_{t}' class='latex' /> (2)</li>
</ul>
<div>Um caminho para a resolução deste sistema é o uso da transformada de Laplace. Comecemos pela equação (1):</div>
<div>
<ul>
<li><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle%5Cmathcal+L%5Cleft%5C%7B%5Cfrac%7BdC_%7Bp%7D%7D%7Bdt%7D%5Cright%5C%7D%3D%5Cmathcal+L+%5C%7Bk_%7B21%7DC_%7Bt%7D%5C%7D-%5Cmathcal+L%5C%7Bk_%7B12%7DC_%7Bp%7D%2Bk+C_%7Bp%7D%5C%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle\mathcal L\left\{\frac{dC_{p}}{dt}\right\}=\mathcal L \{k_{21}C_{t}\}-\mathcal L\{k_{12}C_{p}+k C_{p}\}' title='\displaystyle\mathcal L\left\{\frac{dC_{p}}{dt}\right\}=\mathcal L \{k_{21}C_{t}\}-\mathcal L\{k_{12}C_{p}+k C_{p}\}' class='latex' /></li>
<li><img src='http://l.wordpress.com/latex.php?latex=s%5Cmathcal+L+%5C%7BC_%7Bp%7D%5C%7D-C_%7Bp_%7B0%7D%7D%3Dk_%7B21%7D%5Cmathcal+L%5C%7BC_%7Bt%7D%5C%7D-%28k_%7B12%7D%2Bk%29%5Cmathcal+L%5C%7BC_%7Bp%7D%5C%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='s\mathcal L \{C_{p}\}-C_{p_{0}}=k_{21}\mathcal L\{C_{t}\}-(k_{12}+k)\mathcal L\{C_{p}\}' title='s\mathcal L \{C_{p}\}-C_{p_{0}}=k_{21}\mathcal L\{C_{t}\}-(k_{12}+k)\mathcal L\{C_{p}\}' class='latex' /></li>
<li><img src='http://l.wordpress.com/latex.php?latex=%5Cmathcal+L+%5C%7BC_%7Bp%7D%5C%7D%28s%2Bk_%7B12%7D%2Bk+%29-C_%7Bp_%7B0%7D%7D%3Dk_%7B21%7D%5Cmathcal+L+%5C%7BC_%7Bt%7D%5C%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\mathcal L \{C_{p}\}(s+k_{12}+k )-C_{p_{0}}=k_{21}\mathcal L \{C_{t}\}' title='\mathcal L \{C_{p}\}(s+k_{12}+k )-C_{p_{0}}=k_{21}\mathcal L \{C_{t}\}' class='latex' /></li>
<li><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle%5Cmathcal+L+%5C%7BC_%7Bt%7D%5C%7D%3D%5Cfrac%7B%5Cmathcal+L+%5C%7BC_%7Bp%7D%5C%7D%28s%2Bk_%7B12%7D%2Bk+%29-C_%7Bp_%7B0%7D%7D%7D%7Bk_%7B21%7D%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle\mathcal L \{C_{t}\}=\frac{\mathcal L \{C_{p}\}(s+k_{12}+k )-C_{p_{0}}}{k_{21}}' title='\displaystyle\mathcal L \{C_{t}\}=\frac{\mathcal L \{C_{p}\}(s+k_{12}+k )-C_{p_{0}}}{k_{21}}' class='latex' /> (3)</li>
</ul>
</div>
<p>Apliquemos, então, a transformada de Laplace à equação (2):</p>
<ul>
<li><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle%5Cmathcal+L%5Cleft%5C%7B%5Cfrac%7BdC_%7Bt%7D%7D%7Bdt%7D%5Cright%5C%7D%3D%5Cmathcal+L%5C%7Bk_%7B12%7DC_%7Bp%7D%5C%7D-%5Cmathcal+L%5C%7Bk_%7B21%7DC_%7Bt%7D%5C%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle\mathcal L\left\{\frac{dC_{t}}{dt}\right\}=\mathcal L\{k_{12}C_{p}\}-\mathcal L\{k_{21}C_{t}\}' title='\displaystyle\mathcal L\left\{\frac{dC_{t}}{dt}\right\}=\mathcal L\{k_{12}C_{p}\}-\mathcal L\{k_{21}C_{t}\}' class='latex' /></li>
<li><img src='http://l.wordpress.com/latex.php?latex=s%5Cmathcal+L%5C%7BC_%7Bt%7D%5C%7D-0%3Dk_%7B12%7D%5Cmathcal+L%5C%7BC_%7Bp%7D%5C%7D-k_%7B21%7D%5Cmathcal+L%5C%7BC_%7Bt%7D%5C%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='s\mathcal L\{C_{t}\}-0=k_{12}\mathcal L\{C_{p}\}-k_{21}\mathcal L\{C_{t}\}' title='s\mathcal L\{C_{t}\}-0=k_{12}\mathcal L\{C_{p}\}-k_{21}\mathcal L\{C_{t}\}' class='latex' /></li>
<li><img src='http://l.wordpress.com/latex.php?latex=%5Cmathcal+L%5C%7BC_%7Bt%7D%5C%7D%28s%2Bk_%7B21%7D%29%3Dk_%7B12%7D%5Cmathcal+L%5C%7BC_%7Bp%7D%5C%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\mathcal L\{C_{t}\}(s+k_{21})=k_{12}\mathcal L\{C_{p}\}' title='\mathcal L\{C_{t}\}(s+k_{21})=k_{12}\mathcal L\{C_{p}\}' class='latex' /> (4)</li>
</ul>
<p>Substituindo (3) em (4):</p>
<ul>
<li><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle%5Cfrac%7B%5Cmathcal+L+%5C%7BC_%7Bp%7D%5C%7D%28s%2Bk_%7B12%7D%2Bk+%29-C_%7Bp_%7B0%7D%7D%7D%7Bk_%7B21%7D%7D%28s%2Bk_%7B21%7D%29%3Dk_%7B12%7D%5Cmathcal+L%5C%7BC_%7Bp%7D%5C%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle\frac{\mathcal L \{C_{p}\}(s+k_{12}+k )-C_{p_{0}}}{k_{21}}(s+k_{21})=k_{12}\mathcal L\{C_{p}\}' title='\displaystyle\frac{\mathcal L \{C_{p}\}(s+k_{12}+k )-C_{p_{0}}}{k_{21}}(s+k_{21})=k_{12}\mathcal L\{C_{p}\}' class='latex' /></li>
<li><img src='http://l.wordpress.com/latex.php?latex=%28%5Cmathcal+L+%5C%7BC_%7Bp%7D%5C%7D%28s%2Bk_%7B12%7D%2Bk+%29-C_%7Bp_%7B0%7D%7D%29%28s%2Bk_%7B21%7D%29%3Dk_%7B12%7Dk_%7B21%7D%5Cmathcal+L%5C%7BC_%7Bp%7D%5C%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='(\mathcal L \{C_{p}\}(s+k_{12}+k )-C_{p_{0}})(s+k_{21})=k_{12}k_{21}\mathcal L\{C_{p}\}' title='(\mathcal L \{C_{p}\}(s+k_{12}+k )-C_{p_{0}})(s+k_{21})=k_{12}k_{21}\mathcal L\{C_{p}\}' class='latex' /></li>
<li><img src='http://l.wordpress.com/latex.php?latex=%5Cmathcal+L+%5C%7BC_%7Bp%7D%5C%7D%28%28s%2Bk_%7B12%7D%2Bk+%29%28s%2Bk_%7B21%7D%29-k_%7B12%7Dk_%7B21%7D%29%3DC_%7Bp_%7B0%7D%7D%28s%2Bk_%7B21%7D%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\mathcal L \{C_{p}\}((s+k_{12}+k )(s+k_{21})-k_{12}k_{21})=C_{p_{0}}(s+k_{21})' title='\mathcal L \{C_{p}\}((s+k_{12}+k )(s+k_{21})-k_{12}k_{21})=C_{p_{0}}(s+k_{21})' class='latex' /></li>
<li><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle%5Cmathcal+L+%5C%7BC_%7Bp%7D%5C%7D%3D%5Cfrac%7BC_%7Bp_%7B0%7D%7D%28s%2Bk_%7B21%7D%29%7D%7B%28%28s%2Bk_%7B12%7D%2Bk+%29%28s%2Bk_%7B21%7D%29-k_%7B12%7Dk_%7B21%7D%29%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle\mathcal L \{C_{p}\}=\frac{C_{p_{0}}(s+k_{21})}{((s+k_{12}+k )(s+k_{21})-k_{12}k_{21})}' title='\displaystyle\mathcal L \{C_{p}\}=\frac{C_{p_{0}}(s+k_{21})}{((s+k_{12}+k )(s+k_{21})-k_{12}k_{21})}' class='latex' /></li>
<li><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle%5Cmathcal+L+%5C%7BC_%7Bp%7D%5C%7D%3D%5Cfrac%7BC_%7Bp_%7B0%7D%7D%28s%2Bk_%7B21%7D%29%7D%7Bs%5E%7B2%7D%2B%28k%2Bk_%7B12%7D%2Bk_%7B21%7D%29s%2Bkk_%7B21%7D%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle\mathcal L \{C_{p}\}=\frac{C_{p_{0}}(s+k_{21})}{s^{2}+(k+k_{12}+k_{21})s+kk_{21}}' title='\displaystyle\mathcal L \{C_{p}\}=\frac{C_{p_{0}}(s+k_{21})}{s^{2}+(k+k_{12}+k_{21})s+kk_{21}}' class='latex' /> (5)</li>
</ul>
</div>
<div>Para auxiliar a resolução, fazem-se as seguintes considerações:</div>
<div>
<ul>
<li><img src='http://l.wordpress.com/latex.php?latex=a%2Bb%3Dk%2Bk_%7B12%7D%2Bk_%7B21%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='a+b=k+k_{12}+k_{21}' title='a+b=k+k_{12}+k_{21}' class='latex' /></li>
<li><img src='http://l.wordpress.com/latex.php?latex=ab%3Dkk_%7B21%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='ab=kk_{21}' title='ab=kk_{21}' class='latex' /></li>
</ul>
</div>
<p>Logo, voltando em (5):</p>
<ul>
<li><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle%5Cmathcal+L+%5C%7BC_%7Bp%7D%5C%7D%3D%5Cfrac%7BC_%7Bp_%7B0%7D%7D%28s%2Bk_%7B21%7D%29%7D%7Bs%5E%7B2%7D%2B%28a%2Bb%29s%2Bab%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle\mathcal L \{C_{p}\}=\frac{C_{p_{0}}(s+k_{21})}{s^{2}+(a+b)s+ab}' title='\displaystyle\mathcal L \{C_{p}\}=\frac{C_{p_{0}}(s+k_{21})}{s^{2}+(a+b)s+ab}' class='latex' /></li>
<li><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle%5Cmathcal+L+%5C%7BC_%7Bp%7D%5C%7D%3D%5Cfrac%7BC_%7Bp_%7B0%7D%7D%28s%2Bk_%7B21%7D%29%7D%7B%28s%2Ba%29%28s%2Bb%29%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle\mathcal L \{C_{p}\}=\frac{C_{p_{0}}(s+k_{21})}{(s+a)(s+b)}' title='\displaystyle\mathcal L \{C_{p}\}=\frac{C_{p_{0}}(s+k_{21})}{(s+a)(s+b)}' class='latex' /></li>
</ul>
<div>É possível, desta forma, usar o método das frações parciais:</div>
<div>
<ul>
<li><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle%5Cmathcal+L+%5C%7BC_%7Bp%7D%5C%7D%3D%5Cfrac%7BC_%7Bp_%7B0%7D%7D%28s%2Bk_%7B21%7D%29%7D%7B%28s%2Ba%29%28s%2Bb%29%7D%3D%5Cfrac%7Bx%7D%7B%28s%2Ba%29%7D%2B%5Cfrac%7By%7D%7B%28s%2Bb%29%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle\mathcal L \{C_{p}\}=\frac{C_{p_{0}}(s+k_{21})}{(s+a)(s+b)}=\frac{x}{(s+a)}+\frac{y}{(s+b)}' title='\displaystyle\mathcal L \{C_{p}\}=\frac{C_{p_{0}}(s+k_{21})}{(s+a)(s+b)}=\frac{x}{(s+a)}+\frac{y}{(s+b)}' class='latex' /> (6)</li>
<li><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle%5Cmathcal+L+%5C%7BC_%7Bp%7D%5C%7D%3D%5Cfrac%7BC_%7Bp_%7B0%7D%7D%28s%2Bk_%7B21%7D%29%7D%7B%28s%2Ba%29%28s%2Bb%29%7D%3D%5Cfrac%7Bx%28s%2Bb%29%2By%28s%2Ba%29%7D%7B%28s%2Ba%29%28s%2Bb%29%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle\mathcal L \{C_{p}\}=\frac{C_{p_{0}}(s+k_{21})}{(s+a)(s+b)}=\frac{x(s+b)+y(s+a)}{(s+a)(s+b)}' title='\displaystyle\mathcal L \{C_{p}\}=\frac{C_{p_{0}}(s+k_{21})}{(s+a)(s+b)}=\frac{x(s+b)+y(s+a)}{(s+a)(s+b)}' class='latex' /></li>
<li><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle%5Cmathcal+L+%5C%7BC_%7Bp%7D%5C%7D%3D%5Cfrac%7BC_%7Bp_%7B0%7D%7D%28s%2Bk_%7B21%7D%29%7D%7B%28s%2Ba%29%28s%2Bb%29%7D%3D%5Cfrac%7Bs%28x%2By%29%2Bxb%2Bya%7D%7B%28s%2Ba%29%28s%2Bb%29%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle\mathcal L \{C_{p}\}=\frac{C_{p_{0}}(s+k_{21})}{(s+a)(s+b)}=\frac{s(x+y)+xb+ya}{(s+a)(s+b)}' title='\displaystyle\mathcal L \{C_{p}\}=\frac{C_{p_{0}}(s+k_{21})}{(s+a)(s+b)}=\frac{s(x+y)+xb+ya}{(s+a)(s+b)}' class='latex' /></li>
</ul>
</div>
<div>Portanto, temos que:</div>
<div>
<ul>
<li><img src='http://l.wordpress.com/latex.php?latex=x%2By%3DC_%7Bp_%7B0%7D%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x+y=C_{p_{0}}' title='x+y=C_{p_{0}}' class='latex' /> (7)</li>
<li><img src='http://l.wordpress.com/latex.php?latex=xb%2Bya%3DC_%7Bp_%7B0%7D%7Dk_%7B21%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='xb+ya=C_{p_{0}}k_{21}' title='xb+ya=C_{p_{0}}k_{21}' class='latex' /> (8)</li>
</ul>
</div>
<div>Isolando y em (7):</div>
<div>
<ul>
<li><img src='http://l.wordpress.com/latex.php?latex=y%3DC_%7Bp_%7B0%7D%7D-x&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='y=C_{p_{0}}-x' title='y=C_{p_{0}}-x' class='latex' /> (9)</li>
</ul>
</div>
<div>Substituindo (9) em (8):</div>
<div>
<ul>
<li><img src='http://l.wordpress.com/latex.php?latex=xb%2B%28C_%7Bp_%7B0%7D%7D-x%29a%3DC_%7Bp_%7B0%7D%7Dk_%7B21%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='xb+(C_{p_{0}}-x)a=C_{p_{0}}k_{21}' title='xb+(C_{p_{0}}-x)a=C_{p_{0}}k_{21}' class='latex' /></li>
<li><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle+x%3D%5Cfrac%7BC_%7Bp_%7B0%7D%7D%28k_%7B21%7D-a%29%7D%7Bb-a%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle x=\frac{C_{p_{0}}(k_{21}-a)}{b-a}' title='\displaystyle x=\frac{C_{p_{0}}(k_{21}-a)}{b-a}' class='latex' /> (10)</li>
</ul>
</div>
<div>Substituindo (10) em (9):</div>
<div>
<ul>
<li><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle+y%3DC_%7Bp_%7B0%7D%7D-%5Cfrac%7BC_%7Bp_%7B0%7D%7D%28k_%7B21%7D-a%29%7D%7Bb-a%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle y=C_{p_{0}}-\frac{C_{p_{0}}(k_{21}-a)}{b-a}' title='\displaystyle y=C_{p_{0}}-\frac{C_{p_{0}}(k_{21}-a)}{b-a}' class='latex' /></li>
<li><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle+y%3D%5Cfrac%7BC_%7Bp_%7B0%7D%7D%28b-k_%7B21%7D%29%7D%7Bb-a%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle y=\frac{C_{p_{0}}(b-k_{21})}{b-a}' title='\displaystyle y=\frac{C_{p_{0}}(b-k_{21})}{b-a}' class='latex' /> (11)</li>
</ul>
</div>
<div>Substituindo (10) e (11) em (6):</div>
<div>
<ul>
<li><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle%5Cmathcal+L+%5C%7BC_%7Bp%7D%5C%7D%3D%5Cfrac%7BC_%7Bp_%7B0%7D%7D%28s%2Bk_%7B21%7D%29%7D%7B%28s%2Ba%29%28s%2Bb%29%7D%3D%5Cfrac%7B%5Cfrac%7BC_%7Bp_%7B0%7D%7D%28k_%7B21%7D-a%29%7D%7Bb-a%7D%7D%7B%28s%2Ba%29%7D%2B%5Cfrac%7B%5Cfrac%7BC_%7Bp_%7B0%7D%7D%28b-k_%7B21%7D%29%7D%7Bb-a%7D%7D%7B%28s%2Bb%29%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle\mathcal L \{C_{p}\}=\frac{C_{p_{0}}(s+k_{21})}{(s+a)(s+b)}=\frac{\frac{C_{p_{0}}(k_{21}-a)}{b-a}}{(s+a)}+\frac{\frac{C_{p_{0}}(b-k_{21})}{b-a}}{(s+b)}' title='\displaystyle\mathcal L \{C_{p}\}=\frac{C_{p_{0}}(s+k_{21})}{(s+a)(s+b)}=\frac{\frac{C_{p_{0}}(k_{21}-a)}{b-a}}{(s+a)}+\frac{\frac{C_{p_{0}}(b-k_{21})}{b-a}}{(s+b)}' class='latex' /></li>
</ul>
</div>
<div>Finalmente, consegue-se aplicar a transformada inversa:</div>
<div>
<ul>
<li><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle%5Cmathcal+L%5E%7B-1%7D%5C%7B%5Cmathcal+L+%5C%7BC_%7Bp%7D%5C%7D%5C%7D%3D%5Cmathcal+L%5E%7B-1%7D%5Cleft%5C%7B%5Cfrac%7B%5Cfrac%7BC_%7Bp_%7B0%7D%7D%28k_%7B21%7D-a%29%7D%7Bb-a%7D%7D%7B%28s%2Ba%29%7D%5Cright%5C%7D%2B%5Cmathcal+L%5E%7B-1%7D%5Cleft%5C%7B%5Cfrac%7B%5Cfrac%7BC_%7Bp_%7B0%7D%7D%28b-k_%7B21%7D%29%7D%7Bb-a%7D%7D%7B%28s%2Bb%29%7D%5Cright%5C%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle\mathcal L^{-1}\{\mathcal L \{C_{p}\}\}=\mathcal L^{-1}\left\{\frac{\frac{C_{p_{0}}(k_{21}-a)}{b-a}}{(s+a)}\right\}+\mathcal L^{-1}\left\{\frac{\frac{C_{p_{0}}(b-k_{21})}{b-a}}{(s+b)}\right\}' title='\displaystyle\mathcal L^{-1}\{\mathcal L \{C_{p}\}\}=\mathcal L^{-1}\left\{\frac{\frac{C_{p_{0}}(k_{21}-a)}{b-a}}{(s+a)}\right\}+\mathcal L^{-1}\left\{\frac{\frac{C_{p_{0}}(b-k_{21})}{b-a}}{(s+b)}\right\}' class='latex' /></li>
<li><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle+C_%7Bp%7D%3D%5Cfrac%7BC_%7Bp_%7B0%7D%7D%28k_%7B21%7D-a%29%7D%7Bb-a%7De%5E%7B-at%7D%2B%5Cfrac%7BC_%7Bp_%7B0%7D%7D%28b-k_%7B21%7D%29%7D%7Bb-a%7De%5E%7B-bt%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle C_{p}=\frac{C_{p_{0}}(k_{21}-a)}{b-a}e^{-at}+\frac{C_{p_{0}}(b-k_{21})}{b-a}e^{-bt}' title='\displaystyle C_{p}=\frac{C_{p_{0}}(k_{21}-a)}{b-a}e^{-at}+\frac{C_{p_{0}}(b-k_{21})}{b-a}e^{-bt}' class='latex' /></li>
</ul>
</div>
<div>De uma forma mais simplificada:</div>
<div>
<ul>
<li><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle+A%3D%5Cfrac%7BC_%7Bp_%7B0%7D%7D%28k_%7B21%7D-a%29%7D%7Bb-a%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle A=\frac{C_{p_{0}}(k_{21}-a)}{b-a}' title='\displaystyle A=\frac{C_{p_{0}}(k_{21}-a)}{b-a}' class='latex' /></li>
<li><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle+B%3D%5Cfrac%7BC_%7Bp_%7B0%7D%7D%28b-k_%7B21%7D%29%7D%7Bb-a%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle B=\frac{C_{p_{0}}(b-k_{21})}{b-a}' title='\displaystyle B=\frac{C_{p_{0}}(b-k_{21})}{b-a}' class='latex' /></li>
</ul>
</div>
<div>E finalmente:</div>
<div>
<ul>
<li><img src='http://l.wordpress.com/latex.php?latex=C_%7Bp%7D%3DAe%5E%7B-at%7D%2BBe%5E%7B-bt%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='C_{p}=Ae^{-at}+Be^{-bt}' title='C_{p}=Ae^{-at}+Be^{-bt}' class='latex' /></li>
</ul>
</div>
<div>Referência: Leon Shargel, Susanna Wu-Pong, Andrew B.C. Yu. Applied Biopharmaceutics &amp; Pharmacokinetics. 5th edition, 2005.</div>
</div>
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			<media:title type="html">bolus2</media:title>
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		<title>Administração Intravenosa em bolus &#8211; Modelo de um compartimento</title>
		<link>http://campodearroz.wordpress.com/2008/12/25/administracao-intravenosa-em-bolus-modelo-de-um-compartimento/</link>
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		<pubDate>Thu, 25 Dec 2008 20:02:44 +0000</pubDate>
		<dc:creator>Augusto Maeda</dc:creator>
				<category><![CDATA[Equações diferenciais]]></category>
		<category><![CDATA[Farmacocinética]]></category>
		<category><![CDATA[Farmacologia]]></category>

		<guid isPermaLink="false">http://campodearroz.wordpress.com/?p=87</guid>
		<description><![CDATA[Considerações para o uso deste modelo
Modelo de um compartimento:


Corpo age como um compartimento único;


Pode ser usado nas situações em que a droga no sangue se equilibra rapidamente com tecidos extravasculares;


Administração em bolus:

Droga é administrada de uma vez no compartimento (injeção intravenosa rápida);

Processo de Eliminação:

A eliminação da droga é um processo de primeira ordem

Equação Concentração x Tempo
O [...]<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=campodearroz.wordpress.com&blog=2694588&post=87&subd=campodearroz&ref=&feed=1" />]]></description>
			<content:encoded><![CDATA[<div class='snap_preview'><br /><p class="MsoNormal"><strong>Considerações para o uso deste modelo</strong></p>
<p>Modelo de um compartimento:</p>
<p><!--[endif]--></p>
<ul>
<li>Corpo age como um compartimento único;</li>
</ul>
<ul>
<li><!--[if !supportLists]-->Pode ser usado nas situações em que a droga no sangue se equilibra rapidamente com tecidos extravasculares;<span style="font-family:Symbol;"><span><span style="font-family:&quot;font-style:normal;font-variant:normal;font-weight:normal;font-size:7pt;line-height:normal;"><br />
</span></span></span></li>
</ul>
<p>Administração em bolus:</p>
<ul>
<li><!--[if !supportLists]-->Droga é administrada de uma vez no compartimento (injeção intravenosa rápida);</li>
</ul>
<p>Processo de Eliminação:</p>
<ul>
<li>A eliminação da droga é um processo de primeira ordem</li>
</ul>
<p><strong>Equação Concentração x Tempo</strong></p>
<div id="attachment_60" class="wp-caption alignnone" style="width: 163px"><a><img class="size-full wp-image-60" title="bolus1" src="http://campodearroz.files.wordpress.com/2008/12/bolus1.jpg?w=153&#038;h=145" alt="Adm IV em bolus" width="153" height="145" /></a><p class="wp-caption-text">Adm IV em bolus</p></div>
<p>O modelo compartimental para esta situação leva a seguinte equação diferencial de primeira ordem:</p>
<p class="MsoNormal"><img src='http://l.wordpress.com/latex.php?latex=%5Cfrac%7BdC_%7Bp%7D%7D%7Bdt%7D%3D-kC_%7Bp%7D+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\frac{dC_{p}}{dt}=-kC_{p} ' title='\frac{dC_{p}}{dt}=-kC_{p} ' class='latex' /></p>
<p class="MsoNormal">onde <img src='http://l.wordpress.com/latex.php?latex=C_%7Bp%7D+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='C_{p} ' title='C_{p} ' class='latex' /> é a concentração plasmática do fármaco e k é a constante de eliminação.</p>
<p class="MsoNormal">A equação pode ser resolvida por integração direta, por ser uma equação diferencial separável:</p>
<p class="MsoNormal"><img src='http://l.wordpress.com/latex.php?latex=%5Cfrac%7BdC_%7Bp%7D%7D%7BC_%7Bp%7D%7D%3D-kdt+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\frac{dC_{p}}{C_{p}}=-kdt ' title='\frac{dC_{p}}{C_{p}}=-kdt ' class='latex' /></p>
<p class="MsoNormal">Quando o tempo varia de 0 a t,<span> </span><img src='http://l.wordpress.com/latex.php?latex=C_%7Bp%7D+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='C_{p} ' title='C_{p} ' class='latex' /> varia de <img src='http://l.wordpress.com/latex.php?latex=C_%7Bp_%7B0%7D%7D+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='C_{p_{0}} ' title='C_{p_{0}} ' class='latex' /> a <img src='http://l.wordpress.com/latex.php?latex=C_%7Bp%7D+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='C_{p} ' title='C_{p} ' class='latex' />, onde <img src='http://l.wordpress.com/latex.php?latex=C_%7Bp_%7B0%7D%7D+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='C_{p_{0}} ' title='C_{p_{0}} ' class='latex' /> é a concentração do fármaco no tempo t=0. Logo,</p>
<p class="MsoNormal"><img src='http://l.wordpress.com/latex.php?latex=%5Cint_%7BC_%7Bp_%7B0%7D%7D%7D%5E%7BC_%7Bp%7D%7D+%5Cfrac%7BdC_%7Bp%7D%7D%7BC_%7Bp%7D%7D%3D%5Cint_%7B0%7D%5E%7Bt%7D-+k+dt&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\int_{C_{p_{0}}}^{C_{p}} \frac{dC_{p}}{C_{p}}=\int_{0}^{t}- k dt' title='\int_{C_{p_{0}}}^{C_{p}} \frac{dC_{p}}{C_{p}}=\int_{0}^{t}- k dt' class='latex' /></p>
<p class="MsoNormal"><img src='http://l.wordpress.com/latex.php?latex=%28ln%7CCp%7C%29%5D_%7BC_%7Bp%7D%3DC_%7Bp_%7B0%7D%7D%7D%5E%7BC_%7Bp%7D%3DC_%7Bp%7D%7D%3D%28-kt%29%5D_%7Bt%3D0%7D%5E%7Bt%3Dt%7D+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='(ln|Cp|)]_{C_{p}=C_{p_{0}}}^{C_{p}=C_{p}}=(-kt)]_{t=0}^{t=t} ' title='(ln|Cp|)]_{C_{p}=C_{p_{0}}}^{C_{p}=C_{p}}=(-kt)]_{t=0}^{t=t} ' class='latex' /></p>
<p class="MsoNormal"><img src='http://l.wordpress.com/latex.php?latex=ln%28C_%7Bp%7D%29-ln%28C_%7Bp_%7B0%7D%7D%29%3D-kt+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='ln(C_{p})-ln(C_{p_{0}})=-kt ' title='ln(C_{p})-ln(C_{p_{0}})=-kt ' class='latex' /></p>
<p class="MsoNormal"><img src='http://l.wordpress.com/latex.php?latex=ln+%5Cleft%28%7B%5Cfrac%7BC_%7Bp%7D%7D%7BC_%7Bp_%7B0%7D%7D%7D%7D%5Cright%29+%3D+-kt+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='ln \left({\frac{C_{p}}{C_{p_{0}}}}\right) = -kt ' title='ln \left({\frac{C_{p}}{C_{p_{0}}}}\right) = -kt ' class='latex' /></p>
<p class="MsoNormal"><img src='http://l.wordpress.com/latex.php?latex=%5Cfrac%7BC_%7Bp%7D%7D%7BC_%7Bp_%7B0%7D%7D%7D%3De%5E%7B-kt%7D+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\frac{C_{p}}{C_{p_{0}}}=e^{-kt} ' title='\frac{C_{p}}{C_{p_{0}}}=e^{-kt} ' class='latex' /></p>
<p class="MsoNormal"><img src='http://l.wordpress.com/latex.php?latex=C_%7Bp%7D%3DC_%7Bp_%7B0%7D%7De%5E%7B-kt%7D+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='C_{p}=C_{p_{0}}e^{-kt} ' title='C_{p}=C_{p_{0}}e^{-kt} ' class='latex' /><span> </span>(1)</p>
<p class="MsoNormal">
<p class="MsoNormal"><strong>Determinação de parâmetros farmacocinéticos</strong></p>
<p class="MsoNormal">1) A partir de dados de concentração plasmática</p>
<p class="MsoNormal">1.1) Determinação de k</p>
<p class="MsoNormal">A idéia é transformar a equação (1)<span> </span>em uma equação linear. Para isto, calculamos o logaritmo natural de ambos os lados da equação:</p>
<p class="MsoNormal"><img src='http://l.wordpress.com/latex.php?latex=ln%28C_%7Bp%7D%29%3Dln%28C_%7Bp_%7B0%7D%7De%5E%7B-kt%7D%29+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='ln(C_{p})=ln(C_{p_{0}}e^{-kt}) ' title='ln(C_{p})=ln(C_{p_{0}}e^{-kt}) ' class='latex' /></p>
<p class="MsoNormal"><img src='http://l.wordpress.com/latex.php?latex=ln%28C_%7Bp%7D%29%3Dln%28C_%7Bp_%7B0%7D%7D%29-kt&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='ln(C_{p})=ln(C_{p_{0}})-kt' title='ln(C_{p})=ln(C_{p_{0}})-kt' class='latex' /></p>
<p class="MsoNormal">Vendo <img src='http://l.wordpress.com/latex.php?latex=ln%28C_%7Bp%7D%29+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='ln(C_{p}) ' title='ln(C_{p}) ' class='latex' /> como a variável dependente e t como a variável independente, a equação acima é uma equação do tipo <img src='http://l.wordpress.com/latex.php?latex=y%3D%5Calpha+%2B+%5Cbeta+x+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='y=\alpha + \beta x ' title='y=\alpha + \beta x ' class='latex' />, sendo <img src='http://l.wordpress.com/latex.php?latex=%5Calpha+%3D+ln%28C_%7Bp_%7B0%7D%7D%29+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\alpha = ln(C_{p_{0}}) ' title='\alpha = ln(C_{p_{0}}) ' class='latex' /> e <img src='http://l.wordpress.com/latex.php?latex=%5Cbeta+%3D+-k+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\beta = -k ' title='\beta = -k ' class='latex' />. A partir de dados experimentais, os valores <img src='http://l.wordpress.com/latex.php?latex=%5Calpha&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\alpha' title='\alpha' class='latex' /> e <img src='http://l.wordpress.com/latex.php?latex=%5Cbeta&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\beta' title='\beta' class='latex' /> podem ser determinados com o ajuste dos dados ln(concentração)-tempo a uma reta pelo método dos mínimos quadrados. O valor de k segue imediatamente, pois</p>
<p class="MsoNormal"><img src='http://l.wordpress.com/latex.php?latex=k%3D-%5Cbeta+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='k=-\beta ' title='k=-\beta ' class='latex' /></p>
<p class="MsoNormal">
<p class="MsoNormal">1.2) Determinação de <img src='http://l.wordpress.com/latex.php?latex=V_%7Bd%7D+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='V_{d} ' title='V_{d} ' class='latex' /></p>
<p class="MsoNormal">Primeiro, deve-se determinar a concentração da droga no instante t=0:</p>
<p class="MsoNormal"><img src='http://l.wordpress.com/latex.php?latex=C_%7Bp_%7B0%7D%7D%3De%5E%7B%5Calpha%7D+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='C_{p_{0}}=e^{\alpha} ' title='C_{p_{0}}=e^{\alpha} ' class='latex' /></p>
<p class="MsoNormal">A concentração é a razão dose por volume de distribuição:</p>
<p class="MsoNormal"><img src='http://l.wordpress.com/latex.php?latex=C_%7Bp_%7B0%7D%7D%3D%7B%5Cfrac%7BD_%7BB_%7B0%7D%7D%7D%7BV_%7BD%7D%7D%7D%3De%5E%7B%5Calpha%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='C_{p_{0}}={\frac{D_{B_{0}}}{V_{D}}}=e^{\alpha}' title='C_{p_{0}}={\frac{D_{B_{0}}}{V_{D}}}=e^{\alpha}' class='latex' /></p>
<p class="MsoNormal"><img src='http://l.wordpress.com/latex.php?latex=V_%7BD%7D%3D%7B%5Cfrac%7BD_%7BB_%7B0%7D%7D%7D%7Be%5E%7B%5Calpha%7D%7D%7D+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='V_{D}={\frac{D_{B_{0}}}{e^{\alpha}}} ' title='V_{D}={\frac{D_{B_{0}}}{e^{\alpha}}} ' class='latex' /></p>
<p class="MsoNormal">
<p class="MsoNormal">1.3) Determinação de Cl</p>
<p class="MsoNormal"><span lang="EN-US"><img src='http://l.wordpress.com/latex.php?latex=Cl+%3D+k+%5Ccdot+V_%7Bd%7D+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='Cl = k \cdot V_{d} ' title='Cl = k \cdot V_{d} ' class='latex' /></span></p>
<p class="MsoNormal"><span lang="EN-US"> </span></p>
<p class="MsoNormal">1.4) Determinação de <img src='http://l.wordpress.com/latex.php?latex=t_%7B1%2F2%7D+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='t_{1/2} ' title='t_{1/2} ' class='latex' /></p>
<p class="MsoNormal">Basta considerar dois intervalos de tempo distintos <img src='http://l.wordpress.com/latex.php?latex=t_%7B1%7D+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='t_{1} ' title='t_{1} ' class='latex' /> e <img src='http://l.wordpress.com/latex.php?latex=t_%7B2%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='t_{2}' title='t_{2}' class='latex' />, sendo que:</p>
<p class="MsoNormal"><img src='http://l.wordpress.com/latex.php?latex=C_%7Bp_%7B2%7D%7D%3D%5Cfrac%7BC_%7Bp_%7B1%7D%7D%7D%7B2%7D+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='C_{p_{2}}=\frac{C_{p_{1}}}{2} ' title='C_{p_{2}}=\frac{C_{p_{1}}}{2} ' class='latex' />,</p>
<p class="MsoNormal">onde <img src='http://l.wordpress.com/latex.php?latex=C_%7Bp_%7B2%7D%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='C_{p_{2}}' title='C_{p_{2}}' class='latex' /> é a concentração plasmática em <img src='http://l.wordpress.com/latex.php?latex=t_%7B2%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='t_{2}' title='t_{2}' class='latex' /> e <img src='http://l.wordpress.com/latex.php?latex=C_%7Bp_%7B1%7D%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='C_{p_{1}}' title='C_{p_{1}}' class='latex' /> é a concentração plasmática em <img src='http://l.wordpress.com/latex.php?latex=t_%7B1%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='t_{1}' title='t_{1}' class='latex' />.</p>
<p class="MsoNormal">Temos então que:</p>
<p class="MsoNormal"><img src='http://l.wordpress.com/latex.php?latex=C_%7Bp_%7B1%7D%7D+%3D+C_%7Bp_%7B0%7D%7De%5E%7B-kt_%7B1%7D%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='C_{p_{1}} = C_{p_{0}}e^{-kt_{1}}' title='C_{p_{1}} = C_{p_{0}}e^{-kt_{1}}' class='latex' />,</p>
<p class="MsoNormal"><img src='http://l.wordpress.com/latex.php?latex=C_%7Bp_%7B2%7D%7D%3DC_%7Bp_%7B0%7D%7De%5E%7B-kt_%7B2%7D%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='C_{p_{2}}=C_{p_{0}}e^{-kt_{2}}' title='C_{p_{2}}=C_{p_{0}}e^{-kt_{2}}' class='latex' /></p>
<p class="MsoNormal">e, portanto,</p>
<p class="MsoNormal"><img src='http://l.wordpress.com/latex.php?latex=C_%7Bp_%7B0%7D%7De%5E%7B-kt_%7B2%7D%7D%3D%5Cfrac%7BC_%7Bp_%7B0%7D%7De%5E%7B-kt_%7B1%7D%7D%7D%7B2%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='C_{p_{0}}e^{-kt_{2}}=\frac{C_{p_{0}}e^{-kt_{1}}}{2}' title='C_{p_{0}}e^{-kt_{2}}=\frac{C_{p_{0}}e^{-kt_{1}}}{2}' class='latex' /></p>
<p class="MsoNormal"><img src='http://l.wordpress.com/latex.php?latex=e%5E%7B-kt_%7B1%7D%2Bkt_%7B2%7D%7D%3D2&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='e^{-kt_{1}+kt_{2}}=2' title='e^{-kt_{1}+kt_{2}}=2' class='latex' /></p>
<p class="MsoNormal"><img src='http://l.wordpress.com/latex.php?latex=k%28t_%7B2%7D-t_%7B1%7D%29%3Dln%282%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='k(t_{2}-t_{1})=ln(2)' title='k(t_{2}-t_{1})=ln(2)' class='latex' /></p>
<p class="MsoNormal">Mas como <img src='http://l.wordpress.com/latex.php?latex=t_%7B2%7D-t_%7B1%7D+%3D+t_%7B1%2F2%7D+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='t_{2}-t_{1} = t_{1/2} ' title='t_{2}-t_{1} = t_{1/2} ' class='latex' /></p>
<p class="MsoNormal"><img src='http://l.wordpress.com/latex.php?latex=t_%7B1%2F2%7D%3D%5Cfrac%7Bln%282%29%7D%7Bk%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='t_{1/2}=\frac{ln(2)}{k}' title='t_{1/2}=\frac{ln(2)}{k}' class='latex' /></p>
<p class="MsoNormal">Referência: Leon Shargel, Susanna Wu-Pong, Andrew B.C. Yu. Applied Biopharmaceutics &amp; Pharmacokinetics. 5th edition, 2005.</p>
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		<title>Potenciação de longo prazo</title>
		<link>http://campodearroz.wordpress.com/2008/07/04/potenciacao-de-longo-prazo/</link>
		<comments>http://campodearroz.wordpress.com/2008/07/04/potenciacao-de-longo-prazo/#comments</comments>
		<pubDate>Fri, 04 Jul 2008 19:30:47 +0000</pubDate>
		<dc:creator>Augusto Maeda</dc:creator>
				<category><![CDATA[Farmacodinâmica]]></category>
		<category><![CDATA[Farmacologia]]></category>

		<guid isPermaLink="false">http://campodearroz.wordpress.com/?p=27</guid>
		<description><![CDATA[A potenciação de longo prazo é um fenômeno onde um nível constante de estimulação pré-sináptica é convertido em uma grande saída (output) pós-sináptica. Este fenômeno desempenha papel importante na memória e no aprendizado.
 
Long-Term Potentiation LTP

In normal neuronal communication, input from a presynaptic neuron leads to firing in a postsynaptic neuron. Increased stimulation from the [...]<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=campodearroz.wordpress.com&blog=2694588&post=27&subd=campodearroz&ref=&feed=1" />]]></description>
			<content:encoded><![CDATA[<div class='snap_preview'><br /><p class="MsoNormal" style="text-align:center;"><span style="font-size:10pt;font-family:Arial;">A potenciação de longo prazo é um fenômeno onde um nível constante de estimulação pré-sináptica é convertido em uma grande saída (output) pós-sináptica. Este fenômeno desempenha papel importante na memória e no aprendizado.</span></p>
<p class="MsoNormal"><span style="font-size:10pt;font-family:Arial;"> </span></p>
<p class="MsoNormal" style="text-align:center;" align="center"><span style="font-size:10pt;font-family:Arial;">Long-Term Potentiation LTP</span></p>
<p class="MsoNormal" style="text-align:center;" align="center"><span style="text-align:center; display: block;"><a href="http://campodearroz.wordpress.com/2008/07/04/potenciacao-de-longo-prazo/"><img src="http://img.youtube.com/vi/BwZfLv3Z96A/2.jpg" alt="" /></a></span></p>
<p class="MsoNormal"><span style="font-size:8pt;font-family:Arial;">In normal neuronal communication, input from a presynaptic neuron leads to firing in a postsynaptic neuron. Increased stimulation from the presynaptic neuron causes increased firing of the postsynaptic neuron. But in LTP, the postsynaptic neuron continues to fire at an elevated rate even after the increased stimulation has subsided.</span></p>
<p class="MsoNormal"><span style="font-size:10pt;font-family:Arial;"> </span></p>
<p class="MsoNormal"><span style="font-size:10pt;font-family:Arial;">Os dois vídeos que seguem abaixo, falam sobre modelos que tentam explicar a ocorrência deste fenômeno:</span></p>
<p class="MsoNormal"><span style="font-size:10pt;font-family:Arial;"> </span></p>
<p class="MsoNormal" style="text-align:center;" align="center"><span style="font-size:10pt;font-family:Arial;">LTP mechanisms</span></p>
<p style="text-align:center;"><span style="text-align:center; display: block;"><a href="http://campodearroz.wordpress.com/2008/07/04/potenciacao-de-longo-prazo/"><img src="http://img.youtube.com/vi/GMehTI6DPYI/2.jpg" alt="" /></a></span></p>
<p class="MsoNormal"><span style="font-size:8pt;font-family:Arial;">There are two explanations for what actually occurs at the molecular level during LTP. One hypothesis suggests that the presynaptic neuron fires at abnormal rate, releasing an elevated amount of neurotransmitter. If you release more neurotransmitter, you&#8217;re gonna get a bigger response in the postsynaptic neuron. The other way is actually to, for example, modify the receptor function. So the receptors are more sensitive to the neurotransmitter. In the second model, normal amount of neurotransmitter is released, but the increased sensitivity of the receptor, causes the postsynaptic neuron to fire at an elevated rate.</span></p>
<p class="MsoNormal"><span style="font-size:10pt;font-family:Arial;"> </span></p>
<p class="MsoNormal" style="text-align:center;" align="center"><span style="font-size:10pt;font-family:Arial;">Increased Receptor Sensitivity</span></p>
<p style="text-align:center;"><span style="text-align:center; display: block;"><a href="http://campodearroz.wordpress.com/2008/07/04/potenciacao-de-longo-prazo/"><img src="http://img.youtube.com/vi/Bh-oT-QamOk/2.jpg" alt="" /></a></span></p>
<p class="MsoNormal"><span style="font-size:8pt;font-family:Arial;">When the neurotransmitter glutamate binds with the receptor it opens up a channel through the membrane. Electrically charged ions enter the postsynaptic neuron, which then fires an action potential. But a higher state of sensitivity occurs when an enzyme phosphorylates the protein channel. The next time a neurotransmitter binds with the receptor, more ions stream through the channel. More action potentials fire within the postsynaptic cell, passing a strong signal along to other neurons in the network.</span></p>
<p class="MsoNormal"><span style="font-size:10pt;font-family:Arial;"> </span></p>
<p class="MsoNormal"><span style="font-size:10pt;font-family:Arial;">Segue uma explicação do mecanismo de sensibilização do receptor.</span></p>
<p class="MsoNormal"><span style="font-size:10pt;font-family:Arial;">1 – Moléculas de glutamato, neurotransmissor liberado pelo neurônio pré-sináptico, ligam-se a receptores NMDA na membrana do neurônio pós-sináptico;</span></p>
<p class="MsoNormal"><span style="font-size:10pt;font-family:Arial;">2 – Estes receptores NMDA contêm canais transmembrânicos para Ca<sup>2+</sup>. Assim, a ligação de moléculas de glutamato a estes receptores aumenta a permeabilidade da membrana ao Ca<sup>2+</sup>.</span></p>
<p class="MsoNormal"><span style="font-size:10pt;font-family:Arial;">3 – O influxo de Ca<sup>2+</sup> ativa a proteína cinase II dependente de cálcio-calmodulina (CaMKII). Esta cinase fosforila um segundo tipo de receptor para glutamato, o AMPA. Ao ser fosforilado, o receptor AMPA torna-se mais permeável a íons Na<sup>+</sup>, diminuindo o potencial de repouso da célula. A célula fica, então, mais sensível a impulsos de entrada.</span></p>
<p class="MsoNormal"><span style="font-size:10pt;font-family:Arial;">4 – Além disto, há evidências de que a atividade aumentada da CaMKII aumenta o número de receptores AMPA na sinapse.</span></p>
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		<title>Farmacocinética na Internet</title>
		<link>http://campodearroz.wordpress.com/2008/07/03/farmacocinetica-na-internet/</link>
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		<pubDate>Thu, 03 Jul 2008 20:14:57 +0000</pubDate>
		<dc:creator>Augusto Maeda</dc:creator>
				<category><![CDATA[Farmacocinética]]></category>
		<category><![CDATA[Farmacologia]]></category>
		<category><![CDATA[Add new tag]]></category>

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		<description><![CDATA[Basic Pharmacokinetics: Livro texto para um curso introdutório em farmacocinética.
http://www.wepapers.com/Papers/21306/Makoid_et_al-_Basic_Pharmacokinetics
Biofarmácia e Farmacocinética: Slides de aulas de Maria Deolinda Auxtero
http://www.egasmoniz.edu.pt/ficheiros/alunos/anos_anteriores/biofarmacia/
Pharmacokinetics Tutorial: Introdução de conceitos e farmacocinética aplicada. No final de cada seção, pode-se clicar em um ícone de interrogação para testar os conhecimentos adquiridos na seção.
http://www.rxkinetics.com/pktutorial/1_1.html 
Pharmacokinetics &#8211; Notes &#38; Practice Problems &#8211; University of Alberta
http://www.pharmacy.ualberta.ca/pharm415/contents.htm [...]<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=campodearroz.wordpress.com&blog=2694588&post=18&subd=campodearroz&ref=&feed=1" />]]></description>
			<content:encoded><![CDATA[<div class='snap_preview'><br /><p><strong><span style="font-family:Arial;">Basic Pharmacokinetics:</span></strong><span style="font-family:Arial;"> Livro texto para um curso introdutório em farmacocinética.</span></p>
<p class="MsoNormal"><a href="http://www.wepapers.com/Papers/21306/Makoid_et_al-_Basic_Pharmacokinetics">http://www.wepapers.com/Papers/21306/Makoid_et_al-_Basic_Pharmacokinetics</a></p>
<p class="MsoNormal"><strong><span style="font-family:Arial;">Biofarmácia e Farmacocinética:</span></strong><span style="font-family:Arial;"> Slides de aulas de Maria Deolinda Auxtero</span></p>
<p class="MsoNormal"><a href="http://www.egasmoniz.edu.pt/ficheiros/alunos/anos_anteriores/biofarmacia/">http://www.egasmoniz.edu.pt/ficheiros/alunos/anos_anteriores/biofarmacia/</a></p>
<p class="MsoNormal"><strong><span style="font-family:Arial;">Pharmacokinetics Tutorial:</span></strong><span style="font-family:Arial;"> Introdução de conceitos e farmacocinética aplicada. No final de cada seção, pode-se clicar em um ícone de interrogação para testar os conhecimentos adquiridos na seção.</span></p>
<p class="MsoNormal"><span style="font-family:Arial;"><a href="http://www.rxkinetics.com/pktutorial/1_1.html">http://www.rxkinetics.com/pktutorial/1_1.html</a> </span></p>
<p><strong><span style="font-family:Arial;">Pharmacokinetics &#8211; Notes &amp; Practice Problems &#8211; University of Alberta</span></strong></p>
<p class="MsoNormal"><span style="font-family:Arial;"><a href="http://www.pharmacy.ualberta.ca/pharm415/contents.htm">http://www.pharmacy.ualberta.ca/pharm415/contents.htm</a> </span></p>
<p class="MsoNormal"><strong><span style="font-family:Arial;">A First Course in Pharmacokinetics and Biopharmaceutics </span></strong></p>
<p class="MsoNormal"><span style="font-family:Arial;"><a href="http://www.boomer.org/c/p1/">http://www.boomer.org/c/p1/</a> </span></p>
<p class="MsoNormal"><strong><span style="font-family:Arial;">Pharmacokinetics &#8211; </span></strong><strong><span style="font-family:Arial;">Cornell</span></strong><strong><span style="font-family:Arial;"> </span></strong><strong><span style="font-family:Arial;">College</span></strong><strong><span style="font-family:Arial;"> of Veterinary Medicine</span></strong></p>
<p class="MsoNormal"><strong><span style="font-family:Arial;"><a href="http://web.vet.cornell.edu/public/pharmacokinetics/pharmacokinetics.html"><span style="font-weight:normal;">http://web.vet.cornell.edu/public/pharmacokinetics/pharmacokinetics.html</span></a></span></strong><span style="font-family:Arial;"> </span></p>
<p class="MsoNormal"><strong><span style="font-family:Arial;">Introduction and Review of Basic pharmacokinetics, related responses, and Clinical Applications</span></strong></p>
<p class="MsoNormal"><span style="font-family:Arial;"><a href="http://www.lapk.org/pubsinfo/teaching_topics.php">http://www.lapk.org/pubsinfo/teaching_topics.php</a> </span></p>
<p class="MsoNormal"><strong><span style="font-family:Arial;">Basic Principles of Dose Optimization I – Case Studies</span></strong></p>
<p class="MsoNormal"><span style="font-family:Arial;"><a href="http://www.cop.ufl.edu/safezone/pat/pha5127/previous_case_studies.htm">http://www.cop.ufl.edu/safezone/pat/pha5127/previous_case_studies.htm</a> </span></p>
<p class="MsoNormal"><strong><span style="font-family:Arial;">Farmacocinética</span></strong></p>
<p class="MsoNormal"><span style="font-family:Arial;"><a href="http://www.farmacia.ufmg.br/cespmed/text7.htm">http://www.farmacia.ufmg.br/cespmed/text7.htm</a> </span></p>
<p class="MsoNormal"><strong><span style="font-family:Arial;">General Principles: Pharmacokinetics</span></strong></p>
<p class="MsoNormal"><span style="font-family:Arial;"><a href="http://www.pharmacology2000.com/General/Pharmacokinetics/kinobj1.htm">http://www.pharmacology2000.com/General/Pharmacokinetics/kinobj1.htm</a> </span></p>
<p class="MsoNormal"><strong><span style="font-family:Arial;">Pharmacokinetics and Biopharmaceutics – Lecture Tutorials</span></strong></p>
<p class="MsoNormal"><span style="font-family:Arial;"><a href="http://www.uiowa.edu/~c046138/138tutorialindex.htm">http://www.uiowa.edu/~c046138/138tutorialindex.htm</a> </span></p>
<p class="MsoNormal"><strong><span style="font-family:Arial;">Clinical Pharmacokinetics and T.D.M information on the net</span></strong></p>
<p class="MsoNormal"><span style="font-family:Arial;"><a href="http://www.usal.es/~galenica/clinpkin/marco0.htm">http://www.usal.es/~galenica/clinpkin/marco0.htm</a> </span></p>
<p class="MsoNormal"><span style="font-family:Arial;"> </span></p>
<p class="MsoNormal"><strong><span style="font-family:Arial;">Outros Links: </span></strong></p>
<p class="MsoNormal"><span style="font-family:Arial;">Pharmacokinetics, Metabolism, and Pharmaceutics of Drugs of Abuse</span></p>
<p class="MsoNormal"><span style="font-family:Arial;"><a href="http://www.nida.nih.gov/pdf/monographs/monograph173/download173.html">http://www.nida.nih.gov/pdf/monographs/monograph173/download173.html</a> </span></p>
<p class="MsoNormal"><span style="font-family:Arial;">Merck Manual Professional &#8211; Pharmacokinetics</span></p>
<p class="MsoNormal"><span style="font-family:Arial;"><a href="http://www.merck.com/mmpe/sec20/ch303/ch303a.html">http://www.merck.com/mmpe/sec20/ch303/ch303a.html</a> </span></p>
<p class="MsoNormal"><span style="font-family:Arial;">Pharmacokinetics – an Introduction – Tutorial: Uma explicação rápida sobre alguns conceitos em Farmacocinética.</span></p>
<p class="MsoNormal"><span style="font-family:Arial;"><a href="http://www.4um.com/tutorial/science/pharmak.htm">http://www.4um.com/tutorial/science/pharmak.htm</a> </span></p>
<p class="MsoNormal"><span style="font-family:Arial;">Lista de Software de farmacocinética</span></p>
<p class="MsoNormal"><span style="font-family:Arial;"><a href="http://www.boomer.org/pkin/soft.html">http://www.boomer.org/pkin/soft.html</a> </span></p>
<p class="MsoNormal"><span style="font-family:Arial;">Clinical Pharmacokinetics Preferred Symbols</span></p>
<p class="MsoNormal"><span style="font-family:Arial;"><a href="http://pharmacokinetics.adisonline.com/pt/pt-core/template-adis/cpk/media/CPK_symbols.pdf">http://pharmacokinetics.adisonline.com/pt/pt-core/template-adis/cpk/media/CPK_symbols.pdf</a> </span></p>
<p class="MsoNormal"><span style="font-family:Arial;">PK Solutions 2 Equations – Conjunto de Equações em Farmacocinética</span></p>
<p class="MsoNormal"><span style="font-family:Arial;"><a href="http://www.summitpk.com/files/PKS2Equations.pdf">http://www.summitpk.com/files/PKS2Equations.pdf</a> </span></p>
<p class="MsoNormal"><span style="font-family:Arial;">Periódico &#8211; Drug Metabolism and Pharmacokinetics</span></p>
<p class="MsoNormal"><span style="font-family:Arial;"><a href="http://www.jstage.jst.go.jp/browse/dmpk">http://www.jstage.jst.go.jp/browse/dmpk</a> </span></p>
<p class="MsoNormal"><span style="font-family:Arial;">Guidance for Industry &#8211; Population Pharmacokinetics</span></p>
<p class="MsoNormal"><span style="font-family:Arial;"><a href="http://www.fda.gov/CDER/guidance/1852fnl.pdf">http://www.fda.gov/CDER/guidance/1852fnl.pdf</a> </span></p>
<p class="MsoNormal"><span style="font-family:Arial;">Pharmacokinetics – Wikipedia.</span></p>
<p class="MsoNormal"><span style="font-family:Arial;"><a href="http://en.wikipedia.org/wiki/Pharmacokinetics">http://en.wikipedia.org/wiki/Pharmacokinetics</a></span></p>
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		<title>PPARs</title>
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		<pubDate>Tue, 25 Mar 2008 00:36:35 +0000</pubDate>
		<dc:creator>Augusto Maeda</dc:creator>
				<category><![CDATA[Farmacodinâmica]]></category>
		<category><![CDATA[Farmacologia]]></category>
		<category><![CDATA[RXR PPAR receptores nucleares]]></category>

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		<description><![CDATA[Diferentemente de muitos ligantes que possuem um alvo (receptor) na membrana, os hormônios lipofílicos têm a capacidade de atravessar a membrana plasmática e se ligar a receptores no interior das células. Estes receptores pertencem à superfamília dos receptores nucleares, um conjunto de fatores de transcrição. Esta superfamília compreende os receptores nucleares hormonais (NHRs) e os [...]<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=campodearroz.wordpress.com&blog=2694588&post=12&subd=campodearroz&ref=&feed=1" />]]></description>
			<content:encoded><![CDATA[<div class='snap_preview'><br /><p>Diferentemente de muitos ligantes que possuem um alvo (receptor) na membrana, os hormônios lipofílicos têm a capacidade de atravessar a membrana plasmática e se ligar a receptores no interior das células. Estes receptores pertencem à superfamília dos receptores nucleares, um conjunto de fatores de transcrição. Esta superfamília compreende os receptores nucleares hormonais (NHRs) e os receptores nucleares órfãos. Os NHRs tiveram seus ligantes identificados. Já os receptores órfãos não tiveram seus ligantes conhecidos na época da identificação do receptor.</p>
<p>Os PPARs (<em>Peroxisome proliferator-activated receptor</em>) são NHRs. Eles exercem diversos efeitos no metabolismo de gordura e carboidratos. Segue uma animação mostrando como agem estes receptores.</p>
<p>Primeiro vai a descrição da animação:</p>
<p>1) Um PPAR encontra-se ligado ao DNA. Perceba que há um co-repressor ligado a este receptor.</p>
<p>2) O receptor RXR (<em>Retinoid X Receptor</em>) liga-se ao PPAR, formando um heterodímero.</p>
<p>3) O reconhecimento de um ligante pelo heterodímero faz com que o co-repressor seja desligado do complexo.</p>
<p>4) Uma  mudança conformacional leva ao recrutamento de um co-ativador.</p>
<p>5) Assim, ocorre a transcrição de genes específicos.</p>
<p style="text-align:center;"><span style="text-align:center; display: block;"><a href="http://campodearroz.wordpress.com/2008/03/24/ppars/"><img src="http://img.youtube.com/vi/Wqwlx0LrsVA/2.jpg" alt="" /></a></span></p>
<p>O receptor RXR não está presente apenas na situação acima ilustrada. Ele também forma heterodímeros com vários outros receptores nucleares.</p>
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