por manuoliveira » Qua Fev 27, 2013 18:55
por Russman » Qua Fev 27, 2013 20:15
é a área, 
e 
os lados conhecidos e 
o ângulo entre eles.
, obtemos![\frac{\mathrm{d} S}{\mathrm{d} t} = \frac{1}{2}\frac{\mathrm{d} }{\mathrm{d} t}\left ( ab\sin \left ( \alpha \right ) \right ) = \frac{1}{2}\left [ \frac{\mathrm{d} (ab)}{\mathrm{d} t}\sin \left ( \alpha \right )+\cos \left ( \alpha \right ).\frac{\mathrm{d}\alpha }{\mathrm{d} t}ab \right ]](/latexrender/pictures/b2f0e27d38bc3f13123066341ed9de5a.png "\frac{\mathrm{d} S}{\mathrm{d} t} = \frac{1}{2}\frac{\mathrm{d} }{\mathrm{d} t}\left ( ab\sin \left ( \alpha \right ) \right ) = \frac{1}{2}\left [ \frac{\mathrm{d} (ab)}{\mathrm{d} t}\sin \left ( \alpha \right )+\cos \left ( \alpha \right ).\frac{\mathrm{d}\alpha }{\mathrm{d} t}ab \right ]")
.
. Isolando, então, a taxa de variação de aplicando a regra 
obtemos, finalmente![\frac{\mathrm{d} \alpha }{\mathrm{d} t} = -\tan \left ( \alpha \right )\left [ \frac{1}{b}\frac{\mathrm{d} b}{\mathrm{d} t}+\frac{1}{a}\frac{\mathrm{d} a}{\mathrm{d} t} \right ]](/latexrender/pictures/6df2f19f4d950eab8c1bc78b67930516.png "\frac{\mathrm{d} \alpha }{\mathrm{d} t} = -\tan \left ( \alpha \right )\left [ \frac{1}{b}\frac{\mathrm{d} b}{\mathrm{d} t}+\frac{1}{a}\frac{\mathrm{d} a}{\mathrm{d} t} \right ]")
Voltar para Cálculo: Limites, Derivadas e Integrais
por AlbertoAM » Sáb Mai 21, 2011 14:23
por AlbertoAM » Sáb Mai 28, 2011 15:53
por Pollyanna Moraes » Sáb Out 22, 2011 17:37
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![\frac{\sqrt[]{\sqrt[4]{8}+\sqrt[]{\sqrt[]{2}-1}}-\sqrt[]{\sqrt[4]{8}-\sqrt[]{\sqrt[]{2}-1}}}{\sqrt[]{\sqrt[4]{8}-\sqrt[]{\sqrt[]{2}+1}}}](/latexrender/pictures/981987c7bcdf9f8f498ca4605785636a.png "\frac{\sqrt[]{\sqrt[4]{8}+\sqrt[]{\sqrt[]{2}-1}}-\sqrt[]{\sqrt[4]{8}-\sqrt[]{\sqrt[]{2}-1}}}{\sqrt[]{\sqrt[4]{8}-\sqrt[]{\sqrt[]{2}+1}}}")
![\sqrt[]{2}](/latexrender/pictures/f21662d1cabab6e8b273a4b6f1cd663a.png "\sqrt[]{2}")
(dica : igualar a expressão a 
e elevar ao quadrado os dois lados)