Usuário(a):Nailton Gama/Função-logística-equações

Equação de Verhulst-Pearl

Dedução

${dP \over dt}=r\ P\left(1-{P \over K}\right)\Rightarrow {\int _{P_{\rm {0}}}^{P(t)}}{1 \over {P\left(1-{P \over K}\right)}}\ dP={\int _{\rm {0}}^{t}}r\ dt\Leftrightarrow K\cdot \int _{P_{\rm {0}}}^{P(t)}{1 \over {P(K-P)}}\ dP=r\ t$

$\Leftrightarrow K\ \int _{P_{\rm {0}}}^{P(t)}{1 \over K}\left({1 \over P}+{1 \over {K-P}}\right)dP=r\ t\Leftrightarrow \int _{P_{\rm {0}}}^{P(t)}{1 \over P}\ dP\ +\int _{P_{\rm {0}}}^{P(t)}{1 \over {K-P}}\ dP=r\ t\Leftrightarrow {\ln \left|P\right|}\ {\bigg |}_{P_{\rm {0}}}^{P(t)}-{\ln \left|{K-P}\right|}\ {\bigg |}_{P_{\rm {0}}}^{P(t)}=r\ t$

$\therefore {{{P(t)} \over {K-{P(t)}}}={{{P_{\rm {0}}}\ e^{r\ t}} \over {K-P_{\rm {0}}}}}\therefore P(t)\left(1+{{{P_{\rm {0}}}\ e^{r\ t}} \over {K-{P_{\rm {0}}}}}\right)={{K\ {P_{\rm {0}}}\ e^{r\ t}} \over {K-{P_{\rm {0}}}}}\Leftrightarrow \ln {P \over {K-P}}\ {\Bigg |}_{P_{\rm {0}}}^{P(t)}=r\ t\Leftrightarrow {P \over {K-P}}\ {\Bigg |}_{P_{\rm {0}}}^{P(t)}=e^{r\ t}\therefore$

$P(t)={{K\ {P_{\rm {0}}}\ e^{r\ t}} \over {K+{P_{\rm {0}}}(e^{r\ t}-1)}}\qquad \blacksquare$

Prova de que ${\sqrt {2}}\notin \mathbb {Q}$

Suponha ${\sqrt {2}}\in \mathbb {Q}$

${\rm {\Leftrightarrow \exists !\ p,q\in \mathbb {Z} \vline mdc(p,q)=1\land {\sqrt {2}}={\frac {p}{q}}}}$. Logo,

${\rm {{\sqrt {2}}\in \mathbb {Q} \Rightarrow ({\sqrt {2}})^{2}=\left({\frac {p}{q}}\right)^{2}\Leftrightarrow 2={\frac {p^{2}}{q^{2}}}\Leftrightarrow p^{2}=2q^{2}\Leftrightarrow p^{2}\equiv 0(mod\ 2)\left(\because q^{2}\in \mathbb {Z} \right)\Leftrightarrow p\equiv 0(mod\ 2)\Leftrightarrow {\frac {p}{2}}\in \mathbb {Z} \Rightarrow \left({\frac {p}{2}}\right)^{2}\in \mathbb {Z} .}}$

Além disso,

${\rm {{\sqrt {2}}\in \mathbb {Q} \Leftrightarrow {\frac {\sqrt {2}}{2}}={\frac {\left({\frac {p}{q}}\right)}{2}}\Leftrightarrow {\frac {1}{\sqrt {2}}}={\frac {\left({\frac {p}{2}}\right)}{q}}\Leftrightarrow \left({\frac {1}{\sqrt {2}}}\right)^{2}=\left[{\frac {\left({\frac {p}{2}}\right)}{q}}\right]^{2}\Leftrightarrow {\frac {1}{2}}={\frac {\left({\frac {p}{2}}\right)^{2}}{q^{2}}}\Leftrightarrow q^{2}=2\left({\frac {p}{2}}\right)^{2}\Leftrightarrow q^{2}\equiv 0(mod\ 2)\ \left(\because \left({\frac {p}{2}}\right)^{2}\in \mathbb {Z} \right)\Leftrightarrow q\equiv 0\ (mod\ 2).}}$

Diante disso,

${\begin{cases}{\rm {p\equiv 0\ (mod\ 2)}}\\{\rm {q\equiv 0\ (mod\ 2)}}\end{cases}}{\rm {\Rightarrow mdc(p,q)\geq 2\Leftrightarrow mdc(p,q)\neq 1}}$ ↯.

Conclui-se que ${\sqrt {2}}\notin \mathbb {Q}$

Razão entre os raios iônicos para cada retículo cristalino:

Razão entre os raios iônicos (r+/r–)	Interstício ocupado	Tipo de retículo cristalino
${\frac {r_{+}}{r_{-}}}\geq {\sqrt {3}}-1\approx 0,732$	Cúbico	Tipo CsCl
$0,732\approx {\sqrt {3}}-1>{\frac {r_{+}}{r_{-}}}\geq {\sqrt {2}}-1\approx 0,414$	Octaédrico	Tipo NaCl / sal-gema
$0,414\approx {\sqrt {2}}-1>{\frac {r_{+}}{r_{-}}}\geq {\frac {1}{\sin \left({\frac {\arccos \left(-{\frac {1}{3}}\right)}{2}}\right)}}-1\approx 0,225$	Tetraédrico	Tipo Zns / blenda Tipo CaF2 / fluorita Tipo antifluorita
$0,225\approx {\frac {1}{\sin \left({\frac {\arccos \left(-{\frac {1}{3}}\right)}{2}}\right)}}-1>{\frac {r_{+}}{r_{-}}}\geq {\frac {2{\sqrt {3}}}{3}}-1\approx 0,155$	Trigonal
$0,155\approx {\frac {2{\sqrt {3}}}{3}}-1>{\frac {r_{+}}{r_{-}}}$

${\frac {r_{+}}{r_{-}}}\geq {\sqrt {3}}-1\approx 0,732$

$0,732\approx {\sqrt {3}}-1>{\frac {r_{+}}{r_{-}}}\geq {\sqrt {2}}-1\approx 0,414$

$0,414\approx {\sqrt {2}}-1>{\frac {r_{+}}{r_{-}}}\geq {\frac {1}{\sin \left({\frac {\arccos \left(-{\frac {1}{3}}\right)}{2}}\right)}}-1\approx 0,225$

$0,225\approx {\frac {1}{\sin \left({\frac {\arccos \left(-{\frac {1}{3}}\right)}{2}}\right)}}-1>{\frac {r_{+}}{r_{-}}}\geq {\frac {2{\sqrt {3}}}{3}}-1\approx 0,155$

$0,155\approx {\frac {2{\sqrt {3}}}{3}}-1>{\frac {r_{+}}{r_{-}}}$

${\ce {{TiO2_{(}s)}+{C_{(}s)}+2\ {Cl2_{(}g)}->{TiCl4_{(}s)}+CO2_{(}g)}}$

${\ce {{TiCl4_{(}g)}+2\ Mg_{(}s)->{Ti_{(}s)}+2\ MgCl2_{(}s)}}$