# Proyectos

Páginas: 2 (411 palabras) Publicado: 7 de noviembre de 2011
1)
dx/dt=k x (4000-x)
dx/(x (4000-x))=k dt
∫▒〖dx/(x (4000-x))=〗 ∫▒〖k dt〗
∫▒〖dx/(x (4000-x))=〗 k∫▒〖 dt〗

A/x+B/(4000-x)
(A(x-4000)+Bx)/(x (4000-x))
A(4000-x)+Bx=14000A-Ax+Bx=1
x(B-A)+4000A=1
*B-A=0
*4000A=1

A=1/4000
B=A
B=1/4000
∫▒〖A/x+∫▒B/(4000-x)=〗 k∫▒〖 dt〗
∫▒〖(1/4000)/x+∫▒(1/4000)/(4000-x)=〗 kt+c
∫▒〖(1/4000dx)/x+∫▒(1/4000 dx)/(4000-x)=〗 kt+c
1/4000 ∫▒〖dx/x+1/4000 ∫▒dx/(4000-x)=〗 kt+c
1/4000 ln|x|+1/4000(-ln|4000-x|)=kt+c
1/4000 (ln|x|-ln|4000-x|)=kt+c
1/4000 ln|x/(⁡(4000-x))|=kt+cln|x/(⁡(4000-x))|=4000kt+4000c
e^ln|x/(⁡(4000-x))| =e^(4000kt+4000c)
x/(4000-x)=〖(e〗^4000kt)(e^4000c)
x=4000(e^4000kt.e^4000c )-x(e^4000kt.e^4000c )x+x(e^4000kt.e^4000c )=4000(e^4000kt.e^4000c )
x(1+(e^4000kt.e^4000c ))=4000(e^4000kt.e^4000c )
x=4000(e^4000kt.e^4000c )/(1+(e^4000kt.e^4000c ) )
x (t)=4000(e^4000kt.e^4000c)/(1+(e^4000kt.e^4000c ) )

ln|2/(⁡(4000-2))|=4000k(0)+4000c
ln|2/3998|=4000k(0)+4000c
lln|2/3998|=4000c
ln|2/3998|/4000=c
-0.0019≈c
ln|x/(⁡(4000-x))|=4000kt+4000cln|12/(⁡(4000-12))|≈4000k(10)+4000(-0.0019)
ln|12/3988|≈40000k-7.6
ln|12/3988|+7.6≈40000k
(ln|12/3988|+7.6)/40000≈k
0.000044≈k
a)
x (t)≈4000(e^4000kt.e^4000c )/(1+(e^4000kt.e^4000c) )
x (t)≈4000(e^(4000(0.000044)t).e^(4000(-0.0019)) )/(1+(e^(4000(0.000044)t).e^(4000(-0.0019)) ) )
x (t)≈4000(e^0.176t.e^(-7.6) )/(1+(e^0.176t.e^(-7.6) ) )
x(t)≈4000(e^(0.176(20)).e^(-7.6) )/(1+(e^(0.176(20)).e^(-7.6) ) )
x (t)≈4000(e^3.52.e^(-7.6) )/(1+(e^3.52.e^(-7.6) ) )
x (t)≈66.5054
b)
ln|x/(⁡(4000-x))|=4000kt+4000cln|2000/(⁡(4000-2000))|=4000kt+4000c
ln|2000/2000|≈4000(0.000044)t+4000(-0.0019)
ln|1|≈0.176t-7.6
ln|1|+7.6≈0.176t
0+7.6≈0.176t
7.6≈0.176t
7.6/0.176≈t
43.(18) ̅≈t

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