electro-magnetismo y mecánica-en-ingeniería

Examen de electro-magnetismo:

Principio:

E(x,y,z) = qk·(1/r)^{3}·a·< x^{2},y^{2},z^{2} >

E(yz,zx,xy) = qk·(1/r)^{4}·a^{2}·< (yz)^{2},(zx)^{2},(xy)^{2} >

Ley:

div[ E(x,y,z) ] = ?

Anti-div[ E(yz,zx,xy) ] = ?

Ley:

Anti-Potencial[ E(x,y,z) ] = ?

Potencial[ E(yz,zx,xy) ] = ?

Ley: [ de corrección del examen ]

div[ E(x,y,z) ] = d_{x(yz)}^{2}[ Anti-Potencial[ E(x,y,z) ] ]

Anti-div[ E(yz,zx,xy) ] = d_{x(yz)}^{2}[ Potencial[ E(yz,zx,xy) ] ]

Ley:

R·d_{t}[q(t)]+(-C)·p(t) = W·f(ut)·e^{ut}

p(t) = W·( 1/(uR·d_{ut}[f(ut)]+(-C)·f(ut)) )·f(ut)·e^{ut}

q(t) = W·( ut /o(ut)o/ (uR·f(ut)+(-C)·int[ f(ut) ]d[ut]) ) [o(ut)o] f(ut) [o(ut)o] e^{ut}

Ley:

R·d_{t}[q(t)]+C·p(t) = W·f(ut)·e^{(-1)·ut}

p(t) = W·( 1/((-u)·R·d_{ut}[f(ut)]+C·f(ut)) )·f(ut)·e^{(-1)·ut}

q(t) = W·( ut /o(ut)o/ ((-u)·R·f(ut)+C·int[ f(ut) ]d[ut]) ) [o(ut)o] f(ut) [o(ut)o] e^{(-1)·ut}

Ley:

Sea ( d_{t}[ I_{cx} ] = 0 & d_{t}[ I_{cy} ] = 0 ) ==>

Si d[M_{1}(t)] = (1/2)·mgx·(1/s)^{2}·cos(nw)·d[w] ==>

M_{1}(t) = (1/2)·mg·(x/n)·(1/s)^{2}·sin(nw)

Si d[ d[M_{2}(t)] ] = mg·(1/s)^{2}·sin(nw)·cos(nw)·d[y]d[w] ==>

M_{2}(t) = mg·(y/n)·(1/s)^{2}·(1/2)·( sin(nw) )^{2}

M_{1}(t) = M_{2}(t) <==> ( w(t) = (1/n)·arc-sin( I_{cx}/I_{cy} ) & I_{cx} [< I_{cy} )

Ley:

Sea d_{t}[ I_{c} ] = 0 ==>

Si d[M_{1}(t)] = (1/2)·I_{c}·u^{2}·cos(nw)·d[w] ==>

M_{1}(t) = (1/2)·I_{c}·u^{2}·(1/n)·sin(nw)

Si d[ d[M_{2}(t)] ] = I_{c}·u^{2}·(1/x)·sin(nw)·cos(nw)·d[x]d[w] ==>

M_{2}(t) = I_{c}·u^{2}·ln(ax)·(1/n)·(1/2)·( sin(nw) )^{2}

M_{1}(t) = M_{2}(t) <==> ( w(t) = (1/n)·arc-sin( ( 1/ln( aI_{c}·(1/(md)) ) ) ) & aI_{c} >] md·e )

Este problema es un ejemplo,

de que es divertido estudiar conmigo de profesor,

porque no pongo calcular el ángulo con anti-funciones que es un coñazo.

No buscan su felicidad siguiendo a Dios,

y no entiendo porque no la buscan.

En el sexo no hay la felicidad,

siendo mujer del prójimo hay violación mental.

Ley: [ del calor electro-magnético ]

div[ E_{e}(x,y,z) ] = (-2)·(1/c)·B_{e}(x,y,z)

Deducción:

E_{e}(x,y,z)+int[ B_{e}(x,y,z) ]d[t] = 0 = m·d_{tt}^{2}[ < x,y > ]

x(t) = ct·( cos(w) )^{2}

y(t) = ct·( sin(w) )^{2}

div[ E_{e}(x,y,z) ]+div[ inr[ B_{e}(x,y,z) ]d[t] ] = 0^{2}

div[ int[ B_{e}(x,y,z) ]d[t] ] = ( 1/(d[x]+d[y]) )·(d[x]+d[y]) [o] div[ int[ B_{e}(x,y,z) ]d[t] ]

div[ E_{e}(x,y,z) ]+2·(1/c)·B_{e}(x,y,z) = 0^{2}

div[ E_{e}(x,y,z) ] = (-2)·(1/c)·B_{e}(x,y,z)

Ley: [ del calor gravito-magnético ]

div[ E_{g}(x,y,z) ] = (-2)·(1/c)·B_{g}(x,y,z)

Teorema:

d_{x}[u(x,y,t)]+d_{y}[u(x,y,t)] = (-2)·(1/c)·d_{t}[u(x,y,t)]

u(x,y,0) = H(ax,ay)

u(x,y,t) = sum[k = 1]-[oo][ ( 0·H(ax,ay) || 0 ) ]·e^{ax+ay+(-1)·act || 0}

Teorema:

d_{x}[u(x,y,t)]+d_{y}[u(x,y,t)] = 2·(1/c)·d_{t}[u(x,y,t)]

u(x,y,0) = H(ax,ay)

u(x,y,t) = sum[k = 1]-[oo][ ( 0·H(ax,ay) || 0 ) ]·e^{ax+ay+act || 0}