mecànica teoría

Si m·d_{tt}^{2}[x(t)] = f(t) ==> (1) & (2) & (3)

(1): d_{tt}^{2}[x(t)] = (1/m)·f(t)

(2): d_{t}[x(t)] = (1/m)·int[ f(t) ] d[t]

(3): x(t) = (1/m)·int[ int[ f(t) ] d[t] ] d[t]

Si ( m·d_{tt}^{2}[x(t)] = a·( x(t) )^{n} <==> (m/2)·d_{t}[x(t)]^{2}= ( a/(n+1) )( x(t) )^{n+1} ) ==> ...

... (1) & (2) & (3)

Demostració: ( k+(-2) = kn || 2k+(-2) = k·(n+1) )

(1): x(t) = ...

... ( ( (1+(-n))/(n+1) )^{(1/2)}·( (1+(-n))/2 )^{(1/2)}·(a/m)^{(1/2)}·t )^{2/(1+(-n))}

(2): d_{t}[x(t)] = ...

... ( (1+(-n))/(n+1) )^{(1/2)}·( 2/(1+(-n)) )^{(1/2)}·(a/m)^{(1/2)}·...

... ( ( (1+(-n))/(n+1) )^{(1/2)}·( (1+(-n))/2 )^{(1/2)}·(a/m)^{(1/2)}·t )^{(n+1)/(1+(-n))}

(3): d_{tt}^{2}[x(t)] = ...

... (a/m)·( ( (1+(-n))/(n+1) )^{(1/2)}·( (1+(-n))/2 )^{(1/2)}·(a/m)^{(1/2)}·t )^{(2n)/(1+(-n))}