a)
Given #L = t^2(dotx^2/2-x^6/6)# the movement equations are obtained by doing
#d/dt((partial L)/(partial dot x))-(partial L)/(partial x)=0# giving
#ddot x = -(2dot x/t+x^5)#
b)
The condition for invariance of #int L dt# is given by
#(partial L)/(partial t) tau+(partial L)/(partial x) xi + (partial L)/(partial dot x)((d xi)/(dt)-dot x (d tau)/(dt))+L(d tau)/(dt)=0#
Now expanding the total derivatives #(d xi)/(dt), (d tau)/(dt)# and substituting we arrive at
#(partial L)/(partial t)tau+(partial L)/(partial x)xi+(partial L)/(partial dot x)((partial xi)/(partial t)+(partial xi)/(partial x)dot x-dot x(partial tau)/(partial t)-dot x^2 (partial tau)/(partial x))+L((partial tau)/(partial t)+(partial tau)/(partial x)dot x)=0#
From #L# we have
#(partial L)/(partial t) = t(dot x^2-1/3x^6)#
#(partial L)/(partial x) =-t^2x^5#
#(partial L)/(partial dot x) =t^2 dot x#
Substituting and grouping the coefficients associated to the powers of #dot x# and equating to zero we have
#{
((x tau)/3+t xi + t x/6 (partial tau)/(partial t)=0),
((partial xi)/(partial t)-x^6/6 (partial tau)/(partial x) = 0),
(t tau+t^2(partial xi)/(partial x) - t^2/2 (partial tau)/(partial t) = 0),((partial tau)/(partial x)=0):}#
Now from the last equation, #tau=tau(t)# substituting into the second equation #xi = xi(x)# remaining the system of differential equations
#{
(x/3 tau+t xi + t/6x (d tau)/(dt)=0),
(t tau+t^2 (d xi)/(dx)-t^2/2 (d tau)/(dt)=0):}#
From inspection, the transformation is
#tau=t# and #xi = -x/2# so
#int L dt# is invariant under the transformation group
#t'=t+epsilon t#
#x'=x-1/2epsilon x#
c) Suppose that #L(t,x,dot x)# is variationally invariant on #t_i,t_f# under the former transformation. Then
#(partial L)/(partial dot x) xi+(L-dot x (partial L)/(partial dot x))tau=C^(te)#
Substituting #(partial L)/(partial dot x), L,xi, tau# we get
#(t^2x dot x)/2+(t^3 dot x^2)/2+(t^3x^6)/6 = C^(te)#
