Lagrangian vs conservation of energy When deriving the equations of motions of a system, one can do it using the Lagrangian approach and calculate the kinetic energy $T$ and potential energy $V$. Then one sets $L=T-V$ and applies the Euler-Lagrange equations. Why can't one just set $\frac{d}{dt}(T+V)=0$, because energy has to be conserved, and derive the equations of motion that way? 
 A: That only gives you one equation of motion, which is underconstrained for motion in one dimension. This makes sense: while energy is conserved in a system, it is not the only thing that is conserved. You need to also include momentum conservation, for instance, which the Euler-Lagrange equations do.
A: In general, the conservation of kinetic plus potential energy alone doesn't give a solution to the movement of the system. In a sufficiently complicated mechanical system, e.g. an isolated system of a higher number of mass points, even all conservation laws (energy, linear momentum, angular momentum, etc.) are not sufficient to yield the time and space dependent solutions. These have to be found by equations of motion like the Lagrange or Hamilton equations. 
A: Consider,


*

*The Hamiltonian as $H(q,p) = \frac{p^2}{2m} + V(q)$. 

*If the Hamiltonian is conserved then we expect the Poisson bracket with $H$ to vanish $\{ H, H\} = 0$ meaning $H$ is a first integral. 

*$H$ need not represent the total energy of the system if $\partial _tH \neq 0$ and hence need not be conserved. 

*Given these points we require an approach that is valid in all cases. The correct approach is to insert the Hamiltonian into Hamilton's equations to obtain the equations of motion. 
$$
\dot p _i =- \frac{\partial H}{\partial q^i}, \qquad \dot q^i = \frac{\partial H}{\partial p_i}
$$


*

*We may then construct a Lagrangian by performing a Legendre transform on $H$ if it is regular. In the case that it is not we require a Dirac construction to account for additional constraints. 

A: Consider a particle moving in the $xy$ plane with 
$$
x=r\cos\phi\, ,\qquad y=r\sin\phi
$$
under the influence of a “central” potential $U(r)$.  Then
$$
T=\frac{1}{2}m (\dot r^2+r^2\dot\phi^2)\, ,\qquad 
L=T-U\, .
$$
There are two Euler-Lagrange equations, one for $r$ and the other for $\phi$, neither of which you can recover from $\frac{d}{dt}(T+U)$.
