Distinction between various forms of relativistic energy If I am told to find the total relativistic energy of a particle moving with some velocity v, do I use $E=\gamma mc^2$ or $E^2=p^2c^2+m^2c^4$ and take the square root. I am not sure the distinction between the two.  
 A: If you use $$p=\gamma m \beta c$$ where $\beta = v/c$, then do the algebra for your second equation, you will find out that they are equal.
The distinction for a single particle is usually reduced to whichever one is easier, given the information quickly available.
The advantage and usefulness of the $E^2 = p^2c^2+m^2c^4$ form is that it is useful for systems of several masses.  $E$ will be the total of all the particles in the system:
$$E=\sum_{all~j}\left(K_j+m_jc^2\right)$$
where $K_j$ is the kinetic energy of  particle $j$. The $p^2$ is actually squared magnitude of the vector sum of the momenta of the particles:
$$\vec{p}=\sum_{all~j}\vec{p}_j$$
$$p^2=\vec{p}\cdot \vec{p}.$$
If you then consider re-writing the equation as
$$mc^2=\sqrt{E^2-p^2c^2}$$
you have an Lorentz invariant quantity, the system mass.  Notice that this is not necessarily the sum of the individual masses if you have more than 1 mass.
Consider 2 masses, each with mass $M$ in the center of mass reference frame, both moving toward each other, each with kinetic energy $K=0.5Mc^2$. The total momentum is zero. The total energy is $E=3Mc^2$. This gives us
$$mc^2=\sqrt{9M^2c^4-0}=3Mc^2.$$
The sum of the individual masses is only 2M, but the effective system mass is 3M.
A: 
I am not sure the distinction between the two

Other answers have shown that there is no distinction (for a massive particle).   Let's see how the second expression is derived.  The four-momentum (see Four-vector) $\mathbf{P}$ of a particle has components
$$\mathbf{P} = \left(\frac{E}{c},\vec{p}\right)$$
For a particle of invariant mass $m$, we have the relativistic energy $E$ and momentum $\vec{p}$ are given by
$$E = \gamma m c^2,\quad \vec{p} = \gamma m \vec{v}$$
and so
$$\mathbf{P} = \left(\gamma m c,\gamma m \vec{v}\right)$$
The invariant 'length' of the four-momentum is
$$P_\mu P^\mu \equiv \left(\frac{E}{c}\right)^2 - p^2 = (\gamma m c)^2 - (\gamma m \vec{v})^2 = \left(\gamma m c\right)^2\left(1 - \frac{v^2}{c^2}\right) = \left(mc\right)^2$$
and so
$$E^2 = (pc)^2 + (mc^2)^2$$
$$\Rightarrow (\gamma m c^2)^2 = (pc)^2 + (mc^2)^2$$
or
$$\gamma mc^2 = \sqrt{(pc)^2+(mc^2)^2}$$
A: Both expressions are actually equivalent, and you can use either one at your convenience.
If you start with the definition of relativistic momentum:
$p=m_0\gamma v=\frac{m_0 v}{\sqrt{1-v^2/c^2}}$
Square the expression and multiply it by $c^2$:
$p^2c^2=m_0^2\gamma^2 v^2c^2$
We can conveniently add and substract a term in the expression:
$p^2c^2=m_0^2\gamma^2 v^2c^4-\gamma^2m_0^2c^4+\gamma^2m_0^2c^4$
$p^2c^2=m_0\gamma^2c^4(\frac{v^2}{c^2}-1)+\gamma^2m_0^2c^4$
Where $\gamma^2m_0^2c^4=(\gamma m_0c^2)^2$ and:
$m_0^2c^4\gamma^2(\frac{v^2}{c^2}-1)=m_0^2c^4\frac{\frac{v^2}{c^2}-1}{1-\frac{v^2}{c^2}}=-m_0^2c^4=-(m_0c^2)^2$
Substituting, you find that:
$(\gamma m_0c^2)^2=p^2c^2+(m_0c^2)^2$
And recalling the first definition of $E=\gamma m_0 c^2$, you get:
$E^2=p^2c^2+(m_0c^2)^2$
A: We can verify if the two equation are equivalent:
$$mc^2\gamma  = \sqrt{p^2c^2+m^2c^4}$$
$$mc^2\gamma=mc^2\sqrt{\frac{p^2}{m^2c^2}+1}$$
$$\gamma=\sqrt{\frac{p^2}{m^2c^2}+1}$$
Knowing that $\gamma:=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$, the two equations are the same if $\sqrt{\frac{p^2}{m^2c^2}+1}=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$. Expressing momentum as $p=m\gamma v$, we can obtain:
$$\gamma=\sqrt{\frac{m^2v^2\gamma^2}{m^2c^2}+1}$$
$$\gamma=\sqrt{\frac{v^2\gamma^2}{c^2}+1}$$
$$\gamma^2=\frac{v^2\gamma^2}{c^2}+1$$
$$\gamma^2\left(1-\frac{v^2}{c^2}\right)=1$$
$$\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$
So $\gamma$ is indeed equal to $\sqrt{\frac{p^2}{m^2c^2}+1}$ and the two equations are equivalent.
Notice that for massless particles ($m=0$), $v$ will be equal to $c$ and $mc^2\gamma$ will be undefined ($0\cdot\infty$). But you can see from $E=\sqrt{p^2c^2+m^2c^4}$ that the energy in that case is just: $E=pc$ - this is the case of photons, for example.
A: As long as $v<c$ these two equations are the same. If $v=c$ then the first equation is undefined, but the second equation still holds. 
However, note in the second equation that $p\ne mv $. So you are probably better off to use the first if you are given $v$
