# A partial differential equation for kinetic energy

The kinetic energy of a point particle of mass $m$ and speed $v$ is $K = \frac{1}{2}mv^2$. An elementary mathematics textbook I saw asked one to show that

$$\frac{\partial K}{\partial m}\frac{\partial^2 K}{\partial v^2} = K.$$

While this is a straightforward exercise in partial differentiation, is there supposed to be any physical meaning behind this formula? For example, is there a significance to quantities that satisfy this nonlinear PDE?

• I can't see where this has any utility at all. The point of having any equation, differential or algebraic, is to put constraints on a system. We then solve these equations to obtain an unknown. However, in this case, we already know the answer and the equation gives us no new information. – DJBunk Mar 19 '13 at 15:19
• I am tempted to agree. Feel free to post this as an answer. – Doubt Mar 19 '13 at 18:33
• This isn't proof that this equation is worthless but I've never seen anything like it before in a physics context. I can't think of a principle that would motivate such an equation, and it is not satisfied for the relativistic kinetic energy $E - mc^2 = (\gamma - 1) mc^2$ anyway. – Michael Brown Mar 20 '13 at 6:12
• Making a product ansatz, leads to the equation being solved for all of $K(m,v)=(\frac{m}{2a}+d)(a v^2+bv+c)$. – Nikolaj-K Mar 20 '13 at 9:11

It is not difficult to see physical situations where the kinetic energy of an object depends on variations of both, $m$ and $v$. This is for example the case of a rocket which propels itself by the usual gas emission method. In this case both $m$ and $v$ vary. But a useful quantity to study would be the rate of change of the kinetic energy, which one could right for 1-D motion as
$\frac{dE_k}{dt}= \frac{\partial E_k}{\partial m}\frac{dm}{dt}+\frac{\partial E_k}{\partial v}\frac{dv}{dt}$