# 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?

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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 $K(m,v)=(\frac{m}{2a}+d)(a v^2+bv+c)$. – NikolajK Mar 20 '13 at 9:11

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 subject to the constraint. However in this case, we already know the answer and the equation gives us no new constraints and hence no new info information.

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The given equation does not seem to have application power, as it is an identity in differential form. It assumes the variation of both mass and speed of the object, which from physics point of view it is possible, but not the type of variations suggested by the equation.

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

$\frac{dE_k}{dt}= \frac{\partial E_k}{\partial m}\frac{dm}{dt}+\frac{\partial E_k}{\partial v}\frac{dv}{dt}$

This can relate the rate of change of the kinetic energy to the acceleration of the object, or the force acting on it. I don’t know whether this has any immediate application, but it is a situation that could arise in a rocket motion situation if one is interested in finding propulsion Power for example.

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