>>16919870
It is with respect to velocity. Your image reflects that.
What is your question?

>>16919995
He wrote int p dx. He's clearly too confused to formulate his questions

>>16919995
Not OP, but what's the implication of p being the derivative of KE with respect to v? What does it mean physically?

dKE/dv = p
dKE/dp = v

>>16922136
We'll start with the integral explanation as that's more intuitive.
If you imagine a function which tracks momentum of a point mass as its velocity changes, the total kinetic energy of the function is the accumulation of all that momentum together.
Since velocity is distance over time (as a vector), this accounts both for paths that travel further as well as paths that get there faster.

Understanding KE in this way, the momentum a given point mass has for a given velocity is the quantity you're accumulating to derive KE as that velocity is changing.

>>16922136
>what's the implication of p being the derivative of KE with respect to v
If you know mechanics, it doesn't tell anything new. If you don't know mechanics, it can't be taught in a single 4chan post, but the gist is that from the Lagrangian L(x,v) := KE - PE, classical equations of motion can be gleaned via the Euler-Lagrange equations which say that the infinite dimensional gradient [math] \tfrac{\partial}{\partial x} - \tfrac{d}{dt} \tfrac{\partial}{\partial v} [/math] of L(x,v) should be set to 0 to get the stationary point of the so-called Action[x(t)], or the integral of the Lagrangian over time. Here you see that [math] p(x,v) := \tfrac{\partial}{\partial v} L [/math] makes total sense and the E-L equation is simply Newton's 2nd Law, so the "implication" is more like a definition. That's Lagrangian mechanics.

In Hamiltonian mechanics, which uses the Hamiltonian H(x,p) := p*v(x,p)- L(x,v(x,p)) = KE+PE you get the equivalent equations via a Legendre transform of L(x,v), which is a mathematical transform that has value for convex/concave functions because they got minimums and maximums. Here, the transform takes all the variables v and transforms them to the variables [math] p(x,v) := \tfrac{\partial}{\partial v} L [/math] , and instead of one E-L equation, you get Hamilton's two equations, one of which says [math] \tfrac{dx}{dt} = \tfrac{\partial H}{\partial p} [/math].

So in both cases, it isn't too surprising that >>16922163 is true from the transform as its part of "definition". Here it's interesting only in that KE = KE(v or p) and PE = PE(x)

>>16923509
>What does it mean physically?
I myself couldn't really say it physically means that much per se. If you try and go backwards starting at p(x,v)*dv or p*dv(x,p), it's sorta odd since we only really consider the coordinates (x,v) or (x,p), so it's more [math] \tfrac{\partial L}{\partial v} dv [/math] and the version for H and dp, but again those were "by definition", So it'a little more interesting that H and L are made of terms variable-separable by addition, KE(v or p) and PE(x), or rather, why are H or L are even defined that way in the first place? In special relativity, L = (gamma*mv^2 - gamma*mc^2) - PE. At this point, you'd have to start asking questions like what is KE (or maybe energy related to self) and PE (or maybe energy from interacting with others), or energy in general.

Suppose further that you have a problem where you got some drag/friction where the resistance increases with v, so this isn't some PE(x) we can consider like a spring. Here, energy isn't conserved on the level you'd normally consider, so L and H aren't really applyable, and you gotta start with Newton acceleration instead. Now, you can then work backwards and try to find an L(x,v) and H(x,p) from this, so you get some non-conventional "PE" = PE(v), and H isn't the total energy anymore (btw, math says H is still conserved since L doesn't depend on time explicitly). But if we aren't working with conventional energy to determine motion, it's a little hard to justify why you'd even ask "what is KE" in a conventional way when it isn't even useful here other than maybe "how much conventional energy was lost" into heat. But that's a question to consider when you assume conventional energy is conserved - and it generally is globally - or assuming you can do the conventional L or H at least on some level involving microscopic interactions, which isn't really feasible realistically.