Derivative of velocity with respect to position

Question

Preface: I am a physicist, not a mathematician, but I'm seeking something closer to mathematical rigor.

I'm deriving the expression for kinetic energy from Newton's second law, and it seems to rely on the curious relationship

$$ \frac{dv}{dx} = \frac{a}{v} $$

where $x(t)$ is a function of $t$, $v(t) = dx/dt$, and $a(t) = dv/dt$.

This appears to be a mathematical statement about deriviatves, not a physical definition. Is there (a) a geometric interpretation for this relationship, and/or (b) a way to express this in terms of the fundamental definition of the derivative, i.e. in terms of limits?

Answer 1

This is just a parametric derivative, right? Since $v$ and $x$ are both parameterized by $t$, one has $$\frac{dv}{dx} = \frac{dv(t)/dt}{dx(t)/dt} = \frac{a(t)}{v(t)}=\frac{a}{v}$$ The parameter $t$ is suppressed in the first and last expressions, which is a little sloppy but convenient notationally.

Is this what you're looking for? Said differently, it's basically the chain rule:

$$\underbrace{\frac{dv}{dt}}_a = \frac{dv}{dx}\underbrace{\frac{dx}{dt}}_v$$ which then can be rearranged to get your expression $$\frac{dv}{dx} = \frac{a}{v}$$

Answer 2

Let's start qualitatively. If $a$ is large then our velocity increases a lot if we increase $t$ by a (relatively) small amount. If $v$ is small then a small increase in $t$ results in a (relatively) small increase in position. Putting this together we see that $\frac{dv}{dx}$ will be large when $a$ is large and $v$ is small, so this tracks with your formula. So physically, the formula seems reasonable.

Next, lets do a specific computation in a simple situation. Assume that $v\neq 0$, otherwise the right hand side of your formula is nonsense. For simplicity let's think about the case where $a$ is constant. In this case we get the familiar equations of projectile motion

$$x(t)=x_0+v_0t+1/2at^2,\ \ v(t)=v_0+at,$$ where $x_0$ and $v_0$ are initial positions and velocities. Further, lets assume that we have picked coordinates where $x_0=0$ and $v_0>0$.

Using the quadratic formula we can solve for $t$ in terms of and get $$t=\frac{-v_0\pm \sqrt{v_0^2+2ax}}{a}$$

Plugging this in for $v$ we find that

$$v(x)=\sqrt{v_0^2+2ax},$$ where we are forced to pick the positive square root by our assumption that $v_0=0$. Differentiating we find that

$$\frac{dv}{dx}=\frac{a}{\sqrt{v_0^2+2ax}}=\frac{a}{v}$$

In the case where $a$ is not constant we can follow the same road map but use "Taylor approximations". More precisely, Let $x_0=x(t_0)$, $v_0=v(t_0)=v(x_0)$, and $a_0=a(t_0)$. Our formula for $v(t)$ becomes $$v(t)=v_0+v'(t_0)(t-t_0)+R_1(t)=v_0+a_0(t-t_0)+R_1(t),$$ where $\lim_{t\to t_0}\frac{R_1(t)}{t-t_0}=0$.

Rather than writing down $x(t)$ and solving for $t$ we use the inverse function theorem to write

$$t=t_0+\frac{dt}{dx}\vert_{x=x_0}(x-x_0)+R_2(x)=t_0+\frac{1}{v_0}(x-x_0)+R_2(x),$$ where $\lim_{x\to x_0}\frac{R_2(x)}{x-x_0}=0$.

Note that here we are using the fact that $v_0\neq 0$ and we need to observe that the above formula is only guaranteed to hold for $x$ sufficiently close to $x_0$.

Plugging this into $v$ gives

$$v(x)=v_0+\frac{a_0}{v_0}\left((x-x_0)+R_2(x)\right)+R_1(t(x))$$

If we write down the limit definition we get $\frac{dv}{dx}\vert_{x=x_0}=a_0/v_0$

Morally, speaking, this is all just a more detailed explanation for the heuristic formula $\frac{dv}{dx}=\frac{dv}{dt}\frac{dt}{dx}=\frac{dv/dt}{dx/dt}=a/v$ mentioned in the comments