# Prove a=v*dv/dx

Using calculus, and assuming a particle moving along the x-axis is concerned, prove that $a=v*dv/dx$

~~~~~~~~~~~~ this is what I did, but im not sure it's rigorous enough:

$a=dv/dt$

$t=x/v$

$a=dv/d(x/v)$ (read all x1 as x subscript 1)

$a=\lim_{(x_1 \rightarrow x_0)}\frac {(v_1-v_0)}{(x_1/v_1 - x_0/v_0)}$ Consider the denominator; as $x_1\rightarrow x_0$, $v_1\rightarrow v_0$; so we can rewrite the equation as

$a= \lim_{(x_1 \rightarrow x_0)}\frac {(v_1-v_0)}{((x_1-x_0)/v))}$ (rest of proof flows easily from here)

Now the last line is what im not so sure about; I believe my reasoning is correct, but the proof doesnt seem rigorous to me, at least not in the way it's notated; since we took the limit of denominator v shouldnt be included in the limit, but I dont know how to show it, can I say that $\lim_{(x1 \rightarrow x0)} x_1/v_1 = \frac {\lim_{(x_1 \rightarrow x_0)} x_1)}{(\lim_{ (x_1 \rightarrow x_0)} v_1)}$ would that solve my problem?

• As this is your first post here on MSE, I'd like to present you with this, so that your posts can be more easily read, and thus more quickly answered, in the future. meta.math.stackexchange.com/questions/5020/… Sep 29, 2014 at 20:52

Not sure if this is "calculusy" enough for you:

$$a = \dfrac{dv}{dt} = v \cdot \dfrac{dv}{dt} \cdot \dfrac{1}{v} = v \cdot \dfrac{dv}{dt} \cdot \dfrac{dt}{dx} = v\cdot \dfrac{dv}{dx}$$

• Does the statement hold when moving in $\mathbb R^3$?
– Leo
Apr 13, 2016 at 12:06
• yes, it would be $\vec a=|\vec v|\frac{d\vec v}{ds}$ where $ds=|\vec v|dt$ Dec 1, 2019 at 8:40

Since $$a$$ is defined as the rate of change of velocity with respect to time:

$$a=\frac{dv}{dt}$$ ,

and is identical to $$a=\frac{dv}{dt} .\frac{dx}{dx}$$

where $$\frac{dx}{dt}$$ is velocity, then we are left with: $$a=v\frac{dv}{dx}$$