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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?

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2 Answers 2

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

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

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

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