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If I have a force $F(x)$, can I assume it to be constant in any infinitesimal interval such as $Rd\theta$,$ dy \over cos\phi$, $dz$ etc. or can I assume it to only be constant in the interval $[x,x+dx)$? If yes, can you prove it mathematically and if not, which infinitesimal intervals can I assume a force to be constant in & why?

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  • $\begingroup$ related: impulses and delta functions $\endgroup$
    – user140606
    Feb 19, 2017 at 16:53
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    $\begingroup$ I think you can only do that, in general, if the element is infinitesimal in all dimensions. So $dx\,dy\,dz$ is fine, but $R\,d\theta$ is not. $\endgroup$
    – lemon
    Feb 19, 2017 at 16:57
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    $\begingroup$ @lemon : That looks like an answer, not a comment. $\endgroup$ Feb 19, 2017 at 17:42
  • $\begingroup$ @lemon I disagree. If the force is finite, the change across an infinitesimal interval will always be infinitesimal, so the force can be treated as a constant. The triple infinitesimal is just a much smaller infinitesimal $\endgroup$
    – user126422
    Feb 19, 2017 at 19:55
  • $\begingroup$ @AlbertAspect Consider a large $R$ and the force $f(R)=1/R$. The force is far from constant across $R\,d\theta$... $\endgroup$
    – lemon
    Feb 19, 2017 at 20:08

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What you want to do is a Taylor Expansion. Assuming that $F\in C^1(\mathbb{R}^3)$ you have:

$$ F(\mathbf{x})=F(\mathbf{x}_0)+\nabla F(\mathbf{x})|_{\mathbf{x}_0}\cdot(\mathbf{x}-\mathbf{x}_0) + \mathcal{o}(\mathbf{x}-\mathbf{x}_0)$$

So, you can approximate $F$ to be constant only if the first (and higher) order terms are neglectable for your purposes. Here I've showed you a general expansion, but you can write the gradient in any coordinate system (for example spherical) you wish. If it's legitimate to consider only the zeroth order of the expansion usually depends on the explicit form of $F(\mathbf{x})$ and the particular application in which you want to make this approximation.


Let's now consider a one-variable function $F:\mathbb{R}\to\mathbb{R}$, $F\in C^N(\mathbb{R})$. It's then possible to write it using the Taylor theorem:

$$ F(x)=\sum_{k=0}^{N}\frac{d^kF}{dx^k}|_{x_0}\frac{(x-x_0)^k}{k!} +\mathcal{o}((x-x_0)^k)$$

If we put N=1, we get:

$$ F(x)=F(x_0)+\frac{dF}{dx}|_{x_0}\cdot(x-x_0)+\mathcal{o}(x-x_0)$$

That means that you can legitimately consider $F(x)\simeq F(x_0)=\text{constant}$ in the neighborhood of $x_0$ only if the first (and higher) order derivative of $F(x)$ evaluated in $x=x_0$ times $(x-x_0)$ is small enough to be considered neglectable for your purposes. Here I wrote $F:=F(x)$, but it's exactly the same if $F:=F(\theta)$. Just put $x\mapsto\theta$ (or any other variable you wish) in the above formulas.


Consider now the limit for $\mathbf{x}\to \mathbf{x_0}$, i.e. $\mathbf{x}-\mathbf{x_0}\simeq d\mathbf{x}$. The differential of the function $F(\mathbf{x})$ is its linear increase respect to its variables. It is defined as:

$$dF=\nabla F \cdot d\mathbf{x} =\frac{\partial F}{\partial x}dx+\frac{\partial F}{\partial y}dy+\frac{\partial F}{\partial z}dz$$

So F is not increasing (i.e. $F$ is constant) when moving along a direction only if the partial derivative in the same direction is zero. You can write this in any coordinate system you want. For example in spherical:

$$dF=\frac{\partial F}{\partial r}dr+\frac{1}{r}\frac{\partial F}{\partial \theta}rd\theta+\frac{1}{r\sin\theta}\frac{\partial F}{\partial \phi}r\sin\theta d\phi$$

So just evaluate the partial derivative and see if it's zero. If that's the case, then you can consider the function constant along that direction.

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  • $\begingroup$ I only know single variable calculus(Calc I). Is there any way to do this using those basic tools? $\endgroup$
    – xasthor
    Feb 20, 2017 at 14:54
  • $\begingroup$ Yes, of course. I edited the answer to add the single variable case. Let me know if now it's more clear. $\endgroup$ Feb 20, 2017 at 16:00
  • $\begingroup$ This is essentially telling me when/why I can assume $F(x)$ to be constant in the interval $[x_0, x_0+dx]$ and $F(\theta)$ to be constant in the interval $d\theta$, $f(z)$ to be constant in the interval $dz$ etc. but my question was whether I can assume $f(x)$ to be constant in any infinitesimal interval such as $d\theta$, $dz$, $dy/cos\phi$ etc., not just $dx$ because I've seen it done in my physics classes but I don't understand $why$ I can do it. $\endgroup$
    – xasthor
    Feb 21, 2017 at 9:57
  • $\begingroup$ Well, if you consider $F(\mathbf{x})$ to be a three-dimensional force (i.e. $F:\mathbb{R}^3\to\mathbb{R}^3$), you need to know multi-variable calculus, otherwise you can't be rigorous. Then the answer is the first one i wrote you. If you're thinking about $F(x)$ as having only a single variable (for example if you consider the line $R\theta$ with $R=constant$) the you can simply use the second answer I wrote. If you don't know enough math to understand the above formulas, I'm afraid the only thing you can do is evaluate $\frac{F(x+\Delta x)-F(x)}{F(x)}$ and check if is $\ll 1$. $\endgroup$ Feb 21, 2017 at 10:17
  • $\begingroup$ I edited my answer again, trying to be even more clear and using explicitly infinitesimal quantities. I hope that now you can more easily understand the matter. Let me know. $\endgroup$ Feb 21, 2017 at 10:46

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