# B The meaning of dx

1. Feb 8, 2017

### anonymous24

Hello,
I am slightly confused about the actual meaning of dx. Because I read in a physics textbook, they say something along this line:
"We divide this region into spherical shells of radius r, surface area 4pr 2, thickness dr, and volume dV = 4pr^2 dr."
I don't understand how can we represent dV like this, like it's a small bit of the total V. However, we learnt in mathematics dx represent a small "change" in x. I am not sure whether I should treat dx as an infinite small x or infinite small change in x. Thank you for answering my question in advance!

2. Feb 8, 2017

### mathman

Infinitely small change and infinitely small are the same in this context.

3. Feb 8, 2017

### anonymous24

Thanks for the reply, but why is that? Could you also please give an example where it won't be the same? thank you

4. Feb 8, 2017

### Khashishi

Formally, $dV = 4pr^2 dr$
means
$\frac{dV}{dr} = 4pr^2$

In standard math, a differential only makes any sense if you can rearrange the equation such that one differential sits on top of another. You can treat the differentials as objects that you can divide, but the real definition comes from the epsilon delta definition of the derivative.
(There are some non-standard analysis methods that treats dV as an actual value, but I'm not too familiar with them.)

5. Feb 8, 2017

6. Feb 8, 2017

### davidbenari

Treat it as a little chunk of x. An infinitesimal one. Or in your words, a small change in x.

I hate to say this, but even though a lot of math guys out there try to pretend they're really angry because you're using infinitesimal quantities, you will get pretty far in your math education if you use these infinitesimal quantities. Their anger is kind of bogus too, since I've seen pure math guys use this concept as well. It's just convenient and makes a ton of sense. Also, if you're in physics, you absolutely need this concept since so many equations rely on it.

7. Feb 8, 2017

### anonymous24

Thanks for the reply, so do you mean a small change in x and a small chunk of x is the samething? that's the thing the confuses me.

8. Feb 8, 2017

### Staff: Mentor

I don't think you should spent many thoughts on this answer. The easiest way is in my opinion to look at where it comes from:
$$\frac{dV}{dr} = \lim_{\Delta \, r \rightarrow 0} \frac{\Delta \,V}{\Delta \, r} \ \ \ \ \ \ (1)$$
so in general it's a tiny difference. But a closer look shows us, although $\Delta \,r$ might be a tiny difference in the denominator, it can still result in a big one in the nominator $\Delta \, V$, which makes it context sensitive. I think one should keep this in mind, when dealing with $dr$ as something small. I see @mathman's remark in post #2 as the short form to say: Whether we define it as in $(1)$ or by
$$\frac{dV}{dr} = \lim_{h \rightarrow 0} \frac{V(r+h) - V(r)}{h} \ \ \ \ \ \ (2)$$
doesn't make a difference. In the first case $dr \approx \Delta \, r$ is an infinitesimal small change in $r$ and in the second it is an infinitesimal small piece of $r$, so $dr \approx h$.

As these notations developed, it wasn't clear at the time, which way would be the best to handle them, or even what exactly they mean. So a brief look into history might also be very enlightening. A start could be the corresponding Wikipedia entries:

https://en.wikipedia.org/wiki/Leibniz's_notation
https://en.wikipedia.org/wiki/Notation_for_differentiation#Lagrange.27s_notation
https://en.wikipedia.org/wiki/Hyperreal_number

9. Feb 8, 2017

### Stavros Kiri

The Differential (of a function f=f(x)) in rigorous mathematics is defined as a function, e.g. at a point x0, df|x0(h) = {limx→x0[f(x)-f(x0)]/(x-x0)}•h = f'(x0)•h

By varying h small, you basically get the same results and idea/concept as in physics, or non-rigorous mathematics and/or mathematical physics. I hope that helps. (+ it interprets and unifies some of the above replies* and approaches ...)

*most of them unique, correct and useful in their own ways and perspectives ! ...

P.S. (last edit): CAUTION! (for all) : In trying to understand this post, please be careful [for all] not to be confused ! ... (but it is/(can be) very fruitful and useful if you do in fact understand it ...)
[e.g. you might ask yourself "why/(how come) differential [is] a function? etc." ...]
But the definition works, in [/and covers] all cases ... . Just do the simple math, e.g. f'(x) = df(x)/dx, and plug it in in the definition, to interpret/justify everything, even the symbol/notation df/dx ... . Then think of what a function actually is and does (e.g. allows you to vary the variance h, and make it as small as you want, etc.), and let your mind free ... . The concept tacitly behind it: continuity (of a function, the Reals (i.e. Real numbers set), etc.), ... etc. .
It's all very deep, it works, plus a good intelectual exercise too! [May be rigorous mathematicians aren't useless after all ...] And thanks to the OP (anonymous24), for bringing up the issue.

Last edited: Feb 8, 2017
10. Feb 8, 2017

### Stavros Kiri

You have a [good] point (especially for physicists to pay attention and caution) but not quite true (see previous post above).

Also regarding:

That's the idea, but ... as a function ... not "an actual value" ... (unless you take the value of the differential function df(h) at h=h0 ...)

11. Feb 8, 2017

### davidbenari

I think I see your confusion. You might want to think of it this way:

If you are integrating, then a little chunk of x is a better way to think of it. For example, say you have a density function $\rho(x)$, where $x$ is the coordinate along some rope or something. The mass on a little piece of rope is $\rho(x)dx$ where clearly, I have assumed $dx$ is a little chunk of rope.

If you are differentiating, then thinking of $dx$ as a change in $x$ might be better. For example, say $x$ is the quantity of something, and $F$ is some function of $x$. Then you could think of $\frac{dF}{dx}$ as how much F changes if you input a little change in x.

But this is very wishy washy. Now that I think of it, sometimes you think of $dx$ as a little chunk of something and sometimes as a small change in $x$. I guess it depends on the problem. I think your confusion in this problem might disappear if you do more concrete problems and try to put some meaning into the differential.

12. Feb 8, 2017

### Stavros Kiri

Lol (about the math guys etc.), a bit sarcastic, but mostly or partially true enough!
However see post #9 above ..., it unifies things (but careful [for all] not to be confused ! ...), ...etc.

13. Feb 16, 2017

### laplacianZero

Dx is an infinite small change in x

14. Feb 16, 2017

### Stavros Kiri

This is a vague statement ...
[for example, how do you define 'infinite small'? (or better: 'infinitely small') ... . Limit? Or function with a small variance range? ... etc.]

15. Feb 22, 2017

### laplacianZero

An infinite number of dx that is infinitely small can fill up the segment length of 1.

16. Feb 22, 2017

### Stavros Kiri

That's 01dx = 1, so true enough! ...

17. Feb 23, 2017

### Stavros Kiri

Two more remarks:

1.
Not quite true. We have:
Δx = x-x0 . By taking Δx as small as we want (i.e. in the limit), x0 can still be anything ... . In the h notation though, i.e. x = x0 + Δx = x0+h, the h variable is the one that gets "infinitely small" ... [and x, x0 can still be anything (that meet that small difference) ].

Therefore, since Δx = h, that's the variable that gets infinitely small, not x or x0. {See also fresh_42 ' s post #8 above}
And in that sense, dx is a[n infinitely] small change in x, or small chunk of x, but also a small h ...
So
is good enough.

2. The effect on a (real) function is briefly as follows:
df(x) = f'(x)dx [where f'(x) = df(x)/dx]
So if dx gets "very small", and the derivative f'(x) exists and is finite, then also df(x) gets very small ...

In essence that expresses the known theorem in Calculus that sais that "A differentiable function is necessarilly continuous" (but not vice versa [i.e. not all continuous functions are differentiable] ...).

Also these are relevant to:

Last edited: Feb 23, 2017
18. Feb 25, 2017

### Kumar8434

I don't think $dx$ ever means infinitely small $x$. $dx$ always represents an infinitely small change in the value of $x$. I don't think infinitely small $x$ has any meaning. You can only change $x$ by an infinitely small amount. In calculus, $dx$ is assumed to be smaller than any other change in $x$, such that if an interval has a length $dx$, then even a function, whose value continuously changes can be assumed to be constant over that interval. Let our continuously varying function be $x^2$. At $x=2$, it has a value 4. Even when we change $x$ by a very small, but finite amount, 0.001, then $2.001^2$ comes out to be 4.004001. You see that $x^2$ always changes, no matter how small change in $x$ we bring because it is continuously varying, i.e. it always has a non-constant derivative. $dx$ is assumed to be a change so small, smaller than any other change, that even the value of continuously varying functions remains virtually constant when we change $x$ by that amount. So, the change in the value of $f(x)=x^2$ when $x$ changes from $2$ to $2+dx$ is so small that it can be assumed to be the same at $x=2$ and $x=2+dx$.

Let's try to find surface area of a cylinder. It can be assumed to be made of an infinite number of circles or circumference $2\pi r$ distributed over a height $h$ each of breadth $dx$. The surface area contributed by each of these circles is $2\pi rdx$. So, the surface area must be the 'sum' of these infinite number of circumferences, which is $2\pi rdx+2\pi rdx+2\pi rdx+......$ upto $h$ terms$=2\pi rh$. Let's try to find the volume of a sphere in a similar way. The sphere can be assumed to be made up of an infinite no. of spherical shells each of 'breadth' $dx$. Note that the circumferences of all the circles making up the cylinder were constant but in this case the surface areas of all these shells are not the same. If a shell is at a distance $x$ from the center of the sphere, then its surface area is $4\pi x^2$. Since, $dx$ is infinitely small, so this surface area can be assumed to be constant over this breadth. So, the volume contributed by each spherical shell is $dV=4\pi x^2dx$. Now, the mathematical tool of adding all these non-constant small volume contributions is integration, which gives $V=\int_0^R4\pi x^2dx$

Last edited: Feb 25, 2017
19. Feb 25, 2017

### Stavros Kiri

But rather in the h variable it makes sense (Δx=h can get infinitely small ... [Δx = x-x0 = h can be thought of as a change of variable, for a particular point x0] ).

x is the variable of a real function here. E.g. it can be x=0 or x→0 ...

We sometimes do that , but this is not rigorous calculus (may be mathematical methods in physics, one kind - nothing wrong with that, perhaps). It can get misleading (see ahead). The rigorous concepts behind those things (that we are all trying to describe here) are the notions of limit and continuity (with the regular rigorous ε-δ definition ...) and definitions of derivative (as a limit), differential (see post #9 above) and integral (with the use of Sums and sequences, Sup and Inf and the sequence of partitions/sum/limit ...).
Other than that we are talking about approximations and approximative concepts, which of course can be very useful and enlightening (especially in Physics), but sometimes can be misleading.

For example

... not quite true. That is actually true only for dx=0 or in the limit (dx or h → 0). In fact f(x+dx) ≅ f(x) + f'(x)dx, or more accurately, if the binomial expansion exists:
f(x+dx) = f(x) + (1/1!)f'(x)dx + (1/2!)f''(x)(dx)2 + (1/3!)f'''(x)(dx)3 + ... ...

So what you say is actually true only as approximation (and in fact 0th order, not even 1st order approximation) or in the limit (dx or h → 0). But we use that in proofs, as intermediate step sometimes, to avoid pedantic and tedious ε, δ definitions or nasty partitions sums and limits in the integrals etc.
Thus your comments and post are useful, but I just thought I would make these clarifications.

20. Feb 25, 2017

### Kumar8434

Yeah, I should've said that the derivative of the function can be assumed to be constant when $x$ changes by an amount $dx$. In the sphere example, the volume of the sphere is continuously changing by a rate $4\pi x^2$. So, in the graph of the volume function of a sphere, we can assume the part of the graph between $x$ and $x+dx$ to be a straight line of slope $4\pi x^2$ so that the small change in volume $dV$ contributed by this part can be obtained by directly multiplying the change in $x$ ,i.e. $dx$ with the constant rate of change, i.e. $4\pi x^2$. So, the function is assumed to be changed by a small amount $dV$, and this $dV$ is obtained by assuming that the graph of the function is a straight line between $x$ and $x+dx$ so that its derivative remains constant over that part.
Sorry, I got confused by the 'adding the rectangles' explanation I read in my book. So, I thought that the two sides of the rectangle parallel to the $y-axis$ should be equal and hence the function is assumed to remain constant. But, now I think, if those two sides of rectangle differ by a small amount $dy$, then they're no longer rectangles, then shouldn't we be adding the trapeziums instead?