# Is $dx$ always positive?

When we refer to change in a quantity, we define it to be (final-initial). If it is positive it indicates an increase from the initial value and negative indicates a decrease.

But when we take this to an infinitesimal perspective and denote the change as $$dx$$ for a quantity, do we still follow the definition final-initial? I mean does the quantity $$\text dx$$ carry the sign with itself, or is it always positive?

Recently I came across $$RC$$ circuits and saw that while deriving the equation for discharging of capacitors as a function of time, they took current $$I$$ as $$-\text dq/\text dt$$ and said that the negative sign indicates a decrease in charge of capacitors.

• Would Mathematics be a better home for this question? Jul 25, 2022 at 6:58
• @Qmechanic I dont think so. the word "chage" would confuse them. :) Jul 25, 2022 at 7:03
• I think the negative sign is solving the problem before the algebra is gone through, an over defintion as far as dq goes, Jul 25, 2022 at 7:07
• @annav If I understand the question correctly, your comment is somewhere between right and wrong. There is a classic confusion in this problem. Usually in a circuit problem $dq/dt$ is the magnitude of the current, and the sign is the direction relative to a pre-defined positive direction. But that's not what $dq/dt$ means in this problem. Here $q$ is the quantity of charge on the capacitor, and the sign indicates charging or discharging of the capacitor, not the direction of the current. So the algebra does take care of everything provided the definition of $dq/dt$ is correct Jul 25, 2022 at 12:56

It is best not to think of $$dx$$ as a quantity - it is just part of the notation. The first derivative of a function $$y(x)$$ with respect to $$x$$ is written as $$\frac{dy}{dx}$$ and is defined at a point $$x=a$$ as

$$\frac {dy}{dx} = \lim_{x \rightarrow a} \frac {y(x) - y(a)}{x-a}$$

- always provided this limit actually exists. Note that this definition is completely symmetric; $$x$$ can approach $$a$$ from above, from below, or indeed by bouncing around on different sides of $$a$$. It is also possible to have one-side derivatives, where $$x$$ must approach $$a$$ from above to give a right derivative (or from below to give a left derivative).

The $$\frac {dy}{dx}$$ notation is known as Leibniz's notation. There are alternative notations which do not use $$dx$$ at all. For example, in Lagrange's notation the first derivative is written as $$y'(x)$$; in Euler's notation it is written as $$(Df)(x)$$ which emphasises the derivative as an operator on functions; and in Newton's notation it is written as $$\dot y (x)$$. See this Wikipedia article for more details.

The $$dx$$ can be negative in this and many other cases. In your current example, we have $$dq$$, which is the charge variation. If the charge decreased, then $$dq<0$$, if it increased, then $$dq>0$$.

Now, in the formula that you presented, we have $$I = -\frac{dq}{dt}$$. The minus sign here just tells us that if $$dq$$ decreases, then $$I$$ is positive, and if $$dq$$ increases, then $$I$$ is negative: the sign of $$I$$ tells us the direction the charge current is flowing.

In summary, yes, $$dx$$ can have both signs, but the minus sign in $$I = -\frac{dq}{dt}$$ has no relation to it, as $$dq$$ could still have both signs there.

Your doubt is reasonable, because the authors and the orators often do not clarify such things. If we have a function $$f:A\subset \Bbb R\to \Bbb R$$, which is differentiable at some interior point $$x\in A$$, we acquire its differential $$df(x,Δx):=f'(x)Δx$$, where $$Δx$$ is a real number. So differential of a real function is a real function of two variables. Let us consider $$f(x)=x, x\in A$$.Then we get $$df(x, Δx)=dx(x,Δx)=(x)'Δx$$, so $$dx=Δx$$. This means that the differential of any function can be written as the product of its derivative times $$dx$$, which is any real number: $$df:=dy=f'(x)dx$$. (It will help you to sketch these thoughts; you will see clearly the geometry of a differential.) Very often we suppose that $$dx$$ is "infinitely small" and that "it can be integrated". The first supposition makes no problem to our approach; the second means $$dx$$ as a part of a differential form -which is another task. Finally, precisely about the sign; for $$f(x)=-x$$ we get $$dy=-dx$$. If $$dx$$ is positive, then (the differential) $$dy$$ is negative, which means no problem with the sign.

• Thank you for your help, but the thing is we have not started differentiation in maths in our school so I don't know all the notations you used. I just know the basic aspects of differentiation in physics. it would be kind of you if you could explain it by avoiding notations, in simple language, and finally, do you mean dx represents the change including the sign (+ve / -ve). Jul 25, 2022 at 7:31
• A dx change is a Δx change which can be seen as "very small" and it can be intergrated.
– SK_
Jul 25, 2022 at 8:11

Let $$x$$ represent some scalar quantity: mass, charge, speed, temperature, etc. A change in $$x$$ can be either positive or negative. Typically $$dx$$ represents a small change in $$x$$ and this change can be positive or negative. $$-dx$$ simply means the negative of the change and if $$dx$$ is negative $$-dx$$ is positive.

In an integral the sign of $$dx$$ is controlled by the limits of integration.

Consider the simple function $$y=6$$ for all $$x$$.

Going incrementally from $$x=1$$ to $$x=3$$ the increments $$\Delta x$$ are positive so the area under the graph $$= \sum\limits_{x=1}^{x=3} y \,\Delta x = y(x_{\rm final}-x_{\rm initial}) = 6(3-1) =+12$$ is positive.

However, going incrementally from $$x=3$$ to $$x=1$$ the increments $$\Delta x$$ are negative so the area under the graph $$= \sum\limits_{x=3}^{x=1} y \,\Delta x = y(x_{\rm final}-x_{\rm initial}) = 6(1-3) =-12$$ is negative.

Notice that the form of the summation, $$\sum\limits_{\rm x_{initial}}^{\rm x_{final}} y \,\Delta x\, \left (\displaystyle{\int_{\rm x_{initial}}^{\rm x_{final}} y \,d x }\right )$$, is the same in both cases and the sign of the increment $$\Delta x / dx$$ depending on the limits.

That's the Maths but now let's look at how the differential equations are set up and to do that a sign/labelling convention must be used.
So consider a simple circuit in which a charged capacitor is discharged through a resistor.

Guessing from what has been written in the question the convention used by the OP is that adopted in the MIT 8.02 Electricity and Magnetism course, (Physics), and described in this chapter of the course textbook.
The convention which I shall call A is summarised in the table below.

Using convention A the labelling of the circuit diagram is as follows.

The direction of the current is chosen with the knowledge that in this case the capacitor is discharging.

Applying KVL (Kirchhoff's Voltage Law) for the loop starting at node $$a$$ and going around clockwise gives,

$$\dfrac qC -IR =0$$ and with $$I = - \dfrac {dq}{dt} \Rightarrow \dfrac qC + \dfrac{dq}{dt}R =0$$ which can be solved to give $$q=q_0 \exp (-t/RC)$$ where $$q_0$$ is the initial charge on the capacitor.
Note that here one needs to use the procedure with regard to integration and the limits of integration as explained earlier.

Furthermore since $$I = -\dfrac {dq}{dt} \Rightarrow I(t) = \dfrac {q_0}{RC} \exp (-t/RC)$$ and the voltage across the resistor $$V_{\rm R}(t) = \dfrac {q_0}{C} \exp (-t/RC)$$

Now let's look at the same problem using a different sign/labelling convention which is called the passive sign convention and is used in the edX 6.002 Circuit and Electronics course, (Electrical Engineering).
Here is a table which outlines convention B.

The key parts being that a circuit element is given a $$+$$ and $$-$$ label with the voltage $$V$$ meaning that $$V$$ is the potential of the node labelled $$+$$ relative to the potential of the node labelled $$-$$.
Furthermore a current arrow label always points to the $$+$$ label.

Using convention B the circuit under consideration is labelled as follows,

Applying KVL starting at node $$A$$ and going clockwise gives,

$$V_{\rm C} + V_{\rm R}=0$$ and with $$I = \dfrac {dq}{dt} \Rightarrow \dfrac qC + \dfrac{dq}{dt}R =0$$ which is exactly the same differential equation as found when using convention A.

However, since $$I=\dfrac{dq}{dt}$$, one finds that $$I(t) = \color{red} -\dfrac {q_0}{RC} \exp (-t/RC)$$ and the voltage across the resistor $$V_{\rm R}(t) = \color{red} -\dfrac {q_0}{C} \exp (-t/RC)$$.
Thus the current is actually flowing in the opposite direction to the current arrow label in the diagram and the potential of node $$b$$ is greater than node $$a$$.

The purpose of showing you the two conventions is to try and explain why in convention A the current, $$I$$, is made to equal $$-\dfrac {dq}{dt}$$.
It is no more than for a simple circuit to try and produce a positive value for the current assuming that it is a capacitor discharge situation and thus the direction of current flow is from the label $$q$$ on the capacitor.
That is not to say that in more complex circuits a capacitor assumed to be discharging is actually found to be charging and in such a situation the current will be negative, in the opposite direction to the direction indicated by the current arrow label.

Convention B is probably easier to apply but is more abstract and one does not need to "assume" what will happen, that all comes out in the solution.
Convention B is probably easier to apply for more complex circuits?