# What does the symbol d mean?

What does the symbol "d" mean??

I've seen the symbol "d" many times and in a physics lecture. For example: the professor would write: $$I = \frac {dq}{dt}$$ instead of: $$I = \frac {Q}{t}$$ I think I've also seen it in calculus equations such as derivatives and such. I'm only in Pre-calculus so I haven't not yet gone over anything related to calculus. But I'm interested in finding out what the "d" means in the formulas. Thanks. Related Introductory Physics Homework Help News on Phys.org
mjsd
Homework Helper
"d" means change... in the limit that the change is infinitestimal. in your particular exapmle: $$I=Q/t$$ really means "average" current because that's total change in charge over total change in time: in fact it means
$$I_{av.}=\frac{\delta Q}{\delta t}=\frac{Q_f-Q_i}{t_f-t_i}$$, now when in the limit of very small change.... ie. $$\delta t \rightarrow 0$$ this becomes $$\frac{dQ}{dt}$$, the advantage of this quantity is that you can now specify "I" at any instance.

anyway, calculus means two things in essence: chop things up into small bits or adding small bits togeter.

"d" means change... in the limit that the change is infinitestimal. in your particular exapmle: $$I=Q/t$$ really means "average" current because that's total change in charge over total change in time: in fact it means
$$I_{av.}=\frac{\delta Q}{\delta t}=\frac{Q_f-Q_i}{t_f-t_i}$$, now when in the limit of very small change.... ie. $$\delta t \rightarrow 0$$ this becomes $$\frac{dQ}{dt}$$, the advantage of this quantity is that you can now specify "I" at any instance.

anyway, calculus means two things in essence: chop things up into small bits or adding small bits togeter.
Oh ok, I knew it was related to delta $$\Delta$$ thanks!

cepheid
Staff Emeritus
Gold Member
Yikes, this is not the best way of learning what dq/dt means.

Note that an "infinitesimal change" as a quantity is NOT a well-defined term mathematically. Mjsd's comment about calculus is good as a conceptual way to look at things only. Therefore, although physicists do it all the time, dq and dt really shouldn't be treated as quantities, and dq/dt shouldn't be treated as a ratio. It is the limit of a sequence of such ratios:

$$\frac{dq}{dt} = \lim_{\Delta t \to 0} \frac{\Delta q}{\Delta t} = \lim_{\Delta t \to 0} \frac{q(t + \Delta t) - q(t)}{(t + \Delta t) - t}$$​

To reiterate: the derivative of the function q(t) is given by the limit as $\Delta t \rightarrow 0$ of the above sequence of ratios. A limit IS a well-defined concept in mathematics, and you will learn what it means when you take calculus. It is used to define a derivative rigorously and formally. As a result, d shouldn't be thought of as a symbol, if you want to be mathematically proper. Instead, $$\frac{d}{dt}$$ should be thought of as a symbol that represents the operation of differentiation. When this d/dt acts on a function, the operation of differentiation with respect to time is carried out on that function to produce the first dervative of the function with respect to time.

$$\frac{d}{dt}q(t) = i(t)$$​

In this example, the derivative of the function, denoted by dq/dt, represents the instantaneous rate of change of q(t) (i.e. the instantaneous current, as opposed to the average current over some finite time interval).

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