# 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. ## Answers and Replies

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
Science Advisor
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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