-1

Some days ago I have asked a question about a formula for power, many generous people have answered my question and clarify for me that the correct formula of work is $$\mathrm{d}W= \mathbf{F}\cdot \mathrm{d}\mathbf{r}$$ and not $W= \mathbf{F}\cdot \mathrm{d}\mathbf{r}$.

I am now confused, when I see, for example, $\mathrm{d}A/\mathrm{d}n$. I cannot tell whether the quantity I want to derive is $A=\mathbf{B}\cdot \mathbf{C}$ or $A=\mathbf{B}\cdot \mathrm{d}\mathbf{C}$.

How can I tell which one is the correct one?

This is just trying to generalize my knowledge of my previous question.

A formula for power

Qmechanic
  • 201,751
Sohaib Ali Alburihy
  • 45
  • 2
    Your problem has little to do with dot products. Do you know how to handle differentials, derivatives and integrals ? It seems to me you should check your knowledge regarding these topics. – Hans Wurst May 06 '21 at 15:05
  • What is ${\rm d}A / {\rm d}n$ ? Please explain the terms above. – JAlex May 06 '21 at 19:24
  • I can understand derivatives, and integrals fairly well, but my problem is with the derivative of the dot product of two vector field, more specificaly I can't tell what I am deriving. I will try to restate my question. – Sohaib Ali Alburihy May 06 '21 at 19:29
  • Mr JAlex, let me restate my question, just some moments – Sohaib Ali Alburihy May 06 '21 at 19:32

3 Answers3

4

The differentials $\text{d}r$ and $\text{d}W$ are best thought of as the (infinitesimal) change in those quantities. $\bf{F}\cdot \text{d}\bf{r}$ represents the dot product of the force along an (infinitesimal) change of distance, which is kind of like "scaling up" the (infinitesimal) change. The resulting object should also be infinitesimal, so the correct equation would look like $$\text{d} W = \bf{F} \cdot \text{d} \bf{r}.$$ Even if the change is not infinitesimal, this idea still holds (scaling up a change in a quantity is a change of a different quantity). In this case, it is common to use the symbol $\Delta$ or $\delta$ instead of $\text{d}$.

There is also a geometric explanation for this. If $W$ is a (smooth) function, then $\text{d}W$ is called the differential of $W$, and actually lives in a different vector space than the space of functions. Hence, something like $\text{d}W = Fr$ for functions $F, r$ would not make sense because the product of two functions is again a function, so the two sides of $\text{d}W = Fr$ live in different spaces. However, $\text{d}r$ is also a differential, and so $F\text{d}r$ is also a differential, so the equation $$\text{d}W = F\text{d}r$$ is well-posed.

jsborne
  • 994
  • You have clarified many deep concepts for me, thanks a lot – Sohaib Ali Alburihy May 06 '21 at 19:33
  • I can understand that both sides of the equation must be differentials or functions, well this is an new concept for me, thank you for clarifying.

    But well when I see dP/dt, dW/dt Can I tell am I deriving dW, or dP, just from seeing them, or I must know the formulas of both of them?

     – Sohaib Ali Alburihy May 06 '21 at 19:41
  • The notation is a little ambiguous, but fortunately it is usually fine to treat "fractions" of differentials as literal fractions; meaning if $$\frac{\text{d}W}{\text{d}t} = P,$$ then $$\text{d}W = P\text{d}t.$$ This procedure of "multiplying by a differential on both sides" helps convert differential equations to integral equations. More rigorously, differentials can be seen as linear operators on vectors (which are represented by derivatives). So $$\text{d}W\bigg(\frac{\text{d}}{\text{d}t}\bigg) = \frac{\text{d}W}{\text{d}t}.$$ – jsborne May 06 '21 at 20:01
  • Also, just to answer your question regarding using these concepts to solve problems, it depends on what is given. For example, if $P = P(W)$ is given, then it's better to write $$\frac{\text{d}W}{\text{d}t} = P \implies \int \frac{\text{d}W}{P(W)} = \int \text{d}t,$$ where as if $P = P(t)$ is given, then it's better to write $$\frac{\text{d}W}{\text{d}t} = P \implies \int \text{d}W = \int P(t)\text{d}t.$$ – jsborne May 06 '21 at 20:09
2

I am not able to comment hence I am answering on vague terms.
$$W=\int_a^bF.dr$$ $$or$$ $$dW=F.dr$$
You can't write as   $$W =F.dr (incorrect)$$
It's just for understanding; "the work done to bring about a small displacement is given by....."

Therefore if you see dA/dB=C, You can write it as$$ A= \int C.dB$$ Hope it helps!

hawexrutile
  • 117
0

My question, is this, dW/dt=P In this case we have derived dW with respect to t, and dW is a differential (and infinetsmall amount of W)

But if we derive the power with respect to time, (regardless of physical meaning of this) we will write.

dP/dt, in this case is the derivative of P not dP, with respect to time,

when I see them dW/dt, dP/dt, I see in both (numerators (metaphrically)), dW, dP.

however in the first equation we derived the differential of W, and in the second equation we derived the the function P not its derivative... so can I tell whether I am deriving a function, or a differential just by looking at the derivatives?, or do I have to understand the nature of the quantity which I am deriving (is it defined as differential or a function).

Hopefully you can understand me

Sohaib Ali Alburihy
  • 45
  • I'll try my best to address your confusion (let me know if what I've said so far is unclear). If $P(t)$ is a function of time, then the derivative of $P$ with respect to $t$, written as $$\frac{\text{d}}{\text{d}t} (P) = \frac{\text{d}P}{\text{d}t},$$ is going to be another function of time (in general). Now after applying $\text{d}/\text{d}t$ on $P$, the numerator and denominator can each be treated as a differential, which gives them a completely different meaning! At this point, we can multiply by $\text{d}t$ to move it to the other side, for example, just as you would for a fraction. – jsborne May 06 '21 at 20:14
  • The fact that this works, though, is a coincidence (but an intentional one). So the notation $$\frac{\text{d}P}{\text{d}t}$$ is actually overloaded. On one hand, it represents the derivative operator acting on $P(t)$, but on the other, it represents a "fraction" of differentials. – jsborne May 06 '21 at 20:15
  • One way to see this which might help, is consider the expression $$\int \frac{\text{d}P}{\text{d}t} \text{d}t.$$ On one hand, the integral over $t$ is the inverse operator of the derivative with respect to $t$, so this really just gives $P$, evaluated at the boundary of the integral. On the other hand, the $\text{d}t$'s "cancel": $$\int \frac{\text{d}P}{\text{d}t} \text{d}t = \int \text{d}P = P\vert_\text{boundary},$$ so the two interpretations coincide. – jsborne May 06 '21 at 20:18
  • Thanks a lot for your patience, I think now I can understand – Sohaib Ali Alburihy May 06 '21 at 21:58