I am a high school student who was playing around with some equations, and I derived a formula for which cannot physically imagine.

\begin{align} W & = \vec F \cdot \vec r \\ \frac{dW}{dt} & = \frac{d}{dt}[\vec F \cdot \vec r] = \frac{d\vec F}{dt} \cdot \vec r + \vec F \cdot \frac{d\vec r}{dt} \\ \implies & \boxed{P = \frac{d\vec F}{dt} \cdot \vec r + \vec F \cdot \frac{d\vec r}{dt}} \end{align}

I differentiated Work using its vector form formula $\vec F \cdot \vec r$ So I got this formula by applying the product rule. If in this formula $\frac{d\vec F}{dt}=0$ (Force is constant), than formula just becomes $P = \vec F \cdot \frac{d\vec r}{dt}$ which makes total sense, but this formula also suggests that if $\frac{d\vec r}{dt}=0$ then the formula for power becomes $P =\frac{d\vec F}{dt} \cdot \vec r$, which implies that if the velocity is zero that doesn't necessarily mean that Power of the object will also be zero!

But I don't find this in my high school textbook and I can't think of an example on that top of my head where this situation is true.

From what I have heard and read, if the velocity of the object is zero then power is also zero.

Can someone please clear my supposed misconception or give me an example of the situation where this happens?

  • $\begingroup$ Why do you have $W=\mathbf F\cdot\mathbf v$? Or is that supposed to be $\mathbf r$ instead of $\mathbf v$? $\endgroup$ – BioPhysicist Dec 1 '20 at 18:52
  • $\begingroup$ @BioPhysicist My r looks like a v. I apologize for my handwriting, It's the middle of the night in my country. But could you answer my question? I can assure you that I have used Fr and not Fv. $\endgroup$ – Archit Chhajed Dec 1 '20 at 18:55
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    $\begingroup$ Hi Archit, I edited your post by adding MathJax and removing the attached image. For the future, you can find a tutorial on basic MathJax usage here. $\endgroup$ – J. Murray Dec 1 '20 at 19:14
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    $\begingroup$ Does this answer your question? Why isn't the product rule used in the definition of mechanical work? $\endgroup$ – Brian Drake Dec 2 '20 at 10:37
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    $\begingroup$ @BrianDrake Maybe about integration. But it should be emphasized that the formula for work/power with a constant force is a special case. Students get the idea that's the general formula and have big confusions afterwards. Regarding frames I disagree. Students are usually exposed very early to the intuitive idea that the velocity of something depends on the observer. It doesn't cost much to emphasize that work/power are also observer-dependent. I've seen too many students calculating some work in a frame, then change frame and wrongly use there the work obtained in the previous frame. $\endgroup$ – pglpm Dec 5 '20 at 12:21

The work done by a force is not defined by $W=\mathbf F\cdot\mathbf r$. Work is instead defined in terms of a line integral over a path (your equation just assigns a work for a force and position, which does not match what we mean by the work done by a force). We have

$$W\equiv\int\mathbf F\cdot\text d\mathbf r\to\text dW=\mathbf F\cdot\text d\mathbf r$$

So when we have $P=\text dW/\text dt$ we just have

$$P=\frac{\text dW}{\text dt}=\frac{\mathbf F\cdot\text d\mathbf r}{\text dt}=\mathbf F\cdot\frac{\text d\mathbf r}{\text dt}=\mathbf F\cdot\mathbf v$$

So there is no $\mathbf r\cdot \text d\mathbf F/\text dt$ term in the expression for power. This works out conceptually as well: the power output of a force should not directly depend on the position of the particle (i.e. the location of the origin) in question.

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    $\begingroup$ Oh, now I understand my mistake! Thank you for taking the time. I used the wrong definition of Work. $\endgroup$ – Archit Chhajed Dec 1 '20 at 18:59
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    $\begingroup$ @ArchitChhajed Please remember to upvote all useful answers and to accept an answer if it sufficiently answers your question. $\endgroup$ – BioPhysicist Dec 1 '20 at 19:00
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    $\begingroup$ @MichaelSeifert No, not if $\mathbf r$ represents a position vector as it usually does. I think you mean for constant force $W=\mathbf F\cdot\Delta\mathbf r$. $\endgroup$ – BioPhysicist Dec 1 '20 at 19:15
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    $\begingroup$ @BioPhysicist: Of course. My mistake. $\endgroup$ – Michael Seifert Dec 1 '20 at 19:16
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    $\begingroup$ @Džuris I try to meet the OP where they are at rather than yank them to where I am :) Additionally, I don't think the key point here is missed at all with my notation. Any reader who is familiar with $\delta W$ will understand what is going on here anyway. $\endgroup$ – BioPhysicist Dec 2 '20 at 5:04

Work is defined as $W = \int_{}^{} \vec F \cdot d \vec r = \int_{}^{} \vec F \cdot \vec v \enspace dt$. Power, P, is dW/dt = $\vec F \cdot \vec v$.

Your relationship for work is incorrect, so your relationship for power (boxed-in relationship in your question) is not correct.

  • $\begingroup$ Thanks for the restatement of my answer to help further solidify the point :) $\endgroup$ – BioPhysicist Dec 1 '20 at 19:16
  • $\begingroup$ You are welcome. $\endgroup$ – John Darby Dec 1 '20 at 19:17

As others have already answered, $W = \mathbf F \cdot \Delta \mathbf r$ is a simplification and works only in a special case of constant $\mathbf F$. And so does your formulae.

One way to look at it physically is to recognize that work is not a function of position. Mathematically we usually describe it using the concept of inexact differential:

$$\delta W = \mathbf F \cdot d \mathbf r$$

This notation is used to underline the fact that you can integrate both sides and get the same number, but you may not rearrange this formula and in fact you can not (in general case) express $\mathbf F$ using $W$.

An example of an exact differential and what it allows you to do:

$$d \mathbf r = \mathbf v \, dt \implies \mathbf v = \frac {d \mathbf r} {dt}$$

P.S. There are some special cases where you can write $\mathbf F = \nabla \, W$, in those cases it is said that $\mathbf F$ is a potential force.


When you take derivatives it is of crucial importance having very clear in you mind what is function of what.

In the definition of the work, the force is a function of the position, not of time. This means that although you can certainly move in a force field which varies in time, what matters is the force that you measure at each step in your path regardless of how this force has been in the past or will be in the future.

  • $\begingroup$ The question didn't say anything about force fields. The force may be a function of position, but even if that were true, position is a function of time, so force is still a function of time. Overall, this does post not answer the question. $\endgroup$ – Brian Drake Dec 5 '20 at 9:38
  • $\begingroup$ @BrianDrake I was expecting such a comment. The force is not an explicit function of time. But to formalize this one needs partial derivatives. $\endgroup$ – DarioP Dec 6 '20 at 10:20

The other answers discuss strange things like integrals and differentials. This answer tries to meet the OP where they are: it is targeted at the level of mathematics used in the question and starts with the formula $W = \vec{F} \cdot \vec{r}$.

Presumably the reason you started with this formula is that you did find it in your high school textbook and were taught it at school. That's because the formula is right, unlike what some of the other answers said. But you need to understand two things to apply it correctly:

  1. It requires $\vec{F}$ to be constant.
  2. It requires $\vec{r}$ to be the change in position while the object is subject to the force $\vec{F}$. This would be better written as $\Delta \vec{r}$. [1]

Now let’s look at your problem:

if $\frac{d\vec{r}}{dt} = 0$ then the formula for power becomes $P = \frac{d\vec{F}}{dt} \cdot \vec{r}$, which implies that if the velocity is zero that doesn't necessarily mean that Power of the object will also be zero

This statement fails to account for the two things discussed above:

  1. It does not recognise that $\frac{d\vec{F}}{dt} = 0$.
  2. It does not recognise that $\vec{r}$, which is really $\Delta \vec{r}$, does not really mean anything when velocity is zero. (To address this point properly, we do need integrals – see the other answers.)

[1] For those who know about electricity, this is like how people often writen $V$ when they really mean $\Delta V$.


Assuming we are talking about non-relativistic speeds, then defining $\vec{F}=m\vec{a}$, your proposal (of sorts) is that: $$P=\frac{d\vec{F}}{dt} \cdot \vec{r}+\vec{F} \cdot \frac{d\vec{r}}{dt}=m\left(\vec{r} \cdot \frac{d\vec{a}}{dt}+\vec{a} \cdot \vec{v}\right)$$ with $\vec{v}:=d\vec{r}/dt$.

$d\vec{a}/dt$ is referred to as the "jerk" (or "jolt"), with $\vec{a}$ the acceleration vector. It must occur whenever a force is ramped from $\vec{0}$ to $\vec{F}$. You can learn more about the jerk on its Wikipedia article.

Whether or not the jerk appears in the formula for power depends upon how you define the work: if you use $dW=\vec{F} \cdot d\vec{r}$, then the jerk does not appear. However, if you define $W=\int \vec{F} \cdot d\vec{r}$, then, by applying differentiation under the integral you will have the equation you derived except that $d\vec{F}/dt$ will be replaced by $\partial\vec{F}/\partial t$: According to this differentiation under the integral rule, $$\frac{dW}{dt}=\frac{d}{dt} \int _{\vec{r}(t_0)}^{\vec{r}(t)}\vec{F}(\vec{r},t') \cdot d\vec{r}$$ $$=\vec{F}(\vec{r}(t),t) \cdot \frac{d(\vec{r}(t))}{dt}-\vec{F}(\vec{r}(t_0),t_0) \cdot \frac{d(\vec{r}(t_0))}{dt}+\int _{\vec{r}(t_0)}^{\vec{r}(t)}\frac{\partial\vec{F}}{\partial t} \cdot d\vec{r}$$ $$=:\vec{F} \cdot \vec{v}-\int _{\vec{r}_0}^{\vec{r}}\frac{\partial\vec{F}}{\partial t} \cdot d\vec{r}.$$ If one then assumes that $\frac{\partial\vec{F}}{\partial t}$ is independent of distance (which is not generally the case), then $$\frac{dW}{dt}=:\vec{F} \cdot \vec{v}-\frac{\partial \vec{F}}{\partial t} \cdot \Delta \vec{r}.$$ However, if one considers what would be required for the jolt to be independent of distance (think of the simple example of ramping a force such that the source of the force is always in contact with the object experiencing the force), one can quickly convince oneself that this is highly unusual.

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    $\begingroup$ There is no $(\frac{d}{dt})a$ in the question. There are no integrals, no $dr$, no $dW$. What does it mean for force to be independent of distance, yet depend on time? How does $*$ relate to the dot product (since you say "dot" in some places)? In short, what on Earth are you talking about? $\endgroup$ – Brian Drake Dec 2 '20 at 11:39
  • $\begingroup$ I am using $*$ and dot inter-changeably due to my present lack of knowledge of how to use MathJax. $(d/dt)a$ is meant to be the derivative of acceleration which is (in classical as opposed to special relativity) proportional to the derivative w.r.t. time of the force. The integral comes in if one defines the work by an integral of $dW$ such that the initial work is 0. Typically, the 2 definitions will be numerically equal because the jerk is frequently 0. $\endgroup$ – PrawwarP Dec 3 '20 at 9:52
  • $\begingroup$ By the way, I apologize for my poor knowledge base concerning MathJax. $\endgroup$ – PrawwarP Dec 3 '20 at 9:57
  • $\begingroup$ It looks like you define work as $\int _{\vec{r}(t_0)}^{\vec{r}(t)}\vec{F}(\vec{r},t) \cdot d\vec{r}$ (taking your expression for $dW/dt$ and removing the $d/dt$). This is incorrect, because $\vec{F}$ is always evaluated at time $t$. It should be evaluated at whatever time the object was at that distance. $\endgroup$ – Brian Drake Dec 5 '20 at 13:34
  • $\begingroup$ Sorry, I'll correct my abuse of notation. $\endgroup$ – PrawwarP Dec 5 '20 at 13:43

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