Mathematical Definition of Power 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?
 A: 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.
A: 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.
A: 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.
A: 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:

*

*It requires $\vec{F}$ to be constant.

*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:

*

*It does not recognise that $\frac{d\vec{F}}{dt} = 0$.

*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 $\Delta V$.
A: 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.
A: 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.
