Mechanical power is normally defined as $P = \mathrm{d}W/\mathrm{d}t$, and work is normally defined as $W = \vec F \cdot \vec x$. Today an undergrad pointed out a confusion he had from Griffiths' E&M book where he has the line

$$P = \frac{\mathrm{d}W}{\mathrm{d}t} = \vec F \cdot \vec v$$

Which is a definition of power I had seen before. But the student was confused because it seems like, if you have a time dependent force $\vec F(t)$, the math should work out like:

$$P = \frac{\mathrm{d}W}{\mathrm{d}t} = \frac{\mathrm{d}\vec F(t)}{\mathrm{d}t} \cdot \vec x + \vec F \cdot \frac{\mathrm{d}\vec x(t)}{\mathrm{d}t}$$

(From the product rule.)

However, I've never seen the above formula, so I'm guessing it's wrong. Also, it clashes with the $\vec F \cdot \vec v$ version that I'm pretty sure works completely fine with a time dependent force.

I gave him a pretty weak answer and warned him that it's probably not correct: I told him that if you start with the differential form of work, $dW = \vec F \cdot d \vec x$, it seems to assume that the force stays the same in that tiny $d\vec x$, so it's constant there, and then dividing both by $dt$ gives you the equation we're looking for.

Wikipedia seems to say something similar, basically using the step of $d\vec x = \vec v \ dt$ to get $\vec F \cdot \vec v$, which also assumes that $\vec v$ is constant over $d \vec x$.

So, assuming one of those are correct, I see how they get $\vec F \cdot \vec v$. But what's the flaw in the product rule thing?

  • $\begingroup$ Work is ${\rm d}W = \vec{F} \cdot {\rm d} \vec{x}$ as it applies to small increments of displacement. If you use $W= \vec{F} \cdot \vec{x}$ you are making assumtions of constant force and of linear path. $\endgroup$ Commented Jul 18, 2017 at 16:47
  • $\begingroup$ Note that $\frac{ {\rm d} \vec{F} }{ {\rm d}t} = 0$ is already built in the epxression $W= \vec{F} \cdot \vec{x}$. $\endgroup$ Commented Jul 18, 2017 at 16:54

1 Answer 1


The error in the original reasoning comes from asserting that $W=\vec F\cdot \vec x$. This is only true if the force is constant.

For a particle traveling along a parameterized curve $\vec x(t)$ and under the influence of a force $\vec F(\vec x,t)$ which is explicitly dependent on both position in space and time, the work performed on the particle by this force from a time $t_0$ to a time $t$ is defined as follows: \begin{align} W_{t_0}(t) = \int_{t_0}^{t} \vec F(\vec x(t'),t')\cdot \dot{\vec x}(t') \,dt' \end{align} Note that the expression on the right is often written $\int \vec F\cdot \vec dx$, but this is really schematic, the the mathematically precise definition is what I have written above in terms of a parameterized path with an integral over some range of parameter values. The definition of instantaneous power is then \begin{align} P(t) = \dot W_{t_0}(t) \end{align} Taking the derivative of both sides with respect to $t$, and using the fundamental theorem of calculus, we obtain the desired expression for the power \begin{align} P(t) = F(\vec x(t),t)\cdot \dot{\vec x}(t) \end{align}

  • $\begingroup$ I think $F$ doesn't have to be constant but constant direction. $\endgroup$
    – Shuchang
    Commented Sep 20, 2013 at 0:46
  • 1
    $\begingroup$ @ShuchangZhang That's simply not true. If you want to take the force outside of the integral, then it must not have non-trivial dependence on space or time. $\endgroup$ Commented Sep 20, 2013 at 0:53
  • $\begingroup$ Oh, yes indeed. $\endgroup$
    – Shuchang
    Commented Sep 20, 2013 at 0:56
  • $\begingroup$ $\int\vec F\cdot d\vec x$ is precise if you understand it as symbolic notation for the 1-form $\sum_i F_i dx^i$ $\endgroup$
    – Christoph
    Commented Dec 2, 2013 at 20:09
  • $\begingroup$ @Christoph Agreed, although I hesitate to say such things in an answer like this since I think few who would be confused by the question would also understand differential forms. Thanks for the comment. $\endgroup$ Commented Dec 2, 2013 at 20:19

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