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$\newcommand{\F}{\mathbf{F}}$$\newcommand{\r}{\mathbf{r}}$$\newcommand{\v}{\mathbf{v}}$$\newcommand{\d}[2]{ \frac{ d #1 }{ d #2 }}$ There’s a well-known known formula for the power $P$ exerted on a point mass $P=\F\cdot\v$ where $\F$ is the force doing the work and $\v$ is the velocity with which the point mass moves. The proof my instructor provided back in the day goes as follows:

$$\begin{align} W &= \int\F\cdot d\r \\ W &= \int\F\cdot\d{\r}{t}\,dt \tag{a}\\ \d Wt &= \frac{d}{dt}\int \F\cdot\d{\r}{t}\,dt \tag{b} \\ P &= \frac{d}{dt}\int\F\cdot\v\,dt \tag{c}\\ P &= \F\cdot\v \tag{R}\\ \end{align}$$

There are a few reasons why I find this proof in general less than satisfactory:

  • It makes no reference to limits of integration
  • It does not appear to recognize that $\F$ can vary with $t$ as well as $\r$
  • If does not appear to recognize that $\r$ varies with $t$
  • step $\rm{(a)}$ might be making use of a possibly erroneous “differentials cancel” argument

In an attempt to rigorously derive $\rm{(R)}$, I have carefully defined the following:

  • The interval under consideration ranges from $t=a$ to $t=b$
  • $\r_t(t)$ maps a scalar time to the three-dimensional position vector of the point mass at that time
  • $s_t(\tau)$ maps a scalar time to the scalar distance travelled by the point mass from $t=a$ to $t=\tau$ defined by $s_t(\tau)=\int_{a}^{\tau}\lvert\v_t(t)\rvert\,dt$
  • $\F_\r(t,\r(t))$ maps a scalar time and the three-dimensional position vector of the point mass that corresponds to that time to the three-dimensional vector of the working force
  • $\F_s(t,s(t))$ maps a scalar time and the scalar distance travelled by the point mass to the three-dimensional vector of the working force
  • $F_s(t,s(t))$ maps a scalar time and the scalar distance travelled by the point mass to the scalar magnitude of the working force defined by $F_s(t,s(t))=\lvert\F_s(t,s(t))\rvert$

Assume a more precise definition of work:

$$W_t(a,b)=\int_{\r_t(a)}^{\r_t(b)} \F_\r(t,\r(t))\cdot d(\r_t(t))$$

I’ve made several attempts at this derivation using various methods (which I will not verbosely reproduce here), including:

  • Leibniz integral rule
  • Jacobian determinant
  • Change or variables or $u$-substitution

but to no avail.


Are there any underlying assumptions in saying $P=\F\cdot\v$ that would allow one to derive the equation?

What method could one use to derive $P=\F\cdot\v$?[[

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  • $\begingroup$ In what way do you think this approach ignores time-dependent forces? How would the basic approach change if F did depend on time? I don't see where except in that step where you don't like using the total differential. $\endgroup$ – levitopher Oct 20 '17 at 12:53
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    $\begingroup$ @levitopher Maybe “does not explicitly express” is better than “ignores” $\endgroup$ – let's have a breakdown Oct 20 '17 at 12:55
  • $\begingroup$ en.wikipedia.org/wiki/… $\endgroup$ – Phoenix87 Oct 20 '17 at 14:06
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    $\begingroup$ per khnzhou's answer below, the only real assumption made is that each component of ${\vec r}$ is a piecewise invertible function of $t$. Then, the whole thing is just a change of variables from $r$ to $t$. $\endgroup$ – Jerry Schirmer Oct 20 '17 at 15:11
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Here's a rigorous derivation. The energy is $mv^2/2$, and power is the rate of change of energy, so $$\renewcommand{\v}[1]{\mathbf{#1}} P = \frac{d}{dt} \frac{m \v{v} \cdot \v{v}}{2} = m \v{a} \cdot \v{v} = \v{F} \cdot \v{v}.$$ That's it.


Now let's look at your derivation. The first step $$W = \int \v{F}(\v{r}) \cdot d\v{r} = \int \v{F}(t) \cdot \frac{d\v{r}}{dt} dt$$ is valid, as this is just the chain rule. (Just because it looks like "canceling the differentials" doesn't mean it's wrong! Canceling differentials is completely fine as long as you know what you're doing.) Now rename $t$ to $\tau$ because the name of the integration variable doesn't matter. The next step is $$\frac{dW}{dt} = \frac{d}{dt} \int \v{F}(\tau) \cdot \frac{d\v{r}}{d\tau} d\tau$$ and this is nonsensical because there is no dependence on $t$ to differentiate.

Instead, we should explicitly write integration bounds $t_i$ and $t_f$ where $W(t_i, t_f)$ is the work done from time $t_i$ to time $t_f$. Then $$\frac{dW}{dt_f} = \frac{d}{dt_f} \int_{t_i}^{t_f} \v{F}(\tau) \cdot \frac{d\v{r}}{d\tau} d\tau = \v{F}(t_f) \cdot \v{v}(t_f)$$ where we used the fundamental theorem of calculus. Now the left-hand side is equal to the power at time $t_f$, so $$P(t_f) = \v{F}(t_f) \cdot \v{v}(t_f)$$ which is what we wanted.

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