The derivative of work is $\bf F\cdot v .$ $$P(t)= \frac{\mathrm dW}{\mathrm dt}= \mathbf{F\cdot v}=-\frac{\mathrm dU}{\mathrm dt}\;.$$

But why not $$(\mathrm d(\mathbf F)/\mathrm dx)\cdot \mathbf x + \mathbf{F\cdot v}\;?$$


For a force to do work, it must act upon an object as that object moves some distance (with a component along or opposed to the direction of that force). Holding an object stationary, but carrying the strength of a given force, adds no more work done. Because of this, we say that

$$W=\int_C\vec{F}\cdot d\vec{r}$$

where C is the curve in space along which the object moves. If power is the rate of change of work, then we would expect a product rule to be applied in the derivative, but no power is expended by a force that does no work, and without a displacement, there is no work and no power.


This is because $P(t)$ as stated is the instantaneous power as a function of time and $W =\mathbf F\Delta \mathbf x$ holds only for constant forces. More generally, recall that a definition of work is the integral:

$$W = \int_C\mathbf F(x)\mathrm d\mathbf x$$

Where $C = C(x,t)$ is some curve in space/time. Expressing in terms of time gives:

$$W = \int_t\mathbf F(t)\frac{\mathrm dx}{\mathrm dt}\mathrm dt = \int_t\mathbf F(t)\mathbf v(t)\mathrm dt$$

From the Fundamental Theorem of Calculus, we have:

$$P \equiv \frac{\mathrm dW}{\mathrm dt}= \mathbf F(t)\mathbf v(t)\;.$$

Apologies for sloppy notation, but that's the idea.

  • $\begingroup$ Thanks for answering :) , but i have a follow up question $\endgroup$ – Joe Feb 3 '16 at 6:22
  • $\begingroup$ what if the criteria is given that a body moves under constant power , which means derivative of work with respect to time is constant ,so shouldnt we include in that derivative (according to product rule ) - the derivative (d(F)/dx).x $\endgroup$ – Joe Feb 3 '16 at 6:25
  • $\begingroup$ kb.osu.edu/dspace/bitstream/handle/1811/2458/… this is article im referring to $\endgroup$ – Joe Feb 3 '16 at 6:27
  • $\begingroup$ Hi Joe, the above equation holds whether or not power is constant. The problem is that you only write W = Fd when force is constant (and thus has a vanishing derivative. When force is not constant, you have to resort to the integral definition and in this case, the dF/dx term does not pop up. By the way, thanks for the link. I especially liked the phrase: "... In these days of steam, electric and gasoline transportation..." :) $\endgroup$ – Orko Feb 4 '16 at 6:49

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