# Instantaneous Power problem

1. Apr 14, 2008

### breez

Hey, instantaneous power is computed by F dot v, but this equation should only be valid if F is constant correct?

2. Apr 14, 2008

### Staff: Mentor

Nope. Why do you say that? F and v can be vector functions of time and space and whatever else.

3. Apr 14, 2008

### robertm

It can still be valid because you use instantaneous velocity and instantaneous force, not constant velocity or constant force. It gives the power output for one instant in time regardless of the operations of the system over time.

4. Apr 28, 2008

### breez

I'm talking about in the case for which v is a varying vector.

P = dW/dt = (dF/dt) dot s + F dot v

The above only simplifies to P = F dot v in the case that dF/dt = 0, or in other words, if F is constant.

5. Apr 28, 2008

### rcgldr

The equation should end up as: P(t) = F(t) dot V(t). Note that work is defined as the line intergral of F(s) dot ds.

Last edited: Apr 28, 2008
6. Apr 29, 2008

### breez

Can you show the derivation? Where is the fallacy in my mathematics? I simply used work as a function time while you used work as a function of the arc length.

7. Apr 29, 2008

### rock.freak667

Well here is what I think it is.

W=F.s
When you dot product two vectors you get a scalar.

W=F.s=|F||s|

$$P=\frac{W}{t}=\frac{|F||s|}{t}=|F||v|=F.v$$

8. Apr 29, 2008

### breez

but you assumed F was constant in that...

9. Apr 29, 2008

### Stingray

The work to move something from point x to point y is given by the line integral
$$W(x \rightarrow y) = \int_x^y F(z) \cdot \mathrm{d} z ,$$
where I'm leaving off vector notation. If the path is described parametrically by a function y(t) with y(0) = x, this can be rewritten as
$$W(x \rightarrow y) = \int_0^t F(y(s)) \cdot \frac{\mathrm{d} y(s)}{\mathrm{d} s} \mathrm{d} s .$$

The path parameter here can be anything, but choose it to be time. Differentiating with respect to t only hits the upper limit of integration. The result is that
$$P = \mathrm{d}W/\mathrm{d}t = F(y(t)) \cdot \frac{\mathrm{d} y(t)}{\mathrm{d} t} = F \cdot v .$$
The point here is that the definition of work does not explicitly depend on time. The work required to apply a given force along a given path is identical whether that path is traversed in a second or a year. This sets the definitions as I've given them, and does not allow any t-dependence inside the integrand.

10. Apr 30, 2008

### SimonZ

during time interval dt, force F can be regarded as a constant, the work done in dt is dW = F dot ds, power is P = dW/dt = F dot ds/dt = F dot v. so even though force is not constant, P = F dot v is still valid.

11. Apr 30, 2008

### breez

Ah okay, I see. The function I differentiated didn't make much sense since work isn't F(t) dot s(t), but rather int F(s) dot ds, which can be integrated by switching the parameter to t.