Question

Can someone derive power for me? (F dot product V)

Answer 1

My text book goes from dw/dt to (fdxcostheta)/dt then fcostheta escapes the denominator to become fcostheta(dx/dt)

Answer 2

You're not "deriving" power here.

As I'm sure you know "power" is the rate at which, at time t, work is being done per unit time: \(P(t) = \dfrac{\Delta W(t)}{\Delta t}\)

I think you're really trying to fit the idea of "power" into the maths model you have which, typically, for work, starts life as:

\(dW = \vec F \cdot d \vec x\) where, say, \(\vec x = <x,y,z>\) in 3-D Cartesian, so \(d \vec x = <dx,dy,dz>\)

you can then simplfy, as what you have posted does, using the idea that \(\vec a \cdot \vec b = |\vec a||\vec b|\cos \theta\), so:

\(dW = \vec F \cdot d \vec x = |\vec F| |d \vec x| \cos \theta = F dx \cos \theta\)

And then you can look at "instantaneous" power P as:

\(P(t) = \dfrac{dW }{dt}= \dfrac{ F \color{blue}{dx} \cos \theta}{\color{blue}{dt}} \qquad \triangle\)

Now over the infitessimal space \(<d \vec x , dt>\), \(\vec F\) is constant [ even though \(\vec F = F (\vec x, t)\)]

So \(\triangle\) only makes sense if we join the blue dots, ie

\(P(t) = \dfrac{dW }{dt}= \dfrac{ F \color{blue}{dx} \cos \theta}{\color{blue}{dt}} = F \cos \theta ~ \color{blue}{v}\)

Answer 3

or so i am guessing :-))