Urgent help on a simple instantaneous power problem

[Blkmage]

Homework Statement

The position of a particle on the x-axis is given by x = 2t^2, due to an applied force of F = e^(2t) parallel to the x axis.

a. Write an expression for the power supplied by this force as a function of time.
b. Compute the instantaneous power when the particle is at x = 8.0

Homework Equations

P = dW/dt
P = Fv

The Attempt at a Solution

So I've talked to some of my classmates about this and I did P = dW/dt while some of them did P = Fv. I'm having trouble comprehending which one to use and why the two methods give different answers.

I said W = Fd = 2t^2*e^(2t) and differentiated that using the product rule, resulting in:

P(t) = 4te^(2t) (t + 1)

Meanwhile, using P = Fv, I just get 4te^(2t). Can someone explain why this is? Part b won't be hard for me once I figure out which way is correct in part a

 

[berkeman]

Homework Statement

The position of a particle on the x-axis is given by x = 2t^2, due to an applied force of F = e^(2t) parallel to the x axis.

a. Write an expression for the power supplied by this force as a function of time.
b. Compute the instantaneous power when the particle is at x = 8.0

Homework Equations

P = dW/dt
P = Fv

The Attempt at a Solution

So I've talked to some of my classmates about this and I did P = dW/dt while some of them did P = Fv. I'm having trouble comprehending which one to use and why the two methods give different answers.

I said W = Fd = 2t^2*e^(2t) and differentiated that using the product rule, resulting in:

P(t) = 4te^(2t) (t + 1)

Meanwhile, using P = Fv, I just get 4te^(2t). Can someone explain why this is? Part b won't be hard for me once I figure out which way is correct in part a

When power is changing over time, I think P=dW/dt would be harder to work with. The formulation P(t)=F(t)*v(t) would seem to lend itself better to calculating instaneous power...