Instantaneous power as a function of time

[jdawg]

Homework Statement

If v(t) = t - 4 and i(t) = 3t, find the instantaneous power p(t) as a function of time.

Homework Equations

The Attempt at a Solution

p(t) = ∫ v(t)*i(t) dt
p(t) = ∫ (t-4)*(3t) dt

Is it correct to do this? Or am I supposed to take the derivative of the functions v(t) and i(t) first and then multiply them and take the integral? Thanks!

 

[STEMucator]

Homework Statement

If v(t) = t - 4 and i(t) = 3t, find the instantaneous power p(t) as a function of time.

Homework Equations

The Attempt at a Solution

p(t) = ∫ v(t)*i(t) dt
p(t) = ∫ (t-4)*(3t) dt

Is it correct to do this? Or am I supposed to take the derivative of the functions v(t) and i(t) first and then multiply them and take the integral? Thanks!

Usually power is defined as $p(t) = v(t) i(t)$. In a more general sense $p(t) = \frac{dW(t)}{dt}$ where $W(t)$ is the work function.

 

[phinds]

Why do you want to bring calculus into it? Power is voltage times current. P = V * I, which in this case is P(t) = V(t) * I(t)

EDIT: I see zondrina beat me to it.

 

[jdawg]

So I don't need to integrate? I was just using a formula that I found in my notes.

 

[STEMucator]

So I don't need to integrate? I was just using a formula that I found in my notes.

No integration is required. Perhaps what you are referring to is the change in energy:

$$\Delta W = \int_{t_1}^{t_2} p(t) \space dt = \int_{t_1}^{t_2} v(t) i(t) \space dt$$

 

[phinds]

So I don't need to integrate? I was just using a formula that I found in my notes.

Using forumlae without understanding them is a terrible idea. Forget the forumlae. Focus on the concepts.

In specific answer to your question, re-read post #3

EDIT: HA ... again zondrina beat me to it.

 

[jdawg]

No integration is required. Perhaps what you are referring to is the change in energy:

$$\Delta W = \int_{t_1}^{t_2} p(t) \space dt = \int_{t_1}^{t_2} v(t) i(t) \space dt$$

Yeah that's the one!