Instantaneous power as a function of time

AI Thread Summary
To find the instantaneous power p(t) as a function of time, use the formula p(t) = v(t) * i(t), where v(t) = t - 4 and i(t) = 3t. Integration is not necessary; instead, simply multiply the voltage and current functions directly. The discussion emphasizes the importance of understanding the concepts behind the formulas rather than relying solely on them. The relationship between power, voltage, and current is straightforward and does not require calculus for this specific calculation. Understanding these foundational concepts is crucial for accurate problem-solving in physics.
jdawg
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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!
 
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jdawg said:

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

$$\Delta W = \int_{t_1}^{t_2} p(t) \space dt = \int_{t_1}^{t_2} v(t) i(t) \space dt$$
 
jdawg said:
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.
 
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Zondrina said:
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!
 

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