# Power problem over time using calculus

• smashbrohamme
P = F(t)dt. Then use the Properties of Derivatives to find the derivative of P with respect to t. This will give you the instantaneous power at any point in time. f

#### smashbrohamme

A certain load is characterized by an instantaneous power that increases linearly from zero to 1 kW over a 3-hour period. Determine the total energy in (a) kWh and (b) joules.

p(t) = [dw(t)] / dt

W = F p(t)dt T2 is upper limit, T1 is lower limit.

One of my problems is my calculus is super weak at this point, but I am curious if there is a way to solve this problem without calculus for one, and two...if it truly does take calculus, how do I go about doing it. I did take calculus but I lost the majority of my calculus skills over the course of two years.

I first tried using simple geometry and made a triangle, 0-10,800seconds for the x dimension and 0-1kW for the Y dimension. Of course this lead my answer to 10,800Joules. I know this is completely wrong.

I understand the equation of Energy = Watts x Seconds. This is the instant power equation though, so I am unaware of how to do the function of time equation.

The answer for (a) kWh = 1.5kWh. The answer for (b) is 5.4 x 10^6J.

Ok so I realized the first thing I did wrong was not find the true area of the triangle I made. A = B*H*1/2. When I do this, I actually discovered that I did get the correct Joules value. Something tells me that I should be using calculus to solve this problem though. A = 10800 * 1000W * 1/2 = 5.4 x 10^6J

For your geometric method, if you want a result in kWh you should put kW on the y-axis and Hours on the x-axis. No need to convert to seconds. That makes the units of the area calculation kW x hr = kWh.

To do it by calculus, first write an expression for the power with respect to time. Here its the equation of a line with the appropriate slope. Again, you can use units of hours for time and kW for power.