Power problem over time using calculus

In summary: 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.
  • #1
smashbrohamme
97
1
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.
 
Physics news on Phys.org
  • #2
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
 
  • #3
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.
 

Related to Power problem over time using calculus

What is power problem over time using calculus?

Power problem over time using calculus is a mathematical concept that involves calculating the rate at which power is used or generated over a period of time, using the principles of calculus. It is commonly used in physics and engineering to analyze and predict the behavior of various systems.

How is power problem over time calculated using calculus?

The power problem over time is calculated by taking the derivative of the power function with respect to time. This derivative represents the rate of change of power over time, also known as the power function's instantaneous power. By integrating this derivative over a specific time interval, the total power used or generated over that period can be determined.

What are the key variables involved in power problem over time using calculus?

The key variables involved in power problem over time are power, time, and the power function. Power is the rate at which energy is used or generated, and is typically measured in watts. Time is the duration over which the power is being used or generated, and is typically measured in seconds. The power function is a mathematical representation of how power changes over time, and is typically expressed as P(t) where P is power and t is time.

What are some real-world applications of power problem over time using calculus?

Power problem over time using calculus has many real-world applications, such as in the design and analysis of electrical circuits, the prediction of energy consumption in buildings, and the optimization of power usage in industrial processes. It is also used in fields such as economics and finance to model and predict changes in power consumption or generation over time.

What are some common mistakes to avoid when solving power problem over time using calculus?

Some common mistakes to avoid when solving power problem over time using calculus include not properly defining the variables and units, using incorrect or outdated formulas, and not considering all relevant factors that may affect the power function. It is important to carefully interpret the results and ensure they make sense in the context of the problem being solved.

Similar threads

  • Engineering and Comp Sci Homework Help
Replies
4
Views
6K
  • Engineering and Comp Sci Homework Help
Replies
2
Views
897
  • Introductory Physics Homework Help
Replies
25
Views
1K
Replies
33
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
14
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
892
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Calculus and Beyond Homework Help
Replies
28
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
1
Views
1K
Back
Top