Question

Power and energy are often used interchangeably, but they are quite different. Energy is what makes matter move or heat up. It is measured in units of joules or Calories, where 1 Cal=4184 J. One hour of walking consumes roughly 10⁶J, or 240 Cal. On the other hand, power is the rate at which energy is used, which is measured in watts, where 1 W = 1 J/s. Other useful units of power are kilowatts (1 kW=10³ W) and megawatts (1 MW=10⁶ W). If energy is used at a rate of 1 kW for one hour, the total amount of energy used is 1 kilowatt-hour (1 kWh = 3.6×10⁶ J) Suppose the cumulative energy used in a large building over a 24-hr period is given by E(t)=100t + 4t² − (t³ / 9) kWh where t = 0 corresponds to midnight.
The power is the rate of energy consumption; that is, P(t) = E′(t) Find the power over the interval 0 ≤ t ≤ 24.

Accepted Answer

Step 1: Understand that power is the derivative of energy with respect to time. Given the energy function E(t) = 100t + $$4t^2$$ - $$(t^3 / 9$$), we need to find its derivative to determine the power function P(t). Step 2: Differentiate the term 100t with respect to t. The derivative of 100t is 100, as the derivative of t with respect to t is 1. Step 3: Differentiate the term $$4t^2$$ with respect to t. Using the power rule, the derivative of $$4t^2 is 8t$$, since the power rule states that d/dt [t^n] = n*t^(n-1). Step 4: Differentiate the term $$-(t^3 / 9) $$with respect to t. Again using the power rule, the derivative is $$-(3/9)t^2$$, which simplifies to $$-(1/3)t^2. $$Step 5: Combine the derivatives from Steps 2, 3, and 4 to form the power function P(t). Therefore, P(t) = 100 + 8t - $$(1/3)t^2. $$This function represents the power over the interval 0 ≤ t ≤ 24.

Risposta video verificata per un problema simile:

Concetti chiave

Energy and Power

Derivative in Calculus

Units of Measurement