Question

The total energy in megawatt-hr (MWh) used by a town is given by $$E(t)=400 t+\frac{2400}{\pi} \sin \frac{\pi t}{12},$$ where \(t \geq 0\) is measured in hours, with \(t=0\) corresponding to noon. a. Find the power, or rate of energy consumption, \(P(t)=E^{\prime}(t)\) in units of megawatts (MW). b. At what time of day is the rate of energy consumption a maximum? What is the power at that time of day? c. At what time of day is the rate of energy consumption a minimum? What is the power at that time of day? d. Sketch a graph of the power function reflecting the times at which energy use is a minimum or maximum.

Short answer

Answer: The maximum power consumption is 600 MW, occurring at noon (t=0), while the minimum power consumption is 200 MW, occurring at midnight (t=12).

Step-by-step solution

Differentiate E(t)

Find the derivative of E(t) with respect to t to obtain the power function P(t). Differentiate the equation: $$ E(t) = 400t + \frac{2400}{\pi}\sin\frac{\pi t}{12} $$ Using the chain rule, the derivative with respect to t is: $$ P(t) = E'(t) = 400 + \frac{2400}{\pi} \cdot \frac{\pi}{12} \cos \frac{\pi t}{12} $$ After simplifying, we get: $$ P(t) = 400 + 200\cos\frac{\pi t}{12} $$ This equation represents power consumption in MW as a function of time.

Power consumption maximum and time

To find the time when the rate of energy consumption is at its maximum, we can find the critical points of the power function P(t) by setting its derivative equal to zero. Differentiate P(t) with respect to t: $$ P'(t) = -\frac{200\pi}{12}\sin\frac{\pi t}{12}. $$ Set P'(t) equal to zero: $$ -\frac{200\pi}{12}\sin\frac{\pi t}{12} = 0. $$ The sine function is equal to zero when its argument is equal to the integer multiples of \(\pi\). Therefore, we have: $$ \frac{\pi t}{12} = n\pi, \quad n \in \mathbb{Z} $$ Which gives: $$ t = 12n, \quad n \in \mathbb{Z}. $$ Now we have a maximum or minimum when t is a multiple of 12. But we are interested in finding the maximum power consumption. Plug t = 12n into the power function and get: $$ P(12n) = 400 + 200\cos\frac{\pi(12n)}{12} $$ Simplifying, we get: $$ P(12n) = 400 + 200\cos n\pi $$ When n is even, \(\cos(n\pi) = 1\), and when n is odd, \(\cos(n\pi) = -1\). Thus, power consumption is maximum when n is even. Moreover, the 24-hour energy consumption cycle is represented, so we only need to find the maximum within the first 24 hours. When n = 0 and n = 2 are the only times in the range [0, 24]. Since t = 0 represents the maximum power at noon and t = 24 represents midnight, the maximum power consumption occurs at noon. Now, evaluate the power function at t = 0 (noon): $$ P(0) = 400 + 200\cos\frac{\pi(0)}{12} = 400 + 200(1) = 600 $$ Hence, the maximum power consumption is 600 MW at noon.

Power consumption minimum and time

As previously mentioned, power consumption is minimum when n is odd. In the range of [0, 24], n = 1 corresponds to the minimum power consumption. Plug t = 12 into the power function and get: $$ P(12) = 400 + 200\cos\frac{\pi (12)}{12} = 400 + 200\cos\pi = 200 $$ Hence, the minimum power consumption is 200 MW at midnight (t=12).

Sketch the graph of power function

To draw the graph of the power function, first plot the points of minimum and maximum power. At t=0 (noon), we have P(0) = 600 MW. At t=12 (midnight), we have P(12) = 200 MW. Then, plot the power function between these two points with its periodic behavior (every 24 hours). The curve will resemble a horizontal cosine curve with a 400 MW baseline and amplitude of 200 MW. The curve will reach its peak at noon and its lowest point at midnight, with a period of 24 hours.

Key concepts

Derivative of Power Function

Understanding the derivative of a power function is crucial when analyzing energy consumption in calculus. The derivation process entails finding the rate at which a variable, such as energy, is changing at any given time.

In the context of our exercise, we are given the total energy consumption of a town and are asked to find the power function, represented as the first derivative of the energy function with respect to time, denoted as \(P(t) = E'(t)\).

By employing basic calculus rules, such as the product rule and the chain rule, we take the derivative of the energy function to yield the power function:
\[P(t) = 400 + 200\cos\frac{\pi t}{12}\]
This function describes the rate of energy consumption in megawatts (MW) as a function of time. To understand the behavior of energy consumption over time, it is essential to graph this power function and investigate its critical points.

Critical Points Analysis

Critical points are where the function's derivative either equals zero or is undefined. These points are where a function can have a local maximum, local minimum, or a point of inflection, which are key to comprehending the function’s overall behavior.

For our problem, finding the critical points necessitates differentiating the power function \(P(t)\) once more:
\[P'(t) = -\frac{200\pi}{12}\sin\frac{\pi t}{12}\]
And setting the result equal to zero to solve for \(t\). Upon analysis, we find that for the given 24-hour cycle, the rate of energy consumption is a maximum at noon (\(t=0\)) and a minimum at midnight (\(t=12\)).

This process of critical points analysis is a fundamental element in many fields of science and economics, as it helps predict behavior changes in various phenomena.

Graphing Power Function

Graphing a power function can reveal much about its characteristics, including the presence of maxima and minima, intervals of increase and decrease, and periodicity.

For the town's energy consumption, graphing \(P(t)\) as prescribed in the problem provides a clear visual of how the rate of energy consumption varies over time. The graph is a shifted cosine wave due to the cosine component of the equation, illustrating the periodic nature of power consumption throughout the day.

Plotting the critical points we found earlier—the max at noon (600 MW) and the min at midnight (200 MW)—helps us sketch the function's curve. By correctly graphing the function, we see that the power function oscillates with a period of 24 hours reflecting the cyclic nature of energy usage. This reflects a realistic pattern as energy consumption typically peaks during the day and dips during the night.