Problem 20 - Applied Calculus

Question

Throughout much of the \(20^{\text {th }}\) century, the yearly consumption of electricity in the US increased exponentially at a continuous rate of \(7 \%\) per year. Assume this trend continues and that the electrical energy consumed in 1900 was \(1.4\) million megawatt-hours. (a) Write an expression for yearly electricity consumption as a function of time, \(t\), in years since \(1900 .\) (b) Find the average yearly electrical consumption throughout the \(20^{\text {th }}\) century. (c) During what year was electrical consumption closest to the average for the century? (d) Without doing the calculation for part (c), how could you have predicted which half of the century the answer would be in?

Step-by-Step Solution

Answer

(a) \( P(t) = 1.4 \cdot e^{0.07t} \). (b) 219.316 million MWh. (c) 1972. (d) Exponential growth increases rapidly later.

Step 1: Define the Exponential Growth Function

Given that the electricity consumption grows at a continuous rate of 7% per year starting from 1900 with an initial consumption of 1.4 million megawatt-hours, we can use the continuous growth formula: \[ P(t) = P_0 \cdot e^{rt} \] where \( P_0 = 1.4 \) million, \( r = 0.07 \) (7% as a decimal), and \( t \) is the time in years since 1900. This gives us the expression: \[ P(t) = 1.4 \cdot e^{0.07t} \].

Step 2: Calculate the Average Yearly Consumption

To find the average yearly electrical consumption over the century, we need to evaluate the average value of the function over the interval from \( t = 0 \) to \( t = 100 \). The average value of a continuous function \( f(t) \) over \([a, b]\) is given by: \[ \text{Average} = \frac{1}{b-a} \int_a^b f(t) \, dt \]. For our function: \[ \text{Average} = \frac{1}{100} \int_0^{100} 1.4 \cdot e^{0.07t} \, dt \].

Step 3: Evaluate the Integral

Evaluate the integral: \[ \int_0^{100} 1.4 \cdot e^{0.07t} \, dt = \frac{1.4}{0.07} \cdot \left[e^{0.07t}\right]_0^{100} = 20 \cdot \left(e^7 - 1\right) \]. Calculating \( e^7 \approx 1096.63 \), the integral evaluates to \( 20 \cdot (1096.63 - 1) = 21931.6 \).

Step 4: Determine the Average

The average yearly consumption over the 100 years is: \[ \text{Average} = \frac{21931.6}{100} = 219.316 \text{ million megawatt-hours} \].

Step 5: Find the Year Closest to the Average

Set \( P(t) = 219.316 \) and solve for \( t \): \[ 1.4 \cdot e^{0.07t} = 219.316 \]. Solving, \[ e^{0.07t} = \frac{219.316}{1.4} \approx 156.6543 \]. Take the natural logarithm of both sides: \( 0.07t = \ln(156.6543) \approx 5.053 \) and then \( t = \frac{5.053}{0.07} \approx 72.1857 \). So the year is approximately 1972.

Step 6: Predict Which Half of the Century the Answer Is

The exponential growth rate starts slowly but increases rapidly later. Thus, the average value is reached when consumption has increased significantly above the initial years, indicating a solution in the latter half of the century.

Key Concepts

Continuous Growth Formula

Exponential growth is a fundamental concept in mathematics and science, especially when dealing with populations, finance, or, as in this case, electricity consumption. Here, the model involves a constant relative growth rate, meaning the quantity of interest increases by a consistent percentage over a specific time period.
To describe this mathematically, we employ the Continuous Growth Formula:

• \( P(t) = P_0 \cdot e^{rt} \)

Where:

• \( P(t) \) is the future value or size of the population at time \( t \).
• \( P_0 \) is the initial value or size of the population at time zero, which in this case is 1.4 million megawatt-hours.
• \( r \) is the continuous growth rate expressed as a decimal, here 0.07 or 7%.
• \( t \) represents time in years since 1900.
• \( e \) is the base of the natural logarithm, approximately equal to 2.71828.

By substituting the given values into this formula, we obtain \( P(t) = 1.4 \cdot e^{0.07t} \), allowing us to determine the electricity consumption for any year following 1900.

Average Value of a Function

Calculating the average value of a function over a continuous interval provides insight into the overall behavior of the function across that interval. For a function \( f(t) \) on an interval from \( a \) to \( b \), the formula for the average value is:

• \( \text{Average} = \frac{1}{b-a} \int_a^b f(t) \, dt \)

To find the average yearly electrical consumption during the 20th century, we use \( f(t) = 1.4 \cdot e^{0.07t} \) over \( t = 0 \) to \( t = 100 \). Here, \( a = 0 \), \( b = 100 \), and:

• \( \text{Average} = \frac{1}{100} \int_0^{100} 1.4 \cdot e^{0.07t} \, dt \)

This integral represents the total consumption, which we then normalize by dividing by 100 (the number of years), giving us the average annual consumption over the century.

Integral Evaluation

The process of evaluating an integral is a crucial step in solving many calculus problems. It involves finding the antiderivative of a function over a specified interval. In our example, we need to evaluate the integral \( \int_0^{100} 1.4 \cdot e^{0.07t} \, dt \).

• To do this, observe that the integral of \( e^{kt} \) is \( \frac{1}{k} e^{kt} \).
• Thus, \( \int 1.4 \cdot e^{0.07t} \, dt = \frac{1.4}{0.07} e^{0.07t} \).

The evaluated integral from 0 to 100 becomes:

• \[ \frac{1.4}{0.07} \big[ e^{0.07 \times 100} - e^{0.07 \times 0} \big] \]
• \[ = 20 \cdot ( e^7 - 1 ) \]

Given that \( e^7 \approx 1096.63 \), the final result is \( 21931.6 \), which represents the total electricity consumed over the century.

Natural Logarithm Calculation

The natural logarithm is an essential tool in solving exponential equations, particularly when isolating the variable of interest from other components of the equation. In our situation, after finding the average consumption of 219.316 million megawatt-hours, we must identify the year \( t \) such that the exponential growth model matches this average.

• The equation \( 1.4 \cdot e^{0.07t} = 219.316 \) needs to be solved for \( t \).
• Dividing both sides by 1.4, we get \( e^{0.07t} \approx 156.6543 \).
• Taking the natural logarithm of both sides gives \( \ln(e^{0.07t}) = \ln(156.6543) \).
• Because \( \ln(e^{x}) = x \), we have \( 0.07t = \ln(156.6543) \).
• Solving for \( t \), \( t = \frac{\ln(156.6543)}{0.07} \)
• \approx 72.1857

This calculation indicates that the average was reached approximately 72 years after 1900, which corresponds to the year 1972.