Problem 28 - Applied Calculus

Question

During a surge in the demand for electricity, the rate, \(r\), at which energy is used can be approximated by $$ r=t e^{-a t} $$ where \(t\) is the time in hours and \(a\) is a positive constant. (a) Find the total energy, \(E\), used in the first \(T\) hours. Give your answer as a function of \(a .\) (b) What happens to \(E\) as \(T \rightarrow \infty\) ?

Step-by-Step Solution

Answer

(a) \(E(T) = \frac{1}{a^2}(1 - (aT + 1)e^{-aT})\); (b) \(E \to \frac{1}{a^2}\) as \(T \to \infty\).

Step 1: Understanding the Problem

We need to find the total energy, \(E\), used over a period of time \(T\), given that the rate of energy usage is \(r = t e^{-at}\). After finding the expression for \(E\), we'll also evaluate what happens to \(E\) as \(T\) approaches infinity.

Step 2: Formulating the Integral

The total energy used in the first \(T\) hours is the integral of the rate of usage, \(r(t) = t e^{-at}\), from \(t=0\) to \(t=T\). Therefore, the problem translates to evaluating the following integral: \[ E = \int_{0}^{T} t e^{-at} \, dt. \]

Step 3: Solving the Integral

To solve \( \int t e^{-at} \, dt \), we use integration by parts, where we set \(u = t\) and \(dv = e^{-at} \, dt\). Then, \(du = dt\) and integrating \(dv\), we find \(v = -\frac{1}{a} e^{-at}\). Applying integration by parts: \[ \int t e^{-at} \, dt = -\frac{t}{a} e^{-at} - \int -\frac{1}{a} e^{-at} \, dt. \] This simplifies to: \[ -\frac{t}{a} e^{-at} + \frac{1}{a^2} e^{-at} + C. \]

Step 4: Evaluating the Definite Integral and Simplifying

With the integration complete, we evaluate it from 0 to \(T\): \[ \left[ -\frac{t}{a} e^{-at} + \frac{1}{a^2} e^{-at} \right]_{0}^{T} = \left( -\frac{T}{a} e^{-aT} + \frac{1}{a^2} e^{-aT} \right) - \left( -\frac{0}{a} e^{0} + \frac{1}{a^2} e^{0} \right). \] Simplifying further, we have: \[ E = -\frac{T}{a} e^{-aT} + \frac{1}{a^2} e^{-aT} - \frac{1}{a^2}. \] Combining terms, we get: \[ E = \frac{1}{a^2} \left( 1 - (aT + 1) e^{-aT} \right). \]

Step 5: Considering the Limit as T Approaches Infinity

We now evaluate \( \lim_{T \to \infty} E = \lim_{T \to \infty} \frac{1}{a^2} \left( 1 - (aT + 1) e^{-aT} \right). \) As \(T\) grows large, \((aT + 1) e^{-aT}\) approaches zero, since the exponential function \(e^{-aT}\) declines to zero faster than \(aT\) increases. Thus, the limit becomes: \[ \frac{1}{a^2} (1 - 0) = \frac{1}{a^2}. \]

Key Concepts

Definite Integral

Understanding definite integrals is crucial in calculating the total quantity of a variable that changes with respect to another, such as time. In the electricity usage exercise, the total energy consumed is derived by integrating the rate function, \( r(t) = t e^{-at} \), over a specific time interval \([0, T]\).
The integral, \( \int_{0}^{T} t e^{-at} \, dt \), quantifies the accumulated energy usage over time by considering the continuously changing rate of use at any given moment.

• This integral sums up the infinitesimal amounts of energy consumed at each moment between \( t=0 \) and \( t=T \).
• The evaluation of this integral uses a technique known as integration by parts, often applied when dealing with products of functions, such as \( t \) and \( e^{-at} \).
• The result yields the total energy as a function of \( T \) and \( a \), providing insights into energy consumption patterns over the specified period.

The definite integral thus serves as a powerful tool to aggregate and analyze how quantities accumulate over an interval.

Rate of Change

The concept of rate of change describes how a quantity updates over time or another variable. In the context of energy consumption, it is represented by the function \( r(t) = t e^{-at} \). This function encapsulates how the rate at which energy is used changes as time progresses.

• The function \( r(t) = t e^{-at} \) exhibits both increasing and decreasing behavior over time due to the respective contributions of the linear \( t \) and exponential decay \( e^{-at} \).
• Initially, as time \( t \) increases from zero, the linear term is dominant causing the rate \( r(t) \) to increase.
• However, as \( t \) becomes larger, the exponential decay kicks in significantly, leading the rate \( r(t) \) to decrease as \( t \) continues to grow.

Understanding the rate at which energy is consumed helps in predicting future consumption trends and organizing resources effectively.

Exponential Decay

Exponential decay captures how quantities decrease over time due to a stable rate proportionally reducing the size of the quantity. Within the exercise, the exponential term \( e^{-at} \) portrays how the energy usage rate diminishes over time.
Exponential decay is characterized by:

• A constant proportionality, governed by the decay constant \( a \), influencing how swiftly the value declines.
• As \( t \) increases, the factor \( e^{-at} \) becomes ever smaller, ultimately approaching zero.
• In the energy usage scenario, this means that, even though the immediate demand might surge, long-term energy consumption declines, balancing out over extended periods.

This knowledge of exponential decay is vital for modeling natural processes in physics, chemistry, and related fields, providing critical insights into long-term behavior of diminishing resources.