Problem 27 - 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) \( E(T) = -\frac{T}{a} e^{-aT} + \frac{1}{a^2}(1 - e^{-aT}) \); (b) \( E \rightarrow \frac{1}{a^2} \) as \( T \rightarrow \infty \).

Step 1: Understanding the Problem

We are given a rate function, \( r(t) = t e^{-at} \), and we need to find the total energy \( E \) used during the first \( T \) hours. This requires integrating the rate function from 0 to \( T \).

Step 1: Set Up the Integral for Total Energy

To find the total energy \( E \) used, we set up the integral: \[ E = \int_0^T t e^{-at} \, dt \].

Step 2: Choose the Integration Technique

The integral \( \int t e^{-at} \, dt \) requires integration by parts, where we set \( u = t \) and \( dv = e^{-at} \, dt \). This leads us to differentiate and integrate these parts.

Step 3: Apply Integration by Parts

Using \( u = t \), \( du = dt \), \( v = -\frac{1}{a}e^{-at} \), the integration by parts formula \( \int u \, dv = uv - \int v \, du \) becomes:\[ \int t e^{-at} \, dt = -\frac{t}{a} e^{-at} + \frac{1}{a} \int e^{-at} \, dt \].

Step 4: Solve the Integral \( \int e^{-at} \, dt \)

The integral \( \int e^{-at} \, dt \) is solved as \( -\frac{1}{a}e^{-at} + C \), where \( C \) is the integration constant.

Step 5: Substitute Back and Simplify

Substitute \( \int e^{-at} \, dt \) back into the equation in Step 3:\[-\frac{t}{a} e^{-at} + \frac{1}{a} (-\frac{1}{a} e^{-at}) = -\frac{t}{a} e^{-at} + \frac{1}{a^2} e^{-at} \].

Step 6: Evaluate Definite Integral

Evaluate this expression from 0 to \( T \):\[ \left[ -\frac{t}{a} e^{-at} + \frac{1}{a^2} e^{-at} \right]_0^T = \left( -\frac{T}{a} e^{-aT} + \frac{1}{a^2} e^{-aT} \right) - \left( 0 + \frac{1}{a^2} \right) = -\frac{T}{a} e^{-aT} + \frac{1}{a^2}(1 - e^{-aT}) \].

Step 7: Consider the Limit as \( T \rightarrow \infty \)

As \( T \rightarrow \infty \), the term \( e^{-aT} \rightarrow 0 \), hence the expression simplifies to:\[ E = \frac{1}{a^2} \] when \( T \) approaches infinity.

Key Concepts

Definite Integral

When solving mathematical problems related to rates of change or accumulation over time, the concept of a definite integral is immensely important. The definite integral allows you to find the total quantity of something accumulated over a particular interval. For example, if you have a function that describes the rate at which energy is used over time, integrating this function from one time point to another gives the total energy used during that period.

In the original exercise, the rate function, given as \( r(t) = t \cdot e^{-a t} \), describes how the rate of energy usage changes with time \( t \). To find the total energy \( E \) used over the first \( T \) hours, we define a definite integral:

• Set the integral bounds from 0 to \( T \).
• Integrate the rate function \( r(t) \) using the defined bounds.

The definite integral is \[E = \int_0^T t \cdot e^{-at} \, dt\]This expression must be solved to find the accumulated energy, which requires specific integration techniques like integration by parts.

Exponential Decay

Exponential decay describes a process where the quantity decreases rapidly at first and then slowly over time. This is characterized by a mathematical expression involving an exponential function with a negative exponent. In our problem, the rate of energy usage includes an exponential decay component, represented by \( e^{-a t} \).

Understanding exponential decay is crucial when interpreting how the rate function behaves over time:

• The decay factor \( e^{-a t} \) causes the rate to decrease as \( t \) increases.
• As \( t \) approaches infinity, \( e^{-a t} \) approaches zero.

This behavior signifies that the rate of energy usage decreases significantly as time progresses. When analyzing the energy usage over an infinite time, we consider how this decay influences the total energy. As shown in the solution, the limiting value of the exponential term simplifies the calculations greatly. Ultimately, this results in reaching a finite energy limit as time approaches infinity.

Rate Function

A rate function is a mathematical function that describes how a quantity changes over time. In this exercise, the rate function \( r(t) = t \cdot e^{-a t} \) captures how energy is utilized over time. Recognizing the components and the behavior of the rate function is key to understanding how to approach the problem.

The rate function here is a product of a linear function of time \( t \) and an exponentially decaying function \( e^{-a t} \). Such a combination indicates:

• At the beginning, when \( t \) is small, the usage rate can increase because of the linear term.
• Over time, the exponential decay term dominates, decreasing the overall rate.

To solve the integral involving such a rate function, integration by parts is a suitable technique. This technique helps decompose the integral into simpler parts. This approach facilitates solving complex expressions that arise from the combination of polynomial and exponential components in the rate function.

Ultimately, understanding the rate function allows for accurate prediction and calculation of total energy consumption over a specified period.