Problem 42 - Calculus and its applications

Question

The rate at which electrical energy is used by the Ortiz family, in kilowatt- hours (kW-h) per day, is given by \(K(t)=10 t e^{-t}\) where \(t\) is time, in hours. That is, \(t\) is in the interval [0,24] a) How many kilowatt-hours does the family use in the first \(T\) hours of a day \((t=0\) to \(t=T) ?\) b) How many kilowatt-hours does the family use in the first 4 hours of the day?

Step-by-Step Solution

Answer

a) Energy used is \(10 - 10e^{-T}(T+1)\) kilowatt-hours. b) For the first 4 hours, it is \(10 - 50e^{-4}\) kilowatt-hours.

Step 1: Identify the Problem

We need to determine the total electrical energy usage of the Ortiz family over a given time interval using the function \( K(t) = 10t e^{-t} \) that describes their power consumption rate in kilowatt-hours per day. For part (a), we need to find the energy used from \( t=0 \) to \( t=T \). For part (b), we specifically need to determine the energy used from \( t=0 \) to \( t=4 \).

Step 2: Set Up the Integral for Part (a)

The total energy used from \( t=0 \) to \( t=T \) is given by the definite integral \( \int_{0}^{T} K(t) \, dt \). This calculates the area under the curve of the power usage rate from 0 to \( T \).

Step 3: Evaluate the Integral for Part (a)

Given \( K(t) = 10t e^{-t} \), we evaluate the integral \( \int_{0}^{T} 10t e^{-t} \, dt \). Using integration by parts, let \( u = 10t \) and \( dv = e^{-t} dt \). Then, \( du = 10 \, dt \) and \( v = -e^{-t} \). Apply the integration by parts formula \( \int u \, dv = uv - \int v \, du \).

Step 4: Compute the Antiderivative

Using the integration by parts formula: \[ \int 10t e^{-t} \, dt = -10t e^{-t} - \int -10 e^{-t} \, dt \] This simplifies to:\[ -10t e^{-t} + 10 \int e^{-t} \, dt \] The latter integral evaluates to \( -10e^{-t} \). Thus:\[ -10te^{-t} - 10e^{-t} = -10e^{-t}(t + 1) \] Evaluating from 0 to \( T \) provides:\[ \left[-10e^{-t}(t+1)\right]_{0}^{T} = \left[-10e^{-T}(T+1)\right] - \left[-10e^{0}(0+1)\right] \] This simplifies to \( 10 - 10e^{-T}(T+1) \).

Step 5: Set Up the Integral for Part (b)

For part (b), determine the energy usage from \( t=0 \) to \( t=4 \). Use the integral \( \int_{0}^{4} 10t e^{-t} \, dt \), following similar steps as above.

Step 6: Evaluate the Integral for Part (b)

We use the same antiderivative from part (a):\[ -10e^{-t}(t + 1) \] Evaluate from 0 to 4: \[ \left[-10e^{-t}(t+1)\right]_{0}^{4} = \left[-10e^{-4}(4+1)\right] - \left[-10e^{0}(0+1)\right] \] Simplifies to \( 10 - 50e^{-4} \).

Key Concepts

Integration by Parts

Integration by parts is a useful technique in calculus to solve integrals involving two functions that are multiplied together. A core principle of this technique is the formula \( \int u \, dv = uv - \int v \, du \). Here, "\( u \)" and "\( dv \)" must be chosen carefully to simplify the integration process.

In our example, the power consumption function is \( K(t) = 10t e^{-t} \), where you can see both a polynomial and an exponential component. For these cases, integration by parts works well because of its ability to decompose such multiplicative relationships. We set \( u = 10t \) (making it easy to differentiate), and \( dv = e^{-t} \, dt \) (making it easy to integrate).

By taking the derivative of \( u \) (\( du = 10 \, dt \)) and the antiderivative of \( dv \) (\( v = -e^{-t} \)), we can apply the formula and determine that:

• The integral simplifies to \( -10te^{-t} - 10e^{-t} = -10e^{-t}(t + 1) \).
• This expression helps calculate the definite integral over any interval, as shown in the solution.

Notice how integration by parts essentially reduces a complex integral into a simpler form by strategically selecting functions for differentiation and integration.

Definite Integrals

Definite integrals provide a way to quantify the total accumulation of a quantity, such as area under a curve, over a specific interval. The definite integral of a function \( f(x) \) from \( a \) to \( b \), represented as \( \int_{a}^{b} f(x) \, dx \), is essential for many applications, including physics and engineering.

For the problem at hand, the definite integral \( \int_{0}^{T} 10t e^{-t} \, dt \) gives us the total electrical energy used by the Ortiz family from time \( t=0 \) to \( t=T \). Similarly, evaluating \( \int_{0}^{4} 10t e^{-t} \, dt \) focuses on the energy usage during the first four hours.

• By calculating these integrals, we essentially find the accumulated consumption in terms of kilowatt-hours over a defined period.
• The definite integral involves evaluating the antiderivative at the two endpoints \( a \) and \( b \), and then subtracting the resulting values.

This provides a tangible way to assess how much power is consumed within certain timeframes, reinforcing how definite integrals serve as powerful tools to measure quantities that change continuously over time.

Exponential Functions

Exponential functions are characterized by the constant rate of growth or decay, a feature that makes them prevalent in real-world applications, such as modeling population growth, radioactive decay, or, as in our case, energy consumption over time.

The expression \( e^{-t} \) within our power consumption rate function \( K(t) = 10t e^{-t} \) exemplifies an exponential decay. It reflects how the rate at which energy is used decreases over time, as indicated by the negative exponent.

• Exponential decay implies that the value decreases at a rate proportional to its current value, resulting in a curve that falls sharply initially and levels off as \( t \) increases.
• This behavior is crucial to understand because it impacts the amount of energy consumed, particularly in the early hours versus the later hours of the day.

The exponential function \( e^{-t} \) plays a pivotal role in the integral, specifically affecting how the area under the curve changes over any interval. This highlights the importance of exponential functions in modeling scenarios where decay or growth, informed by a consistent percentage rate, significantly influences the outcome.