Problem 77 - Applied Calculus

Question

On a hot summer afternoon, a city's electricity consumption is \(-3 t^{2}+18 t+10\) units per hour, where \(t\) is the number of hours after noon \((0 \leq t \leq 6) .\) Find the total consumption of electricity between the hours of 1 and \(5 \mathrm{p} \cdot \mathrm{m}\)

Step-by-Step Solution

Answer

The total consumption between 1 pm and 5 pm is 132 units.

Step 1: Identify the Function

The electricity consumption is given by the function \( f(t) = -3t^2 + 18t + 10 \). We need to find the total consumption between 1 pm and 5 pm.

Step 2: Determine the Limits of Integration

Since 1 pm corresponds to \( t = 1 \) and 5 pm corresponds to \( t = 5 \), we will integrate the function from \( t = 1 \) to \( t = 5 \).

Step 3: Set Up the Integral

The expression for the total consumption is the definite integral: \[\int_{1}^{5} (-3t^2 + 18t + 10) \, dt.\]

Step 4: Integrate the Function

Integrate each term separately:\[\int (-3t^2) \, dt = -t^3,\]\[\int 18t \, dt = 9t^2,\]\[\int 10 \, dt = 10t.\]So, the integrated function is \[-t^3 + 9t^2 + 10t.\]

Step 5: Evaluate the Integral

Substitute the limits into the integrated function:\[\left[ -t^3 + 9t^2 + 10t \right]_1^5 = \left( -(5)^3 + 9(5)^2 + 10(5) \right) - \left( -(1)^3 + 9(1)^2 + 10(1) \right).\]

Step 6: Calculate the Result

Evaluate the expressions:\[ -(5)^3 + 9(5)^2 + 10(5) = -125 + 225 + 50 = 150,\]\[-(1)^3 + 9(1)^2 + 10(1) = -1 + 9 + 10 = 18.\]Thus, the total consumption is \(150 - 18 = 132\).

Key Concepts

Electricity Consumption in Summer

Electricity consumption can vary greatly depending on several factors, such as time of day, weather conditions, and human activity levels. For instance, during a hot summer afternoon, the demand for electricity in a city often peaks due to air conditioning usage. In our exercise, electricity consumption is modeled by a polynomial function \( f(t) = -3t^2 + 18t + 10 \).
This function represents the units of electric consumption per hour, with \( t \) being the number of hours after noon. This means, as time progresses from noon onwards, we can observe how consumption changes hourly. By using mathematical modeling, like this polynomial, city planners can anticipate total electricity usage, contributing to efficient energy distribution and management.

Polynomial Integration Simplified

To understand total electricity consumption over a period, integration comes into play. Integration allows us to sum up infinitely small contributions to grasp an entire outcome. In our scenario, we apply polynomial integration to the given function \( -3t^2 + 18t + 10 \).
Each term of the polynomial is integrated separately:

• \( \int (-3t^2) \, dt = -t^3 \)
• \( \int 18t \, dt = 9t^2 \)
• \( \int 10 \, dt = 10t \)

After integrating, we have \(-t^3 + 9t^2 + 10t \).
The process involves reversing differentiation and finding an antiderivative. This approach is crucial for determining the accumulated rate of change over a particular interval, which in our exercise results in understanding the full scope of electricity usage from hour to hour.

Understanding Limits of Integration

The concept of limits in integration is akin to boundaries on a map; they define the start and end of the territory under consideration. For our electricity consumption scenario, the limits of integration are pinpointed between 1 pm and 5 pm. These are represented in the integral as 1 and 5:\[\int_{1}^{5} (-3t^2 + 18t + 10) \, dt.\]
This definite integral covers the period of 1 pm \((t = 1)\) to 5 pm \((t = 5)\). By setting these limits, we aim to calculate the net consumption during this time frame. By substituting these limits into our integrated function \(-t^3 + 9t^2 + 10t\), we compute exact values:

• \(-t^3 + 9t^2 + 10t\) evaluated at \(t = 5\): \(-125 + 225 + 50 = 150\)
• \(-t^3 + 9t^2 + 10t\) evaluated at \(t = 1\): \(-1 + 9 + 10 = 18\)

Finally, to find the total consumption over the specified period, we subtract the evaluated result at 1 pm from the result at 5 pm, yielding 132 units. This method ensures that only the desired time span is calculated, capturing the practical electrical demand within those hours.