Skip to main content
Ch. 6 - Applications of Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 6 - Applications of IntegrationProblem 6.1.53b
Chapter 6, Problem 6.1.53b

Power and energy The terms power and energy are often used interchangeably, but they are quite different. Energy is what makes matter move or heat up and is measured in units of joules (J) or Calories (Cal), where 1 Cal=4184 J. One hour of walking consumes roughly 10⁶ J, or 250 Cal. On the other hand, power is the rate at which energy is used and is measured in watts (W; 1W=1 J/s). Other useful units of power are kilowatts (1 kW=10³ W) and megawatts (1 MW=10⁶ W). If energy is used at a rate of 1 kW for 1 hr, the total amount of energy used is 1 kilowatt-hour (kWh), which is 3.6×10⁶ J. Suppose the power function of a large city over a 24-hr period is given by P(t) = E'(t) = 300 - 200 sin πt/12, where P is measured in megawatts and t=0 corresponds to 6:00 P.M. (see figure).


b. Burning 1 kg of coal produces about 450 kWh of energy. How many kilograms of coal are required to meet the energy needs of the city for 1 day? For 1 year? 

Verified step by step guidance
1
Understand that the power function is given by \(P(t) = E'(t) = 300 - 200 \sin\left(\frac{\pi t}{12}\right)\), where \(P(t)\) is in megawatts (MW) and \(t\) is in hours, with \(t=0\) at 6:00 PM.
To find the total energy used by the city over 1 day (24 hours), integrate the power function over the interval from \(t=0\) to \(t=24\): \(E(24) = \int_0^{24} P(t) \, dt = \int_0^{24} \left(300 - 200 \sin\left(\frac{\pi t}{12}\right)\right) dt\) This will give the total energy in megawatt-hours (MWh).
Evaluate the integral step-by-step: - Integrate the constant term \(300\) over 24 hours. - Integrate the sine term \(-200 \sin\left(\frac{\pi t}{12}\right)\) over 24 hours, using the substitution method if needed.
After finding the total energy in megawatt-hours, convert it to kilowatt-hours (kWh) by multiplying by 1000, since 1 MW = 1000 kW.
Use the given information that burning 1 kg of coal produces 450 kWh of energy. Calculate the kilograms of coal required by dividing the total energy in kWh by 450. For the yearly amount, multiply the daily coal consumption by 365.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
6m
Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Power as the Rate of Energy Consumption

Power represents how quickly energy is used or transferred, measured in watts (W), where 1 W equals 1 joule per second. Understanding power as the derivative of energy with respect to time (P(t) = E'(t)) allows us to calculate total energy consumption by integrating power over a time interval.
Recommended video:
04:16
Intro To Related Rates

Energy Units and Conversion

Energy is measured in joules (J) or kilowatt-hours (kWh), with 1 kWh equal to 3.6×10⁶ J. Converting between these units is essential for practical calculations, such as relating electrical energy consumption to fuel energy content, like the energy produced by burning coal.
Recommended video:
5:10
Evaluate Composite Functions - Values on Unit Circle

Integration to Find Total Energy from Power Function

Given a power function P(t), the total energy used over a time period is found by integrating P(t) with respect to time. This process sums the instantaneous power usage to find the cumulative energy, which is necessary to determine fuel requirements for a given energy demand.
Recommended video:
07:32
Representing Functions as Power Series
Related Practice
Textbook Question

Variable gravity At Earth’s surface, the acceleration due to gravity is approximately g=9.8 m/s² (with local variations). However, the acceleration decreases with distance from the surface according to Newton’s law of gravitation. At a distance of y meters from Earth’s surface, the acceleration is given by a(y) = - g / (1+y/R)², where R=6.4×10⁶ m is the radius of Earth.


b. Use the Chain Rule to show that dv/dt = 1/2 d/dy(v²).

Textbook Question

Displacement and distance from velocity Consider the velocity function shown below of an object moving along a line. Assume time is measured in seconds and distance is measured in meters. The areas of four regions bounded by the velocity curve and the t-axis are also given.

b. What is the displacement of the object over the interval [2, 6]? 

Textbook Question

Displacement and distance from velocity Consider the graph shown in the figure, which gives the velocity of an object moving along a line. Assume time is measured in hours and distance is measured in miles. The areas of three regions bounded by the velocity curve and the t-axis are also given.

b. What is the displacement of the object over the interval [0,3]?

Textbook Question

Filling a tank A 2000-liter cistern is empty when water begins flowing into it (at t=0 at a rate (in L/min) given by Q′(t) = 3√t, where t is measured in minutes.


b. Find the function that gives the amount of water in the tank at any time t≥0.

Textbook Question

21–30. {Use of Tech} Arc length by calculator


b. If necessary, use technology to evaluate or approximate the integral.

y = cos 2x, for 0 ≤ x ≤ π

Textbook Question

Cycling distance A cyclist rides down a long straight road with a velocity (in m/min) given by v(t) = 400−20t, for 0≤t≤10, where t is measured in minutes.


b. How far does the cyclist travel in the first 10 min?