Ch. 6 - Applications of Integration

Briggs - Calculus: Early Transcendentals 3rd Edition

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Briggs 3rd Edition

Ch. 6 - Applications of Integration

Problem 6.1.7b

Chapter 6, Problem 6.1.7b

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]?

Verified step by step guidance

Understand that displacement over a time interval is given by the definite integral of the velocity function over that interval. In this case, displacement from time \(t=0\) to \(t=3\) is the integral of velocity \(v(t)\) from 0 to 3.

Identify the areas under the velocity curve between \(t=0\) and \(t=3\) from the graph. The graph shows two regions: one above the \(t\)-axis from \(t=0\) to \(t=1\) with area 12, and one below the \(t\)-axis from \(t=1\) to \(t=3\) with area 16.

Recall that areas above the \(t\)-axis correspond to positive velocity (positive contribution to displacement), and areas below the \(t\)-axis correspond to negative velocity (negative contribution to displacement).

Calculate the net displacement by subtracting the area below the \(t\)-axis from the area above the \(t\)-axis over the interval \([0,3]\). This means displacement = (area above) - (area below) = 12 - 16.

Express the displacement as the net signed area under the velocity curve from \(t=0\) to \(t=3\), which represents the total change in position of the object during this time.

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Key Concepts

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

Velocity and Displacement

Velocity represents the rate of change of position with respect to time. Displacement over a time interval is the net change in position, calculated as the integral of velocity over that interval. Positive velocity contributes to forward displacement, while negative velocity contributes to backward displacement.

Definite Integral as Area Under the Curve

The definite integral of a velocity function over a time interval corresponds to the net area between the velocity curve and the time axis. Areas above the axis are positive, and areas below are negative, reflecting direction. This integral gives the displacement, not the total distance traveled.

Difference Between Displacement and Distance

Displacement is the net change in position and can be positive, negative, or zero, depending on direction. Distance is the total length traveled, always positive, and equals the sum of the absolute values of areas under the velocity curve. Understanding this distinction is crucial for interpreting velocity graphs.

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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

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?

Textbook Question

Emptying a water trough A water trough has a semicircular cross section with a radius of 0.25 m and a length of 3 m (see figure).

b. If the length is doubled, is the required work doubled? Explain.

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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?

Textbook Question

55–58. Marginal cost Consider the following marginal cost functions.

b. Find the additional cost incurred in dollars when production is increased from 500 units to 550 units.

C′(x) = 300+10x−0.01x²