Ch. 6 - Applications of Integration

Briggs - Calculus: Early Transcendentals 3rd Edition

Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook

Briggs 3rd Edition

Ch. 6 - Applications of Integration

Problem 6.1.8b

Chapter 6, Problem 6.1.8b

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]?

Verified step by step guidance

Identify the interval over which you need to find the displacement, which is from \(t=2\) to \(t=6\) seconds.

Recall that displacement over a time interval is the net change in position, which can be found by integrating the velocity function over that interval: \(\text{Displacement} = \int_{2}^{6} v(t) \, dt\).

Use the given areas under the velocity curve between \(t=2\) and \(t=6\). From the graph, the velocity is positive from \(t=2\) to \(t=4\) with an area of 14, and negative from \(t=4\) to \(t=6\) with an area of 10.

Since the velocity is positive on \([2,4]\), the displacement contribution is +14 meters, and since the velocity is negative on \([4,6]\), the displacement contribution is -10 meters.

Add these signed areas to find the total displacement over \([2,6]\): \(14 + (-10) = 14 - 10\) meters.

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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 is the rate of change of position with respect to time and can be positive or negative depending on direction. Displacement over a time interval is the net change in position, calculated as the integral of velocity over that interval, which corresponds to the signed area under the velocity curve.

Definite Integral as Area Under a Curve

The definite integral of a function over an interval represents the net area between the function's graph and the horizontal axis. For velocity, positive areas indicate movement in one direction, while negative areas indicate movement in the opposite direction, affecting the total displacement.

Interpreting Signed Areas for Displacement

When calculating displacement from velocity, areas above the time-axis contribute positively, and areas below contribute negatively. Summing these signed areas over the interval gives the object's net displacement, reflecting both magnitude and direction of movement.

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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)=200−0.05x

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Work done by a spring A spring on a horizontal surface can be stretched and held 0.5 m from its equilibrium position with a force of 50 N.

b. How much work is done in compressing the spring 0.5 m from its equilibrium position?

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Textbook Question

Winding a chain A 30-m-long chain hangs vertically from a cylinder attached to a winch. Assume there is no friction in the system and the chain has a density of 5kg/m.

b. How much work is required to wind the chain onto the cylinder if a 50-kg block is attached to the end of the chain?

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Textbook Question

13–16. Displacement from velocity Consider an object moving along a line with the given velocity v. Assume time t is measured in seconds and velocities have units of m/s.

b. Find the displacement over the given interval.

v(t) = 50e^−2t on [0, 4]

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Textbook Question

Compressing and stretching a spring Suppose a force of 15 N is required to stretch and hold a spring 0.25 m from its equilibrium position.

b. How much work is required to compress the spring 0.2 m from its equilibrium position?

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Textbook Question

Consider the following curves on the given intervals.

b. Use a calculator or software to approximate the surface area.

y=cos x, for 0≤x≤π/2; about the x-axis

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