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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 6.1.59b
Chapter 6, Problem 6.1.59b

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.


b. When the velocity is positive on an interval, the displacement and the distance traveled on that interval are equal.

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Recall the definitions: displacement on an interval is the net change in position, calculated as \(s(b) - s(a)\) where \(s(t)\) is the position function, and distance traveled is the total length of the path traveled, which accounts for all movement regardless of direction.
Understand that velocity being positive on an interval means the object is moving forward (increasing position) throughout that interval, so \(v(t) > 0\) for all \(t\) in the interval.
Since velocity is positive, the position function \(s(t)\) is strictly increasing on that interval, meaning the object never moves backward.
Because the object only moves forward, the displacement (net change in position) equals the total distance traveled, as there are no backward movements to add extra distance.
Therefore, when velocity is positive on an interval, displacement and distance traveled on that interval are indeed equal.

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Key Concepts

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

Velocity and Its Significance

Velocity is the rate of change of position with respect to time and includes direction, indicated by its sign. A positive velocity means the object is moving in the positive direction, which affects how displacement and distance are calculated over an interval.
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Derivatives Applied To Velocity

Displacement vs. Distance Traveled

Displacement is the net change in position from the start to the end of an interval, considering direction, while distance traveled is the total length of the path covered, always non-negative. They are equal only if the motion does not change direction.
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Using The Acceleration Function Example 1

Relationship Between Velocity and Displacement/Distance

If velocity remains positive over an interval, the object moves continuously in one direction, so displacement equals distance traveled. However, if velocity changes sign, displacement and distance differ, highlighting the importance of velocity's sign in comparing these quantities.
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Derivatives Applied To Velocity
Related Practice
Textbook Question

Emptying a cylindrical tank A cylindrical water tank has height 8 m and radius 2m (see figure).

b. Is it true that it takes half as much work to pump the water out of the tank when it is half full as when it is full? Explain.

Textbook Question

Calculating work for different springs Calculate the work required to stretch the following springs 0.4 m from their equilibrium positions. Assume Hooke’s law is obeyed.

b. A spring that requires 2 J of work to be stretched 0.1 m from its equilibrium position

Textbook Question

Deceleration A car slows down with an acceleration of a(t) = −15 ft/s². Assume v(0)=60 ft/s,s(0)=0, and t is measured in seconds.


b. How far does the car travel in the time it takes to come to rest?

Textbook Question

Emptying a conical tank A water tank is shaped like an inverted cone with height 6 m and base radius 1.5 m (see figure).

b. Is it true that it takes half as much work to pump the water out of the tank when it is filled to half its depth as when it is full? Explain.

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Textbook Question

A torus (doughnut) A torus is formed when a circle of radius 2 centered at (3, 0) is revolved about the y-axis.


b. Use the washer method to write an integral for the volume of the torus.

Textbook Question

Functions from arc length What differentiable functions have an arc length on the interval [a, b] given by the following integrals? Note that the answers are not unique. Give a family of functions that satisfy the conditions.

b. ∫a^b √1+36 cos² 2xdx