Ch. 4 - Applications of Derivatives

Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook

Hass 15th Edition

Ch. 4 - Applications of Derivatives

Problem 51

Chapter 4, Problem 51

Applications

Classical accounts tell us that a 170-oar trireme (ancient Greek or Roman warship) once covered 184 sea miles in 24 hours. Explain why at some point during this feat the trireme’s speed exceeded 7.5 knots (sea or nautical miles per hour).

Verified step by step guidance

Recognize that this problem is an application of the Mean Value Theorem (MVT) from calculus, which states that for a continuous and differentiable function over a closed interval, there exists at least one point where the instantaneous rate of change (derivative) equals the average rate of change over that interval.

Define the function s(t) to represent the distance covered by the trireme at time t, where t is measured in hours and s(t) is in sea miles. The interval of interest is from t = 0 to t = 24 hours.

Calculate the average speed of the trireme over the 24-hour period. The average speed is given by the total distance traveled divided by the total time, which is 184 sea miles / 24 hours.

Apply the Mean Value Theorem: Since s(t) is continuous and differentiable over the interval [0, 24], there exists at least one time c in (0, 24) such that the derivative s'(c), representing the instantaneous speed, equals the average speed.

Compare the average speed to 7.5 knots. If the average speed is less than 7.5 knots, then at some point, the instantaneous speed s'(c) must have exceeded 7.5 knots to compensate for slower speeds at other times, ensuring the average speed is achieved.

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Key Concepts

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

Mean Value Theorem

The Mean Value Theorem (MVT) in calculus states that for a continuous function on a closed interval [a, b] that is differentiable on the open interval (a, b), there exists at least one point c in (a, b) where the instantaneous rate of change (derivative) equals the average rate of change over [a, b]. This theorem is crucial for proving that the trireme's speed exceeded 7.5 knots at some point during its journey.

Average Speed

Average speed is calculated by dividing the total distance traveled by the total time taken. In this scenario, the trireme covered 184 sea miles in 24 hours, resulting in an average speed of 184/24 = 7.67 knots. Understanding average speed helps establish a baseline to apply the Mean Value Theorem, indicating that the speed must have exceeded 7.5 knots at some point.

Instantaneous Speed

Instantaneous speed refers to the speed of an object at a specific moment in time, which can be determined by the derivative of the position function with respect to time. In the context of the trireme, the Mean Value Theorem implies that if the average speed is 7.67 knots, there must be an instant where the instantaneous speed is exactly 7.67 knots, thus exceeding 7.5 knots at some point.

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Related Practice

Textbook Question

In Exercises 9–66, graph the function using appropriate methods from the graphing procedures presented just before Example 9, identifying the coordinates of any local extreme points and inflection points. Then find coordinates of absolute extreme points, if any.

56. y = x² + 2/x

Textbook Question

Each of Exercises 43–48 gives the first derivative of a function y = ƒ(𝓍). (a) At what points, if any, does the graph of ƒ have a local maximum, local minimum, or inflection point? (b) Sketch the general shape of the graph.

y' = 𝓍⁴ ― 2𝓍²

Textbook Question

The range R of a projectile fired from the origin over horizontal ground is the distance from the origin to the point of impact. If the projectile is fired with an initial velocity at an angle with the horizontal, then in Chapter 13 we find that R-v_0^2/g(sin 2α) where g is the downward acceleration due to gravity. Find the angle α for which the range R is the largest possible.

Textbook Question

Sketch the graphs of the rational functions in Exercises 53–60.

y = (x² + 1) / x

Textbook Question

In Exercises 49–52, graph each function. Then use the function’s first derivative to explain what you see.

y = 𝓍²/³ + (𝓍―1)²/³

Textbook Question

Sketch the graphs of the rational functions in Exercises 53–60.

y= (x + 1) / (x - 3)

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