Ch. 4 - Applications of Derivatives

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

Hass 15th Edition

Ch. 4 - Applications of Derivatives

Problem 4.4.106d

Chapter 4, Problem 4.4.106d

106. Motion Along a Line The graphs in Exercises 105 and 106 show the position s=f(t) of an object moving up and down on a coordinate line. At approximately what times is the (d) When is the acceleration positive? Negative?

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To determine when the acceleration is positive or negative, we need to analyze the concavity of the position function s=f(t). Acceleration is the second derivative of the position function, so we are looking for intervals where the graph is concave up (positive acceleration) and concave down (negative acceleration).

Identify the intervals where the graph of s=f(t) is concave up. This occurs when the graph is curving upwards, resembling a U-shape. In these intervals, the second derivative is positive, indicating positive acceleration.

Identify the intervals where the graph of s=f(t) is concave down. This occurs when the graph is curving downwards, resembling an upside-down U-shape. In these intervals, the second derivative is negative, indicating negative acceleration.

Examine the graph closely to find the points of inflection, where the concavity changes. These are the points where the acceleration changes from positive to negative or vice versa.

Using the graph, estimate the time intervals for positive and negative acceleration by observing the changes in concavity. Note these intervals as approximate times where the acceleration is positive or negative.

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Key Concepts

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

Position Function

The position function, denoted as s = f(t), describes the location of an object along a coordinate line at any given time t. It is a fundamental concept in motion analysis, allowing us to understand how the object's position changes over time. The graph of this function visually represents the object's trajectory, with peaks indicating maximum displacement and troughs indicating minimum displacement.

Velocity

Velocity is the rate of change of the position function with respect to time, mathematically expressed as v(t) = f'(t). It indicates both the speed and direction of the object's motion. When analyzing the graph, the velocity is positive when the graph is increasing and negative when it is decreasing, providing insights into the object's movement along the line.

Acceleration

Acceleration is the rate of change of velocity with respect to time, represented as a(t) = v'(t) = f''(t). It indicates how quickly the object's velocity is changing, which can be positive (speeding up) or negative (slowing down). In the context of the graph, acceleration is positive when the slope of the velocity function is increasing and negative when it is decreasing, helping to determine the object's motion dynamics.

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Related Practice

Textbook Question

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In Exercises 51 and 52, give reasons for your answers.

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d. Determine all extrema of f.

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Suppose that f(x) = d/dx (1 − √x) and g(x) = d/dx (x + 2).

Find:

∫[f(x) + g(x)] dx

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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.

y=1-(x+1)^3

Textbook Question

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In Exercises 1–16, find an antiderivative for each function. Do as many as you can mentally. Check your answers by differentiation.

x⁻⁴ + 2x + 3

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Textbook Question

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a. Show that the equation 𝓍⁴ + 2𝓍² ― 2 = 0 has exactly one solution on [0,1] .

[Technology Exercises] b.Find the solution to as many decimal places as you can.

Textbook Question

Finding Antiderivatives

In Exercises 1–16, find an antiderivative for each function. Do as many as you can mentally. Check your answers by differentiation.

2 - 5 / x²