Ch. 3 - Derivatives

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

Hass 15th Edition

Ch. 3 - Derivatives

Problem 3.4.32c

Chapter 3, Problem 3.4.32c

Analyzing Motion Using Graphs

[Technology Exercise] Exercises 31–34 give the position function s = f(t) of an object moving along the s-axis as a function of time t. Graph f together with the velocity function v(t) = ds/dt = f'(t) and the acceleration function a(t) = d²s/dt² = f''(t). Comment on the object’s behavior in relation to the signs and values of v and a. Include in your commentary such topics as the following:

c. When does it change direction?

s = t² - 3t + 2, 0 ≤ t ≤ 5

Verified step by step guidance

Step 1: Identify the position function s(t) = t² - 3t + 2. This function describes the position of the object along the s-axis over time.

Step 2: Find the velocity function v(t) by differentiating the position function s(t) with respect to time t. This gives v(t) = ds/dt = f'(t) = 2t - 3.

Step 3: Determine the acceleration function a(t) by differentiating the velocity function v(t) with respect to time t. This results in a(t) = dv/dt = f''(t) = 2.

Step 4: Analyze when the object changes direction by finding when the velocity function v(t) = 0. Solve the equation 2t - 3 = 0 to find the critical points.

Step 5: Comment on the object's behavior: The object changes direction at the time when v(t) = 0. Since the acceleration a(t) = 2 is constant and positive, it indicates that the object is accelerating in the positive direction throughout the interval 0 ≤ t ≤ 5.

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Key Concepts

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

Position, Velocity, and Acceleration

In calculus, the position function s(t) describes the location of an object over time. The velocity function v(t) is the first derivative of the position function, representing the rate of change of position, or speed and direction. Acceleration a(t) is the second derivative of the position function, indicating the rate of change of velocity. Understanding these relationships is crucial for analyzing motion.

Derivatives and Their Interpretation

The derivative of a function at a point gives the slope of the tangent line to the function at that point, representing the rate of change. For motion, the first derivative of the position function gives velocity, and the second derivative gives acceleration. Analyzing these derivatives helps determine when an object speeds up, slows down, or changes direction.

Sign Analysis of Velocity and Acceleration

The signs of velocity and acceleration provide insights into an object's motion. A positive velocity indicates forward motion, while a negative velocity indicates backward motion. When velocity changes sign, the object changes direction. Acceleration's sign shows whether the object is speeding up or slowing down; positive acceleration increases velocity, while negative acceleration decreases it.

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Derivatives Applied To Acceleration

Related Practice

Textbook Question

Theory and Examples

In Exercises 51–54,

c. For what values of x, if any, is f' positive? Zero? Negative?

y = −x²

Textbook Question

Differentiability and Continuity on an Interval

Each figure in Exercises 45–50 shows the graph of a function over a closed interval D. At what domain points does the function appear to be

c. neither continuous nor differentiable?

Give reasons for your answers.

Textbook Question

Theory and Examples

In Exercises 51–54,

d. Over what intervals of x-values, if any, does the function y = f(x) increase as x increases? Decrease as x increases? How is this related to what you found in part (c)? (We will say more about this relationship in Section 4.3.)

y = −1/x

Textbook Question

The folium of Descartes (See Figure 3.27)

c. Find the coordinates of the point A in Figure 3.29 where the folium has a vertical tangent line.

Textbook Question

By computing the first few derivatives and looking for a pattern, find the following derivatives.

c. d⁷³/dx⁷³ (x sin x)

Textbook Question

Differentiability and Continuity on an Interval

Each figure in Exercises 45–50 shows the graph of a function over a closed interval D. At what domain points does the function appear to be

c. neither continuous nor differentiable?

Give reasons for your answers.