Suppose the position of an object moving horizontally after seconds is given by the function s(t) = 32t - t⁴, where 0 ≤ t ≤ 3 and s is measured in feet, with s \u003e 0 corresponding to positions to the right of the origin. When is the object farthest to the right?

Ch. 4 - Applications of the Derivative

Briggs - Calculus: Early Transcendentals 3rd Edition

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

All textbooks

Briggs 3rd Edition

Ch. 4 - Applications of the Derivative

Problem 4.1.71

Chapter 4, Problem 4.1.71

Suppose the position of an object moving horizontally after seconds is given by the function s(t) = 32t - t⁴, where 0 ≤ t ≤ 3 and s is measured in feet, with s > 0 corresponding to positions to the right of the origin. When is the object farthest to the right?

Verified step by step guidance

To find when the object is farthest to the right, we need to determine the maximum value of the position function s(t) = 32t - t^4 over the interval 0 ≤ t ≤ 3.

First, find the derivative of the position function, s'(t), to determine the critical points. The derivative is s'(t) = 32 - 4t^3.

Set the derivative equal to zero to find the critical points: 32 - 4t^3 = 0. Solve for t to find the critical points.

Evaluate the position function s(t) at the critical points and at the endpoints of the interval, t = 0 and t = 3, to determine which gives the maximum value.

Compare the values of s(t) at these points to determine when the object is farthest to the right. The largest value of s(t) will indicate the time when the object is farthest to the right.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.

Video duration:

Was this helpful?

Key Concepts

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

Position Function

The position function s(t) describes the location of an object at any given time t. In this case, s(t) = 32t - t⁴ represents the position of an object moving horizontally, where t is measured in seconds and s in feet. Understanding this function is crucial for determining the object's position over time and analyzing its motion.

Critical Points

Critical points occur where the derivative of a function is zero or undefined. To find when the object is farthest to the right, we need to calculate the derivative of the position function, s'(t), and set it equal to zero. This helps identify potential maximum positions within the given interval.

Second Derivative Test

The second derivative test is a method used to determine the concavity of a function at critical points. By evaluating the second derivative, s''(t), we can ascertain whether a critical point is a local maximum, minimum, or neither. This is essential for confirming that the object is indeed at its farthest right position when the first derivative is zero.

Recommended video:

06:02

The Second Derivative Test: Finding Local Extrema

Related Practice

Textbook Question

Graphing functions Use the guidelines of this section to make a complete graph of f.

f(x) = x⁴ + 4x³

views

Textbook Question

23–68. Indefinite integrals Determine the following indefinite integrals. Check your work by differentiation.

∫ ((1 + √x)/x)dx

Textbook Question

{Use of Tech} Fixed points An important question about many functions concerns the existence and location of fixed points. A fixed point of f is a value of x that satisfies the equation f(x) = x; it corresponds to a point at which the graph of f intersects the line y = x. Find all the fixed points of the following functions. Use preliminary analysis and graphing to determine good initial approximations.

f(x) = cos x

Textbook Question

Differentials Consider the following functions and express the relationship between a small change in x and the corresponding change in y in the form dy = f'(x)dx.

f(x) = sin² x

Textbook Question

Sketch a continuous function f on some interval that has the properties described. Answers will vary.

The function f satisfies f'(-2) = 2, f'(0) = 0, f'(1) = -3 and f'(4) = 1.

Textbook Question

Suppose f is differentiable on (-∞,∞) and f(5.01) - f(5) = 0.25.Use linear approximation to estimate the value of f'(5).