Ch. 4 - Applications of the Derivative

Briggs - Calculus: Early Transcendentals 3rd Edition

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Briggs 3rd Edition

Ch. 4 - Applications of the Derivative

Problem 4.8.37b

Chapter 4, Problem 4.8.37b

{Use of Tech} A damped oscillator The displacement of an object as it bounces vertically up and down on a spring is given by y(t) = 2.5e⁻ᵗ cos 2t, where the initial displacement is y(0) = 2.5 and y = 0 corresponds to the rest position (see figure). <IMAGE>
b. Find the time and the displacement when the object reaches its lowest point.

Verified step by step guidance

To find the time when the object reaches its lowest point, we need to find the critical points of the function y(t) = 2.5e^(-t) cos(2t). This involves taking the derivative of y(t) with respect to t.

Apply the product rule to differentiate y(t) = 2.5e^(-t) cos(2t). The product rule states that if you have a function h(t) = u(t)v(t), then h'(t) = u'(t)v(t) + u(t)v'(t). Here, u(t) = 2.5e^(-t) and v(t) = cos(2t).

Differentiate u(t) = 2.5e^(-t) to get u'(t) = -2.5e^(-t). Differentiate v(t) = cos(2t) to get v'(t) = -2sin(2t) using the chain rule.

Substitute the derivatives into the product rule: y'(t) = (-2.5e^(-t))cos(2t) + (2.5e^(-t))(-2sin(2t)). Simplify this expression to find y'(t).

Set y'(t) = 0 to find the critical points. Solve the resulting equation for t to find the times when the object reaches its lowest point. Evaluate y(t) at these times to find the corresponding displacements.

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Key Concepts

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

Damped Oscillator

A damped oscillator is a system in which the amplitude of oscillation decreases over time due to energy loss, often from friction or resistance. The displacement function typically includes an exponential decay factor, which represents this loss of energy. In the given equation, the term '2.5e⁻ᵗ' indicates that the oscillation's amplitude diminishes as time progresses.

Cosine Function in Oscillations

The cosine function is fundamental in describing periodic motion, such as oscillations. In the equation y(t) = 2.5e⁻ᵗ cos 2t, the 'cos 2t' part represents the oscillatory behavior of the system, where '2t' indicates the frequency of oscillation. The cosine function oscillates between -1 and 1, determining the position of the object at any given time.

Finding Extrema

To find the lowest point of the oscillation, we need to determine the extrema of the displacement function. This involves taking the derivative of y(t) with respect to time, setting it to zero to find critical points, and then evaluating these points to identify the minimum displacement. The lowest point corresponds to the maximum negative value of the displacement function.

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Related Practice

Textbook Question

{Use of Tech} A damped oscillator The displacement of an object as it bounces vertically up and down on a spring is given by y(t) = 2.5e⁻ᵗ cos 2t, where the initial displacement is y(0) = 2.5 and y = 0 corresponds to the rest position (see figure). <IMAGE>

a. Find the time at which the object first passes the rest position, y = 0.

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Textbook Question

Suppose S = x + 2y is an objective function subject to the constraint xy = 50, for x > 0 and y > 0.

b. Find the absolute minimum value of S subject to the given constraint.

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Textbook Question

Pen problems

b. A rancher plans to make four identical and adjacent rectangular pens against a barn, each with an area of 100 m² (see figure). What are the dimensions of each pen that minimize the amount of fence that must be used? <IMAGE>

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Textbook Question

{Use of Tech} Basketball shot A basketball is shot with an initial velocity of v ft/s at an angle of 45° to the floor. The center of the basketball is 8 ft above the floor at a horizontal distance of 18 feet from the center of the basketball hoop when it is released. The height h (in feet) of the center of the basketball after it has traveled a horizontal distance of x feet is modeled by the function h(x) = 32x² / v² + x + 8 (see figure). <IMAGE>

b. During the flight of the basketball, show that the distance s from the center of the basketball to the front of the hoop is s = √ (x - 17.25)² + ( -(4x² / 81) + x - 2)² (Hint: The diameter of the basketball hoop is 18 inches.)

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Textbook Question

Two poles of heights m and n are separated by a horizontal distance d. A rope is stretched from the top of one pole to the ground and then to the top of the other pole. Show that the configuration that requires the least amount of rope occurs when Θ₁ = Θ₂ (see figure). <IMAGE>

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Textbook Question

{Use of Tech} Every second counts You must get from a point P on the straight shore of a lake to a stranded swimmer who is 50 from a point Q on the shore that is 50 m from you (see figure). Assuming that you can swim at a speed of 2 m/s and run at a speed of 4 m/s, the goal of this exercise is to determine the point along the shore, x meters from Q, where you should stop running and start swimming to reach the swimmer in the minimum time. <IMAGE>

a. Find the function T that gives the travel time as a function of x, where 0 ≤ x ≤ 50.

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