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Ch. 4 - Applications of the Derivative
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 4, Problem 4.8.37d

{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>


d. Find the time and the displacement when the object reaches its high point for the second time.

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Identify the function for displacement: y(t) = 2.5e^(-t) * cos(2t). This function represents a damped harmonic oscillator, where the exponential term 2.5e^(-t) accounts for the damping effect, and cos(2t) represents the oscillatory motion.
To find the high points, we need to determine when the derivative of y(t) with respect to t is zero. This involves finding y'(t) and setting it equal to zero. Use the product rule for differentiation: if u(t) = 2.5e^(-t) and v(t) = cos(2t), then y(t) = u(t)v(t) and y'(t) = u'(t)v(t) + u(t)v'(t).
Calculate the derivatives: u'(t) = -2.5e^(-t) (derivative of 2.5e^(-t)) and v'(t) = -2sin(2t) (derivative of cos(2t)). Substitute these into the product rule formula to find y'(t).
Set y'(t) = 0 to find the critical points. This will give you an equation involving trigonometric and exponential terms. Solve this equation to find the values of t where the displacement reaches a high point.
Determine the second high point by identifying the second positive value of t from the solutions obtained. Substitute this value back into the original displacement function y(t) to find the corresponding displacement at this time.

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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 a resistive force, such as friction or air resistance. In the given equation, the term 'e⁻ᵗ' represents the damping effect, causing the oscillation to gradually lose energy and amplitude as time progresses.
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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' component indicates that the object oscillates with a frequency determined by the coefficient of 't', which affects how quickly the object moves up and down.
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Finding Extrema in Functions

To find the high points (maxima) of the oscillation, one must analyze the function y(t) for critical points. This involves taking the derivative of y(t), setting it to zero, and solving for t to determine when the displacement reaches its maximum value, which corresponds to the high points of the oscillation.
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Related Practice
Textbook Question

107–110. {Use of Tech} Motion with gravity Consider the following descriptions of the vertical motion of an object subject only to the acceleration due to gravity. Begin with the acceleration equation a(t) = v' (t) = -g , where g = 9.8 m/s² .

d. Find the time when the object strikes the ground.

A payload is released at an elevation of 400 m from a hot-air balloon that is rising at a rate of 10 m/s.

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Textbook Question

if ƒ(x) = 1 / (3x⁴ + 5) , it can be shown that ƒ'(x) = 12x³ / (3x⁴ + 5)² and ƒ"(x) = 180x² (x² + 1) (x + 1) (x - 1) / (3x⁴ + 5)³ . Use these functions to complete the following steps.


d. Identify the local extreme values and inflection points of ƒ .

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Textbook Question

Interpreting the derivative The graph of f' on the interval [-3,2] is shown in the figure. <IMAGE>


f. Sketch one possible graph of f.

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Textbook Question

Rectangles in triangles Find the dimensions and area of the rectangle of maximum area that can be inscribed in the following figures.

d. An arbitrary triangle with a given area A (The result applies to any triangle, but first consider triangles for which all the angles are less than or equal to 90° .)

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Textbook Question

Sketch a graph of a function f with the following properties.


f' < 0 and f" < 0, for 8 < x < 10

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Textbook Question

if ƒ(x) = 1 / (3x⁴ + 5) , it can be shown that ƒ'(x) = 12x³ / (3x⁴ + 5)² and ƒ"(x) = 180x² (x² + 1) (x + 1) (x - 1) / (3x⁴ + 5)³ . Use these functions to complete the following steps.


g. Use your work in parts (a) through (f) to sketch a graph of ƒ .

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