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Ch. 3 - Derivatives
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 3 - DerivativesProblem 3.6.55b
Chapter 3, Problem 3.6.55b

{Use of Tech} Spring oscillations A spring hangs from the ceiling at equilibrium with a mass attached to its end. Suppose you pull downward on the mass and release it 10 inches below its equilibrium position with an upward push. The distance x (in inches) of the mass from its equilibrium position after t seconds is given by the function x(t) = 10sin t - 10cos t, where x is positive when the mass is above the equilibrium position. <IMAGE>
b. Find dx/dt and interpret the meaning of this derivative.  

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To find the derivative dx/dt, we need to differentiate the function x(t) = 10sin(t) - 10cos(t) with respect to t. This involves applying the basic rules of differentiation to each term separately.
Differentiate the first term: The derivative of 10sin(t) with respect to t is 10cos(t). This is because the derivative of sin(t) is cos(t), and the constant 10 remains as a coefficient.
Differentiate the second term: The derivative of -10cos(t) with respect to t is 10sin(t). This is because the derivative of cos(t) is -sin(t), and the constant -10 changes sign to become 10.
Combine the derivatives: The derivative dx/dt is the sum of the derivatives of each term, which gives us dx/dt = 10cos(t) + 10sin(t).
Interpret the meaning: The derivative dx/dt represents the rate of change of the distance x with respect to time t. In the context of the spring oscillation, it indicates the velocity of the mass at any given time t. A positive value of dx/dt means the mass is moving upward, while a negative value means it is moving downward.

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Key Concepts

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

Derivatives

A derivative represents the rate of change of a function with respect to a variable. In this context, dx/dt indicates how the position of the mass changes over time. It provides insight into the velocity of the mass as it oscillates, showing whether the mass is moving upward or downward at any given moment.
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Derivatives

Trigonometric Functions

The function x(t) = 10sin(t) - 10cos(t) involves sine and cosine, which are fundamental trigonometric functions. These functions describe periodic motion, such as oscillations, and are essential for modeling the behavior of the mass on the spring. Understanding their properties helps in analyzing the motion's amplitude and frequency.
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Equilibrium Position

The equilibrium position is the point where the forces acting on the mass are balanced, resulting in no net force. In this scenario, it is the position where the spring is neither compressed nor stretched. Understanding this concept is crucial for analyzing the oscillatory motion, as it serves as the reference point for measuring displacement and determining the nature of the oscillations.
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Related Practice
Textbook Question

An object oscillates along a vertical line, and its position in centimeters is given by y(t) = 30(sint - 1), where t ≥ 0 is measured in seconds and y is positive in the upward direction.

Find the velocity of the oscillator, v(t) =y′(t).

Textbook Question

A race Jean and Juan run a one-lap race on a circular track. Their angular positions on the track during the race are given by the functions θ(t) and ϕ(t), respectively, where 0≤t≤4 and t is measured in minutes (see figure). These angles are measured in radians, where θ=ϕ=0 represent the starting position and θ=ϕ=2π represent the finish position. The angular velocities of the runners are θ′(t) and ϕ′(t). <IMAGE>

b. Which runner has the greater average angular velocity?

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Textbook Question

Surface area of a cone The lateral surface area of a cone of radius r and height h (the surface area excluding the base) is A = πr√r²+h².

b. Evaluate this derivative when r=30 and h=40.

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Textbook Question

13-26 Implicit differentiation Carry out the following steps.

b. Find the slope of the curve at the given point.

x = e^y; (2, ln 2)

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Textbook Question

13-26 Implicit differentiation Carry out the following steps.

b. Find the slope of the curve at the given point.

cos y = x; (0, π/2)

Textbook Question

79–82. {Use of Tech} Visualizing tangent and normal lines

b. Graph the tangent and normal lines on the given graph.

(x²+y²)² = 25/3 (x²-y²); (x0,y0) = (2,-1) (lemniscate of Bernoulli)

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