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Ch. 3 - Derivatives
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 3 - DerivativesProblem 54b
Chapter 3, Problem 54b

When the length L of a clock pendulum is held constant by controlling its temperature, the pendulum’s period T depends on the acceleration of gravity g. The period will therefore vary slightly as the clock is moved from place to place on Earth’s surface, depending on the change in g. By keeping track of ΔT, we can estimate the variation in g from the equation T = 2π(L/g)¹/² that relates T, g, and L.


b. If g increases, will T increase or decrease? Will a pendulum clock speed up or slow down? Explain.

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1
First, let's understand the relationship between the period T of a pendulum and the acceleration due to gravity g. The formula given is T = 2π(L/g)¹/², where L is the length of the pendulum.
Notice that T is proportional to the square root of the inverse of g, which means T = 2π√(L/g). This implies that as g increases, the value of √(L/g) decreases because g is in the denominator.
Since T is directly proportional to √(L/g), if g increases, √(L/g) decreases, leading to a decrease in T. Therefore, the period T of the pendulum decreases as g increases.
A decrease in the period T means that the pendulum completes its oscillations more quickly. Thus, the pendulum clock will speed up as g increases.
In summary, if the acceleration due to gravity g increases, the period T of the pendulum decreases, causing the pendulum clock to speed up.

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Key Concepts

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

Pendulum Period Formula

The period T of a pendulum is given by the formula T = 2π(L/g)¹/², where L is the length of the pendulum and g is the acceleration due to gravity. This formula shows that the period is directly related to the square root of the length and inversely related to the square root of gravity. Understanding this relationship is crucial for analyzing how changes in g affect T.
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Introduction to Tangent Graph

Effect of Gravity on Pendulum

Gravity affects the pendulum's period; as gravity increases, the period decreases, meaning the pendulum swings faster. Conversely, if gravity decreases, the period increases, and the pendulum swings slower. This inverse relationship is essential for predicting how a pendulum clock's speed changes with variations in gravity.
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Derivatives Applied To Acceleration Example 2

Pendulum Clock Speed

A pendulum clock's speed is determined by the pendulum's period. If the period decreases due to an increase in gravity, the clock will speed up, as it completes more cycles in a given time. Conversely, if the period increases, the clock will slow down. Understanding this concept helps in determining the clock's behavior when moved to different locations with varying gravity.
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Derivatives Applied To Velocity
Related Practice
Textbook Question

When the length L of a clock pendulum is held constant by controlling its temperature, the pendulum’s period T depends on the acceleration of gravity g. The period will therefore vary slightly as the clock is moved from place to place on Earth’s surface, depending on the change in g. By keeping track of ΔT, we can estimate the variation in g from the equation T = 2π(L/g)¹/² that relates T, g, and L.


a. With L held constant and g as the independent variable, calculate dT and use it to answer parts (b) and (c).

Textbook Question

a. Find an equation for the line that is tangent to the curve y = x³ − 6x² + 5x at the origin.

[Technology Exercise] b. Graph the curve and tangent line together. The tangent line intersects the curve at another point. Use Zoom and Trace to estimate the point’s coordinates.

[Technology Exercise] c. Confirm your estimates of the coordinates of the second intersection point by solving the equations for the curve and tangent line simultaneously.

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Textbook Question

A weight is attached to a spring and reaches its equilibrium position (x = 0). It is then set in motion resulting in a displacement of x = 10 cos t, where x is measured in centimeters and t is measured in seconds. See the accompanying figure.

Find the spring’s displacement when t = 0, t = π/3, and t = 3π/4.

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Textbook Question

In Exercises 51 and 52, find dp/dq.

q = (5p² + 2p)⁻³/²

Textbook Question

Quadratic approximations


[Technology Exercise] c. Graph f(x) = 1/(1 − x) and its quadratic approximation at x = 0. Then zoom in on the two graphs at the point (0,1). Comment on what you see.

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Textbook Question

A weight is attached to a spring and reaches its equilibrium position (x = 0). It is then set in motion resulting in a displacement of x = 10 cos t, where x is measured in centimeters and t is measured in seconds. See the accompanying figure.

b. Find the spring’s velocity when t = 0, t = π/3, and t = 3π/4.