Textbook Question
At all times, the length of a rectangle is twice the width w of the rectangleas the area of the rectangle changes with respect to time t.
a. Find an equation relating A to w.
Ch. 3 - Derivatives
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243
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Table of contents
- Ch. 1 - Functions210
- Ch. 2 - Limits371
- Ch. 3 - Derivatives633
- Ch. 4 - Applications of the Derivative412
- Ch. 5 - Integration328
- Ch. 6 - Applications of Integration331
- Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions177
- Ch. 8 - Integration Techniques394
- Ch. 9 - Differential Equations179
- Ch. 10 - Sequences and Infinite Series339
- Ch. 11 - Power Series218
- Ch.12 - Parametric and Polar Curves215
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 3, Problem 7b
The volume V of a sphere of radius r changes over time t.
b. At what rate is the volume changing if the radius increases at 2 in/min when when the radius is 4 inches?
Verified step by step guidance1
Start by recalling the formula for the volume of a sphere: V = (4/3)πr^3. This formula gives the volume in terms of the radius r.
To find the rate at which the volume changes with respect to time, we need to differentiate the volume formula with respect to time t. This involves using the chain rule since the radius r is a function of time t.
Apply the chain rule: dV/dt = dV/dr * dr/dt. Here, dV/dr is the derivative of the volume with respect to the radius, and dr/dt is the rate at which the radius changes with respect to time.
Calculate dV/dr by differentiating V = (4/3)πr^3 with respect to r. This gives dV/dr = 4πr^2.
Substitute the given values into the differentiated formula: dr/dt = 2 in/min and r = 4 inches. Plug these into dV/dt = 4πr^2 * dr/dt to find the rate at which the volume is changing.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Related Rates
Related rates involve finding the rate at which one quantity changes in relation to another. In this problem, we need to determine how the volume of the sphere changes as the radius changes over time. This requires applying the chain rule of differentiation to relate the rates of change of the radius and the volume.
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Intro To Related Rates
Volume of a Sphere
The volume V of a sphere is given by the formula V = (4/3)πr³, where r is the radius. Understanding this formula is crucial because it allows us to express the volume in terms of the radius, which is necessary for calculating how the volume changes as the radius changes.
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Chain Rule
The chain rule is a fundamental principle in calculus used to differentiate composite functions. In this context, we will use the chain rule to differentiate the volume formula with respect to time, allowing us to relate the rate of change of volume to the rate of change of the radius.
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Related Practice
Textbook Question
The volume V of a sphere of radius r changes over time t.
c. At what rate is the radius changing if the volume increases at 10 in³ when the radius is 5 inches?
Textbook Question
Find the derivative the following ways:
Using the Product Rule or the Quotient Rule. Simplify your result.
f(x) = x(x-1)
Textbook Question
Find the derivative the following ways:
Using the Product Rule or the Quotient Rule. Simplify your result.
g(t) = (t + 1)(t² - t + 1)
Textbook Question
An equation of the line tangent to the graph of f at the point (2,7) is y = 4x−1. Find f(2) and f′(2).
Textbook Question
An equation of the line tangent to the graph of g at x = 3 is y = 5x + 4. Find g(3) and g′(3).