Textbook Question
Find the derivative the following ways:
Using the Product Rule or the Quotient Rule. Simplify your result.
f(x) = x(x-1)
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 7c
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?
Verified step by step guidance1
Start by recalling the formula for the volume of a sphere: .
Differentiate the volume formula with respect to time t to find the rate of change of volume: .
You are given that the rate of change of volume is 10 in³/s when the radius r is 5 inches. Substitute these values into the differentiated equation: .
Solve the equation for to find the rate at which the radius is changing.
Ensure your final answer has the correct units, which should be in inches per second (in/s), as you are finding the rate of change of the radius.
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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 radius of the sphere changes over time as the volume increases. This requires applying the chain rule from calculus to relate the rates of change of volume and radius.
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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, enabling us to differentiate it with respect to time to find the rate of change of the radius.
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Differentiation
Differentiation is a fundamental concept in calculus that involves finding the derivative of a function. In this context, we will differentiate the volume formula with respect to time to relate the change in volume to the change in radius. This process will help us calculate the rate at which the radius is changing when the volume increases.
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Related Practice
Textbook Question
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?
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).
Textbook Question
If h(1) = 2 and h′(1) = 3, find an equation of the line tangent to the graph of h at x = 1.