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Ch. 3 - Derivatives
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Capitolo 3, Problema 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?

Guida verificata passo dopo passo
1
Start by recalling the formula for the volume of a sphere: V = (4/3)πr³.
Differentiate the volume formula with respect to time t to find the rate of change of volume: dV/dt = 4πr²(dr/dt).
You are given that the rate of change of volume dV/dt is 10 in³/s when the radius r is 5 inches. Substitute these values into the differentiated equation: 10 = 4π(5)²(dr/dt).
Solve the equation for dr/dt 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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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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