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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
Tutti i libri di testoBriggs 3rd EditionCh. 3 - DerivativesProblema 112
Capitolo 3, Problema 112

A spherical balloon is inflated at a rate of 10 cm³/min. At what rate is the diameter of the balloon increasing when the balloon has a diameter of 5 cm?

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First, understand that the volume V of a sphere is given by the formula V = (4/3)πr³, where r is the radius of the sphere.
Since the balloon is being inflated, we are given the rate of change of the volume with respect to time, dV/dt = 10 cm³/min.
We need to find the rate of change of the diameter with respect to time, dD/dt, when the diameter is 5 cm. Note that the diameter D is twice the radius, so D = 2r.
Differentiate the volume formula with respect to time t to relate dV/dt and dr/dt. Using the chain rule, we get dV/dt = 4πr²(dr/dt).
Substitute the given dV/dt = 10 cm³/min and the radius r = 2.5 cm (since the diameter is 5 cm) into the differentiated equation to solve for dr/dt. Then, use the relationship dD/dt = 2(dr/dt) to find the rate at which the diameter is increasing.

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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 rate of change of the volume of the balloon relates to the rate of change of its diameter. This concept is essential for solving problems where multiple variables are interdependent.
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Percorso guidato
04:16
Intro To Related Rates

Volume of a Sphere

The volume 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 directly related to the diameter. Since the diameter is twice the radius, we can derive relationships between volume and diameter.
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Example 5: Packaging Design

Chain Rule

The chain rule is a fundamental principle in calculus used to differentiate composite functions. In this context, we will apply the chain rule to relate the rates of change of volume and diameter. By differentiating the volume formula with respect to time, we can find how the diameter changes as the volume increases.
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05:02
Intro to the Chain Rule
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