Ch. 3 - Derivatives

Briggs - Calculus: Early Transcendentals 3rd Edition

Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook

Briggs 3rd Edition

Ch. 3 - Derivatives

Problem 112

Chapter 3, Problem 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?

Verified step by step guidance

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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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 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.

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.

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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Intro to the Chain Rule

Related Practice

Textbook Question

Use the definition of the derivative to evaluate the following limits.

$\lim_{x \to 2} {\frac{5^{x} - 25}{x - 2}}_{}$

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Textbook Question

Suppose the cost of producing x lawn mowers is C(x) = −0.02x²+400x+5000.

a. Determine the average and marginal costs for x = 3000 lawn mowers.

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Textbook Question

Water flows into a conical tank at a rate of 2 ft³/min. If the radius of the top of the tank is 4 ft and the height is 6 ft, determine how quickly the water level is rising when the water is 2 ft deep in the tank.

Textbook Question

Suppose the cost of producing x lawn mowers is C(x) = −0.02x²+400x+5000.

b. Interpret the meaning of your results in part (a).

Textbook Question

A general proof of the Chain Rule Let f and g be differentiable functions with h(x)=f(g(x)). For a given constant a, let u=g(a) and v=g(x), and define H (v) = <1x1 matrix>

c. Show that h′(a) = lim x→a ((H(g(x))+f′(g(a)))⋅g(x)−g(a)/x−a).

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Textbook Question

Derivatives of even and odd functions Recall that f is even if f(−x) = f(x), for all x in the domain of f, and f is odd if f(−x) = −f(x) for all x in the domain of f.

b. If f is a differentiable, odd function on its domain, determine whether f' is even, odd, or neither.

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