Ch. 1 - Functions

Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook

Hass 15th Edition

Ch. 1 - Functions

Problem 2

Chapter 1, Problem 2

Express the radius of a sphere as a function of the sphere’s surface area. Then express the surface area as a function of the volume.

Verified step by step guidance

Start by recalling the formula for the surface area of a sphere, which is given by \( A = 4\pi r^2 \), where \( A \) is the surface area and \( r \) is the radius.

To express the radius \( r \) as a function of the surface area \( A \), solve the surface area formula for \( r \). Begin by dividing both sides by \( 4\pi \) to isolate \( r^2 \): \( r^2 = \frac{A}{4\pi} \).

Take the square root of both sides to solve for \( r \): \( r = \sqrt{\frac{A}{4\pi}} \). This expresses the radius as a function of the surface area.

Next, recall the formula for the volume of a sphere, which is \( V = \frac{4}{3}\pi r^3 \), where \( V \) is the volume.

To express the surface area \( A \) as a function of the volume \( V \), first solve the volume formula for \( r \): \( r = \left(\frac{3V}{4\pi}\right)^{1/3} \). Substitute this expression for \( r \) into the surface area formula \( A = 4\pi r^2 \) to express \( A \) in terms of \( V \).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Surface Area of a Sphere

The surface area of a sphere is calculated using the formula A = 4πr², where A represents the surface area and r is the radius. This relationship shows how the surface area increases with the square of the radius, indicating that even small changes in radius can lead to significant changes in surface area.

Volume of a Sphere

The volume of a sphere is given by the formula V = (4/3)πr³, where V is the volume and r is the radius. This formula illustrates that the volume grows with the cube of the radius, meaning that the volume increases much more rapidly than the surface area as the radius increases.

Function Representation

In mathematics, expressing one quantity as a function of another involves defining a relationship where one variable depends on another. For example, expressing the radius as a function of surface area and then surface area as a function of volume requires rearranging the respective formulas to isolate the desired variable, demonstrating the interconnectedness of geometric properties.

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Properties of Functions

Related Practice

Textbook Question

Composition of Functions

Let f(x) = x − 3, g(x) = √x, h(x) = x³, and j(x) = 2x. Express each of the functions in Exercises 11 and 12 as a composition involving one or more of f, g, h, and j.

f. y = √(x³ − 3)

Textbook Question

Composition of Functions

Evaluate each expression using the functions

f(x) = 2 − x, g(x) = { −x, −2 ≤ x < 0

x − 1, 0 ≤ x ≤ 2

f. f(g(1/2))

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Textbook Question

Combining Functions

Assume that f is an even function, g is an odd function, and both f and g are defined on the entire real line (−∞,∞). Which of the following (where defined) are even? odd?

g. g ∘ f

Textbook Question

Copy and complete the following table of function values. If the function is undefined at a given angle, enter “UND.” Do not use a calculator or tables.

Textbook Question

Graph the functions in Exercises 13–22. What is the period of each function?

sin (x/2)

Textbook Question

Graph the functions in Exercises 13–22. What is the period of each function?

sin (x − π/4) + 1