Table of contents

0. Functions7h 52m
1. Limits and Continuity2h 2m
2. Intro to Derivatives1h 33m
3. Techniques of Differentiation3h 18m
4. Applications of Derivatives2h 38m
5. Graphical Applications of Derivatives6h 2m
6. Derivatives of Inverse, Exponential, & Logarithmic Functions2h 37m
7. Antiderivatives & Indefinite Integrals1h 26m
8. Definite Integrals4h 44m
9. Graphical Applications of Integrals2h 27m
- Area Between Curves
  1h 18m
- Introduction to Volume & Disk Method
  1h 8m
10. Physics Applications of Integrals 3h 16m
- Kinematics
  1h 13m
- Work
  2h 3m
11. Integrals of Inverse, Exponential, & Logarithmic Functions2h 34m
12. Techniques of Integration7h 39m
13. Intro to Differential Equations2h 55m
14. Sequences & Series5h 36m
15. Power Series2h 19m
- Introduction to Power Series
  1h 9m
- Taylor Series & Taylor Polynomials
  1h 10m
16. Parametric Equations & Polar Coordinates7h 58m

0. Functions

Introduction to Functions

Calculus

0. Functions

Introduction to Functions

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2:44 minutes

Problem 2

Textbook Question

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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Related Practice

Textbook Question

In each exercise, a function and an interval of its independent variable are given. The endpoints of the interval are associated with points $P$ and $Q$ on the graph of the function.

a. Sketch a graph of the function and the secant line through $P$ and $Q$ .

b. Find the slope of the secant line in part (a), and interpret your answer in terms of an average rate of change over the interval. Include units in your answer.

After $t$ seconds, an object dropped from rest falls a distance $d=16t^2$ , where $d$ is measured in feet and $2\leq{t}\leq{5}$ .