We are open Saturday and Sunday!

Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 1 : Rate of Change

Study concepts, example questions & explanations for Calculus 1

Create An Account

All Calculus 1 Resources

10 Diagnostic Tests 438 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #241 : Rate Of Change

A spherical balloon is being filled with air. What is ratio of the rate of growth of the volume of the sphere to the rate of growth of the surface area when the radius is one half?

Possible Answers:

$\frac{1}{4}$ $\displaystyle \frac{1}{4}$

$\displaystyle 2$

$\frac{1}{2}$ $\displaystyle \frac{1}{2}$

$\displaystyle 4$

$\displaystyle 1$

Correct answer:

$\frac{1}{4}$ $\displaystyle \frac{1}{4}$

Explanation:

Let's begin by writing the equations for the volume and surface area of a sphere with respect to the sphere's radius:

$V=\frac{4}{3}\pi r^3$ $\displaystyle V=\frac{4}{3}\pi r^3$

$A=4\pi r^2$ $\displaystyle A=4\pi r^2$

The rates of change can be found by taking the derivative of each side of the equation with respect to time:

$\frac{dV}{dt}=4\pi r^2 \frac{dr}{dt}$ $\displaystyle \frac{dV}{dt}=4\pi r^2 \frac{dr}{dt}$

$\frac{dA}{dt}=8\pi r \frac{dr}{dt}$ $\displaystyle \frac{dA}{dt}=8\pi r \frac{dr}{dt}$

The rate of change of the radius is going to be the same for the sphere regardless of the considered parameter. To find the ratio of the rates of changes of the volume and surface area, divide:

$4\pi r^2 \frac{dr}{dt}=(\phi)8\pi r \frac{dr}{dt}$ $\displaystyle 4\pi r^2 \frac{dr}{dt}=(\phi)8\pi r \frac{dr}{dt}$

$\phi =\frac{\frac{1}{2}}{2}$ $\displaystyle \phi =\frac{\frac{1}{2}}{2}$

$\phi =\frac{1}{4}$ $\displaystyle \phi =\frac{1}{4}$

Report an Error

Example Question #241 : Rate Of Change

A spherical balloon is deflating, although it retains a spherical shape. What is ratio of the rate of loss of the volume of the sphere to the rate of loss of the circumference when the radius is 5?

Possible Answers:

$\displaystyle 50$

$\displaystyle 5$

$\displaystyle 15$

$\displaystyle 10$

$\displaystyle 25$

Correct answer:

$\displaystyle 50$

Explanation:

Let's begin by writing the equations for the volume and circumference of a sphere with respect to the sphere's radius:

$V=\frac{4}{3}\pi r^3$ $\displaystyle V=\frac{4}{3}\pi r^3$

$C=2\pi r^$ $\displaystyle C=2\pi r^$

The rates of change can be found by taking the derivative of each side of the equation with respect to time:

$\frac{dV}{dt}=4\pi r^2 \frac{dr}{dt}$ $\displaystyle \frac{dV}{dt}=4\pi r^2 \frac{dr}{dt}$

$\frac{dC}{dt}=2\pi \frac{dr}{dt}$ $\displaystyle \frac{dC}{dt}=2\pi \frac{dr}{dt}$

The rate of change of the radius is going to be the same for the sphere regardless of the considered parameter. To find the ratio of the rates of changes of the volume and circumference, divide:

$4\pi r^2 \frac{dr}{dt}=(\phi)2\pi \frac{dr}{dt}$ $\displaystyle 4\pi r^2 \frac{dr}{dt}=(\phi)2\pi \frac{dr}{dt}$

$\phi =2r^2$ $\displaystyle \phi =2r^2$

$\phi =50$ $\displaystyle \phi =50$

Report an Error

Example Question #243 : Rate Of Change

A spherical balloon is being filled with air. What is ratio of the rate of growth of the volume of the sphere to the rate of growth of the circumference when the radius is 3?

Possible Answers:

$\displaystyle 18$

$\displaystyle 9$

$\displaystyle 3$

$\displaystyle 12$

$\displaystyle 6$

Correct answer:

$\displaystyle 18$

Explanation:

Let's begin by writing the equations for the volume and circumference of a sphere with respect to the sphere's radius:

$V=\frac{4}{3}\pi r^3$ $\displaystyle V=\frac{4}{3}\pi r^3$

$C=2\pi r^$ $\displaystyle C=2\pi r^$

The rates of change can be found by taking the derivative of each side of the equation with respect to time:

$\frac{dV}{dt}=4\pi r^2 \frac{dr}{dt}$ $\displaystyle \frac{dV}{dt}=4\pi r^2 \frac{dr}{dt}$

$\frac{dC}{dt}=2\pi \frac{dr}{dt}$ $\displaystyle \frac{dC}{dt}=2\pi \frac{dr}{dt}$

The rate of change of the radius is going to be the same for the sphere regardless of the considered parameter. To find the ratio of the rates of changes of the volume and circumference, divide:

$4\pi r^2 \frac{dr}{dt}=(\phi)2\pi \frac{dr}{dt}$ $\displaystyle 4\pi r^2 \frac{dr}{dt}=(\phi)2\pi \frac{dr}{dt}$

$\phi =2r^2$ $\displaystyle \phi =2r^2$

$\phi =18$ $\displaystyle \phi =18$

Report an Error

Example Question #2121 : Functions

A spherical balloon is being filled with air. What is ratio of the rate of growth of the surface area of the sphere to the rate of growth of the circumference when the radius is 6?

Possible Answers:

$24\pi$ $\displaystyle 24\pi$

$\displaystyle 12$

$12\pi$ $\displaystyle 12\pi$

$\displaystyle 24$

$6\pi$ $\displaystyle 6\pi$

Correct answer:

$\displaystyle 24$

Explanation:

Let's begin by writing the equations for the surface area and circumference of a sphere with respect to the sphere's radius:

$A=4\pi r^2$ $\displaystyle A=4\pi r^2$

$C=2\pi r$ $\displaystyle C=2\pi r$

The rates of change can be found by taking the derivative of each side of the equation with respect to time:

$\frac{dA}{dt}=8\pi r \frac{dr}{dt}$ $\displaystyle \frac{dA}{dt}=8\pi r \frac{dr}{dt}$

$\frac{dC}{dt}=2\pi \frac{dr}{dt}$ $\displaystyle \frac{dC}{dt}=2\pi \frac{dr}{dt}$

The rate of change of the radius is going to be the same for the sphere regardless of the considered parameter. To find the ratio of the rates of changes of the surface area and circumference, divide:

$8\pi r \frac{dr}{dt}=(\phi)2\pi \frac{dr}{dt}$ $\displaystyle 8\pi r \frac{dr}{dt}=(\phi)2\pi \frac{dr}{dt}$

$\phi =4r$ $\displaystyle \phi =4r$

$\phi =24$ $\displaystyle \phi =24$

Report an Error

Example Question #3152 : Calculus

A spherical balloon is deflating, although it retains a spherical shape. What is ratio of the rate of loss of the surface area of the sphere to the rate of loss of the circumference when the radius is 13?

Possible Answers:

$\displaystyle 26$

$52\pi$ $\displaystyle 52\pi$

$104\pi$ $\displaystyle 104\pi$

$\displaystyle 52$

$26\pi$ $\displaystyle 26\pi$

Correct answer:

$\displaystyle 52$

Explanation:

Let's begin by writing the equations for the surface area and circumference of a sphere with respect to the sphere's radius:

$A=4\pi r^2$ $\displaystyle A=4\pi r^2$

$C=2\pi r$ $\displaystyle C=2\pi r$

The rates of change can be found by taking the derivative of each side of the equation with respect to time:

$\frac{dA}{dt}=8\pi r \frac{dr}{dt}$ $\displaystyle \frac{dA}{dt}=8\pi r \frac{dr}{dt}$

$\frac{dC}{dt}=2\pi \frac{dr}{dt}$ $\displaystyle \frac{dC}{dt}=2\pi \frac{dr}{dt}$

$8\pi r \frac{dr}{dt}=(\phi)2\pi \frac{dr}{dt}$ $\displaystyle 8\pi r \frac{dr}{dt}=(\phi)2\pi \frac{dr}{dt}$

$\phi =4r$ $\displaystyle \phi =4r$

$\phi =52$ $\displaystyle \phi =52$

Report an Error

Example Question #2122 : Functions

A cube is growing in size. What is the ratio of the rate of growth of the cube's volume to the rate of growth of its surface area when its sides have length 16?

Possible Answers:

$\displaystyle 2$

$\displaystyle 4$

$\displaystyle 32$

$\displaystyle 16$

$\displaystyle 8$

Correct answer:

$\displaystyle 4$

Explanation:

Begin by writing the equations for a cube's dimensions. Namely its volume and surface area in terms of the length of its sides:

V=s^3 $\displaystyle V=s^3$

A=6s^2 $\displaystyle A=6s^2$

The rates of change of these can be found by taking the derivative of each side of the equations with respect to time:

$\frac{dV}{dt}=3s^2\frac{ds}{dt}$ $\displaystyle \frac{dV}{dt}=3s^2\frac{ds}{dt}$

$\frac{dA}{dt}=12s\frac{ds}{dt}$ $\displaystyle \frac{dA}{dt}=12s\frac{ds}{dt}$

Now, knowing the length of the sides, simply divide to find the ratio between the rate of change of the volume and surface area:

$3s^2\frac{ds}{dt}=(\phi)12s\frac{ds}{dt}$ $\displaystyle 3s^2\frac{ds}{dt}=(\phi)12s\frac{ds}{dt}$

$\phi=\frac{s}{4}$ $\displaystyle \phi=\frac{s}{4}$

$\phi =4$ $\displaystyle \phi =4$

Report an Error

Example Question #242 : Rate Of Change

A cube is diminishing in size. What is the ratio of the rate of loss of the cube's volume to the rate of loss of its surface area when its sides have length 2?

Possible Answers:

$\displaystyle 4$

$\frac{1}{2}$ $\displaystyle \frac{1}{2}$

$\displaystyle 2$

$\frac{1}{8}$ $\displaystyle \frac{1}{8}$

$\frac{1}{4}$ $\displaystyle \frac{1}{4}$

Correct answer:

$\frac{1}{2}$ $\displaystyle \frac{1}{2}$

Explanation:

Begin by writing the equations for a cube's dimensions. Namely its volume and surface area in terms of the length of its sides:

V=s^3 $\displaystyle V=s^3$

A=6s^2 $\displaystyle A=6s^2$

The rates of change of these can be found by taking the derivative of each side of the equations with respect to time:

$\frac{dV}{dt}=3s^2\frac{ds}{dt}$ $\displaystyle \frac{dV}{dt}=3s^2\frac{ds}{dt}$

$\frac{dA}{dt}=12s\frac{ds}{dt}$ $\displaystyle \frac{dA}{dt}=12s\frac{ds}{dt}$

Now, knowing the length of the sides, simply divide to find the ratio between the rate of change of the volume and surface area:

$3s^2\frac{ds}{dt}=(\phi)12s\frac{ds}{dt}$ $\displaystyle 3s^2\frac{ds}{dt}=(\phi)12s\frac{ds}{dt}$

$\phi=\frac{s}{4}$ $\displaystyle \phi=\frac{s}{4}$

$\phi =\frac{1}{2}$ $\displaystyle \phi =\frac{1}{2}$

Report an Error

Example Question #243 : Rate Of Change

A cube is growing in size. What is the ratio of the rate of growth of the cube's volume to the rate of growth of its diagonal when its sides have length 1?

Possible Answers:

$2\sqrt{3}$ $\displaystyle 2\sqrt{3}$

$\sqrt{3}$ $\displaystyle \sqrt{3}$

$\displaystyle 3$

$\frac{\sqrt{3}}{3}$ $\displaystyle \frac{\sqrt{3}}{3}$

$\frac{1}{3}$ $\displaystyle \frac{1}{3}$

Correct answer:

$\sqrt{3}$ $\displaystyle \sqrt{3}$

Explanation:

Begin by writing the equations for a cube's dimensions. Namely its volume and diagonal in terms of the length of its sides:

V=s^3 $\displaystyle V=s^3$

$d=s\sqrt{3}$ $\displaystyle d=s\sqrt{3}$

The rates of change of these can be found by taking the derivative of each side of the equations with respect to time:

$\frac{dV}{dt}=3s^2\frac{ds}{dt}$ $\displaystyle \frac{dV}{dt}=3s^2\frac{ds}{dt}$

$\frac{dd}{dt}=\sqrt{3}\frac{ds}{dt}$ $\displaystyle \frac{dd}{dt}=\sqrt{3}\frac{ds}{dt}$

Now, knowing the length of the sides, simply divide to find the ratio between the rate of change of the volume and diagonal:

$3s^2\frac{ds}{dt}=(\phi)\sqrt{3}\frac{ds}{dt}$ $\displaystyle 3s^2\frac{ds}{dt}=(\phi)\sqrt{3}\frac{ds}{dt}$

$\phi=s^2\sqrt{3}$ $\displaystyle \phi=s^2\sqrt{3}$

$\phi =\sqrt{3}$ $\displaystyle \phi =\sqrt{3}$

Report an Error

Example Question #244 : Rate Of Change

A cube is diminishing in size. What is the ratio of the rate of loss of the cube's volume to the rate of loss of its diagonal when its sides have length 3?

Possible Answers:

$\frac{\sqrt{3}}{3}$ $\displaystyle \frac{\sqrt{3}}{3}$

$\frac{\sqrt{3}}{9}$ $\displaystyle \frac{\sqrt{3}}{9}$

$\sqrt{3}$ $\displaystyle \sqrt{3}$

$3\sqrt{3}$ $\displaystyle 3\sqrt{3}$

$9\sqrt{3}$ $\displaystyle 9\sqrt{3}$

Correct answer:

$9\sqrt{3}$ $\displaystyle 9\sqrt{3}$

Explanation:

Begin by writing the equations for a cube's dimensions. Namely its volume and diagonal in terms of the length of its sides:

V=s^3 $\displaystyle V=s^3$

$d=s\sqrt{3}$ $\displaystyle d=s\sqrt{3}$

The rates of change of these can be found by taking the derivative of each side of the equations with respect to time:

$\frac{dV}{dt}=3s^2\frac{ds}{dt}$ $\displaystyle \frac{dV}{dt}=3s^2\frac{ds}{dt}$

$\frac{dd}{dt}=\sqrt{3}\frac{ds}{dt}$ $\displaystyle \frac{dd}{dt}=\sqrt{3}\frac{ds}{dt}$

Now, knowing the length of the sides, simply divide to find the ratio between the rate of change of the volume and diagonal:

$3s^2\frac{ds}{dt}=(\phi)\sqrt{3}\frac{ds}{dt}$ $\displaystyle 3s^2\frac{ds}{dt}=(\phi)\sqrt{3}\frac{ds}{dt}$

$\phi=s^2\sqrt{3}$ $\displaystyle \phi=s^2\sqrt{3}$

$\phi =9\sqrt{3}$ $\displaystyle \phi =9\sqrt{3}$

Report an Error

Example Question #245 : Rate Of Change

A cube is growing in size. What is the ratio of the rate of growth of the cube's surface area to the rate of growth of its diagonal when its sides have length 5?

Possible Answers:

$25\sqrt{3}$ $\displaystyle 25\sqrt{3}$

$4\sqrt{3}$ $\displaystyle 4\sqrt{3}$

$5\sqrt{3}$ $\displaystyle 5\sqrt{3}$

$20\sqrt{3}$ $\displaystyle 20\sqrt{3}$

$10\sqrt{3}$ $\displaystyle 10\sqrt{3}$

Correct answer:

$20\sqrt{3}$ $\displaystyle 20\sqrt{3}$

Explanation:

Begin by writing the equations for a cube's dimensions. Namely its surface area and diagonal in terms of the length of its sides:

A=6s^2 $\displaystyle A=6s^2$

$d=s\sqrt{3}$ $\displaystyle d=s\sqrt{3}$

The rates of change of these can be found by taking the derivative of each side of the equations with respect to time:

$\frac{dA}{dt}=12s\frac{ds}{dt}$ $\displaystyle \frac{dA}{dt}=12s\frac{ds}{dt}$

$\frac{dd}{dt}=\sqrt{3}\frac{ds}{dt}$ $\displaystyle \frac{dd}{dt}=\sqrt{3}\frac{ds}{dt}$

Now, knowing the length of the sides, simply divide to find the ratio between the rate of change of the surface area and diagonal:

$12s\frac{ds}{dt}=(\phi)\sqrt{3}\frac{ds}{dt}$ $\displaystyle 12s\frac{ds}{dt}=(\phi)\sqrt{3}\frac{ds}{dt}$

$\phi=4s\sqrt{3}$ $\displaystyle \phi=4s\sqrt{3}$

$\phi =20\sqrt{3}$ $\displaystyle \phi =20\sqrt{3}$

Report an Error

View Calculus Tutors

Collin
Certified Tutor

The University of Texas at Austin, Bachelor of Science, Exercise Science.

View Calculus Tutors

Nnang
Certified Tutor

Columbia University in the City of New York, Bachelor of Science, Economics.

View Calculus Tutors

Tobi
Certified Tutor

Northwestern University, Bachelor of Science, Molecular Biology.

All Calculus 1 Resources

10 Diagnostic Tests 438 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

Reading Tutors in Dallas Fort Worth, Biology Tutors in San Francisco-Bay Area, LSAT Tutors in Washington DC, Algebra Tutors in Chicago, Physics Tutors in Dallas Fort Worth, Calculus Tutors in Phoenix, SSAT Tutors in Chicago, French Tutors in San Francisco-Bay Area, Calculus Tutors in Houston, GRE Tutors in Miami

Popular Courses & Classes

ISEE Courses & Classes in Atlanta, Spanish Courses & Classes in Seattle, GRE Courses & Classes in Seattle, SAT Courses & Classes in Houston, ACT Courses & Classes in Chicago, MCAT Courses & Classes in Dallas Fort Worth, ACT Courses & Classes in Philadelphia, MCAT Courses & Classes in Seattle, ISEE Courses & Classes in Dallas Fort Worth, SSAT Courses & Classes in Atlanta

Popular Test Prep

SSAT Test Prep in Dallas Fort Worth, MCAT Test Prep in Washington DC, ACT Test Prep in Philadelphia, ACT Test Prep in Chicago, SAT Test Prep in Boston, ISEE Test Prep in Chicago, LSAT Test Prep in Denver, MCAT Test Prep in Atlanta, GMAT Test Prep in San Diego, SSAT Test Prep in Denver

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

Calculus 1 : Rate of Change

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #241 : Rate Of Change

Example Question #241 : Rate Of Change

Example Question #243 : Rate Of Change

Example Question #2121 : Functions

Example Question #3152 : Calculus

Example Question #2122 : Functions

Example Question #242 : Rate Of Change

Example Question #243 : Rate Of Change

Example Question #244 : Rate Of Change

Example Question #245 : Rate Of Change

All Calculus 1 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: