We are open Saturday and Sunday!

Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 1 : Calculus

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 #498 : 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 $\frac{1}{10}$ ?

Possible Answers:

$\frac{1}{5}$

$\frac{1}{100}$

$\frac{1}{25}$

$\frac{1}{50}$

$\frac{1}{200}$

Correct answer:

$\frac{1}{50}$

Explanation:

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

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

$C=2\pi r^$

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

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

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

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:

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

$\phi =2r^2$

$\phi =2(\frac{1}{10})^2=\frac{1}{50}$

Report an Error

Example Question #499 : Rate Of Change

Possible Answers:

$\frac{1}{512}$

$\frac{1}{1024}$

$\frac{1}{32}$

$\frac{1}{256}$

$\frac{1}{128}$

Correct answer:

$\frac{1}{128}$

Explanation:

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

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

$C=2\pi r^$

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

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

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

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:

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

$\phi =2r^2$

$\phi =2(\frac{1}{16})^2=\frac{1}{128}$

Report an Error

Example Question #500 : Rate Of Change

Possible Answers:

$\frac{2601}{32}$

4202

2601

$\frac{2601}{4}$

$\frac{2601}{2}$

Correct answer:

$\frac{2601}{2}$

Explanation:

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

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

$C=2\pi r^$

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

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

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

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:

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

$\phi =2r^2$

$\phi =2(\frac{51}{2})^2=\frac{2601}{2}$

Report an Error

Example Question #501 : Rate Of Change

A regular tetrahedron is growing in size. What is the ratio of the rate of change of the volume of the tetrahedron to the rate of change of its surface area when its sides have length $24\sqrt{6}$ ?

Possible Answers:

$\sqrt{6}$

$12\sqrt{6}$

Correct answer:

Explanation:

To solve this problem, define a regular tetrahedron's dimensions, its volume and surface area in terms of the length of its sides:

$V=\frac{s^3}{6\sqrt{2}}$

$A=\sqrt{3}s^2$

Rates of change can then be found by taking the derivative of each property with respect to time:

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

$\frac{dA}{dt}=2\sqrt{3}s\frac{ds}{dt}$

The rate of change of the sides isn't going to vary no matter what dimension of the tetrahedron we're considering; $\frac{ds}{dt}$ is $\frac{ds}{dt}$ . Find the ratio by dividing quantities:

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

$\phi=\frac{s\sqrt{6}}{24}$

$\phi=\frac{(24\sqrt{6})\sqrt{6}}{24}=6$

Report an Error

Example Question #502 : Rate Of Change

A regular tetrahedron is growing in size. What is the ratio of the rate of change of the volume of the tetrahedron to the rate of change of its surface area when its sides have length $12\sqrt{6}$ ?

Possible Answers:

$24\sqrt{6}$

$12\sqrt{3}$

Correct answer:

Explanation:

To solve this problem, define a regular tetrahedron's dimensions, its volume and surface area in terms of the length of its sides:

$V=\frac{s^3}{6\sqrt{2}}$

$A=\sqrt{3}s^2$

Rates of change can then be found by taking the derivative of each property with respect to time:

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

$\frac{dA}{dt}=2\sqrt{3}s\frac{ds}{dt}$

The rate of change of the sides isn't going to vary no matter what dimension of the tetrahedron we're considering; $\frac{ds}{dt}$ is $\frac{ds}{dt}$ . Find the ratio by dividing quantities:

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

$\phi=\frac{s\sqrt{6}}{24}$

$\phi=\frac{(12\sqrt{6})\sqrt{6}}{24}=3$

Report an Error

Example Question #503 : Rate Of Change

A regular tetrahedron is growing in size. What is the ratio of the rate of change of the volume of the tetrahedron to the rate of change of its surface area when its sides have length 14?

Possible Answers:

$\frac{7\sqrt{3}}{12}$

$\frac{7\sqrt{6}}{6}$

$\frac{7\sqrt{6}}{12}$

$\frac{7\sqrt{3}}{4}$

$\frac{7\sqrt{2}}{3}$

Correct answer:

$\frac{7\sqrt{6}}{12}$

Explanation:

To solve this problem, define a regular tetrahedron's dimensions, its volume and surface area in terms of the length of its sides:

$V=\frac{s^3}{6\sqrt{2}}$

$A=\sqrt{3}s^2$

Rates of change can then be found by taking the derivative of each property with respect to time:

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

$\frac{dA}{dt}=2\sqrt{3}s\frac{ds}{dt}$

The rate of change of the sides isn't going to vary no matter what dimension of the tetrahedron we're considering; $\frac{ds}{dt}$ is $\frac{ds}{dt}$ . Find the ratio by dividing quantities:

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

$\phi=\frac{s\sqrt{6}}{24}$

$\phi=\frac{(14)\sqrt{6}}{24}=\frac{7\sqrt{6}}{12}$

Report an Error

Example Question #504 : Rate Of Change

A regular tetrahedron is shrinking in size. What is the ratio of the rate of change of the volume of the tetrahedron to the rate of change of its surface area when its sides have length $48\sqrt{2}$ ?

Possible Answers:

$8\sqrt{6}$

$4\sqrt{3}$

$2\sqrt{6}$

$8\sqrt{3}$

$2\sqrt{3}$

Correct answer:

$4\sqrt{3}$

Explanation:

To solve this problem, define a regular tetrahedron's dimensions, its volume and surface area in terms of the length of its sides:

$V=\frac{s^3}{6\sqrt{2}}$

$A=\sqrt{3}s^2$

Rates of change can then be found by taking the derivative of each property with respect to time:

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

$\frac{dA}{dt}=2\sqrt{3}s\frac{ds}{dt}$

The rate of change of the sides isn't going to vary no matter what dimension of the tetrahedron we're considering; $\frac{ds}{dt}$ is $\frac{ds}{dt}$ . Find the ratio by dividing quantities:

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

$\phi=\frac{s\sqrt{6}}{24}$

$\phi=\frac{(48\sqrt{2})\sqrt{6}}{24}=4\sqrt{3}$

Report an Error

Example Question #505 : Rate Of Change

Possible Answers:

$\frac{3\sqrt{2}}{2}$

$\frac{3\sqrt{3}}{2}$

$9\sqrt{2}$

$\frac{3\sqrt{6}}{2}$

Correct answer:

Explanation:

To solve this problem, define a regular tetrahedron's dimensions, its volume and surface area in terms of the length of its sides:

$V=\frac{s^3}{6\sqrt{2}}$

$A=\sqrt{3}s^2$

Rates of change can then be found by taking the derivative of each property with respect to time:

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

$\frac{dA}{dt}=2\sqrt{3}s\frac{ds}{dt}$

The rate of change of the sides isn't going to vary no matter what dimension of the tetrahedron we're considering; $\frac{ds}{dt}$ is $\frac{ds}{dt}$ . Find the ratio by dividing quantities:

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

$\phi=\frac{s\sqrt{6}}{24}$

$\phi=\frac{(36\sqrt{6})\sqrt{6}}{24}=9$

Report an Error

Example Question #506 : Rate Of Change

Possible Answers:

$12\sqrt{2}$

$8\sqrt{2}$

$12\sqrt{3}$

$8\sqrt{3}$

$4\sqrt{6}$

Correct answer:

$12\sqrt{2}$

Explanation:

To solve this problem, define a regular tetrahedron's dimensions, its volume and surface area in terms of the length of its sides:

$V=\frac{s^3}{6\sqrt{2}}$

$A=\sqrt{3}s^2$

Rates of change can then be found by taking the derivative of each property with respect to time:

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

$\frac{dA}{dt}=2\sqrt{3}s\frac{ds}{dt}$

The rate of change of the sides isn't going to vary no matter what dimension of the tetrahedron we're considering; $\frac{ds}{dt}$ is $\frac{ds}{dt}$ . Find the ratio by dividing quantities:

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

$\phi=\frac{s\sqrt{6}}{24}$

$\phi=\frac{(96\sqrt{3})\sqrt{6}}{24}=12\sqrt{2}$

Report an Error

Example Question #2391 : Functions

A regular tetrahedron is increaseing in size. What is the ratio of the rate of change of the volume of the tetrahedron to the rate of change of its height when its sides have length 6?

Possible Answers:

$9\sqrt{3}$

$3\sqrt{3}$

$9\sqrt{6}$

$18\sqrt{3}$

$18\sqrt{6}$

Correct answer:

$9\sqrt{3}$

Explanation:

To solve this problem, define a regular tetrahedron's dimensions, its volume and height in terms of the length of its sides:

$V=\frac{s^3}{6\sqrt{2}}$

$h=\frac{\sqrt{6}}{3}s$

Rates of change can then be found by taking the derivative of each property with respect to time:

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

$\frac{dh}{dt}=\frac{\sqrt{6}}{3}\frac{ds}{dt}$

The rate of change of the sides isn't going to vary no matter what dimension of the tetrahedron we're considering; $\frac{ds}{dt}$ is $\frac{ds}{dt}$ . Find the ratio by dividing quantities:

$\frac{s^2}{2\sqrt{2}}\frac{ds}{dt}=(\phi)\frac{\sqrt{6}}{3}\frac{ds}{dt}$

$\phi=\frac{\sqrt{3}s^2}{4}$

$\phi=\frac{\sqrt{3}(6)^2}{4}=9\sqrt{3}$

Report an Error

View Calculus Tutors

Michael
Certified Tutor

Carthage College, Bachelor of Science, Finance.

View Calculus Tutors

Trevon
Certified Tutor

Columbia University in the City of New York, Current Undergrad, Chemical Engineering.

View Calculus Tutors

Daisy
Certified Tutor

Northeastern University, Bachelor of Engineering, Chemical Engineering. Northeastern University, Master of Engineering, Chemi...

All Calculus 1 Resources

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

Popular Subjects

GRE Tutors in Denver, SSAT Tutors in Boston, SSAT Tutors in Houston, Statistics Tutors in Boston, GRE Tutors in Phoenix, Biology Tutors in Washington DC, Chemistry Tutors in Seattle, Math Tutors in Miami, Physics Tutors in Houston, LSAT Tutors in Denver

Popular Courses & Classes

ACT Courses & Classes in Denver, LSAT Courses & Classes in Los Angeles, ACT Courses & Classes in Houston, SAT Courses & Classes in Chicago, GRE Courses & Classes in Los Angeles, LSAT Courses & Classes in Denver, SSAT Courses & Classes in Phoenix, SSAT Courses & Classes in Washington DC, ISEE Courses & Classes in New York City, GRE Courses & Classes in Seattle

Popular Test Prep

ACT Test Prep in Phoenix, LSAT Test Prep in Atlanta, LSAT Test Prep in Phoenix, LSAT Test Prep in Chicago, GRE Test Prep in Boston, GRE Test Prep in Atlanta, ACT Test Prep in Chicago, ACT Test Prep in Denver, GRE Test Prep in Houston, LSAT Test Prep in Los Angeles

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 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #498 : Rate Of Change

Example Question #499 : Rate Of Change

Example Question #500 : Rate Of Change

Example Question #501 : Rate Of Change

Example Question #502 : Rate Of Change

Example Question #503 : Rate Of Change

Example Question #504 : Rate Of Change

Example Question #505 : Rate Of Change

Example Question #506 : Rate Of Change

Example Question #2391 : Functions

All Calculus 1 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: