Call Now to Set Up Tutoring:

(704) 350-2687

CLEP Calculus : Functions

Study concepts, example questions & explanations for CLEP Calculus

Create An Account

All CLEP Calculus Resources

2 Practice Tests Question of the Day Flashcards

Example Questions

Example Question #596 : Rate

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:

$18\sqrt{6}$

$3\sqrt{3}$

$18\sqrt{3}$

$9\sqrt{6}$

$9\sqrt{3}$

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

Example Question #597 : Rate

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

Possible Answers:

$\frac{3}{4}$

$\frac{9}{4}$

$\frac{9}{16}$

$\frac{9\sqrt{3}}{16}$

$36\sqrt{3}$

Correct answer:

$\frac{9\sqrt{3}}{16}$

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:

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

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

$\phi=\frac{\sqrt{3}(\frac{\sqrt{3}}{2})^2}{4}=\frac{9\sqrt{3}}{16}$

Report an Error

Example Question #501 : Rate Of Change

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 $4\sqrt{2}$ ?

Possible Answers:

$8\sqrt{6}$

$8\sqrt{3}$

$2\sqrt{6}$

$4\sqrt{6}$

$4\sqrt{3}$

Correct answer:

$8\sqrt{3}$

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:

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

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

$\phi=\frac{\sqrt{3}(4\sqrt{2})^2}{4}=8\sqrt{3}$

Report an Error

Example Question #591 : Rate

A regular tetrahedron is diminishing 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 $\frac{1}{2}$ ?

Possible Answers:

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

$\frac{\sqrt{3}}{32}$

$\frac{\sqrt{3}}{8}$

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

$\frac{\sqrt{3}}{16}$

Correct answer:

$\frac{\sqrt{3}}{16}$

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:

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

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

$\phi=\frac{\sqrt{3}(\frac{1}{2})^2}{4}=\frac{\sqrt{3}}{16}$

Report an Error

Example Question #511 : Rate Of Change

A regular tetrahedron is diminishing 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 $\frac{\sqrt{2}}{2}$ ?

Possible Answers:

$\frac{\sqrt{3}}{8}$

$\frac{\sqrt{3}}{16}$

$\frac{\sqrt{6}}{8}$

$\frac{\sqrt{6}}{4}$

$\frac{\sqrt{3}}{32}$

Correct answer:

$\frac{\sqrt{3}}{8}$

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:

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

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

$\phi=\frac{\sqrt{3}(\frac{\sqrt{2}}{2})^2}{4}=\frac{\sqrt{3}}{8}$

Report an Error

Example Question #511 : Rate Of Change

Possible Answers:

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

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

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

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

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

Correct answer:

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

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:

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

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

$\phi=\frac{\sqrt{3}(\frac{\sqrt{3}}{3})^2}{4}=\frac{\sqrt{3}}{12}$

Report an Error

Example Question #602 : Rate

A regular tetrahedron is burgeoning in size. What is the ratio of the rate of change of the surface area of the tetrahedron to the rate of change of its height when its sides have length 17?

Possible Answers:

$17\sqrt{3}$

$17\sqrt{6}$

$51\sqrt{2}$

$51\sqrt{3}$

$51\sqrt{6}$

Correct answer:

$51\sqrt{2}$

Explanation:

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

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

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

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

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

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

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:

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

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

$\phi=3(17)\sqrt{2}=51\sqrt{2}$

Report an Error

Example Question #603 : Rate

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

Possible Answers:

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

$\sqrt{6}$

$\sqrt{3}$

$\sqrt{2}$

Correct answer:

Explanation:

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

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

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

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

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

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

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:

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

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

$\phi=3(\frac{\sqrt{2}}{2})\sqrt{2}=3$

Report an Error

Example Question #604 : Rate

A regular tetrahedron is burgeoning in size. What is the ratio of the rate of change of the surface area of the tetrahedron to the rate of change of its height when its sides have length 56?

Possible Answers:

$42\sqrt{3}$

$42\sqrt{6}$

$168\sqrt{3}$

$168\sqrt{2}$

$168\sqrt{6}$

Correct answer:

$168\sqrt{2}$

Explanation:

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

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

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

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

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

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

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:

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

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

$\phi=3(56)\sqrt{2}=168\sqrt{2}$

Report an Error

Example Question #605 : Rate

A regular tetrahedron is burgeoning in size. What is the ratio of the rate of change of the surface area of the tetrahedron to the rate of change of its height when its sides have length 25?

Possible Answers:

$75\sqrt{3}$

$75\sqrt{6}$

$15\sqrt{3}$

$15\sqrt{2}$

$75\sqrt{2}$

Correct answer:

$75\sqrt{2}$

Explanation:

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

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

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

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

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

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

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:

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

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

$\phi=3(25)\sqrt{2}=75\sqrt{2}$

Report an Error

View Tutors

Daiven
Certified Tutor

Wofford College, Bachelor of Science, Biology, General.

View Tutors

Graham
Certified Tutor

York Technical College, Associate in Science, Chemical Physics. Clemson University, Bachelor of Science, Chemical and Biomole...

View Tutors

Stephan
Certified Tutor

Richland Community College, Associate in Science, Science, Technology, and Society. Brigham Young University-Provo, Bachelor ...

All CLEP Calculus Resources

2 Practice Tests Question of the Day Flashcards

Popular Subjects

Math Tutors in Miami, GRE Tutors in Phoenix, ACT Tutors in Denver, Computer Science Tutors in Miami, English Tutors in Los Angeles, SSAT Tutors in Los Angeles, Physics Tutors in San Diego, SAT Tutors in Phoenix, MCAT Tutors in Denver, MCAT Tutors in Washington DC

Popular Courses & Classes

MCAT Courses & Classes in San Diego, GRE Courses & Classes in Atlanta, SSAT Courses & Classes in Washington DC, ACT Courses & Classes in Dallas Fort Worth, ISEE Courses & Classes in San Francisco-Bay Area, MCAT Courses & Classes in Miami, Spanish Courses & Classes in Houston, Spanish Courses & Classes in Philadelphia, ACT Courses & Classes in Houston, LSAT Courses & Classes in San Francisco-Bay Area

Popular Test Prep

SAT Test Prep in Phoenix, MCAT Test Prep in Dallas Fort Worth, LSAT Test Prep in Washington DC, ACT Test Prep in Denver, LSAT Test Prep in San Diego, ACT Test Prep in Dallas Fort Worth, MCAT Test Prep in Miami, MCAT Test Prep in San Francisco-Bay Area, LSAT Test Prep in Seattle, ACT Test Prep in Boston

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

CLEP Calculus : Functions

Study concepts, example questions & explanations for CLEP Calculus

All CLEP Calculus Resources

Example Questions

Example Question #596 : Rate

Example Question #597 : Rate

Example Question #501 : Rate Of Change

Example Question #591 : Rate

Example Question #511 : Rate Of Change

Example Question #511 : Rate Of Change

Example Question #602 : Rate

Example Question #603 : Rate

Example Question #604 : Rate

Example Question #605 : Rate

All CLEP Calculus Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: