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 #215 : Rate Of Change

A spherical balloon is being filled with air. What is the circumference of the sphere at the instance the rate of growth of the volume is eleven times the rate of growth of the surface area?

Possible Answers:

$44\pi$

$55\pi$

$96\pi$

$22\pi$

$99\pi$

Correct answer:

$44\pi$

Explanation:

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

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

$A=4\pi r^2$

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

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

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

The rate of change of the radius is going to be the same for the sphere. So given our problem conditions, the rate of growth of the volume is eleven times the rate of growth of the surface area, let's solve for a radius that satisfies it.

$4\pi r^2 \frac{dr}{dt}=(11)8\pi r \frac{dr}{dt}$

r =(11)2

r=22

Circumference is given by

$C=2\pi r$

$C=44\pi$

Report an Error

Example Question #216 : Rate Of Change

A spherical balloon is being filled with air. What is the volume of the sphere at the instance the rate of growth of the volume is an eighth of the rate of growth of the surface area?

Possible Answers:

$\frac{\pi}{48}$

$\frac{\pi}{24}$

$\frac{\pi}{2}$

$\frac{\pi}{6}$

$\frac{\pi}{12}$

Correct answer:

$\frac{\pi}{48}$

Explanation:

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

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

$A=4\pi r^2$

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

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

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

The rate of change of the radius is going to be the same for the sphere. So given our problem conditions, the rate of growth of the volume is an eighth of the rate of growth of the surface area, let's solve for a radius that satisfies it.

$4\pi r^2 \frac{dr}{dt}=(\frac{1}{8})8\pi r \frac{dr}{dt}$

$r =(\frac{1}{8})2$

$r=\frac{1}{4}$

Then to find the volume:

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

$V=\frac{4}{3}\pi (\frac{1}{4})^3$

$V=\frac{\pi}{48}$

Report an Error

Example Question #217 : Rate Of Change

A cube is diminishing in size. What is the surface area the cube at the time that the rate of shrinkage of the cube's volume is equal to a fifth the rate of shrinkage of its surface area?

Possible Answers:

$\frac{96}{25}$

$\frac{24}{5}$

$\frac{16}{25}$

$\frac{4}{5}$

$\frac{96}{5}$

Correct answer:

$\frac{96}{25}$

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

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:

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

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

Now, solve for the length of the side of the cube to satisfy the problem condition, the rate of shrinkage of the cube's volume is equal to a fifth the rate of shrinkage of its surface area:

$3s^2\frac{ds}{dt}=(\frac{1}{5})12s\frac{ds}{dt}$

$s=(\frac{1}{5})4$

$s=\frac{4}{5}$

To find the surface area:

A=6s^2

$A=6(\frac{4}{5})^2$

$A=\frac{96}{25}$

Report an Error

Example Question #221 : Rate Of Change

The rate of change of a cylinder's radius is equal to the rate of change of its height. How does the rate of change of the cylinder's volume compare to the rate of change of its surface area when the radius is five times the height?

Possible Answers:

$\frac{19}{32}r$

$\frac{7}{32}r$

$\frac{13}{32}r$

$\frac{17}{32}r$

$\frac{23}{32}r$

Correct answer:

$\frac{7}{32}r$

Explanation:

To approach this problem, begin by defining the cylinder's volume and surface area in terms of its height and radius:

$V=\pi r^2h$

$A=2\pi r^2 +2\pi rh$

Cylinderdimensions

Rates of change can be found by deriving, then, with respect to time:

$\frac{dV}{dt}=2\pi rh\frac{dr}{dt}+\pi r^2 \frac{dh}{dt}$

$\frac{dA}{dt}=4\pi r \frac{dr}{dt}+2\pi h \frac{dr}{dt}+2\pi r \frac{dh}{dt}$

$\frac{dA}{dt}=(4\pi r+2\pi h )\frac{dr}{dt}+2\pi r \frac{dh}{dt}$

We're told two things:

The rate of change of a cylinder's radius is equal to the rate of change of its height: $\frac{dr}{dt}=\frac{dh}{dt}$

The radius is five times the height r=5h

Using these properties, rewrite the rate equations:

$\frac{dV}{dt}=2\pi r(\frac{1}{5}r)\frac{dr}{dt}+\pi r^2 \frac{dr}{dt}=\frac{7\pi r^2}{5}\frac{dr}{dt}$

$\frac{dA}{dt}=(4\pi r+2\pi (\frac{1}{5}r) )\frac{dr}{dt}+2\pi r \frac{dr}{dt}=\frac{32\pi r}{5}\frac{dr}{dt}$

The comparison between the volume and surface area can be found by taking the ratio of the two:

$\phi = \frac{\frac{7\pi r^2}{5}\frac{dr}{dt}}{\frac{32\pi r}{5}\frac{dr}{dt}}=\frac{7}{32}r$

Report an Error

Example Question #222 : Rate Of Change

The rate of change of a cylinder's radius is equal to half the rate of change of its height. How does the rate of change of the cylinder's volume compare to the rate of change of its surface area when the radius is a third the height?

Possible Answers:

$\frac{1}{7}r$

$\frac{4}{7}r$

$\frac{2}{7}r$

$\frac{7}{2}r$

Correct answer:

$\frac{4}{7}r$

Explanation:

To approach this problem, begin by defining the cylinder's volume and surface area in terms of its height and radius:

$V=\pi r^2h$

$A=2\pi r^2 +2\pi rh$

Cylinderdimensions

Rates of change can be found by deriving, then, with respect to time:

$\frac{dV}{dt}=2\pi rh\frac{dr}{dt}+\pi r^2 \frac{dh}{dt}$

$\frac{dA}{dt}=4\pi r \frac{dr}{dt}+2\pi h \frac{dr}{dt}+2\pi r \frac{dh}{dt}$

$\frac{dA}{dt}=(4\pi r+2\pi h )\frac{dr}{dt}+2\pi r \frac{dh}{dt}$

We're told two things:

The rate of change of a cylinder's radius is equal to half the rate of change of its height: $\frac{dr}{dt}=\frac{1}{2}\frac{dh}{dt}$

The radius is a third the height: $r=\frac{1}{3}h$

Using these properties, rewrite the rate equations:

$\frac{dV}{dt}=2\pi r(3r)\frac{dr}{dt}+\pi r^2 (2\frac{dr}{dt})=8\pi r^2\frac{dr}{dt}$

$\frac{dA}{dt}=(4\pi r+2\pi (3r) )\frac{dr}{dt}+2\pi r (2\frac{dr}{dt})=14\pi r \frac{dr}{dt}$

The comparison between the volume and surface area can be found by taking the ratio of the two:

$\phi = \frac{8\pi r^2\frac{dr}{dt}}{14\pi r \frac{dr}{dt}}=\frac{4}{7}r$

Report an Error

Example Question #223 : Rate Of Change

The rate of change of a cylinder's radius is equal to a fifth of the rate of change of its height. How does the rate of change of the cylinder's volume compare to the rate of change of its surface area when the radius is a fifth of the height?

Possible Answers:

$\frac{5}{8}r$

$\frac{3}{8}r$

$\frac{9}{8}r$

$\frac{1}{8}r$

$\frac{17}{8}r$

Correct answer:

$\frac{5}{8}r$

Explanation:

To approach this problem, begin by defining the cylinder's volume and surface area in terms of its height and radius:

$V=\pi r^2h$

$A=2\pi r^2 +2\pi rh$

Cylinderdimensions

Rates of change can be found by deriving, then, with respect to time:

$\frac{dV}{dt}=2\pi rh\frac{dr}{dt}+\pi r^2 \frac{dh}{dt}$

$\frac{dA}{dt}=4\pi r \frac{dr}{dt}+2\pi h \frac{dr}{dt}+2\pi r \frac{dh}{dt}$

$\frac{dA}{dt}=(4\pi r+2\pi h )\frac{dr}{dt}+2\pi r \frac{dh}{dt}$

We're told two things:

The rate of change of a cylinder's radius is equal to a fifth of the rate of change of its height: $\frac{dr}{dt}=\frac{1}{5}\frac{dh}{dt}$
The radius is a fifth of the height: $r=\frac{1}{5}h$

Using these properties, rewrite the rate equations:

$\frac{dV}{dt}=2\pi r(5r)\frac{dr}{dt}+\pi r^2 (5\frac{dr}{dt})=15\pi r^2 \frac{dr}{dt}$

$\frac{dA}{dt}=(4\pi r+2\pi (5r) )\frac{dr}{dt}+2\pi r (5\frac{dr}{dt})=24\pi r \frac{dr}{dt}$

The comparison between the volume and surface area can be found by taking the ratio of the two:

$\phi = \frac{15\pi r^2 \frac{dr}{dt}}{24\pi r \frac{dr}{dt}}=\frac{5}{8}r$

Report an Error

Example Question #311 : Rate

A regular tetrahedron is growing in size. What is the length of the sides of the tetrahedron at the time the rate of growth of its volume is one thirteenth the rate of growth of its surface area?

Possible Answers:

$\frac{4\sqrt{6}}{13}$

$\frac{2\sqrt{6}}{13}$

$\frac{12}{13}$

$\frac{24}{13}$

$\frac{4}{13}$

Correct answer:

$\frac{4\sqrt{6}}{13}$

Explanation:

To tackle this problem, define a regular tetrahedron's dimensions 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, so given our problem condition, the rate of growth of its volume is one thirteenth the rate of growth of its surface area, solve for the corresponding length of the tetrahedron's sides:

$\frac{s^2}{2\sqrt{2}}\frac{ds}{dt}=(\frac{1}{13})2\sqrt{3}s\frac{ds}{dt}$

$s^2=(\frac{1}{13})4\sqrt{6}s$

$s=\frac{4\sqrt{6}}{13}$

Report an Error

Example Question #312 : Rate

A regular tetrahedron is growing in size. What is the height of the tetrahedron at the time the rate of growth of its volume is ten times the rate of growth of its surface area?

Possible Answers:

$80\sqrt{3}$

$120\sqrt{3}$

$40\sqrt{3}$

Correct answer:

Explanation:

To tackle this problem, define a regular tetrahedron's dimensions 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, so given our problem condition, the rate of growth of its volume is ten times the rate of growth of its surface area, solve for the corresponding length of the tetrahedron's sides:

$\frac{s^2}{2\sqrt{2}}\frac{ds}{dt}=(10)2\sqrt{3}s\frac{ds}{dt}$

$s^2=(10)4\sqrt{6}s$

$s=40\sqrt{6}$

The height of a tetrahedron is given by the equation:

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

$h=\frac{\sqrt{6}}{3}(40\sqrt{6}s)$

h=80

Report an Error

Example Question #226 : Rate Of Change

A regular tetrahedron is growing in size. What is the volume of the tetrahedron at the time the rate of growth of its volume is $\frac{3\sqrt{3}}{2}$ times rate of growth of its surface area?

Possible Answers:

$972\sqrt{2}$

$812\sqrt{2}$

1944

812

$456\sqrt{2}$

Correct answer:

1944

Explanation:

To tackle this problem, define a regular tetrahedron's dimensions 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, so given our problem condition, the rate of growth of its volume is $\frac{3\sqrt{3}}{2}$ times rate of growth of its surface area, solve for the corresponding length of the tetrahedron's sides:

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

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

$s=18\sqrt{2}$

Now find the volume:

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

$V=\frac{(18\sqrt{2})^3}{6\sqrt{2}}$

V=1944

Report an Error

Example Question #313 : Rate

A cube is growing in size. What is the length of the sides of the cube at the time that the rate of growth of the cube's volume is equal to thirteen times the rate of growth of its surface area?

Possible Answers:

$\sqrt{13}$

$2\sqrt{13}$

Correct answer:

Explanation:

Begin by writing s=52 the equations for a cube's dimensions. Namely its volume and surface area in terms of the length of its sides:

V=s^3

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:

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

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

Now, solve for the length of the side of the cube to satisfy the problem condition, the rate of growth of the cube's volume is equal to thirteen times the rate of growth of its surface area:

$3s^2\frac{ds}{dt}=(13)12s\frac{ds}{dt}$

s=(13)4

s=52

Report an Error

View Calculus Tutors

Ben
Certified Tutor

Brown University, Bachelor in Arts, International Relations. Bridge Education Group, Certificate, Teaching French as a Second...

View Calculus Tutors

Rachel
Certified Tutor

University of Arkansas at Little Rock, Bachelor in Arts, Mathematics.

View Calculus Tutors

Aditya
Certified Tutor

University of Minnesota-Twin Cities, Bachelor in Arts, Chemical Engineering.

All Calculus 1 Resources

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

Popular Subjects

SAT Tutors in New York City, Chemistry Tutors in San Diego, GMAT Tutors in Denver, Statistics Tutors in Atlanta, Algebra Tutors in Seattle, Calculus Tutors in Chicago, Spanish Tutors in Philadelphia, French Tutors in Seattle, LSAT Tutors in Denver, ISEE Tutors in Chicago

Popular Courses & Classes

GMAT Courses & Classes in Houston, ISEE Courses & Classes in Boston, SSAT Courses & Classes in Miami, SAT Courses & Classes in New York City, ACT Courses & Classes in Phoenix, ACT Courses & Classes in San Diego, GRE Courses & Classes in San Diego, ISEE Courses & Classes in San Francisco-Bay Area, ACT Courses & Classes in Seattle, Spanish Courses & Classes in Chicago

Popular Test Prep

ISEE Test Prep in Boston, ACT Test Prep in San Diego, ACT Test Prep in Phoenix, SSAT Test Prep in Los Angeles, MCAT Test Prep in Chicago, SSAT Test Prep in Boston, ACT Test Prep in Seattle, GRE Test Prep in San Francisco-Bay Area, SSAT Test Prep in Dallas Fort Worth, MCAT 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 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #215 : Rate Of Change

Example Question #216 : Rate Of Change

Example Question #217 : Rate Of Change

Example Question #221 : Rate Of Change

Example Question #222 : Rate Of Change

Example Question #223 : Rate Of Change

Example Question #311 : Rate

Example Question #312 : Rate

Example Question #226 : Rate Of Change

Example Question #313 : Rate

All Calculus 1 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: