Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 1 : How to find 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 #191 : How To Find Rate Of Change

If the position of an object at time is represented by the function p(t) = 3t^2-12t+9 , when does the object stop moving (i.e. The velocity is zero)?

Possible Answers:

t=3

t=1 or t=3

The velocity is never .

t=1

t=2

Correct answer:

t=2

Explanation:

When the velocity is , that means p'(t)=0 . That is, 6t-12=0 . So t=2 .

Report an Error

Example Question #192 : How To Find Rate Of Change

The width of a rectangular prism increases half as fast as its length and a third as fast as its height. How does the rate of change of the prism's volume compare to that of the rate of change of the width when the length, height, and width are equal?

Possible Answers:

$\frac{1}{4}w^2$

5w^2

6w^2

$\frac{1}{6}w^2$

3w^2

Correct answer:

6w^2

Explanation:

Begin by writing the expression for the volume of a rectangular prism:

V=whl

The rate of change of the volume can be found by taking the derivative of the equation with respect to time:

$\frac{dV}{dt}=wh\frac{dl}{dt}+wl\frac{dh}{dt}+hl\frac{dw}{dt}$

Now, we're given some information:

The width of a rectangular prism increases half as fast as its length and a third as fast as its height: $\frac{dw}{dt}=\frac{1}{2}\frac{dl}{dt},\frac{dw}{dt}=\frac{1}{3}\frac{dh}{dt}$

The width when the length, height, and width are equal: w=l=h

Using this, rewrite the volume equation in terms of width:

$\frac{dV}{dt}=w^2(2\frac{dw}{dt})+w^2(3\frac{dh}{dt})+w^2\frac{dw}{dt}$

$\frac{dV}{dt}=6w^2\frac{dw}{dt}$

The rate of change of the volume is 6w^2 times the rate of change of the rate of change of the width.

Report an Error

Example Question #193 : How To Find Rate Of Change

The width of a rectangular prism increases twice as fast as its length and half as fast as its height. How does the rate of change of the prism's volume compare to that of the rate of change of the width when the width is half the length, which is half the height?

Possible Answers:

20w^2

22w^2

14w^2

9w^2

7w^2

Correct answer:

14w^2

Explanation:

Begin by writing the expression for the volume of a rectangular prism:

V=whl

The rate of change of the volume can be found by taking the derivative of the equation with respect to time:

$\frac{dV}{dt}=wh\frac{dl}{dt}+wl\frac{dh}{dt}+hl\frac{dw}{dt}$

Now, we're given some information:

The width of a rectangular prism increases twice as fast as its length and half as fast as its height: $\frac{dw}{dt}=2\frac{dl}{dt},\frac{dw}{dt}=\frac{1}{2}\frac{dh}{dt}$

The width is half the length, which is half the height: $w=\frac{1}{2}l,l=\frac{1}{2}h\rightarrow w=\frac{1}{4}h$

Using this, rewrite the volume equation in terms of width:

$\frac{dV}{dt}=w(4w)(\frac{1}{2}\frac{dw}{dt})+w(2w)(2\frac{dw}{dt})+(4w)(2w)\frac{dw}{dt}$

$\frac{dV}{dt}=2w^2\frac{dw}{dt}+4w^2\frac{dw}{dt}+8w^2\frac{dw}{dt}$

$\frac{dV}{dt}=14w^2\frac{dw}{dt}$

The rate of change of the volume is 14w^2 times the rate of change of the rate of change of the width.

Report an Error

Example Question #194 : How To Find Rate Of Change

A spherical balloon is deflating, while maintaining its spherical shape. What is the diameter of the sphere at the instance the rate of shrinkage of the volume is equal to the rate of shrinkage of the surface area?

Possible Answers:

$8\pi$

$4\pi$

Correct answer:

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 shrinkage of the volume is equal to the rate of shrinkage of the surface area, let's solve for a radius that satisfies it.

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

r =2

Since the diameter is twice this, d=4

Report an Error

Example Question #195 : How To Find Rate Of Change

A regular tetrahedron is growing in size. What is the height of the tetrahedron at the time the rate of growth of its sides is equal to the rate of growth of its volume?

Possible Answers:

$\frac{\sqrt{3}}{\sqrt[4]{2}}$

$\frac{2\sqrt[4]{2}}{{3}}$

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

$\frac{2\sqrt[4]{2}}{\sqrt{3}}$

$\frac{6\sqrt[4]{2}}{\sqrt{3}}$

Correct answer:

$\frac{2\sqrt[4]{2}}{\sqrt{3}}$

Explanation:

To tackle this problem, define a regular tetrahedron's dimensions in terms of the length of its sides:

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

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

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

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 sides is equal to the rate of growth of its volume, solve for the corresponding length of the tetrahedron's sides:

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

$s^2=2\sqrt{2}=\sqrt{8}$

$s=\sqrt[4]{8}$

The height of a tetrahedron is given by the equation:

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

$h=\frac{\sqrt{6}}{3}\sqrt[4]{8}=\frac{2\sqrt[4]{2}}{\sqrt{3}}$

Report an Error

Example Question #196 : How To Find Rate Of Change

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 surface is equal to 1.5 times the rate of growth of its volume?

Possible Answers:

$\frac{3}{4}$

$\frac{6}{5}$

$\frac{4}{3}$

Correct answer:

$\frac{4}{3}$

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 growth of the cube's surface is equal to 1.5 times the rate of growth of its volume:

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

$s=\frac{12}{9}=\frac{4}{3}$

Report an Error

Example Question #197 : How To Find Rate Of Change

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 twice the rate of growth of its diagonal?

Possible Answers:

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

$\frac{\sqrt{2}}{\sqrt[4]{3}}$

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

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

$\frac{1}{\sqrt[4]{3}}$

Correct answer:

$\frac{\sqrt{2}}{\sqrt[4]{3}}$

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

$d=s\sqrt{3}$

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

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

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

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 twice the rate of growth of its diagonal:

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

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

$s=\frac{\sqrt{2}}{\sqrt[4]{3}}$

Report an Error

Example Question #198 : How To Find Rate Of Change

A cube is growing in size. What is the area of one of the cube's faces at the time that the rate of growth of the cube's volume is equal to twice the rate of growth of its sides?

Possible Answers:

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

$\frac{2}{3}$

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

Correct answer:

$\frac{2}{3}$

Explanation:

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

V=s^3

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

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

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 twice the rate of growth of its sides

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

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

The area of one of those cube's faces is the length of its sides squared:

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

Report an Error

Example Question #197 : How To Find Rate Of Change

A square is growing in area. What is the length of the square's sides at the moment that the rate of growth of the area is the same as the rate of growth of the square's diagonal?

Possible Answers:

$\frac{1}{2}$

$\sqrt{2}$

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

Correct answer:

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

Explanation:

Start by writing the equations for a square's dimensions.

A=s^2

$d=s\sqrt{2}$

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

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

$\frac{dd}{dt}=\sqrt{2}\frac{ds}{dt}$

Now, solve for the length of the side of the square to satisfy the problem condition, the rate of growth of the area is the same as the rate of growth of the square's diagonal:

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

$s=\frac{\sqrt{2}}{2}$

Report an Error

Example Question #198 : How To Find Rate Of Change

Use normal rules of differentiation to find the derrivative f'(0) of the function

$f(x)=2^{5x}+(2x^2+3)^3$

Possible Answers:

$5\ln 2$

$2 \ln 5$

$-5 \ln 2$

$-2 \ln 5$

Correct answer:

$5\ln 2$

Explanation:

Taking the derrivative,

$f'(x)=\frac{d}{dx}[2^{5x}+(2x^2+3)^3]$

We separate the summed terms:

$=\frac{d}{dx}[2^{5x}]+\frac{d}{dx}[(2x^2+3)^3]$

Using the chain rule:

$=2^{5x}\ln 2\frac{d}{dx}[5x]+2(2x^2+3)^2\frac{d}{dx}[(2x^2+3)]$

$=2^{5x}\ln 2*5+3(2x^2+3)^2* 4x$

Simplifying

$=2^{5x}5\ln 2+12x(2x^2+3)^2$

Now, evaluating at f'(0) :

$=2^{5(0)}5\ln 2+12(0)(2(0)^2+3)^2$

$=(1)5 \ln 2+(0)$

$=5 \ln 2$

which is our answer.

Report an Error

View Calculus Tutors

Jack
Certified Tutor

University of Washington-Seattle Campus, Bachelor in Arts, Mathematics. Western Washington University, Master of Science, Mat...

View Calculus Tutors

Celestine
Certified Tutor

University of Ghana, Bachelor of Science, Agricultural Economics. University of Minnesota-Twin Cities, Doctor of Philosophy, ...

View Calculus Tutors

Susana I
Certified Tutor

California State Polytechnic University-Pomona, Bachelor of Science, Geology. The University of Texas at El Paso, Master of S...

All Calculus 1 Resources

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

Popular Subjects

GMAT Tutors in Denver, Computer Science Tutors in Boston, GRE Tutors in Denver, SSAT Tutors in Chicago, Spanish Tutors in Atlanta, LSAT Tutors in Miami, SSAT Tutors in San Francisco-Bay Area, SSAT Tutors in Los Angeles, MCAT Tutors in Atlanta, ISEE Tutors in Miami

Popular Courses & Classes

ISEE Courses & Classes in Los Angeles, GMAT Courses & Classes in Washington DC, SSAT Courses & Classes in Philadelphia, ACT Courses & Classes in Dallas Fort Worth, ACT Courses & Classes in Seattle, LSAT Courses & Classes in Miami, MCAT Courses & Classes in Denver, SAT Courses & Classes in Washington DC, LSAT Courses & Classes in San Diego, SAT Courses & Classes in Denver

Popular Test Prep

GRE Test Prep in New York City, SAT Test Prep in Houston, ACT Test Prep in Denver, SSAT Test Prep in Chicago, MCAT Test Prep in Miami, GMAT Test Prep in Washington DC, ISEE Test Prep in New York City, ISEE Test Prep in Los Angeles, SAT Test Prep in Seattle, MCAT Test Prep in San Francisco-Bay Area

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 : How to find rate of change

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #191 : How To Find Rate Of Change

Example Question #192 : How To Find Rate Of Change

Example Question #193 : How To Find Rate Of Change

Example Question #194 : How To Find Rate Of Change

Example Question #195 : How To Find Rate Of Change

Example Question #196 : How To Find Rate Of Change

Example Question #197 : How To Find Rate Of Change

Example Question #198 : How To Find Rate Of Change

Example Question #197 : How To Find Rate Of Change

Example Question #198 : How To Find 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: