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 #201 : Rate Of Change

Use normal rules of differentiation to find the derrivative $f'(\frac{2 \pi}{3})$ of the function

$f(x)= \tan(\frac{1}{2}x^2)-e^{\cos x}$ .

Possible Answers:

$\approx 10.4568$

$\approx 6.6867$

$\approx 5.2456$

$\approx 4.3333$

$\approx -2.5783$

Correct answer:

$\approx 6.6867$

Explanation:

Taking the derivative

$f'(x)= \frac{d}{dx}[\tan(\frac{1}{2}x^2)-e^{\cos x}]$

Separating the summed terms:

$= \frac{d}{dx}[\tan(\frac{1}{2}x^2)]-\frac{d}{dx}[e^{\cos x}]$

Applying the chain rule:

$= \frac{d}{dx}[\tan(\frac{1}{2}x^2)]-\frac{d}{dx}[e^{\cos x}]$

$= \sec^2 (\frac{1}{2}x^2)\frac{d}{dx}[\frac{1}{2}x^2]-e^{cos x}\frac{d}{dx}[\cos x]$

$= \sec^2 (\frac{1}{2}x^2)*x-e^{cos x}(-\sin x)$

Siimplifying:

$= x\sec^2 (\frac{1}{2}x^2)+\sin x *e^{cos x}$

Evaluating at $x=\frac{2\pi}{3}$ :

$= (\frac{2\pi}{3})\sec^2 (\frac{1}{2}(\frac{2\pi}{3})^2)+\sin (\frac{2\pi}{3}) *e^{cos (\frac{2\pi}{3})}$

$= (\frac{2\pi}{3})\sec^2 (\frac{2\pi^2}{9})+\sin (\frac{2\pi}{3})*e^{cos (\frac{2\pi}{3})}$

$= (\frac{2\pi}{3})\sec^2 (\frac{2\pi^2}{9})+(\frac{\sqrt{3}}{2})*e^{cos (-\frac{1}{2})}$

$\approx 6.6867$

Which is our answer.

Report an Error

Example Question #201 : Rate Of Change

Use implicit differentiation to find the derrivative $\frac{dy}{dx}$ .

Find $\frac{dy}{dx}$ given the equation $\ln (y^2+1)=2xy$ .

Possible Answers:

$\frac{y^3+y}{y-xy^2-x}$

$\frac{1}{y^2+1}-2xy$

$\frac{(y+2)^2}{y+xy^2-xy}$

$\frac{x^3+x}{xy+xy^2-y^2}$

$2x-\frac{2y}{y^2+1}$

Correct answer:

$\frac{y^3+y}{y-xy^2-x}$

Explanation:

We take the derivative of both sides of the equation

$\frac{d}{dx}[\ln (y^2+1)]=\frac{d}{dx}[2xy]$

We apply the chain rule on the left side and the product rule on the right side:

$\frac{1}{(y^2+1)}*\frac{d}{dx}[y^2+1]=\frac{d}{dx}[2x]y+2x\frac{d}{dx}[y]$

$\frac{2y}{(y^2+1)}*\frac{dy}{dx}=2y+2x\frac{dy}{dx}$

$\frac{2y}{(y^2+1)}*\frac{dy}{dx}-2x\frac{dy}{dx}=2y$

$\frac{dy}{dx}[\frac{2y}{(y^2+1)}-2x]=2y$

$\frac{dy}{dx}=\frac{2y}{\frac{2y}{(y^2+1)}-2x}$

Simplifying:

$\frac{dy}{dx}=\frac{2y}{(\frac{2y-2x(y^2+1)}{(y^2+1)})}$

$\frac{dy}{dx}=\frac{2y(y^2+1)}{2y-2x(y^2+1)}$

$\frac{dy}{dx}=\frac{2y(y^2+1)}{2[y-x(y^2+1)]}$

$\frac{dy}{dx}=\frac{y(y^2+1)}{y-x(y^2+1)}$

$\frac{dy}{dx}=\frac{y^3+y}{y-xy^2-x}$

Which is our answer.

Report an Error

Example Question #202 : Rate Of Change

Use implicit differentiation to find the derrivative $\frac{dy}{dx}$ .

Find $\frac{dy}{dx}$ given (x-1)^2+(y+2)^2=36 .

Possible Answers:

$\frac{(x+1)^2}{(y-4)^2}$

$\frac{-2x^2+1}{2y^2+4}-36$

$-\frac{(x+1)^2}{(y-4)^2}$

$\frac{2x+1}{2y-1}$

$\frac{-x+1}{y+2}$

Correct answer:

$\frac{-x+1}{y+2}$

Explanation:

Taking the derivative of both sides of the equation

$\frac{d}{dx}[(x-1)^2+(y+2)^2]=\frac{d}{dx}[36]$

We apply normal rules of differentiation and solve for dy/dx

$2(x-1)+2(y+2)\frac{dy}{dx}=0$

$2(y+2)\frac{dy}{dx}=-2(x-1)$

$\frac{dy}{dx}=\frac{-2(x-1)}{2(y+2)}$

$\frac{dy}{dx}=\frac{-x+1}{y+2}$

Which is our answer.

Report an Error

Example Question #203 : Rate Of Change

Use normal rules of differentiation to find the derrivative $f'(\frac{5\pi}{6})$ of the function $f(x)= \ln(\cos^2(x))$

Possible Answers:

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

$\frac{1}{2}$

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

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

Correct answer:

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

Explanation:

Taking the derivative

$f'(x)=\frac{d}{dx} [\ln(\cos^2(x))]$

Applying the chain rule:

$=\frac{1}{\cos^2 x}*\frac{d}{dx}[\cos^2 x]$

$=\frac{1}{\cos^2 x}*2 \cos x * (- \sin x)$

Simplifying:

$=\frac{-2 \cos x \sin x}{\cos^2 x}$

$=\frac{-2 \sin x}{\cos x}$

$=-2 \tan x$

Evaluating at $x= \frac{5 \pi}{6}$

$=-2 \tan (\frac{5 \pi }{6})$

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

Which is our answer.

Report an Error

Example Question #204 : 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 seven times the rate of growth of the surface area?

Possible Answers:

$14\pi$

$28\pi$

$196\pi$

Correct answer:

$28\pi$

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 seven times the rate of growth of the surface area, let's solve for a radius that satisfies it.

$4\pi r^2 \frac{dr}{dt}=(7)8\pi r \frac{dr}{dt}$

r =(7)2

r=14

Circumference of a sphere is given by the function:

$C=2\pi r$

$C=28\pi$

Report an Error

Example Question #205 : Rate Of Change

A spherical balloon is deflating, while maintaining its spherical shape. What is the volume of the sphere at the instance the rate of shrinkage of the volume is twice the rate of shrinkage of the circumference?

Possible Answers:

$3\pi$

$\frac{1}{3}\pi$

$\frac{2}{3}\pi$

$4\pi$

$\frac{4}{3}\pi$

Correct answer:

$\frac{4}{3}\pi$

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. So given our problem conditions, the rate of shrinkage of the volume is twice the rate of shrinkage of the circumference, let's solve for a radius that satisfies it.

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

r^2 =1

r=1

Then to find the volume at this time:

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

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

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

Report an Error

Example Question #206 : Rate Of Change

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

Possible Answers:

$\frac{8}{7}$

$\frac{1}{7}$

$\frac{2}{7}$

$\frac{16}{7}$

$\frac{4}{7}$

Correct answer:

$\frac{4}{7}$

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 volume is a seventh of the rate of growth of the surface area, let's solve for a radius that satisfies it.

$4\pi r^2 \frac{dr}{dt}=(\frac{1}{7})8\pi r \frac{dr}{dt}$

$r =(\frac{1}{7})2$

$r=\frac{2}{7}$

The diameter is given by the formula:

D=2r

$D=\frac{4}{7}$

Report an Error

Example Question #207 : Rate Of Change

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

Possible Answers:

$\frac{1}{2}$

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

$2\sqrt{3}$

$\sqrt{3}$

Correct answer:

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

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

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 cube's volume is equal to an eigth of the rate of growth of its surface area:

$3s^2\frac{ds}{dt}=(\frac{1}{8})12s\frac{ds}{dt}$

$s=(\frac{1}{8})4$

$s=\frac{1}{2}$

The diagonal of a cube is given by the equation:

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

$d=s\sqrt{3}$

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

Report an Error

Example Question #208 : Rate Of Change

A cube is diminishing in size into nothingness. What is the area of one of the cube's sides at the time that the rate of shrinkage of the cube's volume is equal to thirteen times the rate of shrinkage of its surface area?

Possible Answers:

16224

$\frac{16}{169}$

2704

$\frac{96}{169}$

104

Correct answer:

2704

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 thirteen times the rate of shrinkage of its surface area:

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

s=(13)4

s=52

The area of one of a cube's is given by the equation:

a=s^2

a=52^2

a=2704

Report an Error

Example Question #209 : Rate Of Change

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

Possible Answers:

$96\sqrt{3}$

$48\sqrt{3}$

$384\sqrt{3}$

$84\sqrt{3}$

$192\sqrt{3}$

Correct answer:

$384\sqrt{3}$

Explanation:

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

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

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

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 twice the rate of growth of its surface area, solve for the corresponding length of the tetrahedron's sides:

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

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

$s=8\sqrt{6}$

The surface area of the tetrahedron:

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

$A=\sqrt{3}(8\sqrt{6})^2$

$A=384\sqrt{3}$

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

SAT Tutors in San Diego, Biology Tutors in Los Angeles, SSAT Tutors in Phoenix, MCAT Tutors in Chicago, Chemistry Tutors in Dallas Fort Worth, Physics Tutors in Atlanta, Computer Science Tutors in Washington DC, GRE Tutors in Houston, ACT Tutors in Chicago, LSAT Tutors in Los Angeles

Popular Courses & Classes

LSAT Courses & Classes in Washington DC, MCAT Courses & Classes in Philadelphia, ISEE Courses & Classes in Chicago, GMAT Courses & Classes in San Francisco-Bay Area, GRE Courses & Classes in Seattle, Spanish Courses & Classes in New York City, SSAT Courses & Classes in Washington DC, GRE Courses & Classes in Atlanta, Spanish Courses & Classes in Los Angeles, LSAT Courses & Classes in Dallas Fort Worth

Popular Test Prep

SSAT Test Prep in Dallas Fort Worth, MCAT Test Prep in Houston, GMAT Test Prep in Chicago, SAT Test Prep in Dallas Fort Worth, ISEE Test Prep in San Diego, ISEE Test Prep in Chicago, GMAT Test Prep in San Francisco-Bay Area, SAT Test Prep in Los Angeles, LSAT Test Prep in Dallas Fort Worth, MCAT Test Prep in Philadelphia

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 #201 : Rate Of Change

Example Question #201 : Rate Of Change

Example Question #202 : Rate Of Change

Example Question #203 : Rate Of Change

Example Question #204 : Rate Of Change

Example Question #205 : Rate Of Change

Example Question #206 : Rate Of Change

Example Question #207 : Rate Of Change

Example Question #208 : Rate Of Change

Example Question #209 : 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: