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 #1925 : Functions

Varsity tutors clock

The seconds hand in the clock shown has a length of five inches. Find the velocity of its tip in the and directions at the depicted moment in time.

Possible Answers:

$x:\frac{\pi}{3}sin(\frac{5}{6}\pi)\frac{in}{s}$

$y:\frac{\pi}{3}cos(\frac{5}{6}\pi)\frac{in}{s}$

$x:\frac{\pi}{3}cos(\frac{5}{6}\pi)\frac{in}{s}$

$y:-\frac{\pi}{3}sin(\frac{5}{6}\pi)\frac{in}{s}$

$x:\frac{\pi}{3}sin(\frac{5}{6}\pi) \frac{in}{s}$

$y:\frac{-\pi}{3}cos(\frac{5}{6}\pi)\frac{in}{s}$

$x:\frac{-\pi}{3}cos(\frac{5}{6}\pi)\frac{in}{s}$

$y:\frac{\pi}{3}sin(\frac{5}{6}\pi)\frac{in}{s}$

$x:-\frac{\pi}{3}sin(\frac{5}{6}\pi)\frac{in}{s}$

$y:\frac{\pi}{3}cos(\frac{5}{6}\pi)\frac{in}{s}$

Correct answer:

$x:\frac{\pi}{3}sin(\frac{5}{6}\pi) \frac{in}{s}$

$y:\frac{-\pi}{3}cos(\frac{5}{6}\pi)\frac{in}{s}$

Explanation:

For this problem, it will be useful to work in radians. The angle of the secondshand, between the positive x-axis and itself, is given as:

$\theta=\frac{5}{12}(2\pi)=\frac{5}{6}\pi$

Varsity tutors clock answre

Now, to find the x and y velocities, first write out the form of their positions:

$x=rcos(\theta)$

$y=rsin(\theta)$

Therefore, their respective velocities are given by their respective derivatives:

$\frac{dx}{dt}=\frac{d}{dt}(rcos(\theta))=\frac{dr}{dt}cos(\theta)-r\frac{d\theta}{dt}sin(\theta)$

$\frac{dy}{dt}=\frac{d}{dt}(rsin(\theta))=\frac{dr}{dt}sin(\theta)+r\frac{d\theta}{dt}cos(\theta)$

The secondshand doesn't change in length, so $\frac{dr}{dt}=0$ . As for $\frac{d\theta}{dt}$ , that can be found knowing how long it takes the secondshand to complete the circuit around the clock:

$\frac{d\theta}{dt}=-\frac{2\pi}{180}\frac{360}{60}\frac{rad}{s}=-\frac{\pi}{15}\frac{rad}{s}$

Note that movement in the clockwise direction, when dealing with angular motion, is treated as negative.

Knowing this, we can find the velocities:

$\frac{dx}{dt}=-r\frac{d\theta}{dt}sin(\theta)=\frac{\pi}{3}sin(\frac{5}{6}\pi)$

$\frac{dy}{dt}=r\frac{d\theta}{dt}cos(\theta)=\frac{-\pi}{3}cos(\frac{5}{6}\pi)$

Report an Error

Example Question #41 : How To Find Rate Of Change

A metal cylinder with a radius of five inches and a height of twenty inches is being heated, causing it to expand. If the radius grows at a rate of $0.001\frac{in}{s}$ and the height grows at a rate of $0.002\frac{in}{s}$ , what is the rate of expansion of the cylinder's volume?

Possible Answers:

$0.125\frac{in^3}{s}$

$0.5\pi \frac{in^3}{s}$

$0.25\pi \frac{in^3}{s}$

$0.625\frac{in^3}{s}$

$0.425\pi \frac{in^3}{s}$

Correct answer:

$0.25\pi \frac{in^3}{s}$

Explanation:

The volume of a cylinder is given by the formula:

$V=\pi r^2h$ wherein represents the cylinder's radius, and its height.

Therefore, to find the rate of expansion, derive each side of the equation with respect to time:

$\frac{dV}{dt}=\frac{d}{dt}(\pi r^2h)=2\pi r h\frac{dr}{dt}+\pi r^2 \frac{dh}{dt}$

Therefore, for the values given in the problem statement, the rate of expansion of the volume can be found:

$\frac{dV}{dt}=2\pi r h\frac{dr}{dt}+\pi r^2 \frac{dh}{dt}=2\pi(5in)(20in)\left(0.001\frac{in}{s}\right)+\pi(5in)^2\left(0.002\frac{in}{s}\right)$

$\frac{dV}{dt}=0.25\pi \frac{in^3}{s}$

Report an Error

Example Question #1925 : Functions

The sides of a right triangle are increasing in length. If the shorter side, which has a length of 3in , increases at a rate of $0.3\frac{in}{s}$ , and the longer side, which has a length of 4in , increases at a rate of $0.2\frac{in}{s}$ , what is the rate of growth of the hypotenuse?

Possible Answers:

$0.34\frac{in}{s}$

$1.7\frac{in}{s}$

$\sqrt{0.13}\frac{in}{s}$

$0.68\frac{in}{s}$

$\sqrt{0.65}\frac{in}{s}$

Correct answer:

$0.34\frac{in}{s}$

Explanation:

Since this is a right triangle, the hypotenuse can be related to the sides utilizing the pythagorean theorem:

c^2=a^2+b^2

Although the root of each side may be taken, it would be simpler to take the derivative of each side with respect to time as is:

$2c\frac{dc}{dt}=2a\frac{da}{dt}+2b\frac{db}{dt}$

Therefore, the rate of change of the hypotenuse is:

$\frac{dc}{dt}=\frac{a\frac{da}{dt}+b\frac{db}{dt}}{c}$

Using the Pythagorean Theorem, the hypotenuse is:

$c=\sqrt{3^2+4^2}=5$

Thus:

$\frac{dc}{dt}=\frac{3in(0.3\frac{in}{s})+4in(0.2\frac{in}{s})}{5in}=0.34\frac{in}{s}$

Report an Error

Example Question #1 : Asymptotic And Unbounded Behavior

A cylinder of height 4in and radius 1in is expanding. The radius increases at a rate of $0.1\frac{in}{s}$ and its height increases at a rate of $0.2\frac{in}{s}$ . What is the rate of growth of its surface area?

Possible Answers:

$1.6\pi\frac{in^2}{s}$

$0.4\pi\frac{in^2}{s}$

$1.4\pi\frac{in^2}{s}$

$0.2\pi\frac{in^2}{s}$

$2.8\pi\frac{in^2}{s}$

Correct answer:

$1.6\pi\frac{in^2}{s}$

Explanation:

The surface area of a cylinder is given by the formula:

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

To find the rate of growth over time, take the derivative of each side with respect to time:

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

Therefore, the rate of growth of surface area is:

$\frac{dA}{dt}=4\pi (1in)\left(0.1\frac{in}{s}\right)+2\pi(1in)(4in)\left(0.1\frac{in}{s}\right)+2\pi(1in)\left(0.2\frac{in}{s}\right)$

$\frac{dA}{dt}=(0.4\pi+0.8\pi+0.4\pi)\frac{in^2}{s}=1.6\pi\frac{in^2}{s}$

Report an Error

Example Question #44 : How To Find Rate Of Change

Water is being poured into a cylindrical glass at a rate of $4\frac{in^3}{s}$ . If the cylinder has a diameter of 4in and a height of 8in , what is the rate at which the water rises?

Possible Answers:

$\frac{1}{2\pi}\frac{in}{s}$

$\frac{2}{\pi}\frac{in}{s}$

$\frac{1}{\pi}\frac{in}{s}$

$2\pi\frac{in}{s}$

$\pi\frac{in}{s}$

Correct answer:

$\frac{1}{\pi}\frac{in}{s}$

Explanation:

The volume equation of a cylinder is

$V=\pi r ^2h$

Therefore, the rate of change of each term can be related by taking the derivative of each side of the equation:

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

Treating the glass as solid, the radius of liquid in it will not change, only the height, so $\frac{dr}{dt}=0$ . This simplifies the equation:

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

$\frac{dh}{dt}= \frac{1}{\pi r^2}\frac{dV}{dt}$

Since the radius is half the diameter:

$\frac{dh}{dt}=\frac{1}{\pi}\frac{in}{s}$

Report an Error

Example Question #45 : How To Find Rate Of Change

The position in meters of a particle after seconds is modeled by the equation

$f(t)=2te^{t}$ , where $t\ge0$ .

At what rate, in meters per second, is the position of the particle changing at t = 3 seconds?

Possible Answers:

$8e^{3}$

$2e^{3}$

$2+e^{3}$

$6e^{3}$

Correct answer:

$8e^{3}$

Explanation:

The derivative of a function gives us a new function that describes the rate of change of the original function at every point in its domain. Therefore the derivative of f(t) will produce a new function that describes the rate of change in position with respect to time. Using the product rule, the derivative of f(t) is:

$f'(t)=2e^{t}+2te^{t}$

Next substitute 3 for t to find the rate of change of the particle at exactly 3 seconds.

$f'(3)=2e^{3}+2(3)e^{3} = 8e^{3}$ .

Report an Error

Example Question #46 : How To Find Rate Of Change

What is the rate of change of the function

$y=\frac{x^2}{sin(x)}$ when $x=\frac{\pi}{6}$ ?

Possible Answers:

$\frac{12\pi-\pi^2\sqrt3}{18}$

$\frac{2\pi\sqrt3}{9}$

$\frac{\pi^2-\pi\sqrt2}{2}$

$\frac{\pi\sqrt2}{2}-\frac{\pi^2\sqrt2}{16}$

Correct answer:

$\frac{12\pi-\pi^2\sqrt3}{18}$

Explanation:

The find the rate of change of a function at a particular point, find the derivative of the function, then substitute the point into the function. Using the quotient rule, the derivative is:

$y'=\frac{2xsin(x)-x^2cos(x)}{sin^2(x)}$

Next substitute $\frac{\pi}{6}$ for x.

$y'(\frac{\pi}{6})=\frac{2(\frac{\pi}{6})sin(\frac{\pi}{6})-(\frac{\pi}{6})^2cos(\frac{\pi}{6})}{sin^2(\frac{\pi}{6})} =\frac{(\frac{\pi}{3})(\frac{1}{2})-(\frac{\pi^2}{36})\frac{\sqrt3}{2}}{(\frac{1}{2})^2}$

Now we need to simplify the right hand side.

$\frac{(\frac{\pi}{3})(\frac{1}{2})-(\frac{\pi^2}{36})\frac{\sqrt3}{2}}{(\frac{1}{2})^2} =\frac{(\frac{\pi}{3})(\frac{1}{2})-(\frac{\pi^2}{36})\frac{\sqrt3}{2}}{\frac{1}{4}}$

$=4(\frac{\pi}{3})(\frac{1}{2})-4(\frac{\pi^2}{36})\frac{\sqrt3}{2}$

$=\frac{2\pi}{3}-(\frac{\pi^2}{9})\frac{\sqrt3}{2}$

$=\frac{2\pi}{3}-\frac{\pi^2\sqrt3}{18}$

$=\frac{12\pi-\pi^2\sqrt3}{18}$

Report an Error

Example Question #1931 : Functions

What is the rate of change of a square's sides if it has an area of 144in^2 which is growing at a rate of $1\frac{in^2}{s}$ ?

Possible Answers:

$\frac{1}{24}\frac{in}{s}$

$\frac{1}{6}\frac{in}{s}$

$24\frac{in}{s}$

$12\frac{in}{s}$

$\frac{1}{12}\frac{in}{s}$

Correct answer:

$\frac{1}{24}\frac{in}{s}$

Explanation:

The area of a square in terms of the lengths of its sides is given as:

A=l^2

Therefore the length of each side can be found to be

l^2=144in^2

l=12in

Rates of change can also be related by taking the time derivative of each side:

$\frac{dA}{dt}=2l\frac{dl}{dt}$

The rate of change of the sides, $\frac{dl}{dt}$ , is what is being solved for:

$\frac{dl}{dt}=\frac{1}{2l}\frac{dA}{dt}=\frac{1}{2(12in)}(1\frac{in^2}{s})=\frac{1}{24}\frac{in}{s}$

Report an Error

Example Question #131 : Rate

Find the derivative of the following function through implicit differentiation:

3y^2+2y=x^2+2sin(x)

Possible Answers:

x+cos(x)

$\frac{x+cos(x)}{3y+1}$

{3y+1}

$\frac{-x+cos(x)}{3y+1}$

$\frac{x+cos(x)}{3y+1}$

Correct answer:

$\frac{x+cos(x)}{3y+1}$

Explanation:

3y^2+2y=x^2+2sin(x)

By differentiating both sides:

6ydy+2dy=2xdx+2cos(x)dx

By solving for $\frac{dy}{dx}$ :

$\frac{dy}{dx}=\frac{x+cos(x))}{3y+1}$

Report an Error

Example Question #49 : How To Find Rate Of Change

Air flows out of a spherical balloon at a rate of -12 units^3/min . What is the rate of change of the circumference when the radius of the balloon is units?

Possible Answers:

$\frac{6}{25} units/min$

$\frac{-3}{25\pi} units/min$

$\frac{-6}{25} units/min$

$\frac{3}{25\pi} units/min$

$0\, units/min$

Correct answer:

$\frac{-6}{25} units/min$

Explanation:

It is given that V'=-12 units^3/min .

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

By differentiating both sides with respect to time we can solve this problem:

$V'=4\pi r^2r'$

$r'=\frac{V'}{4\pi r^2}=\frac{-12}{4\pi 5^2} units/min=\frac{-3}{25\pi} units/min$

$C'=2\pi r'=\frac{-2\pi \cdot 3}{25 \pi} unit/min=\frac{-6}{25} units/min$

Report an Error

View Calculus Tutors

Dossevi
Certified Tutor

cole Ste Genevive Versailles, Bachelor of Science, Mathematics. Institut National Polytechnique de Grenoble, Master of Scienc...

View Calculus Tutors

Travis
Certified Tutor

Delaware State University, Bachelor of Science, Mathematics.

View Calculus Tutors

Shiblu
Certified Tutor

University of Central Florida, Doctor of Science, Civil Engineering.

All Calculus 1 Resources

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

Popular Subjects

French Tutors in Dallas Fort Worth, GRE Tutors in San Diego, GMAT Tutors in Phoenix, French Tutors in New York City, Spanish Tutors in Los Angeles, MCAT Tutors in Dallas Fort Worth, Computer Science Tutors in Philadelphia, Statistics Tutors in Houston, English Tutors in Miami, Math Tutors in Washington DC

Popular Courses & Classes

GRE Courses & Classes in Houston, LSAT Courses & Classes in Philadelphia, LSAT Courses & Classes in Boston, LSAT Courses & Classes in New York City, ACT Courses & Classes in Seattle, SAT Courses & Classes in Houston, Spanish Courses & Classes in Dallas Fort Worth, SSAT Courses & Classes in Dallas Fort Worth, ISEE Courses & Classes in Seattle, MCAT Courses & Classes in New York City

Popular Test Prep

GRE Test Prep in Washington DC, GRE Test Prep in Phoenix, ISEE Test Prep in Seattle, ACT Test Prep in Philadelphia, GRE Test Prep in Chicago, SSAT Test Prep in Phoenix, LSAT Test Prep in Dallas Fort Worth, GMAT Test Prep in Boston, LSAT Test Prep in Philadelphia, SAT 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 : How to find rate of change

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #1925 : Functions

Example Question #41 : How To Find Rate Of Change

Example Question #1925 : Functions

Example Question #1 : Asymptotic And Unbounded Behavior

Example Question #44 : How To Find Rate Of Change

Example Question #45 : How To Find Rate Of Change

Example Question #46 : How To Find Rate Of Change

Example Question #1931 : Functions

Example Question #131 : Rate

Example Question #49 : 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: