Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 1 : 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 #1955 : Functions

The circumference of a circle is increasing at a rate of $0.5\pi\frac{in}{s}$ . If the circle has an area of $9\pi in^2$ , what is the rate of growth of the area?

Possible Answers:

$3\frac{in^2}{s}$

$3\pi^2\frac{in^2}{s}$

$2.25\pi\frac{in^2}{s}$

$1.5\pi\frac{in^2}{s}$

$1.5\frac{in^2}{s}$

Correct answer:

$1.5\pi\frac{in^2}{s}$

Explanation:

Begin by finding the radius of the circle:

$A=\pi r^2$

$r=\sqrt{\frac{A}{\pi}}$

$r=\sqrt{\frac{9\pi in^2}{\pi}}$

r=3in

Now, rates of change can be related by taking the time derivative of both sides of an equation:

$\frac{dA}{dt}=2\pi r \frac{dr}{dt}$

However, the rate of change of the radius, $\frac{dr}{dt}$ , is still unknown. It can be found by relating it to the circumference's rate of change:

$C=2\pi r$

$\frac{dC}{dt}=2\pi\frac{dr}{dt}$

$\frac{dr}{dt}=\frac{1}{2\pi}0.5\pi\frac{in}{s}$

$\frac{dr}{dt}=0.25\frac{in}{s}$

Going back to our earlier equation, we can solve for the rate of change of the area:

$\frac{dA}{dt}=2 \pi (3in)(0.25\frac{in}{s})$

$\frac{dA}{dt}=1.5\pi\frac{in^2}{s}$

Report an Error

Example Question #161 : Rate

The radius of a cylinder is given by the function 3+t , while the height at any time is given by the function 12-2t . What is the rate of growth of the cylinder at time t=1 ?

Possible Answers:

$12\pi$

$160\pi$

$-2\pi$

$48\pi$

$108\pi$

Correct answer:

$48\pi$

Explanation:

The volume of a cylinder is given by the equation:

$V=\pi r^2 h$

$V(t)=\pi (3+t)^2(12-2t)$

$V(t)=\pi(t^2+6t+9)(12-2t)$

$V(t)=\pi (108+54t -2t^3)$

From this, the rate of change can be found by taking the derivative with respect to time:

$\frac{dV(t)}{dt}=\pi(54-6t^2)$

$\frac{dV(1)}{dt}=\pi(54-6(1)^2)$

$\frac{dV(1)}{dt}=48\pi$

It may be worth noting that at time t=3 the cylinder actually begins to shrink.

Report an Error

Example Question #1956 : Functions

If $h(x)=\sin(5x^3)$ , what is h'(x) ?

Possible Answers:

$-5x^3\cos(5x^3)$

$15x^2\cos(5x^3)$

$\cos(15x^2)$

$\cos(5x^3)$

Correct answer:

$15x^2\cos(5x^3)$

Explanation:

The given function consists of a function, g(x)=5x^3 , inside another function, $f(x)=\sin(x)$ , such that h(x)=sin(5x^3)=f(g(x)) .

Thus we can use the Chain Rule to find h'(x) .

The Chain Rule says that if

h(x)=f(g(x)) ,

then

$h'(x)=f'(g(x))\cdot g'(x)$ .

Recall (or look up) that the derivative of sine is cosine, so $f'(x)=\cos(x)$ , and use the Power Rule to get g'(x)=15x^2 .

Combining the three functions f'(x) , g(x) , and g'(x) , we have $h'(x)=15x^2\cos(5x^3)$ .

Note: The Power Rule says that for a function

p(x)=kx^n , $p'(x)=knx^{n-1}$ .

Report an Error

Example Question #1957 : Functions

What is the rate of change of the function $g(x)=4x^{\frac{3}{2}}$ when x=25 ?

Possible Answers:

500

750

Correct answer:

Explanation:

The rate of change of the function $g(x)=4x^{\frac{3}{2}}$ at is the value of the derivative

$g'(x)=\frac{dg(x)}{dx}$ at x=25 .

Use the Power Rule to find that

$g'(x)=4\cdot\frac{3}{2}\cdot x^{\frac{3}{2}-1}=6x^{\frac{1}{2}}=6\sqrt{x}$ .

The rate of change at is

$g'(25)=6\sqrt{25}=30$ .

Note: The Power Rule says that for a function

p(x)=kx^n , $p'(x)=knx^{n-1}$ .

Report an Error

Example Question #1958 : Functions

What is $\frac{dy}{dx}$ if $y(x)=\frac{\sin x}{e^{x^2}}$ ?

Possible Answers:

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

$\frac{\sin x-2x\cos x}{4x^2e^{x^2}}$

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

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

Correct answer:

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

Explanation:

Since y(x) is a quotient of two functions $\sin x$ and $e^{x^2}$ , we can use the Quotient Rule, which says that for a function

$h(x)=\frac{f(x)}{g(x)}$ ,

$h'(x)=\frac{g(x)f'(x)-f(x)g'(x)}{(g(x))^2}$ .

$\frac{d}{dx}(\sin x)=\cos x$ and $\frac{d}{dx}(e^{x^2})=2xe^{x^2}$ by the Chain Rule.

Applying the Quotient Rule,

$\\ \frac{dy}{dx}=\frac{e^{x^2}\cdot \frac{d}{dx}(\sin x)-\sin x\cdot \frac{d}{dx}(e^{x^2})}{(e^{x^2})^2}\\ \\=\frac{e^{x^2}\cdot \cos x-\sin x\cdot 2xe^{x^2}}{e^{2x^2}}\\ \\=\frac{e^{x^2}(\cos x-2x\sin x)}{e^{2x^2}}\\ \\=\frac{\cos x-2x\sin x}{e^{x^2}}$ .

Report an Error

Example Question #1959 : Functions

What is the rate of change of $f(x)=\sin^3 x$ when $x=\tfrac{\pi}{4}$ ?

Possible Answers:

$\tfrac{3\sqrt{2}}{2}$

$\tfrac{\sqrt{2}}{2}$

$\tfrac{3\sqrt{2}}{4}$

$\tfrac{\sqrt{2}}{4}$

Correct answer:

$\tfrac{3\sqrt{2}}{4}$

Explanation:

We are looking for $f'(\tfrac{\pi}{4})$ .

The Chain Rule says that for

f(x)=g(h(x)) , f'(x)=g'(h(x))h'(x) .

Applying the Chain Rule,

$f'(x)=\cos x\cdot3(\sin x)^2=3\cos x \sin^2x$ .

So,

$f'(\tfrac{\pi}{4})=3\cos (\tfrac{\pi}{4}) \sin^2 (\tfrac{\pi}{4})=3(\tfrac{\sqrt{2}}{2})(\tfrac{\sqrt2}{2})^2=\tfrac{3\sqrt{2}}{4}$ .

Report an Error

Example Question #71 : How To Find Rate Of Change

A spherical water balloon is filled at a rate of $0.2 \frac{m^3}{s}$ . What is the rate of change of the surface area of the water balloon when the balloon's radius is one meter?

Possible Answers:

$\frac{6}{5}\frac{m^2}{s}$

$\frac{4}{5}\frac{m^2}{s}$

$\frac{1}{20\pi} \frac{m^2}{s}$

$\frac{2}{5}\frac{m^2}{s}$

Correct answer:

$\frac{2}{5}\frac{m^2}{s}$

Explanation:

The rate of change of the surface area is found by relating the volume to the surface area:

$V=\frac{4}{3} \pi r^3$ , $\frac{dV}{dt}=4\pi r^2 \frac{dr}{dt}$

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

We must solve for the rate of change of the radius using the volume equation, and then solve for the rate of change of the surface area by plugging in the given radius and rate of change of the radius:

$\frac{dr}{dt}=\frac{1}{20\pi}=\frac{0.2}{4\pi (1)}$

$\frac{dA}{dt}=8\pi (1)\left(\frac{1}{20 \pi}\right)=\frac{2}{5}$

Report an Error

Example Question #1961 : Functions

The legs of a right triangle are each growing at a rate of $0.6\frac{in}{s}$ . If one leg has a length of five inches and the other a length of twelve inches, what is the rate of growth of hypotenuse?

Possible Answers:

$0.3\frac{in}{s}$

$0.47\frac{in}{s}$

$0.79\frac{in}{s}$

$0.6\frac{in}{s}$

$0.85\frac{in}{s}$

Correct answer:

$0.79\frac{in}{s}$

Explanation:

Begin by finding the hypotenuse using the Pythagorean theorem:

c^2=a^2+b^2

c^2=5^2in^2+12^2in^2

c^2=169in^2

c=13in

Now, to relate the rates of growth, take the time derivative of the Pythagorean Theorem:

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

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

$\frac{dc}{dt}=\frac{5in(0.6\frac{in}{s})+12in(0.6\frac{in}{s})}{13in}$

$\frac{dc}{dt}=0.79\frac{in}{s}$

Report an Error

Example Question #1962 : Functions

The radius of a circle with a square inscribed inside of it (such that each corner touches the edge of the circle) increases at a rate of $0.7\frac{in}{s}$ . What is the rate of growth of the difference in areas of the circle and square if the radius of the circle is $5\sqrt2$ ?

Possible Answers:

$7\sqrt2(2-\pi)\frac{in^2}{s}$

$7\sqrt2(\pi+2)\frac{in^2}{s}$

$14\sqrt2\frac{in^2}{s}$

$\pi7\sqrt2\frac{in^2}{s}$

$7\sqrt2(\pi-2)\frac{in^2}{s}$

Correct answer:

$7\sqrt2(\pi-2)\frac{in^2}{s}$

Explanation:

The areas of the circle and square can be written in terms of the radius of the circle as follows:

$A_{circle}=\pi r^2 = 50\pi in^2$

$A_{square}=(\frac{2r}{\sqrt2})^2=2r^2=50in^2$

The area of the square follows from the fact that the diameter of the circle forms the diagonal of the square, and each side is equal to the diagonal divided by the square root of two. This can be verified via the Pythagorean Theorem (s^2+s^2=d^2) .

Now, the rate of change of areas can be found by taking the derivative of each function with respect to time:

$\frac{dA_{circle}}{dt}=2\pi r \frac{dr}{dt}$

$\frac{dA_{square}}{dt}=4r \frac{dr}{dt}$

Therefore, the rate of growth of the difference in the two areas can be found by finding the difference of these two quantities:

$2r(\pi-2)\frac{dr}{dt}$

$2(5\sqrt2in)(\pi-2)(0.7\frac{in}{s})$

$7\sqrt2(\pi-2)\frac{in^2}{s}$

Report an Error

Example Question #1963 : Functions

The sides of a square have lengths 5+t^2 . What is the rate of growth of the area at time t=2 ?

Possible Answers:

Correct answer:

Explanation:

The area of a square is the square of its sides:

A=s^2

A(t)=(5+t^2)^2

At time t=2

A(2)=(5+4)^2

A=81

Now to find the rate of change of the area as related to the rate of change of the sides, take the derivative of the area equation with respect to time:

$\frac{dA(t)}{dt}=2s\frac{ds}{dt}$

Since s=5+t^2 , we will use the power rule which states,

$x^n\ dx=nx^{n-1}$

therefore

$\frac{ds}{dt}=2t$ , giving:

$\frac{dA(t)}{dt}=2(5+t^2)(2t)$

$\frac{dA(2)}{dt}=2(5+2^2)(2(2))$

$\frac{dA(2)}{dt}=72$

Note that the same result would be found if you took the derivative of the equation wherein the 5+t^2 was substituted in for :

A(t)=(5+t^2)^2

Use the power rule and chain rule to find the derivative of area with respect to time.

Chain Rule: $f(g(x)) dx=f'(g(x))\cdot g'(x)$

$\frac{dA(t)}{dt}=2(2t)(5+t^2)$

Report an Error

View Calculus Tutors

Morgan
Certified Tutor

Catholic University of America, Bachelor of Science, International Finance.

View Calculus Tutors

John
Certified Tutor

The University of Texas at Arlington, Bachelor of Science, Mathematics. Stanford University, Master of Science, Aerospace Eng...

View Calculus Tutors

Muhammad
Certified Tutor

The University of Texas at Dallas, Bachelor of Science, Mechanical Engineering.

All Calculus 1 Resources

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

Popular Subjects

English Tutors in Dallas Fort Worth, French Tutors in Boston, GMAT Tutors in Philadelphia, GRE Tutors in San Diego, GMAT Tutors in Washington DC, French Tutors in Seattle, GRE Tutors in Phoenix, Computer Science Tutors in San Francisco-Bay Area, Calculus Tutors in Los Angeles, Spanish Tutors in Dallas Fort Worth

Popular Courses & Classes

LSAT Courses & Classes in Miami, LSAT Courses & Classes in Chicago, ACT Courses & Classes in Phoenix, ISEE Courses & Classes in Boston, GMAT Courses & Classes in New York City, ISEE Courses & Classes in Philadelphia, ACT Courses & Classes in Boston, ACT Courses & Classes in Atlanta, Spanish Courses & Classes in Washington DC, MCAT Courses & Classes in Philadelphia

Popular Test Prep

LSAT Test Prep in Philadelphia, SAT Test Prep in Philadelphia, ACT Test Prep in Atlanta, ISEE Test Prep in Phoenix, ISEE Test Prep in Los Angeles, ACT Test Prep in San Francisco-Bay Area, ISEE Test Prep in Philadelphia, SAT Test Prep in Boston, SAT Test Prep in Atlanta, LSAT Test Prep in San Diego

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 : Rate of Change

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #1955 : Functions

Example Question #161 : Rate

Example Question #1956 : Functions

Example Question #1957 : Functions

Example Question #1958 : Functions

Example Question #1959 : Functions

Example Question #71 : How To Find Rate Of Change

Example Question #1961 : Functions

Example Question #1962 : Functions

Example Question #1963 : Functions

All Calculus 1 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: