Call Now to Set Up Tutoring:

(888) 888-0446

AP Calculus AB : Derivatives

Study concepts, example questions & explanations for AP Calculus AB

Create An Account

All AP Calculus AB Resources

3 Diagnostic Tests 164 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #4 : Derivative Interpreted As An Instantaneous Rate Of Change

Find the function values and as well as the instantaneous rate of change for the function $g(x)=x\sin(x)$ corresponding to the following values of

$1)\: \: \: \: \: \: \: x =\frac{\pi}{2}$

$2)\: \: \: \: \: \: \: x =\frac{\pi}{4}$

$3)\: \: \: \: \: \: \: x =\pi$

Possible Answers:

B2wrng4

Fixedcorrectanswer

B2wrng3

B2wrng4

Correct answer:

Fixedcorrectanswer

Explanation:

Find the instantaneous rate of change for the function $g(x)=x\sin(x)$ corresponding to the following values of

$1)\: \: \: \: \: \: \: x =\frac{\pi}{2}$

$2)\: \: \: \: \: \: \: x =\frac{\pi}{4}$

$3)\: \: \: \: \: \: \: x =\pi$

Evaluate the function at each value of

$g\left(\frac{\pi}{2} \right )=\frac{\pi}{2}sin\left(\frac{\pi}{2} \right )=\frac{\pi}{2}$

$g\left(\frac{\pi}{4} \right )=\frac{\pi}{4}\sin\left(\frac{\pi}{4} \right )=\frac{\pi}{4}\frac{\sqrt{2}}{2}=\frac{\pi\sqrt{2}}{8}$

$g(\pi)=\pi \sin(\pi)=0$

The instantaneous rate of change at any point (x, g(x)) will be given by the derivative at that point. First compute the derivative of the function:

$g(x)=x\sin(x)$

Apply the product rule:

$g'(x)=x\frac{d}{dx}\sin(x)+\sin(x)\frac{d}{dx}(x)$

$= x\cos(x)+\sin(x)$

Therefore,

$g'(x)=x\cos(x)+\sin(x)$

Now evaluate the derivative for each given value of :

$1)\: \: \: \: \: \: \: x =\frac{\pi}{2}\: \: \: \:\rightarrow\: \: \: \:g'\left(\frac{\pi}{2} \right )\: \: =\: \: \frac{\pi}{2}\cos\left(\frac{\pi}{2} \right )+\sin\left(\frac{\pi}{2} \right )\: \:=\:1$

$2)\: \: \: \: \: \: \: x =\frac{\pi}{4}\: \: \: \:\rightarrow\: \: \: \:g'\left(\frac{\pi}{4} \right )\: \: =\: \: \frac{\pi}{4}\cos\left(\frac{\pi}{4} \right )+\sin\left(\frac{\pi}{4} \right )\: \:=\:\frac{\sqrt{2}}{2}\left(1+\frac{\pi}{4} \right )=\frac{\sqrt{2}}{8}(4+\pi)$

$3)\: \: \: \: \: \: \: x =\pi\: \: \: \:\rightarrow\: \: \: \:g'(\pi)\: \: =\: \: \pi\cos(\pi) +\sin(\pi) \: \:=\:\pi$

Therefore, the instantaneous rate of change of the function g(x) at the corresponding values of are:

Fixedcorrectanswer

Report an Error

Example Question #5 : Derivative Interpreted As An Instantaneous Rate Of Change

Given that v(t) is the velocity of a particle, find the particle's acceleration when t=3.

v(t)=3t^2+6e^x-15t

Possible Answers:

123.5

12.65

Not enough information provided

Correct answer:

123.5

Explanation:

Given that v(t) is the velocity of a particle, find the particle's acceleration when t=3.

v(t)=3t^2+6e^x-15t

We are given velocity and asked to find acceleration. Our first step should be to find the derivative.

v(t)=3t^2+6e^x-15t

We can use our standard power rule for our 1st and 3rd terms, but we need to remember something else for our second term. Namely, that the derivative of e^x is simply e^x

With that in mind, let;s find v'(t)

v(t)=3t^2+6e^x-15t

v'(t)=6t+6e^x-15

Now, for the final push, we need to find the acceleration when t=3. We do this by plugging in 3 for t and simplifying.

v'(t)=6(3)+6e^3-15=18+120.5-15=123.5

So, our answer is 123.5

Report an Error

Example Question #6 : Derivative Interpreted As An Instantaneous Rate Of Change

$\begin{align*}&\text{The position of a particle at a given time is given by the function:}\\&p(t)=6t - sin(\frac{(5\cdot \pi t)}{2}) + 5\\&\text{What is its velocity at time }t=6\text{?}\end{align*}$

Possible Answers:

6.83

13.85

Correct answer:

13.85

Explanation:

$\begin{align*}&\text{Velocity is the time rate of change of position. In lay terms,}\\&\text{it is an instantaneous measure of speed and direction. However,}\\&\text{going back to that first definition, note that it is a rate,}\\&\text{and whenever the term rate is used, derivatives should come}\\&\text{to mind. The rate of change of position with respect to time}\\&\text{is the derivative of position with respect to time:}\\&v(t)=\frac{dp(t)}{dt}\\&\text{Taking the derivative of our position function }6t - sin(\frac{(5\cdot \pi t)}{2}) + 5\\&\text{We find the velocity function: }6 -\frac{ (5\cdot \pi cos(\frac{(5\cdot \pi t)}{2}))}{2}\\&\text{Plugging in values we then get:}\\&v(6)=6 -\frac{ (5\cdot \pi cos(\frac{(5\cdot \pi (6))}{2}))}{2}\\&v(6)=13.85\end{align*}$

Report an Error

Example Question #7 : Derivative Interpreted As An Instantaneous Rate Of Change

$\begin{align*}&\text{The position of a particle at a given time is given by the function:}\\&p(t)=- 3t - cos(5\cdot \pi t) - sin(\frac{(7\cdot \pi t)}{2}) - 5\\&\text{What is its velocity at time }t=3\text{?}\end{align*}$

Possible Answers:

-2.67

-4.67

-125.84

Correct answer:

Explanation:

$\begin{align*}&\text{Velocity is the time rate of change of position. In lay terms,}\\&\text{it is an instantaneous measure of speed and direction. However,}\\&\text{going back to that first definition, note that it is a rate,}\\&\text{and whenever the term rate is used, derivatives should come}\\&\text{to mind. The rate of change of position with respect to time}\\&\text{is the derivative of position with respect to time:}\\&v(t)=\frac{dp(t)}{dt}\\&\text{Taking the derivative of our position function, }- 3t - cos(5\cdot \pi t) - sin(\frac{(7\cdot \pi t)}{2}) - 5\\&\text{We find the velocity function: }5\cdot \pi sin(5\cdot \pi t) -\frac{ (7\cdot \pi cos(\frac{(7\cdot \pi t)}{2}))}{2}- 3\\&\text{Plugging in our value of time we then get:}\\&v(3)=5\cdot \pi sin(5\cdot \pi (3)) -\frac{ (7\cdot \pi cos(\frac{(7\cdot \pi (3))}{2}))}{2}- 3\\&v(3)=-3\end{align*}$

Report an Error

Example Question #8 : Derivative Interpreted As An Instantaneous Rate Of Change

$\begin{align*}&\text{The position of a particle at a given time is given by the function:}\\&p(t)=cos(\frac{(9\cdot \pi t)}{2}) - 2t + cos(\frac{(17\cdot \pi t)}{2}) - 1\\&\text{What is its velocity at time }t=1\text{?}\end{align*}$

Possible Answers:

-42.84

Correct answer:

-42.84

Explanation:

$\begin{align*}&\text{Velocity is the time rate of change of position. In lay terms,}\\&\text{it is an instantaneous measure of speed and direction. However,}\\&\text{going back to that first definition, note that it is a rate,}\\&\text{and whenever the term rate is used, derivatives should come}\\&\text{to mind. The rate of change of position with respect to time}\\&\text{is the derivative of position with respect to time:}\\&v(t)=\frac{dp(t)}{dt}\\&\text{Taking the derivative of our position function, }cos(\frac{(9\cdot \pi t)}{2}) - 2t + cos(\frac{(17\cdot \pi t)}{2}) - 1\\&\text{We find the velocity function: }-\frac{ (9\cdot \pi sin(\frac{(9\cdot \pi t)}{2}))}{2}-\frac{ (17\cdot \pi sin(\frac{(17\cdot \pi t)}{2}))}{2}- 2\\&\text{Plugging in our value of time we then get:}\\&v(1)=-\frac{ (9\cdot \pi sin(\frac{(9\cdot \pi (1))}{2}))}{2}-\frac{ (17\cdot \pi sin(\frac{(17\cdot \pi (1))}{2}))}{2}- 2\\&v(1)=-42.84\end{align*}$

Report an Error

Example Question #9 : Derivative Interpreted As An Instantaneous Rate Of Change

$\begin{align*}&\text{At any given time, a particle has the position :}\\&p(t)=6t - cos(\frac{(13\cdot \pi t)}{2}) + 1\\&\text{Calculate its velocity at time }t=7\end{align*}$

Possible Answers:

-6.14

6.14

-14.42

Correct answer:

-14.42

Explanation:

$\begin{align*}&\text{Velocity is the time rate of change of position. In lay terms,}\\&\text{it is an instantaneous measure of speed and direction. However,}\\&\text{going back to that first definition, note that it is a rate,}\\&\text{and whenever the term rate is used, derivatives should come}\\&\text{to mind. The rate of change of position with respect to time}\\&\text{is the derivative of position with respect to time:}\\&v(t)=\frac{dp(t)}{dt}\\&\text{Taking the derivative of our position function, }6t - cos(\frac{(13\cdot \pi t)}{2}) + 1\\&\text{We find the velocity function: }\frac{(13\cdot \pi sin(\frac{(13\cdot \pi t)}{2}))}{2}+ 6\\&\text{Plugging in our value of time we then get:}\\&v(7)=\frac{(13\cdot \pi sin(\frac{(13\cdot \pi (7))}{2}))}{2}+ 6\\&v(7)=-14.42\end{align*}$

Report an Error

Example Question #10 : Derivative Interpreted As An Instantaneous Rate Of Change

$\begin{align*}&\text{The position of a particle at a given time is given by the function:}\\&p(t)=3t - cos(\frac{(\pi t)}{2}) + sin(4\cdot \pi t) + 5t^{2} + 6t^{3} + 6\\&\text{What is its velocity at time }t=6\text{?}\end{align*}$

Possible Answers:

250.17

723.57

249.33

223.53

Correct answer:

723.57

Explanation:

$\begin{align*}&\text{Velocity is the time rate of change of position. In lay terms,}\\&\text{it is an instantaneous measure of speed and direction. However,}\\&\text{going back to that first definition, note that it is a rate,}\\&\text{and whenever the term rate is used, derivatives should come}\\&\text{to mind. The rate of change of position with respect to time}\\&\text{is the derivative of position with respect to time:}\\&v(t)=\frac{dp(t)}{dt}\\&\text{Taking the derivative of our position function, }3t - cos(\frac{(\pi t)}{2}) + sin(4\cdot \pi t) + 5t^{2} + 6t^{3} + 6\\&\text{We find the velocity function: }10t + 4\cdot \pi cos(4\cdot \pi t) +\frac{ (\pi sin(\frac{(\pi t)}{2}))}{2}+ 18t^{2} + 3\\&\text{Plugging in our value of time we then get:}\\&v(6)=10(6) + 4\cdot \pi cos(4\cdot \pi (6)) +\frac{ (\pi sin(\frac{(\pi (6))}{2}))}{2}+ 18(6)^{2} + 3\\&v(6)=723.57\end{align*}$

Report an Error

Example Question #61 : Concept Of The Derivative

$\begin{align*}&\text{A particle at any point in time has the position :}\\&p(t)=- cos(\frac{(11\cdot \pi t)}{2}) - 6\\&\text{Determine its velocity at time }t=2\end{align*}$

Possible Answers:

-2.5

-298.56

Correct answer:

Explanation:

$\begin{align*}&\text{Velocity is the time rate of change of position. In lay terms,}\\&\text{it is an instantaneous measure of speed and direction. However,}\\&\text{going back to that first definition, note that it is a rate,}\\&\text{and whenever the term rate is used, derivatives should come}\\&\text{to mind. The rate of change of position with respect to time}\\&\text{is the derivative of position with respect to time:}\\&v(t)=\frac{dp(t)}{dt}\\&\text{Taking the derivative of our position function, }- cos(\frac{(11\cdot \pi t)}{2}) - 6\\&\text{We find the velocity function: }\frac{(11\cdot \pi sin(\frac{(11\cdot \pi t)}{2}))}{2}\\&\text{Plugging in our value of time we then get:}\\&v(2)=\frac{(11\cdot \pi sin(\frac{(11\cdot \pi (2))}{2}))}{2}\\&v(2)=0\end{align*}$

Report an Error

Example Question #11 : Derivative Interpreted As An Instantaneous Rate Of Change

$\begin{align*}&\text{The position of a particle at a given time is given by the function:}\\&p(t)=sin(\frac{(5\cdot \pi t)}{2}) + 2\\&\text{What is its velocity at time }t=6\text{?}\end{align*}$

Possible Answers:

0.33

-7.85

Correct answer:

-7.85

Explanation:

$\begin{align*}&\text{Velocity is the time rate of change of position. In lay terms,}\\&\text{it is an instantaneous measure of speed and direction. However,}\\&\text{going back to that first definition, note that it is a rate,}\\&\text{and whenever the term rate is used, derivatives should come}\\&\text{to mind. The rate of change of position with respect to time}\\&\text{is the derivative of position with respect to time:}\\&v(t)=\frac{dp(t)}{dt}\\&\text{Taking the derivative of our position function, }sin(\frac{(5\cdot \pi t)}{2}) + 2\\&\text{We find the velocity function: }\frac{(5\cdot \pi cos(\frac{(5\cdot \pi t)}{2}))}{2}\\&\text{Plugging in our value of time we then get:}\\&v(6)=\frac{(5\cdot \pi cos(\frac{(5\cdot \pi (6))}{2}))}{2}\\&v(6)=-7.85\end{align*}$

Report an Error

Example Question #12 : Derivative Interpreted As An Instantaneous Rate Of Change

$\begin{align*}&\text{The position of a particle at a given time is given by the function:}\\&p(t)=3t + cos(\frac{(7\cdot \pi t)}{2}) + sin(\frac{(9\cdot \pi t)}{2}) - t^{2} - 6t^{3} + 3\\&\text{What is its velocity at time }t=10\text{?}\end{align*}$

Possible Answers:

-606.8

-1831.14

-241.1

-607.2

Correct answer:

-1831.14

Explanation:

$\begin{align*}&\text{Velocity is the time rate of change of position. In lay terms,}\\&\text{it is an instantaneous measure of speed and direction. However,}\\&\text{going back to that first definition, note that it is a rate,}\\&\text{and whenever the term rate is used, derivatives should come}\\&\text{to mind. The rate of change of position with respect to time}\\&\text{is the derivative of position with respect to time:}\\&v(t)=\frac{dp(t)}{dt}\\&\text{Taking the derivative of our position function, }3t + cos(\frac{(7\cdot \pi t)}{2}) + sin(\frac{(9\cdot \pi t)}{2}) - t^{2} - 6t^{3} + 3\\&\text{We find the velocity function: }\frac{(9\cdot \pi cos(\frac{(9\cdot \pi t)}{2}))}{2}- 2t -\frac{ (7\cdot \pi sin(\frac{(7\cdot \pi t)}{2}))}{2}- 18t^{2} + 3\\&\text{Plugging in our value of time we then get:}\\&v(10)=\frac{(9\cdot \pi cos(\frac{(9\cdot \pi (10))}{2}))}{2}- 2(10) -\frac{ (7\cdot \pi sin(\frac{(7\cdot \pi (10))}{2}))}{2}- 18(10)^{2} + 3\\&v(10)=-1831.14\end{align*}$

Report an Error

View AP Calculus AB Tutors

Jeremy
Certified Tutor

University of Illinois at Chicago, Bachelor of Science, Physics.

View AP Calculus AB Tutors

Lyric
Certified Tutor

Spelman College, Bachelor of Science, Mathematics.

View AP Calculus AB Tutors

Abhisek
Certified Tutor

University of Florida, Bachelor of Engineering, Computer Science.

All AP Calculus AB Resources

3 Diagnostic Tests 164 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

Math Tutors in Denver, Math Tutors in Dallas Fort Worth, Biology Tutors in Dallas Fort Worth, ISEE Tutors in San Francisco-Bay Area, Algebra Tutors in Seattle, MCAT Tutors in Los Angeles, Computer Science Tutors in San Francisco-Bay Area, Physics Tutors in Dallas Fort Worth, Spanish Tutors in Denver, French Tutors in San Diego

Popular Courses & Classes

ISEE Courses & Classes in Philadelphia, SSAT Courses & Classes in Chicago, SSAT Courses & Classes in New York City, GRE Courses & Classes in New York City, Spanish Courses & Classes in Atlanta, LSAT Courses & Classes in San Diego, Spanish Courses & Classes in Los Angeles, SSAT Courses & Classes in Phoenix, ACT Courses & Classes in Atlanta, MCAT Courses & Classes in Dallas Fort Worth

Popular Test Prep

GRE Test Prep in Dallas Fort Worth, SSAT Test Prep in Denver, GRE Test Prep in San Diego, ISEE Test Prep in San Francisco-Bay Area, SSAT Test Prep in Atlanta, LSAT Test Prep in Los Angeles, SAT Test Prep in Atlanta, SAT Test Prep in San Diego, SSAT Test Prep in New York City, SAT Test Prep in Chicago

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

AP Calculus AB : Derivatives

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #4 : Derivative Interpreted As An Instantaneous Rate Of Change

Example Question #5 : Derivative Interpreted As An Instantaneous Rate Of Change

Example Question #6 : Derivative Interpreted As An Instantaneous Rate Of Change

Example Question #7 : Derivative Interpreted As An Instantaneous Rate Of Change

Example Question #8 : Derivative Interpreted As An Instantaneous Rate Of Change

Example Question #9 : Derivative Interpreted As An Instantaneous Rate Of Change

Example Question #10 : Derivative Interpreted As An Instantaneous Rate Of Change

Example Question #61 : Concept Of The Derivative

Example Question #11 : Derivative Interpreted As An Instantaneous Rate Of Change

Example Question #12 : Derivative Interpreted As An Instantaneous Rate Of Change

All AP Calculus AB Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: