Call Now to Set Up Tutoring:

(888) 888-0446

AP Calculus AB : Approximate rate of change from graphs and tables of values

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 #31 : Approximate Rate Of Change From Graphs And Tables Of Values

$\begin{align*}&\text{Approximate the acceleration of a particle using}\\&\text{a central difference approach at }t=6\\&\text{Using the following table of positions:}\\&t&p(t)\\&2&26\\&6&-23\\&10&18\\&14&16\\&18&-34\end{align*}$

Possible Answers:

$\frac{45}{8}$

$-\frac{49}{4}$

$-\frac{1}{2}$

Correct answer:

$\frac{45}{8}$

Explanation:

$\begin{align*}&\text{Acceleration is the 2nd time derivative of position. In other}\\&\text{words, you take the derivative of a position function twice}\\&\text{to define it. In this problem, we’re not given a function, but}\\&\text{ a table of times and positions, requiring an approximation,}\\&\text{a 2nd order central difference. This has the form:}\\&f''(x)=\frac{f(x+h)+f(x-h)-2f(x)}{h^2}\\&\text{Where h is the size of the increment between points on the x-axis.}\\&\text{This approximation is actually born from two types of first}\\&\text{order approximations: the forward and backward differences,}\\&\text{which have the forms:}\\&f'(x)_{forward}=\frac{f(x+h)-f(x)}{h}\text{ and }f'(x)_{backward}=\frac{f(x)-f(x-h)}{h}\\&\text{The difference of these two, divided by our increment, gives}\\&\text{a derivative of a derivative.}\\&\text{Applying the formula above, with }h=4\text{, we find:}\\&a(6)\approx\frac{18+(26)-2(-23)}{4^2}\\&a(6)\approx\frac{45}{8}\end{align*}$

Report an Error

Example Question #501 : Derivatives

$\begin{align*}&\text{Approximate the acceleration of a particle using}\\&\text{a central difference approach at }t=5\\&\text{Using the following table of positions:}\\&t&p(t)\\&0&28\\&5&32\\&10&37\end{align*}$

Possible Answers:

$\frac{9}{25}$

$\frac{1}{25}$

$\frac{9}{5}$

$\frac{4}{5}$

Correct answer:

$\frac{1}{25}$

Explanation:

$\begin{align*}&\text{Acceleration is the 2nd time derivative of position. In other}\\&\text{words, you take the derivative of a position function twice}\\&\text{to define it. In this problem, we’re not given a function, but}\\&\text{ a table of times and positions, requiring an approximation,}\\&\text{a 2nd order central difference. This has the form:}\\&f''(x)=\frac{f(x+h)+f(x-h)-2f(x)}{h^2}\\&\text{Where h is the size of the increment between points on the x-axis.}\\&\text{This approximation is actually born from two types of first}\\&\text{order approximations: the forward and backward differences,}\\&\text{which have the forms:}\\&f'(x)_{forward}=\frac{f(x+h)-f(x)}{h}\text{ and }f'(x)_{backward}=\frac{f(x)-f(x-h)}{h}\\&\text{The difference of these two, divided by our increment, gives}\\&\text{a derivative of a derivative.}\\&\text{Applying the formula above, with }h=5\text{, we find:}\\&a''(5)\approx\frac{37+(28)-2(32)}{5^2}\\&a'(5)\approx\frac{1}{25}\end{align*}$

Report an Error

Example Question #861 : Ap Calculus Ab

$\begin{align*}&t&p(t)\\&1&41\\&6&-32\\&11&-24\\&16&-36\\&\text{Using the above table of a particle's positions}\\&\text{ make a central difference approximation of its acceleration at }t=6\end{align*}$

Possible Answers:

$\frac{81}{25}$

$-\frac{73}{5}$

$-\frac{13}{5}$

Correct answer:

$\frac{81}{25}$

Explanation:

$\begin{align*}&\text{Acceleration is the 2nd time derivative of position. In other}\\&\text{words, you take the derivative of a position function twice}\\&\text{to define it. In this problem, we’re not given a function, but}\\&\text{ a table of times and positions, requiring an approximation,}\\&\text{a 2nd order central difference. This has the form:}\\&f''(x)=\frac{f(x+h)+f(x-h)-2f(x)}{h^2}\\&\text{Where h is the size of the increment between points on the x-axis.}\\&\text{This approximation is actually born from two types of first}\\&\text{order approximations: the forward and backward differences,}\\&\text{which have the forms:}\\&f'(x)_{forward}=\frac{f(x+h)-f(x)}{h}\text{ and }f'(x)_{backward}=\frac{f(x)-f(x-h)}{h}\\&\text{The difference of these two, divided by our increment, gives}\\&\text{a derivative of a derivative.}\\&\text{Applying the formula above, with }h=5\text{, we find:}\\&a(6)\approx\frac{-24+(41)-2(-32)}{5^2}\\&a(6)\approx\frac{81}{25}\end{align*}$

Report an Error

Example Question #103 : Derivative At A Point

$\begin{align*}&t&p(t)\\&0&36\\&3&12\\&6&-15\\&9&1\\&12&-10\\&15&-43\\&\text{The table above lists the positions of a particle}\\&\text{At various times. Approximate the acceleration at }t=3\\&\text{Using a central difference approach.}\end{align*}$

Possible Answers:

$-\frac{17}{3}$

$-\frac{1}{3}$

Correct answer:

$-\frac{1}{3}$

Explanation:

$\begin{align*}&\text{Acceleration is the 2nd time derivative of position. In other}\\&\text{words, you take the derivative of a position function twice}\\&\text{to define it. In this problem, we’re not given a function, but}\\&\text{ a table of times and positions, requiring an approximation,}\\&\text{a 2nd order central difference. This has the form:}\\&f''(x)=\frac{f(x+h)+f(x-h)-2f(x)}{h^2}\\&\text{Where h is the size of the increment between points on the x-axis.}\\&\text{This approximation is actually born from two types of first}\\&\text{order approximations: the forward and backward differences,}\\&\text{which have the forms:}\\&f'(x)_{forward}=\frac{f(x+h)-f(x)}{h}\text{ and }f'(x)_{backward}=\frac{f(x)-f(x-h)}{h}\\&\text{The difference of these two, divided by our increment, gives}\\&\text{a derivative of a derivative.}\\&\text{Applying the formula above, with }h=3\text{, we find:}\\&a(3)\approx\frac{-15+(36)-2(12)}{3^2}\\&a(3)\approx-\frac{1}{3}\end{align*}$

Report an Error

Example Question #501 : Derivatives

$\begin{align*}&\text{Approximate the acceleration of a particle using}\\&\text{a central difference approach at }t=3\\&\text{Using the following table of positions:}\\&t&p(t)\\&1&-45\\&3&41\\&5&45\\&7&-1\end{align*}$

Possible Answers:

$\frac{45}{2}$

$-\frac{41}{2}$

Correct answer:

$-\frac{41}{2}$

Explanation:

$\begin{align*}&\text{Acceleration is the 2nd time derivative of position. In other}\\&\text{words, you take the derivative of a position function twice}\\&\text{to define it. In this problem, we’re not given a function, but}\\&\text{ a table of times and positions, requiring an approximation,}\\&\text{a 2nd order central difference. This has the form:}\\&f''(x)=\frac{f(x+h)+f(x-h)-2f(x)}{h^2}\\&\text{Where h is the size of the increment between points on the x-axis.}\\&\text{This approximation is actually born from two types of first}\\&\text{order approximations: the forward and backward differences,}\\&\text{which have the forms:}\\&f'(x)_{forward}=\frac{f(x+h)-f(x)}{h}\text{ and }f'(x)_{backward}=\frac{f(x)-f(x-h)}{h}\\&\text{The difference of these two, divided by our increment, gives}\\&\text{a derivative of a derivative.}\\&\text{Applying the formula above, with }h=2\text{, we find:}\\&a(3)\approx\frac{45+(-45)-2(41)}{2^2}\\&a(3)\approx-\frac{41}{2}\end{align*}$

Report an Error

Example Question #105 : Derivative At A Point

$\begin{align*}&t&p(t)\\&0&28\\&2&-11\\&4&-26\\&6&-10\\&8&-41\\&10&-37\\&12&45\\&14&46\\&\text{The table above lists the positions of a particle}\\&\text{At various times. Approximate the acceleration at }t=8\\&\text{Using a central difference approach.}\end{align*}$

Possible Answers:

$-\frac{31}{2}$

$-\frac{27}{2}$

$\frac{35}{4}$

$-\frac{27}{4}$

Correct answer:

$\frac{35}{4}$

Explanation:

$\begin{align*}&\text{Acceleration is the 2nd time derivative of position. In other}\\&\text{words, you take the derivative of a position function twice}\\&\text{to define it. In this problem, we’re not given a function, but}\\&\text{ a table of times and positions, requiring an approximation,}\\&\text{a 2nd order central difference. This has the form:}\\&f''(x)=\frac{f(x+h)+f(x-h)-2f(x)}{h^2}\\&\text{Where h is the size of the increment between points on the x-axis.}\\&\text{This approximation is actually born from two types of first}\\&\text{order approximations: the forward and backward differences,}\\&\text{which have the forms:}\\&f'(x)_{forward}=\frac{f(x+h)-f(x)}{h}\text{ and }f'(x)_{backward}=\frac{f(x)-f(x-h)}{h}\\&\text{The difference of these two, divided by our increment, gives}\\&\text{a derivative of a derivative.}\\&\text{Applying the formula above, with }h=2\text{, we find:}\\&a(8)\approx\frac{-37+(-10)-2(-41)}{2^2}\\&a(8)\approx\frac{35}{4}\end{align*}$

Report an Error

Example Question #101 : Derivative At A Point

$\begin{align*}&t&p(t)\\&2&24\\&3&6\\&4&-32\\&5&10\\&6&-20\\&7&-37\\&8&-29\\&9&40\\&\text{The above table lists a particle's postion at different times.}\\&\text{Approximate its velocity with a forward difference at }t=2\end{align*}$

Possible Answers:

Correct answer:

Explanation:

$\begin{align*}&\text{Velocity is the time derivative of position. However, since}\\&\text{we do not have a position function, we cannot take the derivative;}\\&\text{we must use an approximation. To find a forward derivative approximation,}\\&\text{we consider function two values: one at the point of interest}\\&\text{and another succeeding it, and finally that distance between}\\&\text{the two. Mathematically, this follows the form:}\\&f'(x)=\frac{f(x+h)-f(x)}{h}\\&\text{Where h is the size of the increment between points on the x-axis.}\\&\text{In approximating a derivative, it is generally prudent to minimize}\\&\text{the distance between function values. In other words, it’s a}\\&\text{good idea to look at two adjacent points. For our problem, that’d}\\&\text{be:}\\&f(2)\text{ and }f(3)\\&\text{Applying the formula above, with }h=1\text{, we find:}\\&v(2)\approx\frac{6-(24)}{1}\\&v(2)\approx-18\end{align*}$

Report an Error

Example Question #31 : Approximate Rate Of Change From Graphs And Tables Of Values

$\begin{align*}&t&p(t)\\&3&-31\\&4&-41\\&5&-19\\&6&-4\\&7&-40\\&\text{The above table lists a particle's postion at different times.}\\&\text{Approximate its velocity with a forward difference at }t=6\end{align*}$

Possible Answers:

Correct answer:

Explanation:

$\begin{align*}&\text{Velocity is the time derivative of position. However, since}\\&\text{we do not have a position function, we cannot take the derivative;}\\&\text{we must use an approximation. To find a forward derivative approximation,}\\&\text{we consider function two values: one at the point of interest}\\&\text{and another succeeding it, and finally that distance between}\\&\text{the two. Mathematically, this follows the form:}\\&f'(x)=\frac{f(x+h)-f(x)}{h}\\&\text{Where h is the size of the increment between points on the x-axis.}\\&\text{In approximating a derivative, it is generally prudent to minimize}\\&\text{the distance between function values. In other words, it’s a}\\&\text{good idea to look at two adjacent points. For our problem, that’d}\\&\text{be:}\\&f(6)\text{ and }f(7)\\&\text{Applying the formula above, with }h=1\text{, we find:}\\&v(6)\approx\frac{-40-(-4)}{1}\\&v(6)\approx-36\end{align*}$

Report an Error

Example Question #32 : Approximate Rate Of Change From Graphs And Tables Of Values

$\begin{align*}&t&p(t)\\&0&33\\&3&-16\\&6&-21\\&9&25\\&12&-49\\&15&-46\\&18&17\\&21&10\\&\text{Using the above table of positions,}\\&\text{make a forward difference velocity approximation at }t=9\end{align*}$

Possible Answers:

$-\frac{37}{3}$

$-\frac{74}{3}$

$\frac{25}{3}$

$\frac{46}{3}$

Correct answer:

$-\frac{74}{3}$

Explanation:

$\begin{align*}&\text{Velocity is the time derivative of position. However, since}\\&\text{we do not have a position function, we cannot take the derivative;}\\&\text{we must use an approximation. To find a forward derivative approximation,}\\&\text{we consider function two values: one at the point of interest}\\&\text{and another succeeding it, and finally that distance between}\\&\text{the two. Mathematically, this follows the form:}\\&f'(x)=\frac{f(x+h)-f(x)}{h}\\&\text{Where h is the size of the increment between points on the x-axis.}\\&\text{In approximating a derivative, it is generally prudent to minimize}\\&\text{the distance between function values. In other words, it’s a}\\&\text{good idea to look at two adjacent points. For our problem, that’d}\\&\text{be:}\\&f(9)\text{ and }f(12)\\&\text{Applying the formula above, with }h=3\text{, we find:}\\&v(9)\approx\frac{-49-(25)}{3}\\&v(9)\approx-\frac{74}{3}\end{align*}$

Report an Error

Example Question #33 : Approximate Rate Of Change From Graphs And Tables Of Values

$\begin{align*}&\text{Approximate a particle's velocity with a forward difference at }t=4\\&\text{Using the following table of positions:}\\&t&p(t)\\&4&6\\&6&-10\\&8&-44\\&10&28\\&12&-16\\&14&11\\&16&24\end{align*}$

Possible Answers:

Correct answer:

Explanation:

$\begin{align*}&\text{Velocity is the time derivative of position. However, since}\\&\text{we do not have a position function, we cannot take the derivative;}\\&\text{we must use an approximation. To find a forward derivative approximation,}\\&\text{we consider function two values: one at the point of interest}\\&\text{and another succeeding it, and finally that distance between}\\&\text{the two. Mathematically, this follows the form:}\\&f'(x)=\frac{f(x+h)-f(x)}{h}\\&\text{Where h is the size of the increment between points on the x-axis.}\\&\text{In approximating a derivative, it is generally prudent to minimize}\\&\text{the distance between function values. In other words, it’s a}\\&\text{good idea to look at two adjacent points. For our problem, that’d}\\&\text{be:}\\&f(4)\text{ and }f(6)\\&\text{Applying the formula above, with }h=2\text{, we find:}\\&v(4)\approx\frac{-10-(6)}{2}\\&v(4)\approx-8\end{align*}$

Report an Error

View AP Calculus AB Tutors

Muhammad
Certified Tutor

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

View AP Calculus AB Tutors

Sharif
Certified Tutor

NCCU, Bachelors, Physics and Mathematics.

View AP Calculus AB Tutors

Megan
Certified Tutor

University of Wisconsin-Madison, Bachelor of Science, Chemical Engineering.

All AP Calculus AB Resources

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

Popular Subjects

Biology Tutors in New York City, LSAT Tutors in San Diego, Chemistry Tutors in Houston, Spanish Tutors in Miami, Statistics Tutors in Los Angeles, GMAT Tutors in Seattle, Biology Tutors in Phoenix, Computer Science Tutors in Dallas Fort Worth, ACT Tutors in Denver, Algebra Tutors in San Francisco-Bay Area

Popular Courses & Classes

ISEE Courses & Classes in Boston, ACT Courses & Classes in Denver, Spanish Courses & Classes in Seattle, ISEE Courses & Classes in Miami, MCAT Courses & Classes in Boston, GRE Courses & Classes in Boston, GMAT Courses & Classes in Los Angeles, MCAT Courses & Classes in Denver, MCAT Courses & Classes in Atlanta, SAT Courses & Classes in Los Angeles

Popular Test Prep

ISEE Test Prep in Dallas Fort Worth, ISEE Test Prep in Washington DC, LSAT Test Prep in Boston, SAT Test Prep in New York City, GMAT Test Prep in Miami, LSAT Test Prep in Philadelphia, SSAT Test Prep in Chicago, MCAT Test Prep in Washington DC, GMAT Test Prep in Washington DC, GRE Test Prep in Boston

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 : Approximate rate of change from graphs and tables of values

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #31 : Approximate Rate Of Change From Graphs And Tables Of Values

Example Question #501 : Derivatives

Example Question #861 : Ap Calculus Ab

Example Question #103 : Derivative At A Point

Example Question #501 : Derivatives

Example Question #105 : Derivative At A Point

Example Question #101 : Derivative At A Point

Example Question #31 : Approximate Rate Of Change From Graphs And Tables Of Values

Example Question #32 : Approximate Rate Of Change From Graphs And Tables Of Values

Example Question #33 : Approximate Rate Of Change From Graphs And Tables Of Values

All AP Calculus AB Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: