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

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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:

$-\frac{49}{4}$

$-\frac{1}{2}$

$\frac{45}{8}$

Correct answer:

$\frac{45}{8}$

Explanation:

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

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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:

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

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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:

$-\frac{13}{5}$

$\frac{81}{25}$

$-\frac{73}{5}$

Correct answer:

$\frac{81}{25}$

Explanation:

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

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Example Question #101 : 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:

$-\frac{1}{3}$

$-\frac{17}{3}$

Correct answer:

$-\frac{1}{3}$

Explanation:

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

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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:

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

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Example Question #861 : Ap Calculus Ab

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

Possible Answers:

$-\frac{31}{2}$

$\frac{35}{4}$

$-\frac{27}{4}$

$-\frac{27}{2}$

Correct answer:

$\frac{35}{4}$

Explanation:

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

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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:

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

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Example Question #862 : Ap Calculus Ab

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

Possible Answers:

Correct answer:

Explanation:

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

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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:

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

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Example Question #101 : Derivative At A Point

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

Possible Answers:

Correct answer:

Explanation:

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

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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 #101 : Derivative At A Point

Example Question #501 : Derivatives

Example Question #861 : Ap Calculus Ab

Example Question #101 : Derivative At A Point

Example Question #862 : Ap Calculus Ab

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

Example Question #101 : Derivative At A Point

All AP Calculus AB Resources

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