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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 #81 : Derivative At A Point

$\begin{align*}&x&f(x)\\&-1&-4\\&1&27\\&3&32\\&5&-40\\&7&-33\\&9&-14\\&11&-45\\&13&2\\&\text{Approximate a backward derivative at }x=3\end{align*}$

Possible Answers:

$\frac{5}{2}$

$\frac{59}{2}$

Correct answer:

$\frac{5}{2}$

Explanation:

$\begin{align*}&\text{To find a backward derivative approximation, we consider function}\\&\text{two values: one at the point of interest and another preceding}\\&\text{it, and finally that distance between the two. Mathematically,}\\&\text{this follows the form:}\\&f'(x)=\frac{f(x)-f(x-h)}{h}\\&\text{Where h is the size of the increment between points on the x-axis.}\\&\text{In approximating a derivative, it is generally prudent to minimize}\\&\text{the distance between function values. In other words, it’s a}\\&\text{good idea to look at two adjacent points. For our problem, that’d}\\&\text{be:}\\&f(3)\text{ and }f(1)\\&\text{Applying the formula above, with }h=2\text{, we find:}\\&f'(3)\approx\frac{32-(27)}{2}\\&f'(3)\approx\frac{5}{2}\end{align*}$

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Example Question #491 : Derivatives

$\begin{align*}&x&f(x)\\&0&42\\&4&-40\\&8&25\\&12&24\\&16&6\\&20&-32\\&24&10\\&28&-20\\&\text{Using the above table, make a backward derivative approximation at }x=8\end{align*}$

Possible Answers:

$-\frac{1}{4}$

$\frac{49}{4}$

$\frac{65}{4}$

$-\frac{15}{4}$

Correct answer:

$\frac{65}{4}$

Explanation:

$\begin{align*}&\text{To find a backward derivative approximation, we consider function}\\&\text{two values: one at the point of interest and another preceding}\\&\text{it, and finally that distance between the two. Mathematically,}\\&\text{this follows the form:}\\&f'(x)=\frac{f(x)-f(x-h)}{h}\\&\text{Where h is the size of the increment between points on the x-axis.}\\&\text{In approximating a derivative, it is generally prudent to minimize}\\&\text{the distance between function values. In other words, it’s a}\\&\text{good idea to look at two adjacent points. For our problem, that’d}\\&\text{be:}\\&f(8)\text{ and }f(4)\\&\text{Applying the formula above, with }h=4\text{, we find:}\\&f'(8)\approx\frac{25-(-40)}{4}\\&f'(8)\approx\frac{65}{4}\end{align*}$

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Example Question #92 : Derivative At A Point

$\begin{align*}&x&f(x)\\&-3&-6\\&-1&-49\\&1&40\\&\text{Approximate a backward derivative at }x=-1\end{align*}$

Possible Answers:

$-\frac{9}{2}$

$-\frac{43}{2}$

$-\frac{55}{2}$

$\frac{89}{2}$

Correct answer:

$-\frac{43}{2}$

Explanation:

$\begin{align*}&\text{To find a backward derivative approximation, we consider function}\\&\text{two values: one at the point of interest and another preceding}\\&\text{it, and finally that distance between the two. Mathematically,}\\&\text{this follows the form:}\\&f'(x)=\frac{f(x)-f(x-h)}{h}\\&\text{Where h is the size of the increment between points on the x-axis.}\\&\text{In approximating a derivative, it is generally prudent to minimize}\\&\text{the distance between function values. In other words, it’s a}\\&\text{good idea to look at two adjacent points. For our problem, that’d}\\&\text{be:}\\&f(-1)\text{ and }f(-3)\\&\text{Applying the formula above, with }h=2\text{, we find:}\\&f'(-1)\approx\frac{-49-(-6)}{2}\\&f'(-1)\approx-\frac{43}{2}\end{align*}$

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Example Question #852 : Ap Calculus Ab

$\begin{align*}&\text{Approximate a backward derivative at }x=7\\&\text{Using the following table:}\\&x&f(x)\\&3&-17\\&4&-20\\&5&-44\\&6&-20\\&7&-46\end{align*}$

Possible Answers:

Correct answer:

Explanation:

$\begin{align*}&\text{To find a backward derivative approximation, we consider function}\\&\text{two values: one at the point of interest and another preceding}\\&\text{it, and finally that distance between the two. Mathematically,}\\&\text{this follows the form:}\\&f'(x)=\frac{f(x)-f(x-h)}{h}\\&\text{Where h is the size of the increment between points on the x-axis.}\\&\text{In approximating a derivative, it is generally prudent to minimize}\\&\text{the distance between function values. In other words, it’s a}\\&\text{good idea to look at two adjacent points. For our problem, that’d}\\&\text{be:}\\&f(7)\text{ and }f(6)\\&\text{Applying the formula above, with }h=1\text{, we find:}\\&f'(7)\approx\frac{-46-(-20)}{1}\\&f'(7)\approx-26\end{align*}$

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Example Question #853 : Ap Calculus Ab

$\begin{align*}&\text{Approximate the acceleration of a particle using}\\&\text{a central difference approach at }t=5\\&\text{Using the following table of positions:}\\&t&p(t)\\&1&-42\\&5&-27\\&9&42\\&13&-35\\&17&33\\&21&4\\&25&50\end{align*}$

Possible Answers:

$\frac{27}{8}$

$\frac{21}{4}$

$\frac{15}{4}$

Correct answer:

$\frac{27}{8}$

Explanation:

$\begin{align*}&\text{Acceleration is the 2nd time derivative of position. In other}\\&\text{words, you take the derivative of a position function twice}\\&\text{to define it. In this problem, we’re not given a function, but}\\&\text{ a table of times and positions, requiring an approximation,}\\&\text{a 2nd order central difference. This has the form:}\\&f''(x)=\frac{f(x+h)+f(x-h)-2f(x)}{h^2}\\&\text{Where h is the size of the increment between points on the x-axis.}\\&\text{This approximation is actually born from two types of first}\\&\text{order approximations: the forward and backward differences,}\\&\text{which have the forms:}\\&f'(x)_{forward}=\frac{f(x+h)-f(x)}{h}\text{ and }f'(x)_{backward}=\frac{f(x)-f(x-h)}{h}\\&\text{The difference of these two, divided by our increment, gives}\\&\text{a derivative of a derivative.}\\&\text{Applying the formula above, with }h=4\text{, we find:}\\&a(5)\approx\frac{42+(-42)-2(-27)}{4^2}\\&a(5)\approx\frac{27}{8}\end{align*}$

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Example Question #95 : Derivative At A Point

Possible Answers:

$\frac{7}{4}$

$\frac{7}{16}$

$\frac{81}{16}$

$-\frac{37}{4}$

Correct answer:

$\frac{81}{16}$

Explanation:

$\begin{align*}&\text{Acceleration is the 2nd time derivative of position. In other}\\&\text{words, you take the derivative of a position function twice}\\&\text{to define it. In this problem, we’re not given a function, but}\\&\text{ a table of times and positions, requiring an approximation,}\\&\text{a 2nd order central difference. This has the form:}\\&f''(x)=\frac{f(x+h)+f(x-h)-2f(x)}{h^2}\\&\text{Where h is the size of the increment between points on the x-axis.}\\&\text{This approximation is actually born from two types of first}\\&\text{order approximations: the forward and backward differences,}\\&\text{which have the forms:}\\&f'(x)_{forward}=\frac{f(x+h)-f(x)}{h}\text{ and }f'(x)_{backward}=\frac{f(x)-f(x-h)}{h}\\&\text{The difference of these two, divided by our increment, gives}\\&\text{a derivative of a derivative.}\\&\text{Applying the formula above, with }h=4\text{, we find:}\\&a(5)\approx\frac{10+(3)-2(-34)}{4^2}\\&a(5)\approx\frac{81}{16}\end{align*}$

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Example Question #96 : Derivative At A Point

$\begin{align*}&\text{Approximate the acceleration of a particle using}\\&\text{a central difference approach at }t=1\\&\text{Using the following table of positions:}\\&t&p(t)\\&0&3\\&1&28\\&2&44\\&3&-37\\&4&7\\&5&-3\end{align*}$

Possible Answers:

Correct answer:

Explanation:

$\begin{align*}&\text{Acceleration is the 2nd time derivative of position. In other}\\&\text{words, you take the derivative of a position function twice}\\&\text{to define it. In this problem, we’re not given a function, but}\\&\text{ a table of times and positions, requiring an approximation,}\\&\text{a 2nd order central difference. This has the form:}\\&f''(x)=\frac{f(x+h)+f(x-h)-2f(x)}{h^2}\\&\text{Where h is the size of the increment between points on the x-axis.}\\&\text{This approximation is actually born from two types of first}\\&\text{order approximations: the forward and backward differences,}\\&\text{which have the forms:}\\&f'(x)_{forward}=\frac{f(x+h)-f(x)}{h}\text{ and }f'(x)_{backward}=\frac{f(x)-f(x-h)}{h}\\&\text{The difference of these two, divided by our increment, gives}\\&\text{a derivative of a derivative.}\\&\text{Applying the formula above, with }h=1\text{, we find:}\\&a(1)\approx\frac{44+(3)-2(28)}{1^2}\\&a(1)\approx-9\end{align*}$

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Example Question #501 : Derivatives

$\begin{align*}&\text{Approximate the acceleration of a particle using}\\&\text{a central difference approach at }t=13\\&\text{Using the following table of positions:}\\&t&p(t)\\&1&33\\&4&9\\&7&5\\&10&42\\&13&-22\\&16&26\end{align*}$

Possible Answers:

$-\frac{16}{3}$

$\frac{112}{9}$

$-\frac{16}{9}$

$-\frac{64}{3}$

Correct answer:

$\frac{112}{9}$

Explanation:

$\begin{align*}&\text{Acceleration is the 2nd time derivative of position. In other}\\&\text{words, you take the derivative of a position function twice}\\&\text{to define it. In this problem, we’re not given a function, but}\\&\text{ a table of times and positions, requiring an approximation,}\\&\text{a 2nd order central difference. This has the form:}\\&f''(x)=\frac{f(x+h)+f(x-h)-2f(x)}{h^2}\\&\text{Where h is the size of the increment between points on the x-axis.}\\&\text{This approximation is actually born from two types of first}\\&\text{order approximations: the forward and backward differences,}\\&\text{which have the forms:}\\&f'(x)_{forward}=\frac{f(x+h)-f(x)}{h}\text{ and }f'(x)_{backward}=\frac{f(x)-f(x-h)}{h}\\&\text{The difference of these two, divided by our increment, gives}\\&\text{a derivative of a derivative.}\\&\text{Applying the formula above, with }h=3\text{, we find:}\\&a(13)\approx\frac{26+(42)-2(-22)}{3^2}\\&a(13)\approx\frac{112}{9}\end{align*}$

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Example Question #851 : Ap Calculus Ab

$\begin{align*}&t&p(t)\\&1&32\\&6&-26\\&11&43\\&16&-15\\&\text{The table above lists the positions of a particle}\\&\text{At various times. Approximate the acceleration at }t=6\\&\text{Using a central difference approach.}\end{align*}$

Possible Answers:

$-\frac{58}{5}$

$\frac{11}{5}$

$\frac{11}{25}$

$\frac{127}{25}$

Correct answer:

$\frac{127}{25}$

Explanation:

$\begin{align*}&\text{Acceleration is the 2nd time derivative of position. In other}\\&\text{words, you take the derivative of a position function twice}\\&\text{to define it. In this problem, we’re not given a function, but}\\&\text{ a table of times and positions, requiring an approximation,}\\&\text{a 2nd order central difference. This has the form:}\\&f''(x)=\frac{f(x+h)+f(x-h)-2f(x)}{h^2}\\&\text{Where h is the size of the increment between points on the x-axis.}\\&\text{This approximation is actually born from two types of first}\\&\text{order approximations: the forward and backward differences,}\\&\text{which have the forms:}\\&f'(x)_{forward}=\frac{f(x+h)-f(x)}{h}\text{ and }f'(x)_{backward}=\frac{f(x)-f(x-h)}{h}\\&\text{The difference of these two, divided by our increment, gives}\\&\text{a derivative of a derivative.}\\&\text{Applying the formula above, with }h=5\text{, we find:}\\&a(6)\approx\frac{43+(32)-2(-26)}{5^2}\\&a(6)\approx\frac{127}{25}\end{align*}$

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Example Question #21 : Approximate Rate Of Change From Graphs And Tables Of Values

$\begin{align*}&t&p(t)\\&2&-28\\&4&25\\&6&-25\\&8&1\\&10&20\\&12&39\\&14&46\\&16&5\\&\text{The table above lists the positions of a particle}\\&\text{At various times. Approximate the acceleration at }t=4\\&\text{Using a central difference approach.}\end{align*}$

Possible Answers:

$\frac{3}{4}$

$\frac{53}{2}$

$-\frac{103}{4}$

$\frac{3}{2}$

Correct answer:

$-\frac{103}{4}$

Explanation:

$\begin{align*}&\text{Acceleration is the 2nd time derivative of position. In other}\\&\text{words, you take the derivative of a position function twice}\\&\text{to define it. In this problem, we’re not given a function, but}\\&\text{ a table of times and positions, requiring an approximation,}\\&\text{a 2nd order central difference. This has the form:}\\&f''(x)=\frac{f(x+h)+f(x-h)-2f(x)}{h^2}\\&\text{Where h is the size of the increment between points on the x-axis.}\\&\text{This approximation is actually born from two types of first}\\&\text{order approximations: the forward and backward differences,}\\&\text{which have the forms:}\\&f'(x)_{forward}=\frac{f(x+h)-f(x)}{h}\text{ and }f'(x)_{backward}=\frac{f(x)-f(x-h)}{h}\\&\text{The difference of these two, divided by our increment, gives}\\&\text{a derivative of a derivative.}\\&\text{Applying the formula above, with }h=2\text{, we find:}\\&a(4)\approx\frac{-25+(-28)-2(25)}{2^2}\\&a(4)\approx-\frac{103}{4}\end{align*}$

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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 #81 : Derivative At A Point

Example Question #491 : Derivatives

Example Question #92 : Derivative At A Point

Example Question #852 : Ap Calculus Ab

Example Question #853 : Ap Calculus Ab

Example Question #95 : Derivative At A Point

Example Question #96 : Derivative At A Point

Example Question #501 : Derivatives

Example Question #851 : Ap Calculus Ab

Example Question #21 : Approximate Rate Of Change From Graphs And Tables Of Values

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