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

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Example Question #41 : Approximate Rate Of Change From Graphs And Tables Of Values

$\begin{align*}&t&p(t)\\&5&24\\&10&-45\\&15&-43\\&20&-42\\&25&30\\&\text{The above table lists a particle's postion at different times.}\\&\text{Approximate its velocity with a forward difference at }t=20\end{align*}$

Possible Answers:

$-\frac{42}{5}$

$\frac{36}{5}$

$\frac{1}{5}$

$\frac{72}{5}$

Correct answer:

$\frac{72}{5}$

Explanation:

$\begin{align*}&\text{Velocity is the time derivative of position. However, since}\\&\text{we do not have a position function, we cannot take the derivative;}\\&\text{we must use an approximation. To find a forward derivative approximation,}\\&\text{we consider function two values: one at the point of interest}\\&\text{and another succeeding it, and finally that distance between}\\&\text{the two. Mathematically, this follows the form:}\\&f'(x)=\frac{f(x+h)-f(x)}{h}\\&\text{Where h is the size of the increment between points on the x-axis.}\\&\text{In approximating a derivative, it is generally prudent to minimize}\\&\text{the distance between function values. In other words, it’s a}\\&\text{good idea to look at two adjacent points. For our problem, that’d}\\&\text{be:}\\&f(20)\text{ and }f(25)\\&\text{Applying the formula above, with }h=5\text{, we find:}\\&v(20)\approx\frac{30-(-42)}{5}\\&v(20)\approx\frac{72}{5}\end{align*}$

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Example Question #111 : Derivative At A Point

$\begin{align*}&t&p(t)\\&0&14\\&1&-17\\&2&16\\&3&25\\&4&8\\&5&24\\&6&-27\\&\text{Using the above table of positions,}\\&\text{make a forward difference velocity approximation at }t=4\end{align*}$

Possible Answers:

Correct answer:

Explanation:

$\begin{align*}&\text{Velocity is the time derivative of position. However, since}\\&\text{we do not have a position function, we cannot take the derivative;}\\&\text{we must use an approximation. To find a forward derivative approximation,}\\&\text{we consider function two values: one at the point of interest}\\&\text{and another succeeding it, and finally that distance between}\\&\text{the two. Mathematically, this follows the form:}\\&f'(x)=\frac{f(x+h)-f(x)}{h}\\&\text{Where h is the size of the increment between points on the x-axis.}\\&\text{In approximating a derivative, it is generally prudent to minimize}\\&\text{the distance between function values. In other words, it’s a}\\&\text{good idea to look at two adjacent points. For our problem, that’d}\\&\text{be:}\\&f(4)\text{ and }f(5)\\&\text{Applying the formula above, with }h=1\text{, we find:}\\&v(4)\approx\frac{24-(8)}{1}\\&v(4)\approx16\end{align*}$

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Example Question #43 : Approximate Rate Of Change From Graphs And Tables Of Values

$\begin{align*}&t&p(t)\\&2&19\\&4&10\\&6&29\\&\text{The above table lists a particle's postion at different times.}\\&\text{Approximate its velocity with a forward difference at }t=2\end{align*}$

Possible Answers:

$-\frac{9}{4}$

$\frac{9}{2}$

$-\frac{9}{2}$

$\frac{9}{4}$

Correct answer:

$-\frac{9}{2}$

Explanation:

$\begin{align*}&\text{Velocity is the time derivative of position. However, since}\\&\text{we do not have a position function, we cannot take the derivative;}\\&\text{we must use an approximation. To find a forward derivative approximation,}\\&\text{we consider function two values: one at the point of interest}\\&\text{and another succeeding it, and finally that distance between}\\&\text{the two. Mathematically, this follows the form:}\\&f'(x)=\frac{f(x+h)-f(x)}{h}\\&\text{Where h is the size of the increment between points on the x-axis.}\\&\text{In approximating a derivative, it is generally prudent to minimize}\\&\text{the distance between function values. In other words, it’s a}\\&\text{good idea to look at two adjacent points. For our problem, that’d}\\&\text{be:}\\&f(2)\text{ and }f(4)\\&\text{Applying the formula above, with }h=2\text{, we find:}\\&v(2)\approx\frac{10-(19)}{2}\\&v(2)\approx-\frac{9}{2}\end{align*}$

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Example Question #41 : Approximate Rate Of Change From Graphs And Tables Of Values

$\begin{align*}&\text{Approximate a particle's velocity with a forward difference at }t=4\\&\text{Using the following table of positions:}\\&t&p(t)\\&2&5\\&4&-27\\&6&14\\&8&-2\\&10&-35\\&12&28\\&14&-40\end{align*}$

Possible Answers:

$\frac{41}{4}$

$-\frac{27}{2}$

$\frac{41}{2}$

Correct answer:

$\frac{41}{2}$

Explanation:

$\begin{align*}&\text{Velocity is the time derivative of position. However, since}\\&\text{we do not have a position function, we cannot take the derivative;}\\&\text{we must use an approximation. To find a forward derivative approximation,}\\&\text{we consider function two values: one at the point of interest}\\&\text{and another succeeding it, and finally that distance between}\\&\text{the two. Mathematically, this follows the form:}\\&f'(x)=\frac{f(x+h)-f(x)}{h}\\&\text{Where h is the size of the increment between points on the x-axis.}\\&\text{In approximating a derivative, it is generally prudent to minimize}\\&\text{the distance between function values. In other words, it’s a}\\&\text{good idea to look at two adjacent points. For our problem, that’d}\\&\text{be:}\\&f(4)\text{ and }f(6)\\&\text{Applying the formula above, with }h=2\text{, we find:}\\&v(4)\approx\frac{14-(-27)}{2}\\&v(4)\approx\frac{41}{2}\end{align*}$

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Example Question #45 : Approximate Rate Of Change From Graphs And Tables Of Values

$\begin{align*}&\text{Approximate a particle's velocity with a forward difference at }t=3\\&\text{Using the following table of positions:}\\&t&p(t)\\&0&-39\\&3&29\\&6&-21\end{align*}$

Possible Answers:

$-\frac{25}{3}$

$-\frac{50}{3}$

$\frac{29}{3}$

$\frac{68}{3}$

Correct answer:

$-\frac{50}{3}$

Explanation:

$\begin{align*}&\text{Velocity is the time derivative of position. However, since}\\&\text{we do not have a position function, we cannot take the derivative;}\\&\text{we must use an approximation. To find a forward derivative approximation,}\\&\text{we consider function two values: one at the point of interest}\\&\text{and another succeeding it, and finally that distance between}\\&\text{the two. Mathematically, this follows the form:}\\&f'(x)=\frac{f(x+h)-f(x)}{h}\\&\text{Where h is the size of the increment between points on the x-axis.}\\&\text{In approximating a derivative, it is generally prudent to minimize}\\&\text{the distance between function values. In other words, it’s a}\\&\text{good idea to look at two adjacent points. For our problem, that’d}\\&\text{be:}\\&f(3)\text{ and }f(6)\\&\text{Applying the formula above, with }h=3\text{, we find:}\\&v(3)\approx\frac{-21-(29)}{3}\\&v(3)\approx-\frac{50}{3}\end{align*}$

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Example Question #46 : Approximate Rate Of Change From Graphs And Tables Of Values

$\begin{align*}&t&p(t)\\&3&16\\&7&-39\\&11&44\\&15&-32\\&19&-24\\&23&30\\&27&-1\\&\text{The above table lists a particle's postion at different times.}\\&\text{Approximate its velocity with a forward difference at }t=19\end{align*}$

Possible Answers:

$\frac{27}{4}$

$\frac{27}{2}$

Correct answer:

$\frac{27}{2}$

Explanation:

$\begin{align*}&\text{Velocity is the time derivative of position. However, since}\\&\text{we do not have a position function, we cannot take the derivative;}\\&\text{we must use an approximation. To find a forward derivative approximation,}\\&\text{we consider function two values: one at the point of interest}\\&\text{and another succeeding it, and finally that distance between}\\&\text{the two. Mathematically, this follows the form:}\\&f'(x)=\frac{f(x+h)-f(x)}{h}\\&\text{Where h is the size of the increment between points on the x-axis.}\\&\text{In approximating a derivative, it is generally prudent to minimize}\\&\text{the distance between function values. In other words, it’s a}\\&\text{good idea to look at two adjacent points. For our problem, that’d}\\&\text{be:}\\&f(19)\text{ and }f(23)\\&\text{Applying the formula above, with }h=4\text{, we find:}\\&v(19)\approx\frac{30-(-24)}{4}\\&v(19)\approx\frac{27}{2}\end{align*}$

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Example Question #47 : Approximate Rate Of Change From Graphs And Tables Of Values

$\begin{align*}&t&p(t)\\&3&-5\\&7&11\\&11&-45\\&\text{Using the above table of positions,}\\&\text{make a forward difference velocity approximation at }t=3\end{align*}$

Possible Answers:

$-\frac{5}{4}$

Correct answer:

Explanation:

$\begin{align*}&\text{Velocity is the time derivative of position. However, since}\\&\text{we do not have a position function, we cannot take the derivative;}\\&\text{we must use an approximation. To find a forward derivative approximation,}\\&\text{we consider function two values: one at the point of interest}\\&\text{and another succeeding it, and finally that distance between}\\&\text{the two. Mathematically, this follows the form:}\\&f'(x)=\frac{f(x+h)-f(x)}{h}\\&\text{Where h is the size of the increment between points on the x-axis.}\\&\text{In approximating a derivative, it is generally prudent to minimize}\\&\text{the distance between function values. In other words, it’s a}\\&\text{good idea to look at two adjacent points. For our problem, that’d}\\&\text{be:}\\&f(3)\text{ and }f(7)\\&\text{Applying the formula above, with }h=4\text{, we find:}\\&v(3)\approx\frac{11-(-5)}{4}\\&v(3)\approx4\end{align*}$

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Example Question #112 : Derivative At A Point

$\begin{align*}&t&p(t)\\&0&-50\\&1&-8\\&2&16\\&\text{The above table lists a particle's postion at different times.}\\&\text{Approximate its velocity with a forward difference at }t=1\end{align*}$

Possible Answers:

Correct answer:

Explanation:

$\begin{align*}&\text{Velocity is the time derivative of position. However, since}\\&\text{we do not have a position function, we cannot take the derivative;}\\&\text{we must use an approximation. To find a forward derivative approximation,}\\&\text{we consider function two values: one at the point of interest}\\&\text{and another succeeding it, and finally that distance between}\\&\text{the two. Mathematically, this follows the form:}\\&f'(x)=\frac{f(x+h)-f(x)}{h}\\&\text{Where h is the size of the increment between points on the x-axis.}\\&\text{In approximating a derivative, it is generally prudent to minimize}\\&\text{the distance between function values. In other words, it’s a}\\&\text{good idea to look at two adjacent points. For our problem, that’d}\\&\text{be:}\\&f(1)\text{ and }f(2)\\&\text{Applying the formula above, with }h=1\text{, we find:}\\&v(1)\approx\frac{16-(-8)}{1}\\&v(1)\approx24\end{align*}$

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

Example Question #111 : Derivative At A Point

Example Question #43 : Approximate Rate Of Change From Graphs And Tables Of Values

Example Question #41 : Approximate Rate Of Change From Graphs And Tables Of Values

Example Question #45 : Approximate Rate Of Change From Graphs And Tables Of Values

Example Question #46 : Approximate Rate Of Change From Graphs And Tables Of Values

Example Question #47 : Approximate Rate Of Change From Graphs And Tables Of Values

Example Question #112 : Derivative At A Point

All AP Calculus AB Resources

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