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AP Calculus AB : AP Calculus AB

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Example Questions

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:

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

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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:

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

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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:

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

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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:

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

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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:

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

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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:

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

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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:

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

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Example Question #13 : Derivative Interpreted As An Instantaneous Rate Of Change

$\begin{align*}&\text{A particle at any point in time has the position :}\\&p(t)=cos(\frac{(\pi t)}{2}) - 4\\&\text{Determine its velocity at time }t=3\end{align*}$

Possible Answers:

-1.33

-0.33

1.57

Correct answer:

1.57

Explanation:

$\begin{align*}&\text{Velocity is the time rate of change of position. In lay terms,}\\&\text{it is an instantaneous measure of speed and direction. However,}\\&\text{going back to that first definition, note that it is a rate,}\\&\text{and whenever the term rate is used, derivatives should come}\\&\text{to mind. The rate of change of position with respect to time}\\&\text{is the derivative of position with respect to time:}\\&v(t)=\frac{dp(t)}{dt}\\&\text{Taking the derivative of our position function, }cos(\frac{(\pi t)}{2}) - 4\\&\text{We find the velocity function: }-\frac{(\pi sin(\frac{(\pi t)}{2}))}{2}\\&\text{Plugging in our value of time we then get:}\\&v(3)=-\frac{(\pi sin(\frac{(\pi (3))}{2}))}{2}\\&v(3)=1.57\end{align*}$

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Example Question #14 : Derivative Interpreted As An Instantaneous Rate Of Change

$\begin{align*}&\text{A particle at any point in time has the position :}\\&p(t)=cos(2\cdot \pi t) - 6t + sin(7\cdot \pi t) + 3t^{2} + 5\\&\text{Determine its velocity at time }t=5\end{align*}$

Possible Answers:

2.01

10.2

-33.48

Correct answer:

2.01

Explanation:

$\begin{align*}&\text{Velocity is the time rate of change of position. In lay terms,}\\&\text{it is an instantaneous measure of speed and direction. However,}\\&\text{going back to that first definition, note that it is a rate,}\\&\text{and whenever the term rate is used, derivatives should come}\\&\text{to mind. The rate of change of position with respect to time}\\&\text{is the derivative of position with respect to time:}\\&v(t)=\frac{dp(t)}{dt}\\&\text{Taking the derivative of our position function, }cos(2\cdot \pi t) - 6t + sin(7\cdot \pi t) + 3t^{2} + 5\\&\text{We find the velocity function: }6t + 7\cdot \pi cos(7\cdot \pi t) - 2\cdot \pi sin(2\cdot \pi t) - 6\\&\text{Plugging in our value of time we then get:}\\&v(5)=6(5) + 7\cdot \pi cos(7\cdot \pi (5)) - 2\cdot \pi sin(2\cdot \pi (5)) - 6\\&v(5)=2.01\end{align*}$

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Example Question #15 : Derivative Interpreted As An Instantaneous Rate Of Change

$\begin{align*}&\text{The position of a particle at a given time is given by the function:}\\&p(t)=- 6t - cos(\frac{(7\cdot \pi t)}{2}) - cos(\frac{(15\cdot \pi t)}{2}) - 4t^{2} - 3\\&\text{What is its velocity at time }t=4\text{?}\end{align*}$

Possible Answers:

-23.25

668.07

Correct answer:

Explanation:

$\begin{align*}&\text{Velocity is the time rate of change of position. In lay terms,}\\&\text{it is an instantaneous measure of speed and direction. However,}\\&\text{going back to that first definition, note that it is a rate,}\\&\text{and whenever the term rate is used, derivatives should come}\\&\text{to mind. The rate of change of position with respect to time}\\&\text{is the derivative of position with respect to time:}\\&v(t)=\frac{dp(t)}{dt}\\&\text{Taking the derivative of our position function, }- 6t - cos(\frac{(7\cdot \pi t)}{2}) - cos(\frac{(15\cdot \pi t)}{2}) - 4t^{2} - 3\\&\text{We find the velocity function: }\frac{(7\cdot \pi sin(\frac{(7\cdot \pi t)}{2}))}{2}- 8t +\frac{ (15\cdot \pi sin(\frac{(15\cdot \pi t)}{2}))}{2}- 6\\&\text{Plugging in our value of time we then get:}\\&v(4)=\frac{(7\cdot \pi sin(\frac{(7\cdot \pi (4))}{2}))}{2}- 8(4) +\frac{ (15\cdot \pi sin(\frac{(15\cdot \pi (4))}{2}))}{2}- 6\\&v(4)=-38\end{align*}$

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AP Calculus AB : AP Calculus AB

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

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

Example Question #13 : Derivative Interpreted As An Instantaneous Rate Of Change

Example Question #14 : Derivative Interpreted As An Instantaneous Rate Of Change

Example Question #15 : Derivative Interpreted As An Instantaneous Rate Of Change

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