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AP Calculus AB : Derivative interpreted as an instantaneous rate of change

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Example Questions

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:

-7.85

0.33

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 #61 : Concept Of The Derivative

$\begin{align*}&\text{The position of a particle at a given time is given by the function:}\\&p(t)=3t + cos(\frac{(7\cdot \pi t)}{2}) + sin(\frac{(9\cdot \pi t)}{2}) - t^{2} - 6t^{3} + 3\\&\text{What is its velocity at time }t=10\text{?}\end{align*}$

Possible Answers:

-607.2

-606.8

-1831.14

-241.1

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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Example Question #61 : Concept Of The Derivative

$\begin{align*}&\text{A particle at any point in time has the position :}\\&p(t)=sin(8\cdot \pi t) + 2\\&\text{Determine its velocity at time }t=4\end{align*}$

Possible Answers:

10.5

0.5

25.13

Correct answer:

25.13

Explanation:

$\begin{align*}&\text{Velocity is the time rate of change of position. In lay terms,}\\&\text{it is an instantaneous measure of speed and direction. However,}\\&\text{going back to that first definition, note that it is a rate,}\\&\text{and whenever the term rate is used, derivatives should come}\\&\text{to mind. The rate of change of position with respect to time}\\&\text{is the derivative of position with respect to time:}\\&v(t)=\frac{dp(t)}{dt}\\&\text{Taking the derivative of our position function, }sin(8\cdot \pi t) + 2\\&\text{We find the velocity function: }8\cdot \pi cos(8\cdot \pi t)\\&\text{Plugging in our value of time we then get:}\\&v(4)=8\cdot \pi cos(8\cdot \pi (4))\\&v(4)=25.13\end{align*}$

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Example Question #11 : Derivative Interpreted As An Instantaneous Rate Of Change

$\begin{align*}&\text{At any given time, a particle has the position :}\\&p(t)=cos(4\cdot \pi t) - 2t - sin(2\cdot \pi t) - 3\\&\text{Calculate its acceleration at time }t=4\end{align*}$

Possible Answers:

-0.63

-8.28

-157.91

Correct answer:

-157.91

Explanation:

$\begin{align*}&\text{Just as velocity is the rate of change of position with respect}\\&\text{to time, acceleration is the rate of change of velocity with}\\&\text{respect to time. That is to say, acceleration is a measure of}\\&\text{how quickly velocity is changing. As velocity is the time derivative}\\&\text{of position, acceleration is the time derivative of velocity:}\\&a(t)=\frac{dv(t)}{dt}=\frac{d^2p(t)}{dt^2}\\&\text{Taking the derivative of our position function, }cos(4\cdot \pi t) - 2t - sin(2\cdot \pi t) - 3\\&\text{We find the velocity function: }- 2\cdot \pi cos(2\cdot \pi t) - 4\cdot \pi sin(4\cdot \pi t) - 2\\&\text{Taking the derivative of velocity gives, in turn, acceleration:}4\cdot \pi ^{2}sin(2\cdot \pi t) - 16\cdot \pi ^{2}cos(4\cdot \pi t)\\&\text{Now, to find the acceleration at our time of interest:}\\&a(4)=4\cdot \pi ^{2}sin(2\cdot \pi (4)) - 16\cdot \pi ^{2}cos(4\cdot \pi (4))\\&a(4)=-157.91\end{align*}$

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Example Question #69 : Concept Of The Derivative

$\begin{align*}&\text{A particle at any point in time has the position :}\\&p(t)=5t + sin(3\cdot \pi t) - 6t^{2} + 1\\&\text{Determine its acceleration at time }t=3\end{align*}$

Possible Answers:

-4.22

-40.42

-18.28

Correct answer:

Explanation:

$\begin{align*}&\text{Just as velocity is the rate of change of position with respect}\\&\text{to time, acceleration is the rate of change of velocity with}\\&\text{respect to time. That is to say, acceleration is a measure of}\\&\text{how quickly velocity is changing. As velocity is the time derivative}\\&\text{of position, acceleration is the time derivative of velocity:}\\&a(t)=\frac{dv(t)}{dt}=\frac{d^2p(t)}{dt^2}\\&\text{Taking the derivative of our position function, }5t + sin(3\cdot \pi t) - 6t^{2} + 1\\&\text{We find the velocity function: }3\cdot \pi cos(3\cdot \pi t) - 12t + 5\\&\text{Taking the derivative of velocity gives, in turn, acceleration:}- 9\cdot \pi ^{2}sin(3\cdot \pi t) - 12\\&\text{Now, to find the acceleration at our time of interest:}\\&a(3)=- 9\cdot \pi ^{2}sin(3\cdot \pi (3)) - 12\\&a(3)=-12\end{align*}$

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Example Question #11 : Derivative Interpreted As An Instantaneous Rate Of Change

$\begin{align*}&\text{At any given time, a particle has the position :}\\&p(t)=3t^{2} - sin(9\cdot \pi t) - 4t + 2\\&\text{Calculate its acceleration at time }t=6\end{align*}$

Possible Answers:

3.73

2.39

Correct answer:

Explanation:

$\begin{align*}&\text{Just as velocity is the rate of change of position with respect}\\&\text{to time, acceleration is the rate of change of velocity with}\\&\text{respect to time. That is to say, acceleration is a measure of}\\&\text{how quickly velocity is changing. As velocity is the time derivative}\\&\text{of position, acceleration is the time derivative of velocity:}\\&a(t)=\frac{dv(t)}{dt}=\frac{d^2p(t)}{dt^2}\\&\text{Taking the derivative of our position function, }3t^{2} - sin(9\cdot \pi t) - 4t + 2\\&\text{We find the velocity function: }6t - 9\cdot \pi cos(9\cdot \pi t) - 4\\&\text{Taking the derivative of velocity gives, in turn, acceleration:}81\cdot \pi ^{2}sin(9\cdot \pi t) + 6\\&\text{Now, to find the acceleration at our time of interest:}\\&a(6)=81\cdot \pi ^{2}sin(9\cdot \pi (6)) + 6\\&a(6)=6\end{align*}$

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AP Calculus AB : Derivative interpreted as an instantaneous rate of change

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #61 : Concept Of The Derivative

Example Question #11 : Derivative Interpreted As An Instantaneous Rate Of Change

Example Question #61 : Concept Of The Derivative

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

Example Question #61 : Concept Of The Derivative

Example Question #11 : Derivative Interpreted As An Instantaneous Rate Of Change

Example Question #69 : Concept Of The Derivative

Example Question #11 : Derivative Interpreted As An Instantaneous Rate Of Change

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