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AP Calculus AB : Concept of the Derivative

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Example Questions

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Example Question #20 : 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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Example Question #21 : Derivative Interpreted As An Instantaneous Rate Of Change

$\begin{align*}&\text{A particle at any point in time has the position :}\\&p(t)=cos(\frac{(15\cdot \pi t)}{2}) - sin(\frac{(7\cdot \pi t)}{2}) + 5\\&\text{Determine its acceleration at time }t=1\end{align*}$

Possible Answers:

-120.9

23.56

34.56

Correct answer:

-120.9

Explanation:

$\begin{align*}\\&a(t)=\frac{dv(t)}{dt}=\frac{d^2p(t)}{dt^2}\\&\text{Taking the derivative of our position function, }cos(\frac{(15\cdot \pi t)}{2}) - sin(\frac{(7\cdot \pi t)}{2}) + 5\\&\text{We find the velocity function: }-\frac{ (7\cdot \pi cos(\frac{(7\cdot \pi t)}{2}))}{2}-\frac{ (15\cdot \pi sin(\frac{(15\cdot \pi t)}{2}))}{2}\\&\text{Taking the derivative of velocity gives, in turn, acceleration: }\frac{(49\cdot \pi ^{2}sin(\frac{(7\cdot \pi t)}{2}))}{4}-\frac{ (225\cdot \pi ^{2}cos(\frac{(15\cdot \pi t)}{2}))}{4}\\&\text{Now, to find the acceleration at our time of interest:}\\&a(1)=\frac{(49\cdot \pi ^{2}sin(\frac{(7\cdot \pi (1))}{2}))}{4}-\frac{ (225\cdot \pi ^{2}cos(\frac{(15\cdot \pi (1))}{2}))}{4}\\&a(1)=-120.9\end{align*}$

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Example Question #22 : Derivative Interpreted As An Instantaneous Rate Of Change

$\begin{align*}&\text{A particle at any point in time has the position :}\\&p(t)=5t^{3} - cos(\pi t) - 6t^{2} - 4t + 5\\&\text{Determine its acceleration at time }t=1\end{align*}$

Possible Answers:

8.13

Correct answer:

8.13

Explanation:

$\begin{align*}&\text{Just as velocity is the rate of change of position with respect}\\&\text{to time, acceleration is the rate of change of velocity with}\\&\text{respect to time. That is to say, acceleration is a measure of}\\&\text{how quickly velocity is changing. As velocity is the time derivative}\\&\text{of position, acceleration is the time derivative of velocity:}\\&a(t)=\frac{dv(t)}{dt}=\frac{d^2p(t)}{dt^2}\\&\text{Taking the derivative of our position function, }5t^{3} - cos(\pi t) - 6t^{2} - 4t + 5\\&\text{We find the velocity function: }\pi sin(\pi t) - 12t + 15t^{2} - 4\\&\text{Taking the derivative of velocity gives, in turn, acceleration: }30t + \pi ^{2}cos(\pi t) - 12\\&\text{Now, to find the acceleration at our time of interest:}\\&a(1)=30(1) + \pi ^{2}cos(\pi (1)) - 12\\&a(1)=8.13\end{align*}$

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Example Question #23 : Derivative Interpreted As An Instantaneous Rate Of Change

$\begin{align*}&\text{A particle at any point in time has the position :}\\&p(t)=sin(2\cdot \pi t) - 3t - 3t^{2} + 4t^{3} + 5\\&\text{Determine its acceleration at time }t=9\end{align*}$

Possible Answers:

921.28

210

102

32.73

Correct answer:

210

Explanation:

$\begin{align*}&\text{Just as velocity is the rate of change of position with respect}\\&\text{to time, acceleration is the rate of change of velocity with}\\&\text{respect to time. That is to say, acceleration is a measure of}\\&\text{how quickly velocity is changing. As velocity is the time derivative}\\&\text{of position, acceleration is the time derivative of velocity:}\\&a(t)=\frac{dv(t)}{dt}=\frac{d^2p(t)}{dt^2}\\&\text{Taking the derivative of our position function, }sin(2\cdot \pi t) - 3t - 3t^{2} + 4t^{3} + 5\\&\text{We find the velocity function: }2\cdot \pi cos(2\cdot \pi t) - 6t + 12t^{2} - 3\\&\text{Taking the derivative of velocity gives, in turn, acceleration: }24t - 4\cdot \pi ^{2}sin(2\cdot \pi t) - 6\\&\text{Now, to find the acceleration at our time of interest:}\\&a(9)=24(9) - 4\cdot \pi ^{2}sin(2\cdot \pi (9)) - 6\\&a(9)=210\end{align*}$

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Example Question #24 : Derivative Interpreted As An Instantaneous Rate Of Change

$\begin{align*}&\text{At any given time, a particle has the position :}\\&p(t)=cos(2\cdot \pi t) - 4t - cos(5\cdot \pi t) + 5t^{2} + 3t^{3} + 2\\&\text{Calculate its acceleration at time }t=4\end{align*}$

Possible Answers:

289.26

16.13

180

Correct answer:

289.26

Explanation:

$\begin{align*}&\text{Just as velocity is the rate of change of position with respect}\\&\text{to time, acceleration is the rate of change of velocity with}\\&\text{respect to time. That is to say, acceleration is a measure of}\\&\text{how quickly velocity is changing. As velocity is the time derivative}\\&\text{of position, acceleration is the time derivative of velocity:}\\&a(t)=\frac{dv(t)}{dt}=\frac{d^2p(t)}{dt^2}\\&\text{Taking the derivative of our position function, }cos(2\cdot \pi t) - 4t - cos(5\cdot \pi t) + 5t^{2} + 3t^{3} + 2\\&\text{We find the velocity function: }10t - 2\cdot \pi sin(2\cdot \pi t) + 5\cdot \pi sin(5\cdot \pi t) + 9t^{2} - 4\\&\text{Taking the derivative of velocity gives, in turn, acceleration: }18t - 4\cdot \pi ^{2}cos(2\cdot \pi t) + 25\cdot \pi ^{2}cos(5\cdot \pi t) + 10\\&\text{Now, to find the acceleration at our time of interest:}\\&a(4)=18(4) - 4\cdot \pi ^{2}cos(2\cdot \pi (4)) + 25\cdot \pi ^{2}cos(5\cdot \pi (4)) + 10\\&a(4)=289.26\end{align*}$

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Example Question #25 : Derivative Interpreted As An Instantaneous Rate Of Change

$\begin{align*}&\text{A particle at any point in time has the position :}\\&p(t)=1 - cos(\frac{(13\cdot \pi t)}{2})\\&\text{Determine its acceleration at time }t=6\end{align*}$

Possible Answers:

108.6

-416.99

0.06

Correct answer:

-416.99

Explanation:

$\begin{align*}&\text{Just as velocity is the rate of change of position with respect}\\&\text{to time, acceleration is the rate of change of velocity with}\\&\text{respect to time. That is to say, acceleration is a measure of}\\&\text{how quickly velocity is changing. As velocity is the time derivative}\\&\text{of position, acceleration is the time derivative of velocity:}\\&a(t)=\frac{dv(t)}{dt}=\frac{d^2p(t)}{dt^2}\\&\text{Taking the derivative of our position function, }1 - cos(\frac{(13\cdot \pi t)}{2})\\&\text{We find the velocity function: }\frac{(13\cdot \pi sin(\frac{(13\cdot \pi t)}{2}))}{2}\\&\text{Taking the derivative of velocity gives, in turn, acceleration: }\frac{(169\cdot \pi ^{2}cos(\frac{(13\cdot \pi t)}{2}))}{4}\\&\text{Now, to find the acceleration at our time of interest:}\\&a(6)=\frac{(169\cdot \pi ^{2}cos(\frac{(13\cdot \pi (6))}{2}))}{4}\\&a(6)=-416.99\end{align*}$

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Example Question #26 : Derivative Interpreted As An Instantaneous Rate Of Change

$\begin{align*}&\text{The position of a particle at a given time is given by the function:}\\&p(t)=3t^{2} - cos(\frac{(13\cdot \pi t)}{2}) - 4t + 2\\&\text{What is its acceleration at time }t=7\text{?}\end{align*}$

Possible Answers:

3.08

2.47

17.58

Correct answer:

Explanation:

$\begin{align*}&\text{Just as velocity is the rate of change of position with respect}\\&\text{to time, acceleration is the rate of change of velocity with}\\&\text{respect to time. That is to say, acceleration is a measure of}\\&\text{how quickly velocity is changing. As velocity is the time derivative}\\&\text{of position, acceleration is the time derivative of velocity:}\\&a(t)=\frac{dv(t)}{dt}=\frac{d^2p(t)}{dt^2}\\&\text{Taking the derivative of our position function, }3t^{2} - cos(\frac{(13\cdot \pi t)}{2}) - 4t + 2\\&\text{We find the velocity function: }6t +\frac{ (13\cdot \pi sin(\frac{(13\cdot \pi t)}{2}))}{2}- 4\\&\text{Taking the derivative of velocity gives, in turn, acceleration: }\frac{(169\cdot \pi ^{2}cos(\frac{(13\cdot \pi t)}{2}))}{4}+ 6\\&\text{Now, to find the acceleration at our time of interest:}\\&a(7)=\frac{(169\cdot \pi ^{2}cos(\frac{(13\cdot \pi (7))}{2}))}{4}+ 6\\&a(7)=6\end{align*}$

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Example Question #1271 : Ap Calculus Ab

$\begin{align*}&\text{At any given time, a particle has the position :}\\&p(t)=sin(5\cdot \pi t) - 4t - 2t^{2} - 2t^{3} - 4\\&\text{Calculate its acceleration at time }t=2\end{align*}$

Possible Answers:

-20.29

Correct answer:

Explanation:

$\begin{align*}&\text{Just as velocity is the rate of change of position with respect}\\&\text{to time, acceleration is the rate of change of velocity with}\\&\text{respect to time. That is to say, acceleration is a measure of}\\&\text{how quickly velocity is changing. As velocity is the time derivative}\\&\text{of position, acceleration is the time derivative of velocity:}\\&a(t)=\frac{dv(t)}{dt}=\frac{d^2p(t)}{dt^2}\\&\text{Taking the derivative of our position function, }sin(5\cdot \pi t) - 4t - 2t^{2} - 2t^{3} - 4\\&\text{We find the velocity function: }5\cdot \pi cos(5\cdot \pi t) - 4t - 6t^{2} - 4\\&\text{Taking the derivative of velocity gives, in turn, acceleration: }- 12t - 25\cdot \pi ^{2}sin(5\cdot \pi t) - 4\\&\text{Now, to find the acceleration at our time of interest:}\\&a(2)=- 12(2) - 25\cdot \pi ^{2}sin(5\cdot \pi (2)) - 4\\&a(2)=-28\end{align*}$

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Example Question #1271 : Ap Calculus Ab

$\begin{align*}&\text{A particle at any point in time has the position :}\\&p(t)=- t - sin(\frac{(13\cdot \pi t)}{2}) - 5\\&\text{Determine its acceleration at time }t=1\end{align*}$

Possible Answers:

416.99

20.42

Correct answer:

416.99

Explanation:

$\begin{align*}&\text{Just as velocity is the rate of change of position with respect}\\&\text{to time, acceleration is the rate of change of velocity with}\\&\text{respect to time. That is to say, acceleration is a measure of}\\&\text{how quickly velocity is changing. As velocity is the time derivative}\\&\text{of position, acceleration is the time derivative of velocity:}\\&a(t)=\frac{dv(t)}{dt}=\frac{d^2p(t)}{dt^2}\\&\text{Taking the derivative of our position function, }- t - sin(\frac{(13\cdot \pi t)}{2}) - 5\\&\text{We find the velocity function: }-\frac{ (13\cdot \pi cos(\frac{(13\cdot \pi t)}{2}))}{2}- 1\\&\text{Taking the derivative of velocity gives, in turn, acceleration: }\frac{(169\cdot \pi ^{2}sin(\frac{(13\cdot \pi t)}{2}))}{4}\\&\text{Now, to find the acceleration at our time of interest:}\\&a(1)=\frac{(169\cdot \pi ^{2}sin(\frac{(13\cdot \pi (1))}{2}))}{4}\\&a(1)=416.99\end{align*}$

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Example Question #1271 : Ap Calculus Ab

$\begin{align*}&\text{A particle at any point in time has the position :}\\&p(t)=3t + sin(4\cdot \pi t) - 2t^{2} + 3t^{3} - 2\\&\text{Determine its acceleration at time }t=10\end{align*}$

Possible Answers:

28.28

176

875.57

Correct answer:

176

Explanation:

$\begin{align*}&\text{Just as velocity is the rate of change of position with respect}\\&\text{to time, acceleration is the rate of change of velocity with}\\&\text{respect to time. That is to say, acceleration is a measure of}\\&\text{how quickly velocity is changing. As velocity is the time derivative}\\&\text{of position, acceleration is the time derivative of velocity:}\\&a(t)=\frac{dv(t)}{dt}=\frac{d^2p(t)}{dt^2}\\&\text{Taking the derivative of our position function, }3t + sin(4\cdot \pi t) - 2t^{2} + 3t^{3} - 2\\&\text{We find the velocity function: }4\cdot \pi cos(4\cdot \pi t) - 4t + 9t^{2} + 3\\&\text{Taking the derivative of velocity gives, in turn, acceleration: }18t - 16\cdot \pi ^{2}sin(4\cdot \pi t) - 4\\&\text{Now, to find the acceleration at our time of interest:}\\&a(10)=18(10) - 16\cdot \pi ^{2}sin(4\cdot \pi (10)) - 4\\&a(10)=176\end{align*}$

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AP Calculus AB : Concept of the Derivative

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #20 : Derivative Interpreted As An Instantaneous Rate Of Change

Example Question #21 : Derivative Interpreted As An Instantaneous Rate Of Change

Example Question #22 : Derivative Interpreted As An Instantaneous Rate Of Change

Example Question #23 : Derivative Interpreted As An Instantaneous Rate Of Change

Example Question #24 : Derivative Interpreted As An Instantaneous Rate Of Change

Example Question #25 : Derivative Interpreted As An Instantaneous Rate Of Change

Example Question #26 : Derivative Interpreted As An Instantaneous Rate Of Change

Example Question #1271 : Ap Calculus Ab

Example Question #1271 : Ap Calculus Ab

Example Question #1271 : Ap Calculus Ab

All AP Calculus AB Resources

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