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AP Calculus AB : Derivatives

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Example Questions

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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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Example Question #711 : Derivatives

The motion of a particle on Mars at any time can be represented by the following equation:

p(t)=3t^2+1

Find the velocity of the particle at t=5

Possible Answers:

v(5)=15

v(5)=76

v(5)=31

v(5)=18

Correct answer:

v(5)=31

Explanation:

We can represent the velocity as the instantaneous rate of change of a particles motion, that is $v(t)=\frac{\mathrm{d}p }{\mathrm{d} t}$ . We solve the problem by taking the derivative of the function that describes the particles motion with respect to time

v(t)=p'(t)=6t+1

We then evaluate the velocity equation at the specified time, which in this case was t=5

v(5)=6*5+1=31

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Example Question #1271 : Ap Calculus Ab

A particles position at any point in time can be modeled by the following equation

p(t)=2t^2+e^t

Find the velocity of the particle when t=2

Possible Answers:

v(2)=8+2e^2

v(2)=4-e

v(2)=8+e^2

$v(2)=\frac{3}{2}+e^2$

Correct answer:

v(2)=8+e^2

Explanation:

The velocity is the instantaneous rate of change of the particles position with respect to time ( $\frac{\mathrm{d}p }{\mathrm{d} t}$ ), so by taking the derivative of the function we can find the particles velocity ant any point in time

p'(t)=v(t)=4t+e^t

Evaluating at the time we specified in the problem statement, we get

v(2)=8+e^2

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Example Question #1272 : Ap Calculus Ab

The amount of bacteria in a dish (as a function of time) is modeled according to the following expression:

$b(t)=10e^{2t}-2t$

At what time will the amount of bacteria be growing at a rate of 38?

Possible Answers:

$\frac{\ln2}{2}$

None of the other answers

$\frac{40}{e^{2}}$

$\ln 2$

Correct answer:

$\frac{\ln2}{2}$

Explanation:

To determine the rate of change of the amount of bacteria at a specific time, we must take the derivative of the function:

$b'(t)=20e^{2t}-2$

The derivative was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u\frac{\mathrm{d} u}{\mathrm{d} x}$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x}ax=a$

Now, we are asked to find the time corresponding to a certain rate of growth of the bacteria. This is the output of the derivative function, from a certain time input into the function. So, we must solve for the time that gets us 38 as our output:

$b'(t)=20e^{2t}-2=38$

$20e^{2t}=40$

$e^{2t}=2$

$2t=\ln 2$

$t= \frac{\ln 2}{2}$

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Example Question #1273 : Ap Calculus Ab

An electron in motion can be described by the following equation, which gives its position at any point in time

p(t)=5t^2+2t+1

Find the speed of the electron when t=4

Possible Answers:

Correct answer:

Explanation:

To find the velocity of the electron, we take the derivative of its position equation. Velocity is the instantaneous rate of change ofposition with respect to time, which we will evaluate at a specified time

$p'(t)=v(t)=\frac{\mathrm{d} }{\mathrm{d} t}(5t^2+2t+1)=10t+2$

Evaluating when t=4

v(4)=10(4)+2=42

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AP Calculus AB : Derivatives

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

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

Example Question #711 : Derivatives

Example Question #1271 : Ap Calculus Ab

Example Question #1272 : Ap Calculus Ab

Example Question #1273 : Ap Calculus Ab

All AP Calculus AB Resources

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