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AP Calculus AB : AP Calculus AB

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3 Diagnostic Tests 164 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #27 : Derivative Interpreted As An Instantaneous Rate Of Change

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

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 #711 : Derivatives

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

Possible Answers:

20.42

416.99

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 #29 : Derivative Interpreted As An Instantaneous Rate Of Change

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

Possible Answers:

875.57

176

28.28

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 #21 : Derivative Interpreted As An Instantaneous Rate Of Change

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)=18

v(5)=31

v(5)=76

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:

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

v(2)=8+2e^2

v(2)=4-e

v(2)=8+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 #712 : Derivatives

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:

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

None of the other answers

$\ln 2$

$\frac{\ln2}{2}$

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 #711 : Derivatives

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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Example Question #31 : Derivative Interpreted As An Instantaneous Rate Of Change

A car going up a hill can be described by the following equation:

p(t)=5t^2+3

How fast is the car moving at the time t=3

Possible Answers:

v(3)=33

v(3)=25

v(3)=48

v(3)=30

Correct answer:

v(3)=30

Explanation:

The cars velocity is the instantaneous rate of change of the position with respect to time, $\frac{\mathrm{dp} }{\mathrm{d} t}$ , so by taking the derivative of the function we can get the velocity equation which can be evaluated for any time

$v(t)=\frac{\mathrm{dp} }{\mathrm{d} t}=\frac{\mathrm{d} }{\mathrm{d} t}(5t^2+3)=10t$

Evaluated at t=3 , we get

v(3)=10*3=30

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Example Question #82 : Concept Of The Derivative

Given that h(t) represents the height of a frisbee as a function of time, find the frisbee's velocity when t=1.5 .

h(t)=12t^5-45t^4+13t^2+14

Possible Answers:

v=-264.75

v=3.75

v=-344.25

v=26.75

Correct answer:

v=-264.75

Explanation:

Given that h(t) represents the height of a frisbee as a function of time, find the frisbee's velocity when t=1.5 .

h(t)=12t^5-45t^4+13t^2+14

To find the velocity function from a height or position function, we simply need to find the first derivative of the function. Then, to find the velocity at a given value of t, simply plug the value of t into the velocity function.

Recall the power rule:

$\frac{dy}{dx}x^n=nx^{n-1}$

This states that we can find the derivative of any polynomial by taking each term, multiplying by its exponent, and then subtracting one from the exponent.

Doing so yields:

h(t)=12t^5-45t^4+13t^2+14

h'(t)=(5)12t^4-(4)45t^3+(2)13t

h'(t)=60t^4-180t^3+26t

Notice that our 14 dropped out. Any constant terms will drop our while differentiating.

Change it to v(t) and we are almost there.

v(t)=60t^4-180t^3+26t

We need v(1.5)

v(1.5)=60(1.5)^4-180(1.5)^3+26(1.5)=303.75-607.5+39=-264.75

So, we get -264.75. This might seem like a strange velocity for a frisbee, but we don't know the units, so it might make sense within the relevant system.

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Example Question #83 : Concept Of The Derivative

Given that h(t) represents the height of a frisbee as a function of time, find the frisbee's acceleration when t=-1 .

h(t)=12t^5-45t^4+13t^2+14

Possible Answers:

a=233.65

a=75.44

a=-754

Correct answer:

a=-754

Explanation:

Given that h(t) represents the height of a frisbee as a function of time, find the frisbee's acceleration when t=-1 .

h(t)=12t^5-45t^4+13t^2+14

To find the acceleration function, we need to recall that velocity is the first derivative of position, and that acceleration is the first derivative of velocity. In other words, we need to find our second derivative. Once we have that, we need to plug in -1 for t.

Recall the power rule:

$\frac{dy}{dx}x^n=nx^{n-1}$

This states that we can find the derivative of any polynomial by taking each term, multiplying by its exponent, and then subtracting one from the exponent.

Doing so yields:

h(t)=12t^5-45t^4+13t^2+14

h'(t)=(5)12t^4-(4)45t^3+(2)13t

h'(t)=60t^4-180t^3+26t

Notice that our 14 dropped out. Any constant terms will drop our while differentiating.

Change it to v(t) and we are almost there.

v(t)=60t^4-180t^3+26t

Next, find the derivative of v(t), again with the power rule.

a(t)=v'(t)=(4)60t^3-(3)180t^2+26

Making our acceleration function:

a(t)=240t^3-540t^2+26

Now, finally, plug in -1 for t and solve.

a(t)=240(-1)^3-540(-1)^2+26=-240-540+26=-754

So, our acceleration is:

-754

We have no units, so although this number may seem a little strange, we can;t really say much about it.

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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 #27 : Derivative Interpreted As An Instantaneous Rate Of Change

Example Question #711 : Derivatives

Example Question #29 : Derivative Interpreted As An Instantaneous Rate Of Change

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

Example Question #1271 : Ap Calculus Ab

Example Question #712 : Derivatives

Example Question #711 : Derivatives

Example Question #31 : Derivative Interpreted As An Instantaneous Rate Of Change

Example Question #82 : Concept Of The Derivative

Example Question #83 : Concept Of The Derivative

All AP Calculus AB Resources

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