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

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Example Questions

Example Question #1 : Derivative Interpreted As An Instantaneous Rate Of Change

Evaluate the following limit:

$\lim_{x\rightarrow 0} (\frac{1-cos2x}{sinx})$

Possible Answers:

$\infty$

Correct answer:

Explanation:

When approaches 0 both sinx and (1-cos2x) will approach . Therefore, L’Hopital’s Rule can be applied here. Take the derivatives of the numerator and denominator and try the limit again:

$\lim_{x\rightarrow 0} (\frac{1-cos2x}{sinx})= \lim_{x\rightarrow 0} (\frac{2sin2x}{cosx})=\frac{0}{1}=0$

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Example Question #2 : Derivative Interpreted As An Instantaneous Rate Of Change

Find the instantaneous rate of change for the function,

y = 2x^3 - x-4

at the point (-1,-7) .

Possible Answers:

293

712

Correct answer:

Explanation:

Find the instantaneous rate of change for the function,

y = 2x^3 - x-4

at the point (-1,-7) .

1) First compute the derivative of the function, since this will give us the instantaneous rate of change of the function as a function of .

$\frac{dy}{dx}=\frac{d}{dx}(2x^3-x-4)$

$\frac{dy}{dx}=\frac{d}{dx}(2x^3)-\frac{d}{dx}(x)-\frac{d}{dx}(4)$

$\frac{dy}{dx}= 6x^2-1$

2) Now evaluate the derivative at the value x = -1 ,

$\frac{dy}{dx}_{|x=-1}= 6(-1)^2-1 = 5$

Therefore, is the instantaneous rate of change of the function

y = 2x^3 - x-4 at the point (-1,-7) .

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Example Question #3 : Derivative Interpreted As An Instantaneous Rate Of Change

A particle is traveling in a straight line along the x-axis with position function $x(t) = t+\sin(t)$ . What is the instantaneous rate of change in the particle's position at time $\pi$ seconds?

Possible Answers:

$\pi+1$

$\pi$

$\pi-1$

Correct answer:

Explanation:

To find the instantaneous rate of change of the particle at time $t=\pi$ , we have to find the derivative of x(t) and plug $\pi$ into it.

$x'(t) :=v(t) = 1+\cos(t)$ .

And

$v(\pi) = 1+\cos(\pi) = 1+-1 = 0$ .

Hence the instantaneous rate of change in position (or just 'velocity') of the particle at $t=\pi$ is . (At that very instant, the particle is not moving.)

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Example Question #4 : Derivative Interpreted As An Instantaneous Rate Of Change

Find the function values and as well as the instantaneous rate of change for the function $g(x)=x\sin(x)$ corresponding to the following values of

$1)\: \: \: \: \: \: \: x =\frac{\pi}{2}$

$2)\: \: \: \: \: \: \: x =\frac{\pi}{4}$

$3)\: \: \: \: \: \: \: x =\pi$

Possible Answers:

B2wrng4

Fixedcorrectanswer

B2wrng3

B2wrng4

Correct answer:

Fixedcorrectanswer

Explanation:

Find the instantaneous rate of change for the function $g(x)=x\sin(x)$ corresponding to the following values of

$1)\: \: \: \: \: \: \: x =\frac{\pi}{2}$

$2)\: \: \: \: \: \: \: x =\frac{\pi}{4}$

$3)\: \: \: \: \: \: \: x =\pi$

Evaluate the function at each value of

$g\left(\frac{\pi}{2} \right )=\frac{\pi}{2}sin\left(\frac{\pi}{2} \right )=\frac{\pi}{2}$

$g\left(\frac{\pi}{4} \right )=\frac{\pi}{4}\sin\left(\frac{\pi}{4} \right )=\frac{\pi}{4}\frac{\sqrt{2}}{2}=\frac{\pi\sqrt{2}}{8}$

$g(\pi)=\pi \sin(\pi)=0$

The instantaneous rate of change at any point (x, g(x)) will be given by the derivative at that point. First compute the derivative of the function:

$g(x)=x\sin(x)$

Apply the product rule:

$g'(x)=x\frac{d}{dx}\sin(x)+\sin(x)\frac{d}{dx}(x)$

$= x\cos(x)+\sin(x)$

Therefore,

$g'(x)=x\cos(x)+\sin(x)$

Now evaluate the derivative for each given value of :

$1)\: \: \: \: \: \: \: x =\frac{\pi}{2}\: \: \: \:\rightarrow\: \: \: \:g'\left(\frac{\pi}{2} \right )\: \: =\: \: \frac{\pi}{2}\cos\left(\frac{\pi}{2} \right )+\sin\left(\frac{\pi}{2} \right )\: \:=\:1$

$2)\: \: \: \: \: \: \: x =\frac{\pi}{4}\: \: \: \:\rightarrow\: \: \: \:g'\left(\frac{\pi}{4} \right )\: \: =\: \: \frac{\pi}{4}\cos\left(\frac{\pi}{4} \right )+\sin\left(\frac{\pi}{4} \right )\: \:=\:\frac{\sqrt{2}}{2}\left(1+\frac{\pi}{4} \right )=\frac{\sqrt{2}}{8}(4+\pi)$

$3)\: \: \: \: \: \: \: x =\pi\: \: \: \:\rightarrow\: \: \: \:g'(\pi)\: \: =\: \: \pi\cos(\pi) +\sin(\pi) \: \:=\:\pi$

Therefore, the instantaneous rate of change of the function g(x) at the corresponding values of are:

Fixedcorrectanswer

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Example Question #5 : Derivative Interpreted As An Instantaneous Rate Of Change

Given that v(t) is the velocity of a particle, find the particle's acceleration when t=3.

v(t)=3t^2+6e^x-15t

Possible Answers:

123.5

12.65

Not enough information provided

Correct answer:

123.5

Explanation:

Given that v(t) is the velocity of a particle, find the particle's acceleration when t=3.

v(t)=3t^2+6e^x-15t

We are given velocity and asked to find acceleration. Our first step should be to find the derivative.

v(t)=3t^2+6e^x-15t

We can use our standard power rule for our 1st and 3rd terms, but we need to remember something else for our second term. Namely, that the derivative of e^x is simply e^x

With that in mind, let;s find v'(t)

v(t)=3t^2+6e^x-15t

v'(t)=6t+6e^x-15

Now, for the final push, we need to find the acceleration when t=3. We do this by plugging in 3 for t and simplifying.

v'(t)=6(3)+6e^3-15=18+120.5-15=123.5

So, our answer is 123.5

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Example Question #1 : Derivative Interpreted As An Instantaneous Rate Of Change

$\begin{align*}&\text{The position of a particle at a given time is given by the function:}\\&p(t)=6t - sin(\frac{(5\cdot \pi t)}{2}) + 5\\&\text{What is its velocity at time }t=6\text{?}\end{align*}$

Possible Answers:

13.85

6.83

Correct answer:

13.85

Explanation:

$\begin{align*}&\text{Velocity is the time rate of change of position. In lay terms,}\\&\text{it is an instantaneous measure of speed and direction. However,}\\&\text{going back to that first definition, note that it is a rate,}\\&\text{and whenever the term rate is used, derivatives should come}\\&\text{to mind. The rate of change of position with respect to time}\\&\text{is the derivative of position with respect to time:}\\&v(t)=\frac{dp(t)}{dt}\\&\text{Taking the derivative of our position function }6t - sin(\frac{(5\cdot \pi t)}{2}) + 5\\&\text{We find the velocity function: }6 -\frac{ (5\cdot \pi cos(\frac{(5\cdot \pi t)}{2}))}{2}\\&\text{Plugging in values we then get:}\\&v(6)=6 -\frac{ (5\cdot \pi cos(\frac{(5\cdot \pi (6))}{2}))}{2}\\&v(6)=13.85\end{align*}$

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

Study concepts, example questions & explanations for AP Calculus AB

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Example Questions

Example Question #1 : Derivative Interpreted As An Instantaneous Rate Of Change

Example Question #2 : Derivative Interpreted As An Instantaneous Rate Of Change

Example Question #3 : Derivative Interpreted As An Instantaneous Rate Of Change

Example Question #4 : Derivative Interpreted As An Instantaneous Rate Of Change

Example Question #5 : Derivative Interpreted As An Instantaneous Rate Of Change

Example Question #1 : Derivative Interpreted As An Instantaneous Rate Of Change

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 #2 : Derivative Interpreted As An Instantaneous Rate Of Change

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