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

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Question

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

Answer

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

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:

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

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

We need v(1.5)

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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Question

Evaluate the following limit:

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

Answer

When approaches 0 both and 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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Question

Find the instantaneous rate of change for the function,

at the point .

Answer

Find the instantaneous rate of change for the function,

at the point .

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 ,

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

Therefore, is the instantaneous rate of change of the function

at the point .

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Question

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?

Answer

To find the instantaneous rate of change of the particle at time $t=\pi$ , we have to find the derivative of 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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Question

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

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

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

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

Answer

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 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 at the corresponding values of are:

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Question

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

Answer

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

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

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 is simply

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

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.

So, our answer is 123.5

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Question

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

Answer

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

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Question

$\begin{align}&\text{The position of a particle at a given time is given by the function:}\&p(t)=- 3t - cos(5\cdot \pi t) - sin(\frac{(7\cdot \pi t)}{2}) - 5\&\text{What is its velocity at time }t=3\text{?}\end{align}$

Answer

$\begin{align}&\text{Velocity is the time rate of change of position. In lay terms,}\&\text{it is an instantaneous measure of speed and direction. However,}\&\text{going back to that first definition, note that it is a rate,}\&\text{and whenever the term rate is used, derivatives should come}\&\text{to mind. The rate of change of position with respect to time}\&\text{is the derivative of position with respect to time:}\&v(t)=\frac{dp(t)}{dt}\&\text{Taking the derivative of our position function, }- 3t - cos(5\cdot \pi t) - sin(\frac{(7\cdot \pi t)}{2}) - 5\&\text{We find the velocity function: }5\cdot \pi sin(5\cdot \pi t) -\frac{ (7\cdot \pi cos(\frac{(7\cdot \pi t)}{2}))}{2}- 3\&\text{Plugging in our value of time we then get:}\&v(3)=5\cdot \pi sin(5\cdot \pi (3)) -\frac{ (7\cdot \pi cos(\frac{(7\cdot \pi (3))}{2}))}{2}- 3\&v(3)=-3\end{align}$

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Question

$\begin{align}&\text{The position of a particle at a given time is given by the function:}\&p(t)=cos(\frac{(9\cdot \pi t)}{2}) - 2t + cos(\frac{(17\cdot \pi t)}{2}) - 1\&\text{What is its velocity at time }t=1\text{?}\end{align}$

Answer

$\begin{align}&\text{Velocity is the time rate of change of position. In lay terms,}\&\text{it is an instantaneous measure of speed and direction. However,}\&\text{going back to that first definition, note that it is a rate,}\&\text{and whenever the term rate is used, derivatives should come}\&\text{to mind. The rate of change of position with respect to time}\&\text{is the derivative of position with respect to time:}\&v(t)=\frac{dp(t)}{dt}\&\text{Taking the derivative of our position function, }cos(\frac{(9\cdot \pi t)}{2}) - 2t + cos(\frac{(17\cdot \pi t)}{2}) - 1\&\text{We find the velocity function: }-\frac{ (9\cdot \pi sin(\frac{(9\cdot \pi t)}{2}))}{2}-\frac{ (17\cdot \pi sin(\frac{(17\cdot \pi t)}{2}))}{2}- 2\&\text{Plugging in our value of time we then get:}\&v(1)=-\frac{ (9\cdot \pi sin(\frac{(9\cdot \pi (1))}{2}))}{2}-\frac{ (17\cdot \pi sin(\frac{(17\cdot \pi (1))}{2}))}{2}- 2\&v(1)=-42.84\end{align}$

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Question

$\begin{align}&\text{At any given time, a particle has the position :}\&p(t)=6t - cos(\frac{(13\cdot \pi t)}{2}) + 1\&\text{Calculate its velocity at time }t=7\end{align}$

Answer

$\begin{align}&\text{Velocity is the time rate of change of position. In lay terms,}\&\text{it is an instantaneous measure of speed and direction. However,}\&\text{going back to that first definition, note that it is a rate,}\&\text{and whenever the term rate is used, derivatives should come}\&\text{to mind. The rate of change of position with respect to time}\&\text{is the derivative of position with respect to time:}\&v(t)=\frac{dp(t)}{dt}\&\text{Taking the derivative of our position function, }6t - cos(\frac{(13\cdot \pi t)}{2}) + 1\&\text{We find the velocity function: }\frac{(13\cdot \pi sin(\frac{(13\cdot \pi t)}{2}))}{2}+ 6\&\text{Plugging in our value of time we then get:}\&v(7)=\frac{(13\cdot \pi sin(\frac{(13\cdot \pi (7))}{2}))}{2}+ 6\&v(7)=-14.42\end{align}$

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Question

$\begin{align}&\text{The position of a particle at a given time is given by the function:}\&p(t)=3t - cos(\frac{(\pi t)}{2}) + sin(4\cdot \pi t) + 5t^{2} + 6t^{3} + 6\&\text{What is its velocity at time }t=6\text{?}\end{align}$

Answer

$\begin{align}&\text{Velocity is the time rate of change of position. In lay terms,}\&\text{it is an instantaneous measure of speed and direction. However,}\&\text{going back to that first definition, note that it is a rate,}\&\text{and whenever the term rate is used, derivatives should come}\&\text{to mind. The rate of change of position with respect to time}\&\text{is the derivative of position with respect to time:}\&v(t)=\frac{dp(t)}{dt}\&\text{Taking the derivative of our position function, }3t - cos(\frac{(\pi t)}{2}) + sin(4\cdot \pi t) + 5t^{2} + 6t^{3} + 6\&\text{We find the velocity function: }10t + 4\cdot \pi cos(4\cdot \pi t) +\frac{ (\pi sin(\frac{(\pi t)}{2}))}{2}+ 18t^{2} + 3\&\text{Plugging in our value of time we then get:}\&v(6)=10(6) + 4\cdot \pi cos(4\cdot \pi (6)) +\frac{ (\pi sin(\frac{(\pi (6))}{2}))}{2}+ 18(6)^{2} + 3\&v(6)=723.57\end{align}$

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Question

$\begin{align}&\text{A particle at any point in time has the position :}\&p(t)=- cos(\frac{(11\cdot \pi t)}{2}) - 6\&\text{Determine its velocity at time }t=2\end{align}$

Answer

$\begin{align}&\text{Velocity is the time rate of change of position. In lay terms,}\&\text{it is an instantaneous measure of speed and direction. However,}\&\text{going back to that first definition, note that it is a rate,}\&\text{and whenever the term rate is used, derivatives should come}\&\text{to mind. The rate of change of position with respect to time}\&\text{is the derivative of position with respect to time:}\&v(t)=\frac{dp(t)}{dt}\&\text{Taking the derivative of our position function, }- cos(\frac{(11\cdot \pi t)}{2}) - 6\&\text{We find the velocity function: }\frac{(11\cdot \pi sin(\frac{(11\cdot \pi t)}{2}))}{2}\&\text{Plugging in our value of time we then get:}\&v(2)=\frac{(11\cdot \pi sin(\frac{(11\cdot \pi (2))}{2}))}{2}\&v(2)=0\end{align}$

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Question

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

Answer

$\begin{align}&\text{Velocity is the time rate of change of position. In lay terms,}\&\text{it is an instantaneous measure of speed and direction. However,}\&\text{going back to that first definition, note that it is a rate,}\&\text{and whenever the term rate is used, derivatives should come}\&\text{to mind. The rate of change of position with respect to time}\&\text{is the derivative of position with respect to time:}\&v(t)=\frac{dp(t)}{dt}\&\text{Taking the derivative of our position function, }sin(\frac{(5\cdot \pi t)}{2}) + 2\&\text{We find the velocity function: }\frac{(5\cdot \pi cos(\frac{(5\cdot \pi t)}{2}))}{2}\&\text{Plugging in our value of time we then get:}\&v(6)=\frac{(5\cdot \pi cos(\frac{(5\cdot \pi (6))}{2}))}{2}\&v(6)=-7.85\end{align}$

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Question

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

Answer

$\begin{align}&\text{Velocity is the time rate of change of position. In lay terms,}\&\text{it is an instantaneous measure of speed and direction. However,}\&\text{going back to that first definition, note that it is a rate,}\&\text{and whenever the term rate is used, derivatives should come}\&\text{to mind. The rate of change of position with respect to time}\&\text{is the derivative of position with respect to time:}\&v(t)=\frac{dp(t)}{dt}\&\text{Taking the derivative of our position function, }3t + cos(\frac{(7\cdot \pi t)}{2}) + sin(\frac{(9\cdot \pi t)}{2}) - t^{2} - 6t^{3} + 3\&\text{We find the velocity function: }\frac{(9\cdot \pi cos(\frac{(9\cdot \pi t)}{2}))}{2}- 2t -\frac{ (7\cdot \pi sin(\frac{(7\cdot \pi t)}{2}))}{2}- 18t^{2} + 3\&\text{Plugging in our value of time we then get:}\&v(10)=\frac{(9\cdot \pi cos(\frac{(9\cdot \pi (10))}{2}))}{2}- 2(10) -\frac{ (7\cdot \pi sin(\frac{(7\cdot \pi (10))}{2}))}{2}- 18(10)^{2} + 3\&v(10)=-1831.14\end{align}$

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Question

$\begin{align}&\text{A particle at any point in time has the position :}\&p(t)=cos(\frac{(\pi t)}{2}) - 4\&\text{Determine its velocity at time }t=3\end{align}$

Answer

$\begin{align}&\text{Velocity is the time rate of change of position. In lay terms,}\&\text{it is an instantaneous measure of speed and direction. However,}\&\text{going back to that first definition, note that it is a rate,}\&\text{and whenever the term rate is used, derivatives should come}\&\text{to mind. The rate of change of position with respect to time}\&\text{is the derivative of position with respect to time:}\&v(t)=\frac{dp(t)}{dt}\&\text{Taking the derivative of our position function, }cos(\frac{(\pi t)}{2}) - 4\&\text{We find the velocity function: }-\frac{(\pi sin(\frac{(\pi t)}{2}))}{2}\&\text{Plugging in our value of time we then get:}\&v(3)=-\frac{(\pi sin(\frac{(\pi (3))}{2}))}{2}\&v(3)=1.57\end{align}$

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Question

$\begin{align}&\text{A particle at any point in time has the position :}\&p(t)=cos(2\cdot \pi t) - 6t + sin(7\cdot \pi t) + 3t^{2} + 5\&\text{Determine its velocity at time }t=5\end{align}$

Answer

$\begin{align}&\text{Velocity is the time rate of change of position. In lay terms,}\&\text{it is an instantaneous measure of speed and direction. However,}\&\text{going back to that first definition, note that it is a rate,}\&\text{and whenever the term rate is used, derivatives should come}\&\text{to mind. The rate of change of position with respect to time}\&\text{is the derivative of position with respect to time:}\&v(t)=\frac{dp(t)}{dt}\&\text{Taking the derivative of our position function, }cos(2\cdot \pi t) - 6t + sin(7\cdot \pi t) + 3t^{2} + 5\&\text{We find the velocity function: }6t + 7\cdot \pi cos(7\cdot \pi t) - 2\cdot \pi sin(2\cdot \pi t) - 6\&\text{Plugging in our value of time we then get:}\&v(5)=6(5) + 7\cdot \pi cos(7\cdot \pi (5)) - 2\cdot \pi sin(2\cdot \pi (5)) - 6\&v(5)=2.01\end{align}$

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Question

$\begin{align}&\text{The position of a particle at a given time is given by the function:}\&p(t)=- 6t - cos(\frac{(7\cdot \pi t)}{2}) - cos(\frac{(15\cdot \pi t)}{2}) - 4t^{2} - 3\&\text{What is its velocity at time }t=4\text{?}\end{align}$

Answer

$\begin{align}&\text{Velocity is the time rate of change of position. In lay terms,}\&\text{it is an instantaneous measure of speed and direction. However,}\&\text{going back to that first definition, note that it is a rate,}\&\text{and whenever the term rate is used, derivatives should come}\&\text{to mind. The rate of change of position with respect to time}\&\text{is the derivative of position with respect to time:}\&v(t)=\frac{dp(t)}{dt}\&\text{Taking the derivative of our position function, }- 6t - cos(\frac{(7\cdot \pi t)}{2}) - cos(\frac{(15\cdot \pi t)}{2}) - 4t^{2} - 3\&\text{We find the velocity function: }\frac{(7\cdot \pi sin(\frac{(7\cdot \pi t)}{2}))}{2}- 8t +\frac{ (15\cdot \pi sin(\frac{(15\cdot \pi t)}{2}))}{2}- 6\&\text{Plugging in our value of time we then get:}\&v(4)=\frac{(7\cdot \pi sin(\frac{(7\cdot \pi (4))}{2}))}{2}- 8(4) +\frac{ (15\cdot \pi sin(\frac{(15\cdot \pi (4))}{2}))}{2}- 6\&v(4)=-38\end{align}$

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Question

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

Answer

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

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Question

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

Answer

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

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Question

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

Answer

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

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