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

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Question

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

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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

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

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

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Question

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

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

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Question

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

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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

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

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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

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

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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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Question

$\begin{align}&\text{At any given time, a particle has the position :}\&p(t)=3t^{2} - sin(9\cdot \pi t) - 4t + 2\&\text{Calculate its acceleration at time }t=6\end{align}$

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Answer

$\begin{align}&\text{Just as velocity is the rate of change of position with respect}\&\text{to time, acceleration is the rate of change of velocity with}\&\text{respect to time. That is to say, acceleration is a measure of}\&\text{how quickly velocity is changing. As velocity is the time derivative}\&\text{of position, acceleration is the time derivative of velocity:}\&a(t)=\frac{dv(t)}{dt}=\frac{d^2p(t)}{dt^2}\&\text{Taking the derivative of our position function, }3t^{2} - sin(9\cdot \pi t) - 4t + 2\&\text{We find the velocity function: }6t - 9\cdot \pi cos(9\cdot \pi t) - 4\&\text{Taking the derivative of velocity gives, in turn, acceleration:}81\cdot \pi ^{2}sin(9\cdot \pi t) + 6\&\text{Now, to find the acceleration at our time of interest:}\&a(6)=81\cdot \pi ^{2}sin(9\cdot \pi (6)) + 6\&a(6)=6\end{align}$

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Question

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

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Answer

$\begin{align}&\text{Just as velocity is the rate of change of position with respect}\&\text{to time, acceleration is the rate of change of velocity with}\&\text{respect to time. That is to say, acceleration is a measure of}\&\text{how quickly velocity is changing. As velocity is the time derivative}\&\text{of position, acceleration is the time derivative of velocity:}\end{align}$

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

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Question

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

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Answer

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

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Question

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

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Answer

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

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Question

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

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Answer

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

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Question

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

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Answer

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

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Question

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

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Answer

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

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Question

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

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Answer

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

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Question

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

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Answer

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

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Question

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

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Answer

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

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Question

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

Find the velocity of the particle at

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Answer

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

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

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Question

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

Find the velocity of the particle when

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Answer

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

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

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Question

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?

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Answer

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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Question

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

Find the speed of the electron when

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Answer

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

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