### 15.3.1 Instantaneous Rate of Change

This subsection is not completed yet, but in the meantime here are some practice questions.

When graphing the derivative of a graph, take note of the slope. Where $f(x)$ is increasing, the derivative is positive, and where $f(x)$ is decreasing, the derivative is negative. Horizontal points on the graph have a slope (derivative) of zero. Points of inflection have a local maximum or minimum slope.
Using this information, you can graph the derivative:
If you’re going the opposite way and graphing $f(x)$ knowing what the graph of $f’(x)$ looks like, remember that where $f’(x)$ is negative, $f(x)$ is decreasing; where $f’(x)=0$, $f(x)$ is stationary; and where $f’(x)$ is positive, $f(x)$ is increasing. It doesn’t matter what value of $f(x)$ you choose to start with – your graph may be shifted up or down compared to one drawn by someone else.

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Graph the derivative of the function shown below:

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Graph the derivative of the function shown below:

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Graph the derivative of the function shown below:

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Graph the function whose derivative is shown below:

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(Your graph may be shifted up or down compared to this one.)

Graph the function whose derivative is shown below:

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(Your graph may be shifted up or down compared to this one.)

Graph the function whose derivative is shown below:

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(Your graph may be shifted up or down compared to this one.)

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The derivative of displacement with respect to time is velocity. Speed is the magnitude of velocity. Acceleration is the derivative of velocity with respect to time (which makes it the second derivative of displacement with respect to time).
\begin{align*}
v &= \frac{\dif x}{\dif t} = \dot x\\
\mathrm{speed} &= \abs{v}\\
a &= \frac{\dif v}{\dif t} = \dot v \\
&= \frac{\dif^2 x}{\dif t^2} = \ddot x
\end{align*}

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What is the relationship between velocity and displacement?

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Velocity is the derivative of displacement with respect to time. \[v = \dot x = \frac{\dif x}{\dif t}\]

What is the relationship between velocity and acceleration?

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Acceleration is the derivative of velocity with respect to time. \[a = \dot v = \frac{\dif v}{\dif t}\]

What is the relationship between velocity and speed?

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Speed is the magnitude of velocity.

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##### Question 15.3.1

If $\dot x = -3\,\mathrm{m/s}$ and $\ddot x = 1\,\mathrm{m/s^2}$, is the speed increasing or decreasing?
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The velocity is $-3\,\mathrm{m/s}$, which means the speed is $3\,\mathrm{m/s}$. Since the acceleration is positive, the velocity is increasing, but does this mean that the speed is increasing?

Consider what happens after $0.01\,\mathrm{s}$. The velocity will increase to approximately $-2.99\,\mathrm{m/s}$, but the speed will go from $3\,\mathrm{m/s}$ down to $2.99\,\mathrm{m/s}$. Therefore the speed is decreasing.

You could think of it in terms of ‘getting closer to zero’ rather than ‘increasing’ or ‘decreasing’. The velocity is increasing, but because it’s negative, it’s getting closer to zero, which means the speed is also getting closer to zero, and is therefore decreasing.