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Chapter 5 Average and instantaneous rates

Section 5.1 Averaging variation over an extended time period

Example 5.1.1. Average rainfall.

A rainfall gauge is essentially a graduated cylinder that collects rain as it falls. But instead of being marked with volume measurements, it is marked with height measurements.
With rain in the forecast, you decide to set out a rain gauge at 10 o’clock in the evening, even though it is not currently raining. You check on the rain gauge at 8 o’clock the next morning, and find that it has collected an amount of water so that the gauge reads 4 mm. This measurement means that if the ground was perfectly flat and impermeable so that the rain could neither run off nor be absorbed by the ground, you’d be standing in a large puddle 4 mm deep.
So the average rate of accumulation over those 10 hours was 0.4 mm⁄h.
The average rate of accumulation measurement in Example 5.1.1 does not provide any information about the intensity of the weather that produced the rain. Was the accumulation caused by a light drizzle for most of the night, or by a sudden downpour that only lasted a few minutes at some point? There is no way to tell because this was not observed. Average rate of accumulation replaces all the minute-to-minute or second-to-second variation in the actual rate of accumulation with a single, constant rate of accumulation.
Calculating average rate of accumulation from a quantity model is simple — simply divide the accumulation amount by the time duration. However, we won’t always use \(t = 0 \) as the start of the time interval, nor will it always be the case that the initial quantity is assumed to be \(0 \text{,}\) but the formula from Definition 1.3.3 takes that it account: for model \(q(t) \) representing the amount of some quantity, the average rate of accumulation over the time interval \(t_0 \le t \le t_1 \) is
\begin{equation} \avg{r} = \slope{q}{t} = \frac{ q(t_1) - q(t_0) }{ t_1 - t_0 }\text{.}\tag{✶} \end{equation}
We explored a couple examples of calculating average rate in Section 1.3; let’s do another here involving a quantity function.

Example 5.1.2. Calculating average rate from a quantity model.

From physics we know that an object launched upward with initial velocity 30 m⁄s, from the surface of an airless planet where acceleration due to gravity is 10 m⁄s2 will experience parabolic motion, so that its height above the surface at time \(t \) can be expressed as
\begin{equation*} h(t) = -\frac{10}{2} t^2 + 30 t = - 5 t^2 + 30 t \text{,} \end{equation*}
with \(t \) measured in seconds. In this scenario, with both positive (upward) and negative (downward) directions of travel possible, an average rate of change represents an average velocity rather than an average speed as in previous motion examples.

Aside: A look ahead.

Suppose we would like to calculate the average velocity over the time domain \(1 \le t \le 3 \text{.}\) The endpoint heights are
\begin{align*} h(1) \amp = -5 \cdot 1^2 + 30 \cdot 1 \amp h(3) \amp = -5 \cdot 3^2 + 30 \cdot 3\\ \amp = 25 \text{,} \amp \amp = 45 \text{.} \end{align*}
Using Definition 1.3.3, we have
\begin{align*} \avg{v} \amp = \slope{h}{t} \\ \amp = \frac{ h(3) - h(1) }{ 3 - 1 } \\ \amp = \frac{ 45 - 25 }{ 3 - 1 } \\ \amp = \frac{ 20 }{ 2 } \\ \amp = 10 \end{align*}
metres per second.

Section 5.2 Graphical interpretations of average rate of accumulation

On the graph of a quantity function \(q(t) \text{,}\) the formula
\begin{equation*} \avg{r} = \slope{q}{t} = \frac{ q(t_1) - q(t_0) }{ t_1 - t_0 } \end{equation*}
has a particular interpretation. The data points \(\bbrac{t_0, q(t_0)} \) and \(\bbrac{t_1, q(t_1)} \) used in the average rate calculation represent points on the graph of the quantity function, the line segment connecting them is called a secant line for the graph, and the formula for \(\avg{r} \) above is just the rise-over-run formula for the slope of that secant line.
Interpreting an average rate graphically as a slope of a secant line.
Figure 5.2.2. Average rate of accumulation as a slope on the quantity graph.

Example 5.2.3. Average velocity.

Let’s return to the situation of Example 5.1.2. According to Pattern 5.2.1, the average velocity of 10 m⁄s that we calculated represents the slope of a secant line on the position (height-versus-time) graph.
An average velocity as a secant-line slope on a height-versus-time graph.
Figure 5.2.4. An average velocity as a secant-line slope on a height-versus-time graph.

Example 5.2.5. Average rate for a linear quantity model.

Suppose we have a linear quantity model
\begin{equation*} q(t) = m t + q_0 \text{,} \end{equation*}
where \(q_0 = q(0) \) is the initial quantity. Over any time domain \(t_0 \le t \le t_1 \text{,}\) the average rate of accumulation is
\begin{align*} \avg{r} \amp = \slope{q}{t} \\ \amp = \frac{ q(t_1) - q(t_0) }{ t_1 - t_0 } \\ \amp = \frac{ (m t_1 + q_0) - (m t_0 + q_0) }{ t_1 - t_0 } \\ \amp = \frac{ m (t_1 - t_0) }{ t_1 - t_0 } \\ \amp = m \text{.} \end{align*}
The average rate of accumulation is the same, no matter what time domain is used, and is equal to the slope of the quantity’s graph, so that our model could be expressed as
\begin{equation*} q(t) = \avg{r} t + q_0 \text{.} \end{equation*}
A graph depicting the fact that a secant on a linear quantity graph is parallel to the graph.
Figure 5.2.6. A secant on a linear quantity graph is parallel to the graph.
According to Pattern 1.4.3, a linear quantity model arises from a constant rate model. So it is not surprising that the average rate of change is always the same value for a linear quantity model — the average of a constant is that constant.

Section 5.3 Approximating instantaneous rate of accumulation

Observing that your rain gauge has collected 4 mm of water over 10 h says nothing about how intense the rainfall was at any given moment during that time. We’ll need more data to be more specific.

Example 5.3.1. A multi-day road trip.

You drive from Edmonton to Toronto, a total distance of approximately 3375 km. The trip takes a total of 35 h of driving time split over three days, for a total-trip average speed of approximately
\begin{equation*} \frac{3375}{35} = 96.4 \end{equation*}
kilometres per hour.
However, on the first day you are driving on straight, flat, Prairie highways with higher speed limits, and cover 1305 km over 13 h of driving, for a first-day average speed of approximately
\begin{equation*} \frac{1305}{13} = 100.4 \end{equation*}
kilometres per hour.
The second day is much more of a slog through never-ending Northwestern Ontario on two-lane highways with lower speed limits, through hilly terrain with lots of transport trucks and camper-trailers but few opportunities to pass. You cover 1385 km, but it takes you 15 h of driving, for a second-day average speed of approximately
\begin{equation*} \frac{1385}{15} = 92.3 \end{equation*}
kilometres per hour.
The final stretch is better — still a lot of hilly terrain until you get into Southern Ontario, but many more stretches of divided highway with two lanes in each direction and higher speed limits. The last 685 km takes you 7 h, for a final-day average speed of approximately
\begin{equation*} \frac{685}{7} = 97.9 \end{equation*}
kilometres per hour.
Example 5.3.1 demonstrates that we can be more precise about rate behaviour by “zooming in” from “global” data to more “local” data. In Example 5.3.1, this “zooming in” is from whole-trip to day-by-day. We start to see patterns of variation in the rate of accumulation if we consider shorter time durations. What if we take this idea further?

Example 5.3.2. Designing a basic speedometer.

One of your hobbies is building remote-controlled (RC) vehicles. You would like your current model to wirelessly relay data to a communication app on your phone that will let you monitor its speed as close to real-time as possible. A tachometer is a device that counts revolutions of a rotating object. You install one on the rear axle of your RC vehicle, and interface it to the Bluetooth module on the control board. Your phone app now receives axle revolution counts every 100 ms. (Assume that the tachometer is reasonably high-precision, so that fractional revolutions are reported.) How can you use this data to get an idea of the vehicle’s speed?
Let \(q(t) \) represent the quantity of axle revolutions reported at time \(t \text{,}\) where \(t \) is measured in milliseconds. Then
\begin{equation*} \change{q} = q(t) - q(t - 100) \end{equation*}
represents the number of revolutions in a given 100 ms interval, The rear wheels on the RC vehicle measure 108 mm in diameter, so that the distance travelled in a 100 ms interval is
\begin{equation*} \change{s} = 108 \pi \change{q} \text{.} \end{equation*}
So you program the app on your phone to display the result of calculating
\begin{equation*} \slope{s}{t} = \frac{ 108 \pi \bbrac{ q(t) - q(t - 100) } }{100} \cdot \frac{3600}{1000} \text{,} \end{equation*}
where
  • \(q(t) \) represents the most recent tachometer reading
  • \(q(t - 100) \) represents the previous tachometer reading
  • the extra factor of \(3600 / 1000 \) is to convert from millimetres per millisecond to kilometres per hour.
Both Example 5.3.1 and Example 5.3.2 demonstrate the same pattern — we can break the time domain up into smaller subdomains of length \(\change{t} \) to get more precise information. On a short enough time domain, as in Example 5.3.2, our average rate calculations give us a pretty good idea of the instantaneous rate of accumulation.

Principle 5.3.3. Average rate of accumulation over short domains approximates instantaneous rate of accumulation.

In general, given a quantity function \(q(t) \) over a time domain \(t_0 \le t \le t_1 \) so that
\begin{equation*} \change{t} = t_1 - t_0 \end{equation*}
is sufficiently small, it is reasonable to say that average rate of accumulation over that time domain is approximately equal to the instantaneous rate of accumulation at \(t \text{.}\)

Remark 5.3.4.

The meaning of “sufficiently small” is relative to the particular quantity function \(q(t) \text{,}\) the time domain considered, and what one is willing to accept as “reasonable” in the approximation.
The reason that we need \(\change{t} \) to be small for the approximation to be “reasonable” is that if \(\change{t} \) is too long, we will capture information that is irrelevant to the current rate of accumulation. Here is an extreme example to illustrate this.

Example 5.3.5. Bailing a tub by the bucket-full.

You are slowly bailing out a 50 L tub using a 10 L pail. You are able to fill and empty a pail-full every 5 s. (Let’s assume you are able to get a full pail even when there is only 10 L left in the tub.)
A graph of volume versus time for the process of bailing water out of a tub.
Figure 5.3.6. Quantity of water in tub during emptying.
If we compute \(\avg{r} \) over the full time domain \(0 \le t \le 25 \text{,}\) we obtain
\begin{align*} \avg{r} \amp = \slope{q}{t} \\ \amp = \frac{ q(25) - q(0) }{ 25 - 0 } \\ \amp = \frac{ 0 - 50 }{ 25 } \\ \amp = -2 \end{align*}
in litres per second, where we have obtained a negative result because the tub was losing water rather than accumulating water. This value represents the slope of the red line segment in Figure 5.3.7.
But clearly the rate of accumulation (or, really, the rate of loss) of water in the tub is 0 L⁄s at any instant between scoops with the pail. That is, we expect \(r(t) = 0 \) for every value of \(t \) outside of the times that the pail is being filled from the tub. We can better approximate these instantaneous rate of accumulation values with an average rate of accumulation calculation over a shorter time domain, so that we focus only on “local” data rather than using the entire “global” collection of data. For example, over the time domain \(6 \le t \le 9 \text{,}\) we have
\begin{align*} \avg{r} \amp = \slope{q}{t} \\ \amp = \frac{ q(9) - q(6) }{ 9 - 6 } \\ \amp = \frac{ 30 - 30 }{ 3 } \\ \amp = 0 \text{.} \end{align*}
This value represents the slope of the green line segment in Figure 5.3.7.
A graph of volume versus time for the process of bailing water out of a tub, including two secant lines.
Figure 5.3.7. Two secant lines on the quantity graph.

Principle 5.3.8. Shorter time domain yields better approximation.

For a “well-behaved” quantity function \(q(t) \text{,}\) an average rate calculation over a shorter time domain will yield a better approximation of the actual rate of change, compared to an average rate calculation over a longer time domain.

Remark 5.3.9.

It is possible to come up with “wild” examples where Principle 5.3.8 does not hold, which is why we have included the (admittedly vague) “well-behaved” condition.

Section 5.4 Calculating instantaneous rate of accumulation

Suppose we have a quantity model \(q(t) \) and we want to calculate instantaneous rate of accumulation at a specific instant \(t^\ast \) exactly. The idea in Principle 5.3.8 is precisely the situation that we developed the idea of limit to handle: can we find some pattern in the values of
\begin{equation*} \avg{r} = \slope{q}{t} \end{equation*}
as we take \(\change{t} \to 0 \text{?}\) To see a pattern of increasing accuracy in a collection of approximations, we need a more systematic approach than just “compute \(\avg{r} \) over a short time domain containing \(t^\ast \text{.}\)” Here are three such approaches.

Definition 5.4.1. Three standard difference approximations to rate.

Assume a small value of \(\change{t} \) has been chosen. The following are three different average rate calculations, each over a time domain containing a particular point of interest \(t = t^\ast \) and of length \(\change{t} \text{,}\) that can be used as approximations to the instantaneous rate of accumulation of quantity function \(q(t) \) at \(t = t^\ast \text{.}\)
  • Forward difference approximation at \(t^\ast \).
    Over time domain
    \begin{equation*} t^\ast \le t \le t^\ast + \change{t} \text{,} \end{equation*}
    average rate
    \begin{equation*} \frac{ q(t^\ast + \change{t}) - q(t^\ast) }{ \change{t} } \text{.} \end{equation*}
  • Backward difference approximation at \(t^\ast \).
    Over time domain
    \begin{equation*} t^\ast - \change{t} \le t \le t^\ast \text{,} \end{equation*}
    average rate
    \begin{equation*} \frac{ q(t^\ast) - q(t^\ast - \change{t}) }{ \change{t} } \text{.} \end{equation*}
  • Central difference approximation at \(t^\ast \).
    Over time domain
    \begin{equation*} t^\ast - c \le t \le t^\ast + c \quad \text{for} \quad c = \frac{\change{t}}{2} \text{,} \end{equation*}
    average rate
    \begin{equation*} \frac{ q(t^\ast + c) - q(t^\ast - c) }{ \change{t} } \text{.} \end{equation*}
Secant lines on a graph whose slopes represent the forward, backward, and central approximations to rate.
Figure 5.4.2. Secant lines whose slopes represent the forward (green), backward (red), and central (blue) approximations to rate.

Remark 5.4.3.

  • Since
    \begin{equation*} \frac{ q(t^\ast) - q(t^\ast - \change{t}) }{ \change{t} } = \frac{ q(t^\ast - \change{t}) - q(t^\ast) }{ - \change{t} } \text{,} \end{equation*}
    the forward and backward approximations can be unified into one by using the formula for the forward difference approximation but allowing \(\change{t} \) to be chosen as either a positive or negative value.
  • The central difference formula could also be written as
    \begin{equation*} \frac{ q(t^\ast + c) - q(t^\ast - c) }{ 2 c } \text{.} \end{equation*}
Let’s practise using these formulas.

Example 5.4.4. Approximate instantaneous rate for a cubic quantity model.

Suppose we have quantity model \(q(t) = t^3 \text{.}\) Let’s approximate the instantaneous rate of accumulation of the quantity at time \(t^\ast = 2 \) using the three average-rate approximations in Definition 5.4.1, using \(\change{t} = 0.1 \text{.}\)

Forward difference approximation.

Over time domain \(2 \le t \le 2.1 \text{,}\)
\begin{align*} \avg{r} \amp = \frac{ q(2 + \change{t}) - q(2) }{ \change{t} } \\ \amp = \frac{ (2.1)^3 - 2^3 }{ 0.1 } \\ \amp = 12.61 \text{.} \end{align*}

Backward difference approximation.

Over time domain \(1.9 \le t \le 2 \text{,}\)
\begin{align*} \avg{r} \amp = \frac{ q(2) - q(2 - \change{t}) }{ \change{t} } \\ \amp = \frac{ 2^3 - (1.9)^3 }{ 0.1 } \\ \amp = 11.41 \text{.} \end{align*}

Central difference approximation.

Over time domain \(1.95 \le t \le 2.05 \text{,}\)
\begin{align*} \avg{r} \amp = \frac{ q(2 + \change{t}/2) - q(2 - \change{t}/2) }{ \change{t} } \\ \amp = \frac{ (2.05)^3 - (1.95)^3 }{ 0.1 } \\ \amp = 12.0025 \text{.} \end{align*}
In Example 5.4.4, all three average rate values appear to be in the “neighbourhood” of \(12 \text{.}\) Is this the exact instantaneous rate value? Let’s explore further in the next example.

Example 5.4.5. Instantaneous rate for a cubic quantity model.

Continuing Example 5.4.4, for \(q(t) = t^3 \) let’s attempt to get closer to the true instantaneous rate value at \(t^\ast = 2 \) by calculating approximations using smaller values of \(\change{t} \text{.}\) We’ll use Sage to speed up the calculations and allow us to easily compute new approximations for smaller and smaller values of \(\change{t} \text{.}\) Since Sage doesn’t like Greek letters very much, we’ll write dt in place of \(\change{t} \text{.}\) We’ll also write u in place of \(t^\ast \text{.}\)
Examine the formulas above for forward, backward, and central in the Sage cell above, and make sure that you see how they correspond to the formulas in Definition 5.4.1. We’ve set up the initial value for dt so that you can have Sage verify the results of Example 5.4.4. After doing so by evaluating the Sage cell, change the value of dt to something smaller and re-evaluate. Repeat with smaller and smaller values of dt until the forward, backward, and central difference approximations all seem to be “converging” to a single value. Is it the value that we predicted following Example 5.4.4?

Section 5.5 The derivative

In many cases we can use average-rate approximations with smaller and smaller \(\change{t} \) and observe an obvious pattern toward some specific value. As noted in the previous section, this is the precise reason we introduced the concept of limit, and now that we have set up several ways to approach the instantaneous rate of change, we are ready to put that concept to work for us.

Definition 5.5.1. Derivative value.

Given a function \(q(t) \text{,}\) if
\begin{equation*} \lim_{\change{t} \to 0} \frac{ q(t^\ast + \change{t}) - q(t^\ast) }{ \change{t} } \end{equation*}
exists, then we say that \(q(t) \) is differentiable at \(t = t^\ast \text{,}\) and we call the value of the limit the derivative value of \(q(t) \) at \(t = t^\ast \). We write either \(q'(t^\ast) \) or
\begin{equation*} \funceval{\ddt{q}}{t = t^\ast} \end{equation*}
to represent the value of this limit.

Remark 5.5.2.

Note that we have chosen the unified forward/backward difference approximation formula as the basis for this limit. When evaluating a limit where \(\change{t} \to 0 \text{,}\) we consider \(\change{t} \approx 0 \) to potentially be either positive or negative, so both the forward and backward difference approximations are represented here. But the central difference approximation has been left out. It turns out that this omission doesn’t matter, since if \(q(t) \) is differentiable at \(t = t^\ast \text{,}\) the limit as \(\change{t} \to 0 \) of the central difference approximation formula would evaluate to the same derivative value.
Notationally, we could write
\begin{equation*} \slope{q}{t} \to \ddt{q} \qquad \text{as} \qquad \change{t} \to 0 \text{.} \end{equation*}
The idea is that while \(\change{t} \) represents an actual time interval length, \(dt \) represents an “infinitesimally” small time interval length, at the “end” of the limiting process, and \(dq \) represents the corresponding “infinitesimally” small change in the value of \(q(t) \text{.}\)
Recall that the definition of limit involves the idea that the values of the expression in the limit can be made arbitrarily close to the proposed limiting value. In the next example, we will test that idea for a limit that represents a derivative value.

Example 5.5.3. Approximating instantaneous rate to within a given constraint.

Let’s revisit Example 5.4.5. Hopefully you found that the three average-rate approximations to instantaneous rate at \(t = 2 \) all became close to \(d = 12 \) as \(\change{t} \) (dt in the Sage code) was made smaller.
Below is some Sage code to test an arbitrarily chosen range containing \(12 \text{,}\)
\begin{equation*} 12 - \epsilon \lt \slope{q}{t} \lt 12 + \epsilon \text{.} \end{equation*}
In the Sage code we use e in place of the Greek letter \(\epsilon \text{,}\) and start with the specific value \(\epsilon = 0.1 \text{.}\) Try to determine a small enough \(\change{t} \) value (again, represented as dt in the Sage code) so that all three of the forward, backward, and central approximations fall within the desired range.
Once you’ve done this, try changing e to a smaller value, and see if the previous value of dt still works, or whether you need to make it smaller.

Section 5.6 Graphical interpretation of the derivative

Subsection 5.6.1 Tangent lines

Consider what happens on the quantity graph when we form the forward, backward, and central approximations to the quantity’s rate of accumulation at a specific time \(t = t^\ast \) for several iterations of smaller \(\change{t} \) values. For emphasis, we will extend the associated secant line segments to actual lines.
Secant lines on a graph whose slopes represent the forward, backward, and central approximations to rate, using a somewhat long time subdomain.
(a) Somewhat large \(\change{t} \text{.}\)
Secant lines on a graph whose slopes represent the forward, backward, and central approximations to rate, using a shorter time subdomain.
(b) Smaller \(\change{t} \text{.}\)
Secant lines on a graph whose slopes represent the forward, backward, and central approximations to rate, using a very short time subdomain.
(c) Small \(\change{t} \text{.}\)
Figure 5.6.1. Three iterations of secant lines whose slopes represent the forward (green), backward (red), and central (blue) approximations to rate, using decreasing values of \(\change{t} \text{.}\)
When \(q(t) \) is differentiable at \(t = t^\ast \text{,}\) by definition the slopes of the three lines in each of the subfigures of Figure 5.6.1 must become arbitrarily close to some specific value as \(\change{t} \) is made smaller, hence arbitrarily close to each other, and lines with the same slope are parallel. As you can see in the sequence in Figure 5.6.1, not only do these three lines become closer to being parallel as \(\change{t} \) becomes smaller, it appears that if we continued to make \(\change{t} \) smaller they would “converge” toward becoming the exact same line. We could think of this one line that the three are converging toward as the case of \(\change{t} \) becoming \(dt \) — a secant line over a subdomain containing \(t^\ast \) of length so small that the endpoints of the subdomain are essentially at \(t^\ast \) as well.
A "secant line" on a graph over an infinitesimally short time subdomain.
Figure 5.6.2. A “secant” line over a subdomain of “infinitesimal length” \(dt \text{.}\)
The line in Figure 5.6.2 has a special name.

Definition 5.6.3. Tangent line.

If \(q(t) \) is differentiable at \(t = t^\ast \text{,}\) then the line that passes through the point \(\bbrac{ t^\ast, q(t^\ast) } \) with slope
\begin{equation*} \funceval{\dqdt}{t = t^\ast} \end{equation*}
is called the tangent line to the graph of \(q(t) \) at \(t = t^\ast \text{.}\)

Example 5.6.5. Determining a tangent line equation.

Once again, let’s revisit the quantity function \(q(t) = t^3 \) previously analysed in Example 5.4.5 and Example 5.5.3, where we found that
\begin{equation*} \funceval{\dqdt}{t = 2} = 12 \text{.} \end{equation*}
So the tangent line through the point \((2, 8) \) with slope \(m = 12 \) is
\begin{align*} y \amp = 12 t + (8 - 12 \cdot 2) \\ \amp = 12 t - 16 \end{align*}
A tangent line on the cubic graph.
Figure 5.6.6. A tangent line to the graph of \(q(t) = t^3 \text{.}\)

Subsection 5.6.2 Local linearity

Another way to think of both the derivative value and the tangent line is through the concept of local linearization. Graphically, shrinking \(\change{t} \) is similar to “zooming in” on the point \(\bbrac{ t^\ast, q(t^\ast) } \) on the graph of \(q(t) \text{.}\) In Figure 5.6.7, a blue box of width \(\change{t} \) is drawn around the point \(\bbrac{ t^\ast, q(t^\ast) } \text{,}\) and then that square is used as the frame for the next zoom level with a smaller \(\change{t} \text{.}\)
Graph with a box that will be zoomed into next.
(a) Focussing on a point of interest.
Graph with a box that will be zoomed into next.
(b) A closer look, along with a smaller window to be investigated next.
Graph with a box that will be zoomed into next.
(c) Still closer around the point of interest.
Graph with a box that will be zoomed into next.
(d) And closer.
Graph with a box that will be zoomed into next.
(e) And closer still.
Figure 5.6.7. Zooming in to see the local linearity of a graph. For each subsequent graph, the outer bounding box is the same as the small box around the point of interest in the previous graph.
As you can see in Figure 5.6.7, zooming in around a point excludes more and more of the variation in the graph, until the graph becomes essentially linear through the point being considered. Not every point on every graph will exhibit this kind of pattern when zooming in, but it will always be present at differentiable points, with the derivative value describing the slope of that essentially linear piece of the graph.

Section 5.7 Some non-differentiable examples

Subsection 5.7.1 Cusps

Example 5.7.1. Measuring distance rather than displacement.

A sensor is placed to measure the distance of some object from the sensor. For a period of time, the object moves directly toward the sensor at a constant rate, then passes by it and continues on the same path at the same rate away from the sensor. A typical plot of the data capture by the sensor appears in Figure 5.7.2.
A graph of an object’s distance from a sensor as the object travels toward and then away from the sensor, with accompanying secant lines.
Figure 5.7.2. Graph of distance from a sensor of an object in motion (blue), with backward (red), forward (green), and central (purple) secant lines.
The distance graph is a V-shape, as in Figure 5.7.2, where \(t = t^\ast \) is the instant that the object passes the sensor. We can see that no matter how far we zoom in around \(t = t^\ast \text{,}\) the graph will always have a V-shape there, and will never by locally linear around \(t = t^\ast \text{.}\) So we expect the function to not be differentiable at \(t = t^\ast \text{.}\) Indeed, if we compute forward, backward, and central average-rate approximations to the rate at \(t = t^\ast \text{,}\) we will always find that the forward average rate is \(r \) (the slope of the right “arm” of the V-shape), the backward average rate is \(- r \) (the slope of the left “arm” of the V-shape), and the central average rate is \(0 \text{,}\) where \(r \) is the constant rate of travel of the object. If we choose a small enough range around any one of those three average rate values, it will be impossible to shrink \(\change{t} \) so that all three approximations are within that range.
The sharp corner at \(t = t^\ast \) in Figure 5.7.2 is called a cusp, and functions with such a feature are never differentiable at that point.

Subsection 5.7.2 Jump discontinuities

Example 5.7.3. Differentiability of a “step” function.

Returning to Example 5.3.5, it would be non-trivial to devise a model for the changing water level during the short periods of time where a full pail is being lifted out of the tub. The model graphed in Figure 5.3.6 is a crude attempt to make the graph appear to be realistic, but often when modelling it is more expedient to not attempt to realistically model every aspect of the system, using a simplified model instead. In this case, it is very easy to create a simple model for the amount of water in the tank by using what is called a step function, because it “steps” abruptly from output value to output value.
Aside: A look ahead.
A graph of a step function as volume versus time modelling the process of bailing water out of a tub.
Figure 5.7.4. Quantity of water in tub during emptying as a step function.
Consider the point on the graph at \(t = 5 \text{.}\) If we “zoomed in” around this point, we would see only a line segment on the right, but nothing on the left. Since we do not see a linear graph through the point being considered, we would not call this graph locally linear at \(t = 5 \text{,}\) and so we should expect \(q(t) \) to not be differentiable at \(t = 5 \text{.}\)
Also consider what would happen if we calculated some average rates around \(t = 5 \) for small \(\change{t} \text{.}\) Backward and central average rates would be large negative values, since the corresponding secant lines would have left endpoints on the upper step at output level \(40 \) and right endpoints on the lower step at output level \(30 \text{.}\) But forward average rates would always be \(0 \text{,}\) since the corresponding secant lines would have both endpoints on the lower step. There is no possibility that these average rates could all be made arbitarily close to some proposed limiting value, so the function is not differentiable at \(t = 5 \text{.}\)
And the same would be true at all times where the step function “steps”.

Subsection 5.7.3 Singularities

Example 5.7.5.

Consider a model function
\begin{equation*} q(t) = \frac{1}{(t-2)^2} \text{.} \end{equation*}
This function has a singularity at \(t = 2 \text{,}\) where \(q(t) \to \infty \) for both \(t \to 2^- \) and \(t \to 2^+ \text{.}\)
Consider some average rate calculations around \(t = 2 \text{.}\) Note that the definition of Derivative value does specifically reference the Three standard difference approximations to rate — for a given maximum \(\change{t} \text{,}\) we should consider all average-rate approximations of rate that involve that \(\change{t} \) or smaller.
Average rates near a singularity as secant-line slopes on the quantity graph.
Figure 5.7.6. Average rates near a singularity as secant-line slopes.
From Figure 5.7.6, hopefully it is obvious that no matter how tightly around \(t = 2 \) we restrict ourselves, we can always create average rate calculations with results that are arbitrarily large and positive (for example, by sliding the right endpoint of the green secant line upwards), others that are arbitrarily large and negative (for example, by sliding the left endpoint of the red secant line upwards), and others that are exactly \(0 \) (for example, the purple secant line, or any similar horizontal secant line above it). Even worse, it is not possible to calculate forward or backward average-rate approximations, since \(q(2) \) is undefined.
For these reasons, this function is not differentiable at the singularity at \(t = 2 \text{.}\)

Subsection 5.7.4 Vertical tangents

Example 5.7.7. Vertical tangent.

Now consider a shifted cube-root function,
\begin{equation*} q(t) = 1 + \sqrt[3]{t - 1} \text{,} \end{equation*}
and picture a secant line about \(t = 1 \text{.}\)
Average rate as a secant-line slope near an instantaneously vertical point.
Figure 5.7.8. Average rate as a secant-line slope near an instantaneously vertical point.
As the red endpoints of the secant line in Figure 5.7.8 are drawn closer to \(t = 1 \) to shrink \(\change{t} \) in the average rate calculation, the line segment will become more steep and its slope (representing the average rate) will become larger without bound. We could imagine that the result of the secant-lines-become-tangent-line process from Subsection 5.6.1 is a vertical line at \(t = 1 \) running through the graph at the point \((1,1) \text{,}\) but we can’t say that the function is differentiable at \(t = 1 \) because a vertical line has undefined slope, and the increasing-without-bound average rate values do not display a pattern that converge to a specific number.
In Subsection 5.7.2 and Subsection 5.7.3, we observed that two different kinds of function discontinuities lead to non-differentiability, and the same is true in general.
Note that, among other types of discontinuities, Pattern 5.7.9 may be applied to all values of \(t \) that are not in the domain of the function, since at such points we would not be able to form either the forward or backward average-rate approximations to rate.

Warning 5.7.10. Functions can be non-differentiable even if continuous.

Functions are always non-differentiable at discontinuities, but that is not the only they can fail to be differentiable, as we observed in Subsection 5.7.1 and Subsection 5.7.4.
Considering Pattern 5.7.9 in reverse, if a function is differentiable at a particular value of \(t \text{,}\) then it must be continuous there, since otherwise Pattern 5.7.9 would imply that it wouldn’t have been differentiable in the first place.
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