CCM Average and instantaneous rates

Chapter 11 Average and instantaneous rates

Section 11.1 Averaging variation over an extended time period

Example 11.1.1. 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 11.1.1 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, similarly to our use of step functions in Chapter 5.

Definition 11.1.2. Average rate of accumulation.

The constant rate of accumulation that produces the same accumulation amount over a given time period.

Calculating average rate of accumulation from a quantity model is simple — simply divide the accumulation amount by the time duration.

Example 11.1.3. Average speed.

Consider again our Camrose-Wetaskiwin trip from Example 5.2.9, also discussed in Subsection 6.1.1. The actual total distance travelled according to the “more realistic” rate model in Figure 6.1.1 is approximately 50.5 km, and the trip lasts a total of 39 min, for an average speed of

\begin{equation*} \frac{50.5}{39/60} \approx 77.7 \end{equation*}

in kilometres per hour. If it were possible to go from parked to full speed instantaneously, maintain that full speed constantly over the full 39 min, and then go from full speed to parked again instantaneously at the destination, then the exact same distance of 50.5 km would be traversed over such a trip. In Figure 11.1.4, this equality of distances is represented by equality of the red and blue areas.

Graph comparing accumulation as area between constant average rate and variable rate in a rate-of-travel model. — Figure 11.1.4. Maintaining a constant speed at the average speed value traverses the same distance.

We won’t always use $t = 0$ as the start of the time domain, nor will it always be the case that the initial quantity is assumed to be $0\text{,}$ so here is a formula that takes both those possibilities into account.

Pattern 11.1.5. Average rate of accumulation formula.

Given 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{.} \end{equation*}

Remark 11.1.6.

The formula in Pattern 11.1.5 is an example of what is called a difference quotient, as it is literally a “quotient of two differences.”

We can interpret the formula in Pattern 11.1.5 in words as

\begin{equation*} \text{avg rate} = \frac{\text{accumulation}}{\text{duration}} \text{,} \end{equation*}

where

\begin{equation*} \text{accumulation} = (\text{final quantity}) - (\text{initial quantity}) \text{.} \end{equation*}

Example 11.1.7. Calculating average rate of accumulation from quantity.

You check your investment account and find a balance of $\$10,385.92\text{.}$ Four weeks later, you check again and find a balance of $\$10,593.64\text{.}$ So over that time period, your account has accumulated

\begin{equation*} 10,593.64 - 10,385.92 = 207.72 \end{equation*}

dollars, at an average rate of

\begin{equation*} \frac{207.72}{4} = 51.93 \end{equation*}

dollars per week.

Example 11.1.8. Calculating average rate of accumulation from a quantity model.

Consider 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. According to the analysis Example 7.3.4, the objects height above the surface at any time $t \ge 0$ can be determined using the model

\begin{equation*} h(t) = -5 t^2 + 30 t \text{,} \end{equation*}

with $t$ measured in seconds. (In Example 7.3.4 we used the traditional physics notation $s(t)$ to represent the position function, but here we will use $h(t)$ in its place to avoid confusion with the use of the letter s to represent seconds as the unit of time.)

The “quantity” being measured here is height above the surface, and we consider motion to be an accumulation (or loss) of height. In this case, “average rate of accumulation” really means “average velocity”.

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 Pattern 11.1.5, 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 11.2 Graphical interpretations of average rate of accumulation

We can interpret average rate of accumulation graphically in two ways — on the rate graph, and on the quantity graph.

On the rate graph, the interpretation is precisely just a graphical restatement of the definition (Definition 11.1.2), as accumulation is equal to total (oriented) area under the rate graph (Principle 6.3.5).

Principle 11.2.1. Graphical interpretation of an average rate on the rate graph.

The average rate of accumulation is the constant rate value that will create the same (oriented) area amount over the same domain.

Interpreting an average rate graphically as bounding an equivalent area. — Figure 11.2.2. Interpreting accumulation as the net oriented area under the rate graph over time domain $t_0 \le t \le t_1 \text{,}$ the rectangle of height $\avg{r}$ bounds an equivalent area over that domain, so that the rectangular green area is exactly equal to the total of the blue area and the (negative) red area.

But on the quantity graph, the formula

\begin{equation*} \avg{r} = \slope{q}{t} = \frac{ q(t_1) - q(t_0) }{ t_1 - t_0 } \end{equation*}

leads to a completely different 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.

Principle 11.2.3. Graphical interpretation of an average rate on the quantity graph.

The average rate of accumulation is the slope of the secant line connecting the points on the quantity graph at the endpoints of the domain.

Interpreting an average rate graphically as a slope of a secant line. — Figure 11.2.4. Average rate of accumulation as a slope on the quantity graph.

Example 11.2.5. Average velocity.

Let’s return to the situation of Example 11.1.8. We have previously plotted both position and velocity graphs for this parabolic motion in Figure 7.3.5. According to Principle 11.2.3, the average velocity of 10 ^m⁄_s that we calculated represents the slope of a secant line on the position graph.

Figure 11.2.6. An average velocity as a secant-line slope on the position graph.

At the same time, if we use the calculated $\avg{v}$ as a constant velocity over the same time domain, we achieve the same accumulation of height. According to Principle 11.2.1, this means that the green and red areas in Figure 11.2.7 are the same.

An average velocity as an "equivalent" constant velocity on the velocity graph. — Figure 11.2.7. An average velocity as an “equivalent” constant velocity on the velocity graph.

Example 11.2.8. 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 11.2.9. A secant on a linear quantity graph is parallel to the graph.

According to Pattern 7.2.2, 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 11.3 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 11.3.1. 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 11.3.1 demonstrates that we can be more precise about rate behaviour by “zooming in” from “global” data to more “local” data. In Example 11.3.1, 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 11.3.2. 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 11.3.1 and Example 11.3.2 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 11.3.2, our average rate calculations give us a pretty good idea of the actual rate.

Principle 11.3.3. Average rate of accumulation over short domains approximates actual 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 approximate

\begin{equation*} r(t) \approx \avg{r} \end{equation*}

for all $t$ in that time domain.

Remark 11.3.4.

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 11.3.5. 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 11.3.6. 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 11.3.7.

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 11.3.7.

A graph of volume versus time for the process of bailing water out of a tub, including two secant lines. — Figure 11.3.7. Two secant lines on the quantity graph.

Principle 11.3.8. Shorter time domain yields better approximation.

In general, given a 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 11.3.9.

Compare Principle 11.3.8 with Principle 6.1.5 — both involve obtaining a better approximation by using smaller values of $\change{t}\text{.}$
It is possible to come up with “wild” examples where Principle 11.3.8 does not hold, which is why we have the proviso “In general.”

Section 11.4 Calculating instantaneous rate of accumulation

Suppose we have a quantity model $q(t)$ and we want to calculate a specific rate value $r(t^\ast)$ exactly. Just as we computed exact accumulation values from approximations of (hopefully) increasing accuracy, we take the same approach here. But 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 11.4.1. Three standard difference approximations to rate.

Assume a small value of $\change{t}$ has been chosen

Forward difference approximation at $t^\ast$.

The average-rate approximation to $r(t^\ast)$ over the time domain

\begin{equation*} t^\ast \le t \le t^\ast + \change{t} \text{,} \end{equation*}

so that

\begin{equation*} r(t^\ast) \approx \frac{ q(t^\ast + \change{t}) - q(t^\ast) }{ \change{t} } \text{.} \end{equation*}
Backward difference approximation at $t^\ast$.

The average-rate approximation to $r(t^\ast)$ over the time domain

\begin{equation*} t^\ast - \change{t} \le t \le t^\ast \text{,} \end{equation*}

so that

\begin{equation*} r(t^\ast) \approx \frac{ q(t^\ast) - q(t^\ast - \change{t}) }{ \change{t} } \text{.} \end{equation*}
Central difference approximation at $t^\ast$.

The average-rate approximation to $r(t^\ast)$ over the time domain

\begin{equation*} t^\ast - c \le t \le t^\ast + c \end{equation*}

for

\begin{equation*} c = \frac{\change{t}}{2} \text{,} \end{equation*}

so that

\begin{equation*} r(t^\ast) \approx \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 11.4.2. Secant lines whose slopes represent the forward (green), backward (red), and central (blue) approximations to rate.

Remark 11.4.3.

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*}

Just as we took the number of rectangles $n$ to be larger and larger in upper and lower sum approximations to accumulation and looked for a pattern, here we want to take $\change{t}$ to be smaller and smaller and look for a pattern in the above approximations.

Example 11.4.4. Approximate instantaneous rate for a cubic quantity model.

Suppose we have quantity model $q(t) = t^3\text{.}$ Let’s calculate average-rate approximations to the actual rate $r(t^\ast)$ for $t^\ast = 2\text{,}$ using $\change{t} = 0.1\text{.}$

Forward difference approximation.

\begin{align*} r(2) \amp \approx \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.

\begin{align*} r(2) \amp \approx \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.

\begin{align*} r(2) \amp \approx \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*}

We seem to be obtaining approximate rates in the “neighbourhood” of $r(2) \approx 12\text{.}$ Is this the actual value? Let’s explore further in the next example.

Example 11.4.5. Instantaneous rate for a cubic quantity model.

Continuing Example 11.4.4, for $q(t) = t^3$ let’s attempt to get closer to the true value of $r(t^\ast)$ for $t^\ast = 2$ 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 11.4.1. We’ve set up the initial value for dt so that you can have Sage verify the results of Example 11.4.4. 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 in Example 11.4.4?

Section 11.5 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. When approximating accumulation with Riemann sums, we obtained the definite integral value by observing the upper and lower sums “squeezing” together toward a specific value. Unfortunately, there is no systematic way an “upper average rate” or “lower average rate” to squeeze together, so we will have to be a little more subtle.

Definition 11.5.1. Derivative value.

Given a function $q(t)$ on domain $a \le t \le b$ and a specific time $t^\ast$ strictly within that domain (so that $a \lt t^\ast \lt b$), suppose there exists one unique number $d$ so that all values of average rate calculations

\begin{equation*} \slope{q}{t} \end{equation*}

over subdomains containing $t^\ast$ can be made arbitrarily close to $d$ by taking $\change{t}$ sufficiently small.

In this case, we say that $q(t)$ is differentiable at $t = t^\ast\text{,}$ call the value $d$ the derivative of $q(t)$ at $t = t^\ast$, and write

\begin{equation*} \funceval{\dqdt}{t = t^\ast} \end{equation*}

to represent the value of $d\text{.}$

Remark 11.5.2. Derivative versus rate.

Given that we use average rate calculations over short time domains to approximate actual rate, we expect that

\begin{equation*} \funceval{\dqdt}{t = t^\ast} = r(t^\ast) \end{equation*}

will be true, and in fact we will be able to theoretically justify this in later chapters.

Compare the notation

\begin{equation*} \slope{q}{t} \to \dqdt \end{equation*}

with the previous notation

\begin{equation*} \sum_{k = 1}^{n} r(t_k^\ast) \change{t_k} \to \ccmint{a}{b}{r(t)}{t} \text{.} \end{equation*}

In both situations, we are computing approximations that (we hope) are more and more accurate, until they are so close we can’t tell the difference. And in both notational choices, $\change{t}$ becomes $dt\text{,}$ signally to the reader that we have let our approximations become “infinitely close” to the “actual” value by considering time subdomain(s) of “infinitesimal” length.

Now, what does arbitrarily close and sufficiently small mean? We have encountered these terms before in Chapter 3, and they mean the same thing here. Here is a reminder of what they mean in the context of Definition 11.5.1. Arbitrarily close means that we may choose any small range around the proposed value $d$ that we like, and in response we will be able to set a “maximum” $\change{t}$ so that the result of every average rate calculation over a time subdomain containing $t^\ast$ and of that maximum $\change{t}$ length or smaller will fall within our proposed range around $d\text{.}$ That is, given a small positive $\epsilon$ we expect to always find that

\begin{equation*} d - \epsilon \lt \slope{q}{t} \lt d + \epsilon \end{equation*}

for all average rate calculations calculated over a domain containing $t^\ast$ and with $\change{t}$ smaller than some particular maximum. (Making the range around $d$ smaller by changing the value of $\epsilon$ will likely require a different maximum $\change{t}$ value — in general, we expect a smaller range around $d$ to require a smaller maximum $\change{t}$ value.)

Remark 11.5.3.

It was not necessary to have this sort of “restrict to a small range around $d$” reasoning in the definition of the Definite integral, as the upper and lower sums effectively acted as boundaries of a small range around the definite integral value.

Example 11.5.4. Determining rate within a given constraint.

Let’s revisit Example 11.4.5. Hopefully you found that the three average-rate approximations to $r(2)$ 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 around $12\text{,}$

\begin{equation*} 12 - \epsilon \lt \slope{q}{t} \lt 12 + \epsilon \text{.} \end{equation*}

(We use e in place of the Greek letter $\epsilon\text{.}$) 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 are no more than the value of e away from $12\text{.}$

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 11.6 Graphical interpretation of the derivative

Subsection 11.6.1 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. — (a) Somewhat large $\change{t}\text{.}$

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 11.6.1 must become arbitrarily close to some specific value as $\change{t}$ is made smaller, hence arbitrarily close to each other. As you can see in the sequence in Figure 11.6.1, not only do these three lines become closer to being parallel as $\change{t}$ becomes smaller, it is appears that if we continued to make $\change{t}$ smaller they would “converge” to 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 11.6.2. A “secant” line over a subdomain of “infinitesimal length” $dt\text{.}$

The line in Figure 11.6.2 has a special name.

Definition 11.6.3. 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 $t = t^\ast\text{.}$

Pattern 11.6.4. Equation of a tangent line.

If $q(t)$ is differentiable at $t = t^\ast\text{,}$ then the equation of the tangent line to the graph of $q(t)$ at $t = t^\ast$ is

\begin{equation*} y = m t + b \end{equation*}

where

\begin{align*} m \amp = \funceval{\dqdt}{t = t^\ast} \text{,} \amp b \amp = q(t^\ast) - m t^\ast \text{.} \end{align*}

Example 11.6.5. Determining a tangent line equation.

Once again, let’s revisit the quantity function $q(t) = t^3$ previously analysed in Example 11.4.5 and Example 11.5.4, 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 11.6.6. A tangent line to the graph of $q(t) = t^3\text{.}$

Subsection 11.6.2 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 11.6.7, 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)

As you can see in Figure 11.6.7, 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 11.7 Some non-differentiable examples

Subsection 11.7.1 Cusps

Example 11.7.1. 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 11.7.2.

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 11.7.2. 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 11.7.2, 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 11.7.2 is called a cusp, and functions with such a feature are never differentiable at that point.

Subsection 11.7.2 Jump discontinuities

Example 11.7.3. Differentiability of step functions.

Returning to Example 11.3.5, 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 11.3.6 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 a step function.

A graph of a step function as volume versus time modelling the process of bailing water out of a tub. — Figure 11.7.4. 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 close enough to each other to fit in an arbitrarily small range around some proposed rate 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 11.7.3 Singularities

Example 11.7.5.

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 $t \to 2^\plus$ and $t \to 2^\minus\text{.}$

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 11.7.6. Average rates near a singularity as secant-line slopes.

From Figure 11.7.6, 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 11.7.4 Vertical tangents

Example 11.7.7. 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 $t = 1\text{.}$

Figure 11.7.8. Average rate as a secant-line slope near an instantaneously vertical point.

As the red endpoints of the secant line in Figure 11.7.8 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 11.6.1 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 11.7.2 and Subsection 11.7.3, we observed that two different kinds of function discontinuities lead to non-differentiability, and the same is true in general.

Pattern 11.7.9. Discontinuous implies non-differentiable.

If a function is discontinuous at some value of $t\text{,}$ then it cannot be differentiable there.

Note that, among other types of discontinuities, Pattern 11.7.9 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 11.7.10. 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 11.7.1 and Subsection 11.7.4.

Considering Pattern 11.7.9 in reverse, if a function is differentiable at a particular value of $t\text{,}$ then it must be continuous there, since otherwise Pattern 11.7.9 would imply that it wouldn’t have been differentiable in the first place.

Pattern 11.7.11. Differentiable implies continuous.

If a function is differentiable at some value of $t\text{,}$ then it must be continuous there.

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