CCM Accumulation trends

Chapter 18 Accumulation trends

Often we are more interested in general accumulation trends rather than absolute values. Will water levels rise or recede? Will profit increase or decrease? When will they do one, and when will they do the other? When they increase, will they increase slowly or quickly? When they decrease, will they decrease slowly or quickly?

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We can answer these questions with the derivative function and the second derivative function.

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Section 18.1 Domains of increase and decrease

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It’s easy to measure whether a quantity has increased or decreased from one time to a later time.

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Definition 18.1.1. Increase and decrease.

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We say that a quantity

q (t)

is increasing on the domain

a < t < b

Δ q

is always positive between any two points in that domain.

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We say that a quantity

q (t)

is decreasing on the domain

a < t < b

Δ q

is always negative between any two points in that domain.

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Note 18.1.2.

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In Definition 18.1.1, we are always looking for increase or decrease as time increases, so “Between any two points in that domain” requires that we are looking at

Δ q = q (t_{2}) - q (t_{1})

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for

t_{1}, t_{2}

in the stated domain with positive

Δ t = t_{2} - t_{1} .

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Remark 18.1.3.

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We can also apply the descriptors increasing or decreasing to a quantity on a domain that includes one or both of its endpoints.

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Example 18.1.4. An upright parabola decreases then increases.

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On any domain

0 \leq t \leq b

we can say that the quantity function

q (t) = t^{2}

is increasing, because a number that is larger in magnitude will produce a larger square:

\begin{aligned} t_{1} & < t_{2} & ⟹ & t_{1}^{2} & < t_{2}^{2} . \end{aligned}

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That is,

\begin{aligned} Δ t = t_{2} - t_{1} & > 0 & ⟹ & Δ q = q_{2} - q_{1} = t_{2}^{2} - t_{1}^{2} & > 0 . \end{aligned}

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On a domain of the form

a \leq t \leq 0,

the same principle applies, but for negative numbers satisfying

t_{1} < t_{2}

it will be

t_{1}

that is greater in magnitude. So our quantity function

q (t) = t^{2}

is decreasing on such a domain, since in that case

\begin{aligned} t_{1} & < t_{2} & ⟹ & t_{1}^{2} & > t_{2}^{2} . \end{aligned}

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That is,

\begin{aligned} Δ t = t_{2} - t_{1} & > 0 & ⟹ & Δ q = q_{2} - q_{1} = t_{2}^{2} - t_{1}^{2} & < 0 . \end{aligned}

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Δ q

is always positive between a pair of points in some domain (with positive

Δ t

as well), then the average rate

Δ q / Δ t

must always be positive as well. Similarly, domains of decrease will produce negative average rates. Since derivative values are approximated by average rate values, we can connect domains of increase and decrease to the sign of the derivative.

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Pattern 18.1.5. Derivative values versus domains of increase and decrease.

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1. If $q (t)$ is both differentiable and increasing on the domain $a < t < b,$ then $q^{'} (t)$ cannot be negative at any point in that domain.
2. Conversely, if $q (t)$ is differentiable on the domain $a < t < b,$ with $q^{'} (t)$ positive at all points in that domain, then $q (t)$ must be increasing on that domain.
1. If $q (t)$ is both differentiable and decreasing on the domain $a < t < b,$ then $q^{'} (t)$ cannot be positive at any point in that domain.
2. Conversely, if $q (t)$ is differentiable on the domain $a < t < b,$ with $q^{'} (t)$ negative at all points in that domain, then $q (t)$ must be decreasing on that domain.

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Justification.

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First, if

q (t)

is increasing on

a < t < b,

then

Δ q

calculated between any pair of points in that domain will always be positive. Since

Δ t

is always positive, this makes every average rate calculation

Δ q / Δ t

positive. But if

q

is differentiable, then at any given point

t = t^{*}

the average rate approximations to the derivative value

q^{'} (t^{*})

can be made arbitrarily close to it by taking sufficiently small

Δ t .

Since positive average rates cannot be made arbitrarily close to a negative value,

q^{'} (t^{*})

cannot be negative. However,

q^{'} (t^{*}) = 0

is possible, since it could be the case that all of these average rate calculations around

t = t^{*}

are all positive but very small in magnitude.

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Now suppose it is known that

q^{'} (t)

is positive on all of

a < t < b .

Then near any given value

t = t^{*}

in that domain, there is a small subdomain (by taking

Δ t

sufficiently small) around that point where all of the average rate calculations

Δ q / Δ t

are close to that positive derivative value

q^{'} (t^{*})

and hence must be positive themselves. Since

Δ t

is always positive, we must have that all of these

Δ q

values of positive, so

q

is at least increasing on this small subdomain around

t = t^{*} .

By “stitching” together all of these small subdomains, we find that

q

must be increasing on all of the full domain

a < t < b .

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Finally, similar arguments can be made to confirm the claims in Statement 2 of this pattern.

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Remark 18.1.6.

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Notice that we have been careful to say “

q^{'} (t)

cannot be negative” in the first statement and “

q^{'} (t)

positive” in the second statement. If we start with the assumption that

q

is increasing, then, as described in the justification, it is still possible for

q^{'} (t^{*}) = 0

to be true at some points in that domain.

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On the other hand, if we started with the assumption

q^{'} (t) \geq 0

on some domain, then we couldn’t say for sure that

q

is always increasing on that domain, since there could be an extended period where

q^{'} (t) = 0 .

On such a period, the graph of

q (t)

would be a horizontal line, and

q

would be neither increasing nor decreasing.

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Warning 18.1.7.

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Even if

q^{'} (t^{*}) > 0

at some particular point

t = t^{*},

it is never appropriate to say “

q

is increasing at

t = t^{*} .

” The adjectives increasing and decreasing are comparative, so you need at least two data points in order to be able to say that

q

increased or decreased from this point to that point. In general, we will only apply the descriptors increasing and decreasing to a function over an entire domain such as

a \leq t \leq b .

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Example 18.1.8. The natural logarithm and its derivative.

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The natural logarithm function

\ln (t)

has domain

t > 0

and derivative function

r (t) = \ln^{'} (t) = 1 / t .

Since we have

\begin{aligned} t & > 0 & ⟹ & \frac{1}{t} & > 0, \end{aligned}

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Pattern 18.1.5 says that

\ln (t)

should be always increasing on its domain. We easily confirm this using our knowledge of the graph of the logarithm function. (See Figure 8.1.2.)

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On the other hand, consider the rate function

r (t) = 1 / t

itself. Its derivative is

r^{'} (t) = - \frac{1}{t^{2}},

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and that derivative function is always negative on the domain

t > 0 .

Pattern 18.1.5 then says that

r (t)

should be decreasing on that same domain, and we know from its graph that this is so. But consider what this means for the logarithm function: since

$r (t)$ is the rate of variation of $\ln (t)$
the variation in $\ln (t)$ is that it is increasing on its domain
the rate function $r (t)$ is decreasing on that same domain,

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we can say that the natural increases on domain

t > 0,

but it does so at a decreasing rate.

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Example 18.1.9. The natural exponential and its derivative.

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The natural exponential function

q (t) = \exp (t)

has domain the full real number line, and is its own derivative function:

r (t) = q^{'} (t) = \exp (t) .

Since

\exp (t) > 0

for all

t,

we can say that

q (t)

is always increasing.

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But we also have

r^{'} (t) = \exp (t) > 0

for all

t,

so the rate function

r (t)

is also always increasing. This is what makes exponential growth effectively “runaway” growth: not only is the quantity function

q (t)

increasing, it is increasing at an increasing rate.

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Example 18.1.10. Determining domains of increase and decrease.

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Consider the function

q (t) = \frac{2 t^{2}}{t^{2} - 1} .

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Let’s ask Sage to compute its derivative. (We could also turn the denominator into a factor with a negative exponent, and then use a combination of the Product Rule and the Chain Rule.


    
        
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var('t')
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q(t) = 2 * t^2 / (t^2 - 1)
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qp = q.derivative()
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print(qp.simplify_full())

    
    
    
    
        
            
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That denominator can be factored:

q^{'} (t) = \frac{- 4 t}{(t^{2} - 1)^{2}} .

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To use Pattern 18.1.5 to determine the domains of increase and decrease, we would like to solve the inequalities

\begin{aligned} q^{'} (t) & > 0 & q^{'} (t) & < 0, \end{aligned}

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though solving one will effectively solve the other. Because of the outer square, the denominator is positive for all

t \neq \pm 1 .

Because of the extra negative in the numerator, the numerator is negative for

t > 0

and positive for

t < 0 .

\begin{aligned} q^{'} (t) & > 0 & for & t & < 0, t \neq \pm 1 \\ q^{'} (t) & < 0 & for & t & > 0, t \neq \pm 1, \end{aligned}

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and we conclude that

q

is increasing on the domains

t < - 1

and

- 1 < t < 0,

and decreasing on the domains

0 < t < 1

and

t > 1 .

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Example 18.1.11. Joining two domains of increase.

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Consider the function

q (t) = t^{3} .

We have

q^{'} (t) = 3 t^{2},

and so except for

q^{'} (0) = 0

we have

q^{'} (t) > 0

for all

t .

At this point we can say that

q

increases on the domain

t < 0,

and again on the domain

t > 0 .

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What about at

t = 0 ?

We should never talk about increasing or decreasing at one single point. But suppose we compare the output

q (0) = 0

to other nearby outputs. For example, if we use

t_{1} = 0

and

t_{2} > 0,

then

Δ q = q (t_{2}) - q (t_{1}) = t_{2}^{3} - 0 > 0 .

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On the other side, if we use

t_{1} < 0

and

t_{2} = 0,

then

Δ q = q (t_{2}) - q (t_{1}) = 0 - t_{2}^{3} > 0,

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because

t_{2}^{3}

itself will be negative for

t_{2} < 0 .

So it is appropriate to say that

q

increases through

t = 0,

and we can join our two domains of increase together to say that

q (t) = t^{3}

is always increasing.

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Why didn’t we do the same thing in Example 18.1.10, and join the two adjacent domains

t < - 1

and

- 1 < t < 0

together to say that that particular function was increasing on the domain

t < 0 ?

Because in that case

q (- 1)

was undefined, and so there is no point there for the function to increase through. Even though

q

is increasing on each of those two adjacent domains, we cannot join them together into one domain of increase.

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Section 18.2 Critical points and local extremes

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An important feature in Example 18.1.10 is the boundary between the domain

- 1 < t < 0,

where the function is increasing, and the domain

0 < t < 1,

where the function is decreasing. Notice that

q^{'} (0) = 0

at that boundary. We can think of this feature in several different ways.

When increasing, the quantity is accumulating, and when decreasing, the quantity is dissipating. At a boundary between these two behaviours, there must be an instant where the quantity is neither accumulating nor dissipating; that is, where $r (t) = 0 .$
If the function represents the position of an object in motion, the derivative function represents velocity. When the velocity is positive, the object’s position is increasing in the “positive direction”, and so we think of the object as moving “forward.” When the velocity is negative, the object’s position is increasing in the “negative direction”, and so we think of the object as moving “backward.” At a boundary between these two boundaries, in order to switch direction of movement there must be an instant where the object comes to a stop; that is, where $v (t) = 0 .$
Graphically, if a function increases and then decreases, it must have come to a “peak” at that boundary point between the two behaviours. On the increasing side of the peak we will see positive slopes, and on the decreasing side we will see negative slopes. So if the function is differentiable at that boundary point, then we should see a slope $q^{'} (t) = 0,$ where the tangent line is horizontal.

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Pattern 18.2.1. Changes in behaviour.

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At a boundary point

t = t^{*}

between a domain of increase and a domain of decrease, we must see at least one of the following:

$q (t^{*})$ is undefined
$q^{'} (t^{*})$ is undefined
$q^{'} (t^{*}) = 0 .$

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Justification.

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q (t^{*})

is undefined, then one of the listed occurrences has indeed occurred. If

q (t^{*})

is defined but

q^{'} (t^{*})

is not, then again one of the listed occurrences has indeed occurred. So we really only need to concern ourselves with the case that both

q (t^{*})

and

q^{'} (t^{*})

are defined, and try to determine the value of

q^{'} (t^{*})

in that case.

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Suppose that

q (t)

is increasing on some domain

a < t \leq t^{*}

and decreasing on some domain

t^{*} \leq t < b .

(The argument in the case that

q

is decreasing on the first domain and increasing on the second is essentially the same.) In that case, every forward difference approximation to the instantaneous rate will be negative (since

Δ q

is negative when

q

is decreasing), and every backward difference approximation to the instantaneous rate will be positive (since

Δ q

is positive when

q

is increasing). However, if

q^{'} (t^{*})

is defined, then by Definition 11.5.1, all of those approximations, both positive and negative, for sufficiently small

Δ t

must be close to the derivative value

q^{'} (t^{*}) .

The only number that can simultaneously be arbitrarily close to both positive and negative values is

0 .

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Ignoring the first of the three situations in Pattern 18.2.1 for now, points on the graph of

q (t)

where one of the other two situations occur are important, because they signal the potential for changes in behaviour.

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Definition 18.2.2. Critical points.

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A point

t = t^{*}

in the domain of

q

where

q^{'} (t^{*})

is either undefined or equal to

0

is called a critical point for the function.

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We have already seen an example of a critical point with

q^{'} (t^{*}) = 0

in Example 18.1.10, for

t^{*} = 0 .

Here is an example of the other variety of critical point.

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Example 18.2.3. A critical point can be a cusp.

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Consider

q (t) = t^{2 / 3} .

Then the derivative function

q^{'} (t) = - \frac{2}{3 t^{1 / 3}}

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is undefined at

t = 0,

even though the original function

q

is defined there.

Graph with a cusp as a critical point. — Figure 18.2.4. The cusp at $t = 0$ is a critical point.

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On the graph we can see that the function does change from decreasing to increasing at

t = 0,

but it does with an instantaneous change in behaviour instead of a gradual one, and leaves behind a cusp at that boundary between behaviours.

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Warning 18.2.5.

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A critical point is not always a boundary between different behaviours, as the example of

q (t) = t^{3}

demonstrates, where

t = 0

is a critical point but the function increases on its entire domain.

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When a ball is thrown up into the air, it increases in height until it comes to a momentary stop, when it begins to descend again. That critical point occurs at the moment of maximum height.

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Definition 18.2.6. Local extremes.

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A function $q (t)$ is said to have a local minimum at $t = t^{*}$ if there is a domain containing $t = t^{*}$ where $Δ q$ is always negative to the left of that point and always positive to the right of that point. In this case, $q (t^{*})$ called a local minimum value of $q$ .
A function $q (t)$ is said to have a local maximum at $t = t^{*}$ if there is a domain containing $t = t^{*}$ where $Δ q$ is always positive to the left of that point and always negative to the right of that point. In this case, $q (t^{*})$ called a local maximum value of $q$ .

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In general, we say that

q

has a local extreme at

t = t^{*}

if it has either a local minimum or a local maximum there.

Graph of a function with a local minimum. — (a) A function with a local minimum.

Graph of a function with a local maximum. — (a) A function with a local minimum.

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Remark 18.2.8.

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These are referred to as local minimum and maximum values, since they will only be the lowest or highest values on some (possibly small) domain around them. The function could then go on to achieve lower or higher values on a larger domain, as in Figure 18.2.7.

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Example 18.2.9. Repeating local extremes.

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Because of its periodic nature, the cosine function has an infinite number of local extreme locations: a local maximum at each point

t = n π

with

n

an even integer, and a local minimum at each point

t = n π

with

n

an odd integer. (See Figure 10.3.2.)

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Hopefully it is obvious from our definition that a local extreme can only occur at a critical point.

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Pattern 18.2.10. Location of local extremes.

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If a function has a local extreme at

t = t^{*},

then it must have a critical point there.

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Justification.

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We will discuss only the case of a local minimum — the case of a local maximum is essentially the same.

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q (t)

has a local minimum at

t = t^{*},

then by definition

Δ q

is always negative to the left and positive to the right of that point, on some suitably small domain around it. This means that

q

must be decreasing down to that minimum value, and then increasing away from it. Since we are assuming that

t = t^{*}

is in the domain of

q,

then Pattern 18.2.1 essentially says that

q

must have a critical point there.

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But again, as the cube example demonstrates, a critical point is not always a local extreme. Here is a way that we can determine if

q

has a local extreme at a particular critical point, and what kind if it does.

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Pattern 18.2.11. First Derivative Test.

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Suppose that

q

has a critical point at

t = t^{*} .

If $q$ changes in behaviour from increasing to decreasing at $t = t^{*},$ then $q$ has a local maximum there.
If $q$ changes in behaviour from decreasing to increasing at $t = t^{*},$ then $q$ has a local minimum there.
If $q$ does not change behaviour at $t = t^{*},$ then $q$ has neither a local minimum nor a local maximum there.

Graph of a function that changes from increasing to decreasing at a local maximum. — (a) Change from increasing to decreasing at a local maximum.

Graph of a function that changes from decreasing to increasing at a local minimum. — (a) Change from increasing to decreasing at a local maximum.

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Remark 18.2.13.

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A critical point in the third category of Pattern 18.2.11 is sometimes called a saddle point for

q .

Notice that such a critical point can still have a horizontal tangent there if

q^{'} (t^{*}),

but in this case the tangent passes through the curve at the saddle point.

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Example 18.2.14. Applying the First Derivative Test.

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Consider the accumulation function

A (t) = \int_{0}^{t} (u - 2) (u - 4)^{2} d_{} u .

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When does the accumulated quantity come to a momentary high or low value? By the Theorem 13.3.1, the derivative function is

A^{'} (t) = \frac{d}{d t} \int_{0}^{t} (u - 2) (u - 4)^{2} d_{} u = (t - 2) (t - 4)^{2} .

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So there are two critical points, at

t = 2

and

t = 4 .

Except for at

t = 4,

the squared factor in the derivative function is always positive, so increasing/decreasing behaviour is completely determined by the

(t - 2)

factor:

decreasing for $t < 2$
increasing for $t > 2 .$

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So the quantity was initially dissipating, reached a local minimum value at

t = 2,

and then began to accumulate forever more after that. In particular, the critical point at

t = 4

is only a saddle point and not a local extreme.

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Let’s ask Sage to compute that momentary low quantity for us.


    
        
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integrate( (u - 2) * (u - 4)^2, u, 0, 2 );

    
    
    
    
        
            
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So the quantity will bottom out at

68 / 3

below whatever initial quantity there was at

t = 0,

and then start to recover.

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Section 18.3 The second derivative

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r (t) = q^{'} (t)

is the rate of variation of quantity

q,

then

r^{'} (t)

is the rate of variation of that rate of variation. Since this derivative function is the derivative of the derivative of

q,

we usually refer to it as the second derivative of

q,

and write

q^{″} (t) or \frac{d^{2} q}{{d t}^{2}}

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to represent it.

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The second derivative tells about the variation of the rate function

r (t)

— essentially it gives us information about whether a quantity is “speeding up” or “slowing down” in its variation. For example, we’ve already seen in Example 18.1.8 and Example 18.1.9 that both the natural logarithm and exponential functions are increasing on their respective domains, but the logarithm function increases at a decreasing rate, while the exponential function just keeps increasing faster and faster.

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However, when using the idea of “speeding up” and “slowing down,” we’ve got to be careful to take “direction” into account.

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Example 18.3.1. Speeding up versus slowing down.

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Suppose an object exhibits linear motion, with position function

\begin{aligned} s (t) & = t^{3} - 12 t^{2} + 36 t & t & \geq 0 . \end{aligned}

🔗

The derivative function represents velocity:

\begin{aligned} v (t) & = s^{'} (t) \\ = 3 t^{2} - 24 t + 36 \\ = 3 (t - 2) (t - 6) . \end{aligned}

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Variation in velocity is called acceleration:

a (t) = v^{'} (t) = s^{″} (t) = 6 t - 24 = 6 (t - 4) .

🔗

As described at the beginning of this chapter, when

v (t) > 0,

position is increasing and we think of the object as moving forward. When

v (t) < 0,

position is decreasing and we think of the object as moving backward. In this case, the object is

moving forward on $0 \leq t < 2$ and again on $t > 6$
moving backward on $2 < t < 6 .$

🔗

From our acceleration function, we can say that velocity is

decreasing on $0 \leq t < 4$
increasing on $t > 4 .$

🔗

However, velocity increasing/decreasing is not the same as speeding up / slowing down, since velocity takes direction into account but speed does not. In particular, on

2 < t < 4

we have both negative velocity and negative acceleration. If velocity is decreasing from an already negative value, then velocity is becoming more negative. In other words, the magnitude of velocity (speed) is actually increasing, and the object is speeding up on that time domain.

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In general, an object in motion is speeding up when velocity and acceleration have the same “direction,” and slowing down otherwise. For this object, we have

moving forward but slowing down on $0 \leq t < 2$
momentarily stopped at $t = 2$
moving backward and speeding up on $2 < t < 4$
moving backward but slowing down on $4 < t < 6$
momentarily stopped at $t = 6$
moving forward and speeding up on $t > 6 .$

🔗

Here are some representative graphs for the different combinations of increasing/decreasing function and increasing/decreasing rate.

Typical graph shape when quantity and rate are both increasing. — (a) Quantity and rate both increasing.

Typical graph shape when quantity decreases then increases, but rate is always increasing. — (a) Quantity and rate both increasing.

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Notice that these representative graphs are split into two groups based on a common feature, depending on whether the rate is increasing or decreasing: each graph with an increasing rate exhibits a curve that opens upward, whereas each graph with an increasing rate exhibits a curve that opens downward.

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Definition 18.3.3. Concavity and inflection points.

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Suppose

q (t)

is a twice-differentiable function on domain

a < t < b .

If $q^{″} (t) > 0$ at all points on that domain, then we say that $q$ is concave up there.
If $q^{″} (t) < 0$ at all points on that domain, then we say that $q$ is concave down there.
If $t = t^{*}$ is in the domain of $q$ and at a boundary between concave up and down domains, then we say that $q$ has an inflection point there.

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Remark 18.3.4.

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Since the second derivative of

q

is the first derivative of

q^{'},

domains of concavity for

q

coincide with domains of increase/decrease for

q^{'},

and inflection points for

q

must be critical points for

q^{'} .

That means that to determine inflection points, we should look for points where $q^{″} (t)$ is either $0$ or undefined. However, just as with critical points versus local extremes, note that having one of those two conditions does not necessarily mean that

q

will have an inflection point there.

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Warning 18.3.5.

🔗

A common mistake for beginning students is to use critical points for

q

in their analysis of concavity and inflection points. While some functions will coincidentally have

q^{″} (t) = 0

at some of the same

t

-values where

q^{'} (t) = 0

occurs, in general only points with

q^{″} (t) = 0

(or

q^{″} (t)

undefined) should be considered as potential inflection points.

🔗

Example 18.3.6. Determining concavity.

🔗

Let’s continue Example 18.1.10, where we’ve already determined the domains of increase and decrease for

q (t) = \frac{2 t^{2}}{t^{2} - 1} .

🔗

Let’s now ask Sage to calculate the second derivative for us.


    
        
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var('t')
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q(t) = 2 * t^2 / (t^2 - 1)
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qpp = q.derivative().derivative()
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print(qpp.simplify_full())

    
    
    
    
        
            
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🔗

That denominator is ugly, let’s have Sage factor it for us.


    
        
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var('t')
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factor(t^6 - 3*t^4 + 3*t^2 - 1)

    
    
    
    
        
            
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So now we have

q^{″} (t) = \frac{4 (3 t^{2} + 1)}{{(t + 1)}^{3} {(t - 1)}^{3}} .

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Note that the numerator is always positive, so

q^{″} (t) = 0

is not possible. This function is undefined at

t = \pm 1,

but so is the original function

q (t),

so those cannot be inflection points. However, they could still be boundaries between different concavities for

q .

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Since the numerator is always positive, to solve the inequalities

\begin{aligned} q^{″} (t) & > 0 & q^{″} (t) & < 0 \end{aligned}

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we only need to look at the denominator. We see:

${(t + 1)}^{3}$ is negative on $t < - 1$ and positive on $t > - 1$
${(t - 1)}^{3}$ is negative on $t < 1$ and positive on $t > 1 .$

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Putting these together, we have that

q

concave up on $t < - 1$ and on $t > 1$
concave up down on $- 1 < t < 1 .$

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Section 18.4 Graph trends

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Computers are great at drawing graphs for us, but they have some draw backs.

A restricted viewport means you never know if you are seeing all the important features and behaviour of the the graph, no matter how far you zoom out.
Computer-generated graphs do not perform well with functions whose values “explode” quickly, especially near singularities.

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Essentially, the problem is that a computer can only show you what you ask it to show you, and is unforgiving in its accuracy. As humans we can sacrifice some accuracy and use what we have learned in this and previous chapters to sketch a total graphical picture of the behaviour of a function’s graph, one that we can be sure captures all of the important features of the graph. To do this, we don’t plot a whole bunch of points and blindly connect the dots. Instead, we only plot important points (for example, intercepts, critical points, inflection points) and then connect these using our knowledge of the graph’s behaviour between them (increasing/decreasing, concave up/down, singularities, long-term/ancient behaviour).

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Example 18.4.1. Sketching a graph.

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Let’s finish analyzing the function from Example 18.1.10 and Example 18.3.6:

q (t) = \frac{2 t^{2}}{t^{2} - 1} .

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We already know:

increasing on $t < - 1$ and $- 1 < t < 0$
decreasing on $0 < t < 1$ and $t > 1$
concave up on $t < - 1$ and on $t > 1$
concave up down on $- 1 < t < 1 .$

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To this we can add:

intercepts both axes at the origin, since $q (0) = 0,$ and there are no other axis intercepts
singularities at $t = - 1$ and $t = 1$
long-term behaviour

$\frac{2 t^{2}}{t^{2} - 1} \to 2 as t \to \infty$
ancient behaviour

$\frac{2 t^{2}}{t^{2} - 1} \to 2 as t \to - \infty .$

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We can also combine some of these pieces of information.

Since the graph increases up to the intercept at $q (0) = 0$ and then decreases away, there must be a local maximum there.
Since $q$ increases on $t < - 1,$ the graph must be above the horizontal asymptote $y = 2$ at the far left of the graph, and must increase away from it so that eventually $y \to \infty$ as $t = - 1^{-}$ on the left of the vertical asymptote at $t = - 1 .$
Since $q$ also increases on $- 1 < t < 0,$ the graph must increase away from the vertical asymptote at $t = - 1,$ so $y \to - \infty$ as $t \to - 1^{+} .$
With similar reasoning we can conclude
$\begin{matrix} y \to - \infty as t \to 1^{-} \\ y \to \infty as t \to 1^{+} \end{matrix}$
and that the graph must decrease down to the horizontal asymptote at $y = 2$ from above at the far right of the graph.

Graph of an analyzed function. — Figure 18.4.2. Graph of $q (t) = 2 t^{2} / (t^{2} - 1) .$

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Here is a general procedure one can follow to sketch a graph that contains all the important features of the function. The steps do not necessarily all need to be completed, nor do they necessarily need to be completed in the stated order.

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Procedure 18.4.3. Curve-sketching procedure.

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Determine the domain of the function, and sketch in any vertical asymptotes.
Determine the long-term and ancient behaviour, and sketch in any horizontal or slant asymptotes.
Determine and plot any axis intercepts.
Determine any critical points, calculate the output values at those points, and plot their approximate locations.
Determine the domains of increase and decrease. Use the First Derivative Test to classify each critical point as a local minimum, local maximum, or a saddle point.
Determine the domains of concavity. Determine any inflection points, calculate the output values at those points, and plot their approximate locations.
Finally, sketch the graph between the asymptotes and plotted points, being sure that the sketch reflects the increase/decrease/concavity information on each subdomain.

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Here is one final example. Verifying the provided information is left to you, the reader.

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Example 18.4.4. Using the curve-sketching procedure.

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For function

q (t) = \frac{t^{3}}{t^{2} + 1},

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we have derivative and second derivative

\begin{aligned} q^{'} (t) & = \frac{t^{2} (t^{2} + 3)}{(x^{2} + 1)^{3}} & q^{″} (t) & = \frac{2 t (3 - t^{2})}{(t^{2} + 1)^{3}} . \end{aligned}

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Here is the information requested by the Curve-sketching procedure.

Domain.
The whole real number line. No singularities.
Long-term/ancient behaviour.

$\begin{aligned} q (t) & \to \infty & as & t \to \infty & q (t) & \to - \infty & as & t \to - \infty \end{aligned}$

So no horizontal asymptotes, but we can divide

$\frac{t^{3}}{t^{2} + 1} = t - \frac{t}{t^{2} + 1},$

with

$\frac{t}{t^{2} + 1} \to 0 as t \to \pm \infty,$

so we have a slant asymptote of $y = t .$
Intercepts.
Only at the origin, as $q (0) = 0 .$
Critical points.
Only at $q^{'} (0) = 0,$ and we already know that $q (0) = 0 .$
Increase/decrease.
Increasing on the entire domain, so the critical point at the origin is a saddle point.
Concavity.
- Concave up on $t < - \sqrt{3}$ and again on $0 < t < \sqrt{3} .$
- Concave down on $- \sqrt{3} < t < 0$ and again on $t > \sqrt{3} .$
- Inflection points at
  $\begin{aligned} q (- \sqrt{3}) & = - \frac{3 \sqrt{3}}{4} & q (0) & = 0 & q (\sqrt{3}) = \frac{3 \sqrt{3}}{4} . \end{aligned}$

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And here is the graph. Being a computer-generated graph, it’s a bit difficult to see the change in concavity at inflection points

t = \pm \sqrt{3} .

But the graph must change concavity at these points, because it is impossible for the graph to approach the slant asymptote from below at the far right of the graph with a concave-up orientation, and similarly for the approach to the slant asymptote at the far left of the graph.

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