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Calculus Concepts and Modelling: An introduction to the concepts of calculus

Jeremy Sylvestre

Chapter 7 Polynomial growth

Section 7.1 Power and polynomial functions

Subsection 7.1.1 Definitions

A polynomial is the result of performing algebra in one variable when you restrict yourself to the operations of addition, subtraction, and multiplication. (Keep in mind that a power with a positive integer exponent just represents repeated multiplication.) The generic form of a polynomial in the variable \(t\) is
\begin{equation*} p(t) = a_0 + a_1 t + a_2 t^2 + a_3 t^3 + \dotsb + a_{n-1} t^{n-1} + a_n t^n \end{equation*}
where we assume \(a_n \neq 0\) (since otherwise we would have omitted that term and terminated the polynomial at a lower power of \(t\)). The constants \(a_0, a_1, \dotsc, a_n\) are called the coefficients of the polynomial, and the highest exponent \(n\) is called the degree.
A polynomial that consists of just a single term,
\begin{equation*} p(t) = a_n t^n \text{,} \end{equation*}
is called a power function. For \(n \ge 2\) and even, the graph of a power function is essentially a parabola, while for \(n \ge 3\) and odd, the graph of a power function is essentially a parabola with one arm flipped vertically. In both cases, higher values of \(n\) flatten out the bottom and steepen the sides.
Graphs of power functions of even degree.
(a) Graphs of power functions of even degree.
Graphs of power functions of odd degree.
(b) Graphs of power functions of odd degree.
Figure 7.1.1. Graphs of basic power functions.
Solving \(p(t) = 0\) for \(p\) a polynomial is of fundamental importance in algebra. A solution to \(p(t) = 0\) is called a root of \(p\text{,}\) and graphically these solutions coincide with horizontal intercepts of the polynomial’s graph. The meaning of the term \(\nth\) root of a number is tied to the roots of polynomials, as a square root \(\sqrt{a}\) means the (positive) root of the equation
\begin{equation*} t^2 - a = 0\text{,} \end{equation*}
a cube root \(\sqrt[3]{a}\) means the unique real root of the equation
\begin{equation*} t^3 - a = 0\text{,} \end{equation*}
and so on.

Subsection 7.1.2 Polynomials of small degree

Since we often factor polynomials into factors that are polynomials of smaller degree, polynomials of small degree are of special interest.
Subsubsection Constant polynomials
A polynomial of degree \(0\) is just a a constant function, and its graph is a horizontal line. The zero function \(p(t) = 0\) can be considered a polynomial, though, depending on the context, it may be appropriate to say that the zero polynomial has no degree to distinguish it from non-zero constant polynomials. However, we will just lump the zero polynomial in with the constant polynomials.
Subsubsection Linear polynomials
A polynomial of degree \(1\text{,}\)
\begin{equation*} p(t) = a_0 + a_1 t \text{,} \end{equation*}
is called a linear polynomial because its graph is a line, with vertical intercept \(a_0\) and slope \(a_1\text{,}\) as usual. The root of a linear polynomial is easy to determine:
\begin{align*} a_0 + a_1 t \amp = 0 \amp \amp\implies \amp t \amp = - \frac{a_0}{a_1} \text{.} \end{align*}
Subsubsection Quadratic polynomials
A polynomial of degree \(2\text{,}\)
\begin{equation*} p(t) = a_0 + a_1 t + a_2 t^2 \text{,} \end{equation*}
is called a quadratic polynomial. Its graph is always a parabola, though it may be shifted and/or reflected. If the parabola opens upward we say that it is concave up, and if it opens downward we say that it is concave down. Which of these two shapes the parabola exhibits is determined by \(a_2 \gt 0\) or \(a_2 \lt 0\text{,}\) respectively. The vertex of a parabola is the point where a concave up parabola reaches its minimum height, or where a concave down parabola reaches its maximum height.
A power function of degree \(2\text{,}\) \(p(t) = a_2 t^2 \text{,}\) has its vertex at the origin. For other degree-\(2\) polynomials, we can determine the location of the vertex of its graph by completing the square.
Example 7.1.2. Completing the square.
Consider polynomial
\begin{equation*} p(t) = 6 + 10 t - 2 t^2 \text{.} \end{equation*}
First, we note that this parabola is concave down because of the negative coefficient on the \(t^2\) term. After factoring out that coefficient, we complete the square by manipulating the constant term to create a perfect square.
\begin{align*} p(t) \amp = - 2 t^2 + 10 t - 6 \\ \amp = - 2 (t^2 - 5 t + 3) \\ \amp = - 2 \left( t^2 - 5 t \: {\color{red} + {\left(\frac{5}{2}\right)}^2 - {\left(\frac{5}{2}\right)}^2} + 3 \right) \\ \amp = - 2 \left[ {\left(t - \frac{5}{2}\right)}^2 - \frac{25}{4} + 3 \right] \\ \amp = - 2 \left[ {\left(t - \frac{5}{2}\right)}^2 - \frac{13}{4} \right] \\ \amp = - 2 {\left(t - \frac{5}{2}\right)}^2 + \frac{13}{2} \end{align*}
Using our knowledge of transformations of graphs from Section 2.5, we interpret the last formula above as a combination of transformations of the standard parabola \(f(t) = t^2\text{,}\) in the order listed:
  • horizontal shift by \(5/2\) to the right
  • vertical stretch by a factor of \(2\)
  • vertical reflection
  • vertical shift by \(13/2\) up.
The two shifts put the vertex at the point \((5/2,13/2)\text{,}\) and the vertical reflection reminds us that the parabola is concave down.
Graph of a parabola whose shape and location were determined by completing the square.
Figure 7.1.3. Graph of a parabola whose shape and location were determined by completing the square.
By considering different shifts and vertical reflections of parabolas, we can see that a quadratic polynomial can have no roots, one unique root, or two distinct roots. If we write our quadratic polynomial as
\begin{equation*} a t^2 + b t + c \text{,} \end{equation*}
then whether this polynomial has none, one, or two roots is determined by the discriminant
\begin{equation*} b^2 - 4 a c \text{.} \end{equation*}
no roots
when \(b^2 - 4 a c \lt 0\text{;}\) a concave up parabola with vertex above the horizontal axis or a concave down parabola with vertex below the horizontal axis
one unique root
when \(b^2 - 4 a c = 0\text{;}\) a parabola with vertex on the horizontal axis
two distinct roots
when \(b^2 - 4 a c \gt 0\text{;}\) a concave up parabola with vertex below the horizontal axis or a concave down parabola with vertex above the horizontal axis
In any case, there exists a formula for the roots of quadratic polynomial.
Subsubsection Cubic polynomials
The graph of a cubic polynomial often follows an “up-down-up” or “down-up-down” pattern, as in Figure 7.1.5.
Graph of a cubic that follows an "up-down-up" pattern.
(a) Graph of a cubic that follows an “up-down-up” pattern.
Graph of a cubic that follows an "down-up-down" pattern.
(b) Graph of a cubic that follows an “down-up-down” pattern.
Figure 7.1.5. Typical graphs of cubic functions.
The axes have been intentionally omitted from the graphs in Figure 7.1.5, so that you can imagine different vertical shifts of these graphs. In particular, the dashed lines indicate different possible locations of the horizontal axis, demonstrating the possibility for one, two, or three roots for a cubic polynomial.

Subsection 7.1.3 Roots and factoring

The shape of graphs and roots of higher degree polynomials is obviously more complicated. There exist formulas for the roots of cubic and quartic (degree-\(4\)) polynomials similar to the Pattern 7.1.4, but already there is (famously) no such general formula possible for quintic (degree-\(5\)) polynomials. However, we can say something about the number of roots of a polynomial.
The reason for this is that solving for roots and factoring are equivalent tasks.
Pattern 7.1.7 is the reason the Fundamental Theorem of Algebra is true: if every root means we can factor out a linear factor, then \(n\) roots means \(n\) linear factors, at which point we will have reached the degree of the polynomial and can factor no further.
Example 7.1.8. Factoring out a root.
Clearly \(t = -1\) is a root of \(p(t) = t^4 - 1\text{.}\) This means that we can factor
\begin{equation*} \bbrac{t - (-1)} = (t + 1) \end{equation*}
out of \(p\text{.}\)
\(\color{red} t^3\) \(\color{blue} -\) \(\color{blue} t^2\) \(\color{green} +\) \(\color{green} t\) \(\color{purple} -\) \(\color{purple} 1\)
\(t + 1\) \(t^4\) \(+\) \(0 t^3\) \(+\) \(0 t^2\) \(+\) \(0 t\) \(-\) \(1\)
\(-\) \(\color{red} t^4\) \(\color{red} +\) \(\color{red} t^3\) \(\downarrow\) \(\downarrow\) \(\downarrow\)
\(- t^3\) \(+\) \(0 t^2\)
\(-\) \(\color{blue} - t^3\) \(\color{blue} -\) \(\color{blue} t^2\)
\(t^2\) \(+\) \(0 t\)
\(-\) \(\color{green} t^2\) \(\color{green} +\) \(\color{green} t\)
\(- t\) \(-\) \(1\)
\(\color{purple} - t\) \(\color{purple} -\) \(\color{purple} 1\)
\(0\)
The remainder of \(0\) tells us that \(t + 1\) does indeed divide evenly into \(t^4 - 1\text{,}\) and the quotient result at the top says
\begin{equation*} t^4 - 1 = (t + 1) (t^3 - t^2 + t - 1) \text{.} \end{equation*}

Section 7.2 Constant rate of variation

The simplest type of polynomial is a constant function, which can be considered a degree-0 polynomial. We have already seen in Pattern 1.4.3 that a constant rate of variation always corresponds to a linear quantity function.

Example 7.2.1. Constant acceleration.

Variation in the velocity of an objection in motion is called acceleration. A small object dropped over the surface of an airless planet will experience constant acceleration toward the planet’s surface. We usually denote this constant acceleration as \(g\text{,}\) since its cause is gravity. If we measure velocity in metres per second, then variation of velocity measures how many metres per second change we have per second, so that \(g\) is measured in metres per second per second (or “metres per second-squared”).
If the object is dropped with initial velocity 0 m⁄s, then we can use either Pattern 6.4.7 (just as in Example 6.4.10) or Pattern 1.4.3 to compute its velocity at any point in time after that, up until the time that the object hits the surface of the planet:
\begin{equation*} v(t) = 0 + \ccmint{0}{t}{g}{u} = g t \text{.} \end{equation*}
Pattern 7.2.2 can be interpreted in reverse as well.
While Pattern 7.2.2 is just a restatement of Pattern 1.4.3, Pattern 7.2.3 is new, though it should have been expected. We will verify this new statement mathematically in Chapter 12.

Section 7.3 Linear rate of variation

Next we consider a linear rate of variation.

Example 7.3.1. Linear velocity.

In Example 7.2.1, we saw that a constant acceleration model led to a linear velocity model. But velocity itself is the rate of variation of position. So we can integrate velocity to obtain a model of the accumulation of distance from the starting position, otherwise known as displacement.
To simplify things, let us label the starting position as “position 0” so that displacement is the same as position. Our rate function is \(v(t) = g t\) over a time domain beginning at \(t = 0\) and ending at the variable time \(t = z\text{.}\) Since our graph is linear, we can easily compute the accumulation for this rate function as an area under the graph, similarly to Example 6.3.7.
Graph of a velocity function with bounded oriented area as accumulation of displacement.
Figure 7.3.2. Accumulation of displacement as net oriented area under a velocity graph.
The triangle in Figure 7.3.2 has height \(v(t)\) and base length \(t\text{.}\) So, using the formula for area of a triangle, we obtain
\begin{equation*} \ccmint{0}{t}{v(u)}{u} = \frac{v(t) \cdot t}{2} = \frac{g t^2}{2} \text{.} \end{equation*}
Let’s ask Sage to confirm this for us.
So it seems that a linear rate function leads to parabolic growth/decay.
We can combine the pattern of the result of Example 7.3.1 with Pattern 7.2.2 to deal with all linear rate functions, since Property 3 of Pattern 6.3.11 says that
\begin{equation*} \ccmint{0}{t}{ m u + b }{u} = \ccmint{0}{t}{ m u }{u} + \ccmint{0}{t}{ b }{u} \text{.} \end{equation*}

Example 7.3.4. Parabolic motion.

An object is thrown upwards from the surface of an airless planet with an initial velocity of \(v_0\) metres per second. We would like to model the height of the object above the planet’s surface as a function in time. We start as before with the acceleration function \(a(t)\text{.}\) As the object is initially moving upwards and the gravity of the planet pulls the object downwards, this time we will take the constant acceleration to be a negative function:
\begin{equation*} a(t) = - g \text{,} \end{equation*}
where \(g\) represents the acceleration due to gravity in metres per second per second. Since acceleration is the rate of variation of velocity, Pattern 7.2.2 tells us that the velocity of the object can be modelled as
\begin{equation*} v(t) = -g t + v_0 \text{.} \end{equation*}
And velocity is the rate of variation of position, so from Pattern 7.2.2, the position of the object (as a height above the surface) can be modelled as
\begin{equation*} s(t) = -\frac{g}{2} t^2 + v_0 t \end{equation*}
if we take the initial position to be \(s(0) = 0\) at the surface of the planet.
There are some natural questions about the object’s motion that arise.
  • When does the object return to the surface?
    To answer this, we want to solve for \(t\) in the instances where \(s(t) = 0\text{.}\) We can do this by factoring; for convenience we will also factor out the fraction.
    \begin{equation*} s(t) = -\frac{g}{2} t^2 + v_0 t = \frac{t}{2} ( - g t + 2 v_0 ) \end{equation*}
    From the factored form, we can see that the object will be at the position of the surface when \(t = 0\) (as we assumed) and when
    \begin{gather*} - g t + 2 v_0 = 0 \\ \implies \quad t = \frac{2 v_0}{g} \text{.} \end{gather*}
  • With what velocity does it strike the surface?
    We just need to substitute the time of return into the velocity model:
    \begin{equation*} v(2 v_0 / g) = - g (2 v_0 / g) + v_0 = - v_0 \text{.} \end{equation*}
    Considering the symmetry of the parabolic position function (see Figure 7.3.5) it makes sense that the object returns to the surface at the same speed that it left the surface, but oppositely directed.
  • When does the object reach its maximum height?
    From the symmetry of the parabolic position function (again see Figure 7.3.5), we expect that the object will reach its maximum height halfway between its departure from and return to the surface, at
    \begin{equation*} t = \frac{v_0}{g} \text{.} \end{equation*}
  • What is the maximum height achieved? What is the velocity at the instant the maximum height is reached?
    Substituting the “halfway” time into the position and velocity functions, we have
    \begin{gather*} \text{max height} = s(v_0 / g) = \frac{v_0^2}{2 g} \\ v(v_0 / g) = 0 \text{.} \end{gather*}
    The fact that the velocity is \(0\) at the maximum height should not be a surprise — when the object is thrown upwards, it immediately begins to slow down because of gravity pulling back down, until it momentarily stops at its maximum height before beginning to descend.
A position and velocity graph for parabolic motion due to gravity.
Figure 7.3.5. A position and velocity graph for parabolic motion due to gravity.

Section 7.4 Polynomial rate of variation

A pattern is beginning to emerge:
  • a constant rate model leads to a linear quantity model
  • a linear rate model leads to a quadratic quantity model.
We might guess that a quadratic rate model should lead to a cubic quantity model. Let’s ask Sage.
It’s true! (With some work, this could be verified using lower and upper Riemann sums.) Because of Property 2 and Property 3 of Pattern 6.3.11, we only need a pattern for rate functions that are a power of \(t\) in order to be able to handle any polynomial rate function, and that pattern appears to be
\begin{gather} \ccmint{0}{t}{u^m}{u} = \frac{t^{m + 1}}{m + 1} \text{.}\tag{✶} \end{gather}

Example 7.4.1. Obtaining a quantity model from a polynomial rate function.

For rate function
\begin{equation*} r(t) = 3 t^5 - 37 t^4 + 22 t^3 + 7 t^2 - t + 56 \text{,} \end{equation*}
the associated quantity model is
\begin{equation*} q(t) = 3 \cdot \frac{t^6}{6} - 37 \cdot \frac{t^5}{5} + 22 \cdot \frac{t^4}{4} + 7 \cdot \frac{t^3}{3} - \frac{t^2}{2} + 56 t + q_0 \text{,} \end{equation*}
where \(q_0 = q(0)\) is the initial quantity.
In fact, the pattern of (✶) is valid for all constant exponents \(m\text{,}\) with one important exception that we state below. But because negative powers of \(t\) exhibit singular behaviour at \(t = 0\text{,}\) we’ll have to choose another time to act as the “initial” time. The next logical choice for a “standard” initial time is \(t = 1\) for domains of positive time. Let’s ask Sage about accumulation for rate function \(r(t) = t^{-2}\) on a domain starting at \(t = 1\) and ending at a variable time. Note the extra assume command in this code block, where we tell Sage a little extra information about our variable domain end time so that Sage knows we are avoiding the singularity at \(t = 0\)
If we rewrite the Sage result as
\begin{equation*} \frac{t^{-1}}{-1} + 1 \text{,} \end{equation*}
then it essentially fits the pattern of (✶) for \(m = -2\text{,}\) and we can think of the extra plus-\(1\) as an “initial value correction term” from the fact that we have shifted our “initial” time.
The case of \(m = -1\) where the power integral becomes
\begin{equation*} \ccmint{1}{t}{\frac{1}{u}}{u} \end{equation*}
is a special one that we will explore in the next chapter.

Remark 7.4.3.

  • Why two formulas?
    It’s usually convenient to base our integral at \(t = 0\text{,}\) as that is the typical initial time for an observation period. However, for a negative exponent this is not possible because in that case the rate function is undefined at that initial input value.
    We can use Property 5 of Pattern 6.3.11 to verify that the second formula agrees with the first in the case of a positive exponent. Assuming that the first formula is correct, we have
    \begin{gather*} \ccmint{0}{t}{u^m}{u} = \ccmint{0}{1}{u^m}{u} + \ccmint{1}{t}{u^m}{u} \\ \frac{t^{m + 1}}{m + 1} = \frac{1^{m + 1}}{m + 1} + \ccmint{1}{t}{u^m}{u} \\ \frac{t^{m + 1}}{m + 1} = \frac{1}{m + 1} + \ccmint{1}{t}{u^m}{u} \\ \frac{t^{m + 1}}{m + 1} - \frac{1}{m + 1} = \ccmint{1}{t}{u^m}{u} \end{gather*}
    which agrees with the second formula.
  • Negative domains.
    The results of Pattern 7.4.2 hold over negative domains as well. For positive exponents, both formulas are valid as-is for \(t \lt 0\text{,}\) as long as the rate function \(r(t) = t^m\) is defined there. For example, the formulas would not be valid on \(t \lt 0\) for
    \begin{equation*} r(t) = t^{1/2} = \sqrt{t} \text{,} \end{equation*}
    since that rate function is only defined for \(t \ge 0\text{.}\) For negative exponents, a very similar formula to the one based at \(t = 1\) can be obtained for \(t \lt 0\) using a lower bound of \(t = -1\) instead. (Again, this can only be done for those exponents where a negative domain makes sense for the rate function.)
    Also note that for a power of \(t\) whose domain extends into the negatives, its graph will always exhibit either even or odd symmetry, so accumulation over a negative domain can always be determined from the accumulation over the “mirror image” positive domain, and then taking into account the symmetry of those two accumulation areas under the rate graph.
Finally, let’s record the pattern for polynomials that we used in Example 7.4.1.

Example 7.4.5. Using Pattern 7.4.4.

Suppose we want to compute
\begin{equation*} \ccmint{2}{5}{ 4 - 5 t^2 + 9 t^3 }{t} \text{.} \end{equation*}
Pattern 7.4.4 tells us how to compute a polynomial integral based at \(t = 0\text{,}\) but our integral is based at \(t = 2\text{.}\) However, we may use Pattern 6.3.14 to re-base at \(t = 0\text{,}\) and then apply Pattern 7.4.4:
\begin{align*} \ccmint{2}{5}{ 4 - 5 t^2 + 9 t^3 }{t} \amp = \ccmint{0}{5}{ 4 - 5 t^2 + 9 t^3 }{t} - \ccmint{0}{2}{ 4 - 5 t^2 + 9 t^3 }{t}\\ \amp = \left( 4 \cdot 5 - \frac{5}{3} \cdot 5^3 + \frac{9}{4} \cdot 5^4 \right) - \left( 4 \cdot 2 - \frac{5}{3} \cdot 2^3 + \frac{9}{4} \cdot 2^4 \right)\\ \amp = \frac{4749}{4} \text{.} \end{align*}

Example 7.4.6. Using Pattern 7.4.2.

Suppose we want to compute
\begin{equation*} \ccmint{2}{5}{ 6 t^2 + \frac{7}{t^4} }{t} \text{.} \end{equation*}
This time, since the rate function involves a negative exponent, we’ll use Pattern 6.3.14 to re-base the integral at \(t = 1\text{,}\) and then we’ll be able to apply Formula 2 from Pattern 7.4.2 to both powers of \(t\) in the rate function.
\begin{align*} \amp \ccmint{2}{5}{ 6 t^2 + \frac{7}{t^4} }{t} \\ \amp = \ccmint{1}{5}{ 6 t^2 + \frac{7}{t^4} }{t} - \ccmint{1}{2}{ 6 t^2 + \frac{7}{t^4} }{t}\\ \amp = \left(6 \ccmint{1}{5}{ t^2 }{t} + 7 \ccmint{1}{5}{ t^{-4} }{t} \right) - \left(6 \ccmint{1}{2}{ t^2 }{t} + 7 \ccmint{1}{2}{ t^{-4} }{t} \right)\\ \amp = \left( 6 \cdot \frac{5^3 - 1}{3} + 7 \cdot \frac{5^{-3} - 1}{-3} \right) - \left( 6 \cdot \frac{2^3 - 1}{3} + 7 \cdot \frac{2^{-3} - 1}{-3} \right)\\ \amp = \frac{234273}{1000} \text{.} \end{align*}
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