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

Section 7.1 Power and polynomial functions

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Subsection 7.1.1 Definitions

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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

$$p(t)=a_0+a_{1}t+a_2{t}^2+a_3{t}^3+\cdots +a_{n-1}{t}^{n-1}+a_n{t}^n$$

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where we assume $a_n\ne 0$ (since otherwise we would have omitted that term and terminated the polynomial at a lower power of $t$). The constants $a_0,a_1,\dots ,a_n$ are called the coefficients of the polynomial, and the highest exponent $n$ is called the degree.

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A polynomial that consists of just a single term,

$$p(t)=a_n{t}^n\text{,}$$

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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.

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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 $n^{\mathrm{th}}$ root of a number is tied to the roots of polynomials, as a square root $\sqrt{a}$ means the (positive) root of the equation

$$t^2-a=0\text{,}$$

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a cube root $\sqrt[3]{a}$ means the unique real root of the equation

$$t^3-a=0\text{,}$$

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and so on.

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Subsection 7.1.2 Polynomials of small degree

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Since we often factor polynomials into factors that are polynomials of smaller degree, polynomials of small degree are of special interest.

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Subsubsection Constant polynomials

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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.

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Subsubsection Linear polynomials

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A polynomial of degree $1\text{,}$

$$p(t)=a_0+a_{1}t\text{,}$$

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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:

$$\begin{array}{rlrlrl}{a}_0+a_{1}t& =0& & ⟹& t& =-\frac{a_0}{a_1}\text{.}\end{array}$$

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Subsubsection Quadratic polynomials

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A polynomial of degree $2\text{,}$

$$p(t)=a_0+a_{1}t+a_2{t}^2\text{,}$$

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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 $a_2>0$ or $a_2<0\text{,}$ 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.

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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.

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Example 7.1.2. Completing the square.

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Consider polynomial

$$p(t)=6+10t-2t^2\text{.}$$

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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.

$$\begin{array}{rl}p(t)& =-2t^2+10t-6\\ & =-2(t^2-5t+3)\\ & =-2(t^2-5t+{\left(\frac{5}{2}\right)}^2-{\left(\frac{5}{2}\right)}^2+3)\\ & =-2[{(t-\frac{5}{2})}^2-\frac{25}{4}+3]\\ & =-2[{(t-\frac{5}{2})}^2-\frac{13}{4}]\\ & =-2{(t-\frac{5}{2})}^2+\frac{13}{2}\end{array}$$

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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.

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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.

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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

$$a{t}^2+bt+c\text{,}$$

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then whether this polynomial has none, one, or two roots is determined by the discriminant

$$b^2-4ac\text{.}$$

no roots

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when $b^2-4ac<0\text{;}$ a concave up parabola with vertex above the horizontal axis or a concave down parabola with vertex below the horizontal axis

one unique root

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when $b^2-4ac=0\text{;}$ a parabola with vertex on the horizontal axis

two distinct roots

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when $b^2-4ac>0\text{;}$ a concave up parabola with vertex below the horizontal axis or a concave down parabola with vertex above the horizontal axis

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In any case, there exists a formula for the roots of quadratic polynomial.

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Pattern 7.1.4. Quadratic formula.

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If the discriminant satisfies $b^2-4ac\ge 0\text{,}$ then the roots of

$$a{t}^2+bt+c=0$$

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are

$$\begin{array}{rlrl}t& =\frac{-b-\sqrt{b^2-4ac}}{2a}& t& =\frac{-b+\sqrt{b^2-4ac}}{2a}\text{,}\end{array}$$

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where these two roots are one and the same in the case that $b^2-4ac=0\text{.}$

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Subsubsection Cubic polynomials

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The graph of a cubic polynomial often follows an “up-down-up” or “down-up-down” pattern, as in Figure 7.1.5.

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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.

Section 7.2 Constant rate of variation

Section 7.3 Linear rate of variation

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Example 7.3.1. Linear velocity.

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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.

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To simplify things, let us label the starting position as “position 0” so that displacement is the same as position. Our rate function is $v(t)=gt$ 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.

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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

$${\int }_0^{t}v(u)d_{}u=\frac{v(t)\cdot t}{2}=\frac{g{t}^2}{2}\text{.}$$

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Let’s ask Sage to confirm this for us.

    

        
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var('g,t,u')
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v(u) = g * u
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v.integrate(u, 0, t)

    
    
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So it seems that a linear rate function leads to parabolic growth/decay.

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Example 7.3.4. Parabolic motion.

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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:

$$a(t)=-g\text{,}$$

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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

$$v(t)=-gt+v_0\text{.}$$

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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

$$s(t)=-\frac{g}{2}{t}^2+v_{0}t$$

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if we take the initial position to be $s(0)=0$ at the surface of the planet.

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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.

  $$s(t)=-\frac{g}{2}{t}^2+v_{0}t=\frac{t}{2}(-gt+2v_0)$$

  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

  $$\begin{array}{c}-gt+2v_0=0\\ ⟹t=\frac{2v_0}{g}\text{.}\end{array}$$
• With what velocity does it strike the surface?
  We just need to substitute the time of return into the velocity model:

  $$v(2v_0/g)=-g(2v_0/g)+v_0=-v_0\text{.}$$

  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

  $$t=\frac{v_0}{g}\text{.}$$
• 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

  $$\begin{array}{c}\text{max height}=s(v_0/g)=\frac{v_0^2}{2g}\\ v(v_0/g)=0\text{.}\end{array}$$

  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.

Section 7.4 Polynomial rate of variation

    

        
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var('t')
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r(x) = x^2
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r.integrate(x, 0, t)

    
    
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1
var('t')
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assume(t > 1)
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r(x) = x^(-2)
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r.integrate(x, 1, t)

    
    
    Evaluate (Sage)
    

        

            
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