Skip to main content

Chapter 10 Polynomials

Section 10.1 Power and polynomial functions

Subsection 10.1.1 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 10.1.1. 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 10.1.2 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 10.1.2. 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 2.4, 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 10.1.3. 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 10.1.5.
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 10.1.5. Typical graphs of cubic functions.
The axes have been intentionally omitted from the graphs in Figure 10.1.5, 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 10.1.3 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 10.1.4, 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 10.1.7 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 10.1.8. 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 10.2 Limits and continuity of polynomial functions

Repeated application of the Limit laws (Pattern 3.3.1) allows us to handle limits of power functions, polynomial functions, and most rational functions of polynomials.
The patterns above can be summarized simply as: a limit of a power function, a polynomial function, or a rational function of polynomials can be evaluated by evaluating the function (with the caveat that one cannot divide by zero). In other words, these types of functions are continuous (Definition 3.5.1).

Section 10.3 Derivatives of polynomial quantity functions

The simplest type of polynomial is a constant function, which can be considered a degree-0 polynomial. We have already established that the derivative of a constant is uniformly \(0 \) in Pattern 6.3.1.
A degree-\(1 \) polynomial is a linear function, and we have already established that the derivative of a linear function is its slope in Pattern 6.3.11.
For higher-degree polynomials, we should first establish the derivatives of power functions. We’ll start with some examples of low degree.

Example 10.3.1. Derivative of the square function.

Consider \(q(t) = t^2 \text{.}\) From Definition 6.1.1, we have
\begin{equation*} \dqdt = \lim_{\change{t} \to 0} \slope{q}{t} \text{.} \end{equation*}
Let’s work in steps to simplify the difference quotient in the limit.
\begin{align*} q(t + \change{t}) \amp = (t + \change{t})^2 \\ \amp = t^2 + 2 t \change{t} + {(\change{t})}^2 \end{align*}
\begin{align*} \implies \qquad \change{q} \amp = q(t + \change{t}) - q(t) \\ \amp = \bbrac{\bcancel{t^2} + 2 t \change{t} + {(\change{t})}^2} - \bcancel{t^2} \\ \amp = 2 t \change{t} + {(\change{t})}^2 \\ \amp = (2 t + \change{t}) \change{t} \end{align*}
\begin{align*} \implies \qquad \slope{q}{t} \amp = \frac{ q(t + \change{t}) - q(t) }{ \change{t} } \\ \amp = \frac{ (2 t + \change{t}) \bcancel{\change{t}} }{ \bcancel{\change{t}} } \\ \amp = 2 t + \change{t} \end{align*}
Note that when we take the limit as \(\change{t} \to 0 \text{,}\) we assume \(\change{t} \neq 0 \text{,}\) so the cancellation of \(\change{t} \) through division above is valid.
Now consider the limit of the expression that resulted from simplifying the original difference quotient. Clearly
\begin{equation*} \slope{q}{t} \to 2 t \qquad \text{as} \qquad \change{t} \to 0 \text{,} \end{equation*}
and so
\begin{equation*} \ddt{q} = 2 t \text{.} \end{equation*}

Example 10.3.2. Derivative of the cube function.

For \(q(t) = t^3 \text{,}\) we will again work to simplify the difference quotient \(\inlineslope{q}{t} \text{,}\) just as we did in Example 10.3.1. This time, we will have to deal with
\begin{equation*} q(t + \change{t}) = (t + \change{t})^3 \text{.} \end{equation*}
This sum-inside-a-power formula is an example of a binomial expression, and there is a formula for expanding it:
\begin{equation*} (t + \change{t})^3 = t^3 + 3 t^2 \change{t} + 3 t {(\change{t})}^2 + {(\change{t})}^3 \text{.} \end{equation*}
So we have
\begin{align*} \change{q} \amp = q(t + \change{t}) - q(t) \\ \amp = \bbrac{ \bcancel{t^3} + 3 t^2 \change{t} + 3 t {(\change{t})}^2 + {(\change{t})}^3 } - \bcancel{t^3} \\ \amp = 3 t^2 \change{t} + 3 t {(\change{t})}^2 + {(\change{t})}^3 \\ \amp = \bbrac{ 3 t^2 + 3 t \change{t} + {(\change{t})}^2 } \change{t} \text{,} \end{align*}
leading to
\begin{align*} \slope{q}{t} \amp = \frac{ q(t + \change{t}) - q(t) }{ \change{t} } \\ \amp = \frac{ \bbrac{ 3 t^2 + 3 t \change{t} + {(\change{t})}^2 } \bcancel{\change{t}} }{ \bcancel{\change{t}} } \\ \amp = 3 t^2 + 3 t \change{t} + {(\change{t})}^2 \text{.} \end{align*}
Clearly
\begin{equation*} \slope{q}{t} \to 3 t^2 \qquad \text{as} \qquad \change{t} \to 0 \text{,} \end{equation*}
and therefore
\begin{equation*} \ddt{q} = 3 t^2 \text{.} \end{equation*}
We have the beginnings of a pattern here:
\begin{align*} \ddt{t^2} \amp = 2 t^1 \amp \ddt{t^3} \amp = 3 t^2 \text{.} \end{align*}
So far the pattern appears to be: to obtain the derivative function for a power of \(t \text{,}\) multiply by the exponent and then decrease the exponent. And this pattern does in fact hold for higher-degree power functions.

Justification for positive, integer \(n \).

We follow the template of Example 10.3.2. For quantity function
\begin{equation*} q(t) = t^n \text{,} \end{equation*}
we have
\begin{align*} q(t + \change{t}) \amp = (t + \change{t})^n \\ \amp = t^n + b_1 t^{n - 1} \change{t} + b_2 t^{n - 2} {(\change{t})}^2 + \dotsb + b_{n - 1} t {(\change{t})}^{n - 1} + {(\change{t})}^n \text{,} \end{align*}
where \(b_1,b_2,\dotsc,b_{n-1} \) are the so-called binomial coefficients for the expansion of a binomial expression of the form \((x+y)^n \text{.}\) To compute \(\change{q} \text{,}\) we need to subtract \(q(t) = t^n \) from \(q(t + \change{t}) \text{.}\) But there is a \(t^n \) term at the beginning of our expanded expression for \(q(t + \change{t}) \) above that will cancel in this subtraction, leaving
\begin{align*} \change{q} \amp = b_1 t^{n - 1} \change{t} + b_2 t^{n - 2} {(\change{t})}^2 + \dotsb + b_{n - 1} t {(\change{t})}^{n - 1} + {(\change{t})}^n \\ \amp = \bbrac{ b_1 t^{n - 1} + b_2 t^{n - 2} \change{t} + \dotsb + b_{n - 1} t {(\change{t})}^{n - 2} + {(\change{t})}^{n - 1} } \change{t} \text{.} \end{align*}
Therefore,
\begin{align*} \slope{q}{t} \amp = \frac{ q(t + \change{t}) - q(t) }{ \change{t} } \\ \amp = \frac{ \bbrac{ b_1 t^{n - 1} + b_2 t^{n - 2} \change{t} + \dotsb + b_{n - 1} t {(\change{t})}^{n - 2} + {(\change{t})}^{n - 1} } \bcancel{\change{t}} }{ \bcancel{\change{t}} } \\ \amp = b_1 t^{n - 1} + b_2 t^{n - 2} \change{t} + \dotsb + b_{n - 1} t {(\change{t})}^{n - 2} + {(\change{t})}^{n - 1} \text{.} \end{align*}
All terms above that contain a factor of \(\change{t} \) tend to \(0 \) as \(\change{t} \to 0 \text{,}\) so we have
\begin{equation*} \slope{q}{t} \to b_1 t^{n - 1} \qquad \text{as} \qquad \change{t} \to 0 \text{,} \end{equation*}
and therefore
\begin{equation*} \ddt{q} = b_1 t^{n - 1} \text{.} \end{equation*}
Finally, it turns out that the first binomial coefficient \(b_1 \) for the expansion of a binomial expression of the form \((x+y)^n \) is \(b_1 = n \) (refer to the Binomial Theorem), so that
\begin{equation*} \ddt{(t^n)} = n t^{n - 1} \text{,} \end{equation*}
as claimed.

Remark 10.3.4. Non-positive and/or non-integer exponents.

While our justification is only valid for positive, integer exponents \(n \text{,}\) the Power Rule (Pattern 10.3.3) itself is valid for all exponents, and we will justify this in later chapters after we have a few more techniques and patterns available to us. In the meantime, recall that for negative exponents we cannot attach a derivative value to the quantity function at the singularity at \(t = 0 \) (see Subsection 5.7.3), which is the reason for including the phrase “everywhere that it is defined” in the statement of the pattern. We could also consider the pattern to be true for \(n = 0 \text{,}\) since in that case the quantity function satisfies
\begin{align*} q(t) \amp = t^0 = 1 \amp \text{for } t \amp \gt 0 \text{,} \end{align*}
while the Power Rule gives
\begin{equation*} \ddt{(t^0)} = 0 t^{-1} = 0 \end{equation*}
(again assuming \(t \gt 0 \)). So in some sense the Power Rule “contains” our rule for the derivative of a constant (Pattern 6.3.1). And, for that matter, it also “contains” our rule for the derivative of the independent variable (Pattern 6.3.2) when we take \(n = 1 \text{.}\)
Here are a couple examples of using the Power Rule.

Example 10.3.5. Derivative of a power with a negative exponent.

\begin{align*} q(t) \amp = t^{-3} \amp \amp\implies \amp \ddt{q} \amp = -3 t^{-4} \end{align*}

Note 10.3.6.

While we might describe part of the Power Rule as decrease the exponent, this does not mean decrease the magnitude of the exponent. For example, a common mistake in applying the Power Rule is to use \(t^{-2} \) in the derivative formula for \(t^{-3} \text{,}\) instead of \(t^{-4} \) as in Example 10.3.5.

Example 10.3.7. Derivative of a power with an irrational exponent.

\begin{align*} q(t) \amp = t^\pi \amp \amp\implies \amp \ddt{q} \amp = \pi t^{\pi - 1} \end{align*}
Finally, we can combine Power Rule with the Constant Multiple Rule and the Sum Rule to obtain a general formula for the derivative function of any polynomial. We will leave the justification up to you, the reader.

Example 10.3.9. Derivative of a polynomial.

\begin{align*} q(t) \amp = 3 t^5 - 4 t^3 + t^2 + 7 t + 6 \\ \implies \quad q'(t) \amp = 15 t^4 - 12 t^2 + 2 t + 7 \end{align*}

Example 10.3.10. Parabolic motion.

Let’s return again to situation of Example 6.2.2, where we considered parabolic motion due to gravity:
\begin{equation*} h(t) = -5 t^2 + 30 t \text{.} \end{equation*}
We can now use Pattern 10.3.8 to obtain a general formula for the velocity of this object at arbitrary time \(t \text{:}\)
\begin{align*} v(t) = h'(t) \amp = -5 \cdot 2 t + 30 \\ \amp = -10 t + 30 \text{.} \end{align*}
Keep in mind that with height measured as a distance above the surface of the fictional planet, a positive velocity means upward motion and a negative velocity means downward motion.
There are some natural questions about the object’s motion that arise.
  • With what velocity was the object launched from the surface?
    We can answer this by simply computing the initial velocity
    \begin{equation*} v(0) = -10 \cdot 0 + 30 = 30 \text{,} \end{equation*}
    which agrees with the initial velocity we used to construct the height model in the first place in Example 5.1.2.
  • When does the object return to the surface?
    To answer this, we want to solve for \(t \) in the instances where \(h(t) = 0 \text{.}\) We can do this by factoring.
    \begin{gather*} h(t) = 0 \\ -5 t^2 + 30 t = 0 \\ -5 t (t - 6) = 0 \text{.} \end{gather*}
    From the factored form on the left, we can see that the object will be at the position of the surface when \(t = 0 \) (as we assumed, since the object was initially launched from the surface) and when
    \begin{gather*} t - 6 = 0 \\ t = 6 \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(6) = -10 \cdot 6 + 30 = - 30 \text{.} \end{equation*}
    Considering the symmetry of the parabolic position function (see Figure 10.3.11 below) 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 10.3.11 below), we expect that the object will reach its maximum height halfway between its departure from and return to the surface, at \(t = 3 \text{.}\)
  • 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{gather*} \text{max height} = h(3) = -5 \cdot 3^2 + 30 \cdot 3 = 45 \\ v(3) = -10 \cdot 3 + 30 = 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 10.3.11. A position and velocity graph for parabolic motion due to gravity.

Section 10.4 Integrals of polynomial rate functions

Subsection 10.4.1 Constant rate function

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 10.4.1. Constant acceleration.

Variation in the velocity of an objection in motion is called acceleration. If \(v(t) \) is a function that represents the velocity of the object at time \(t \text{,}\) then the associated acceleration function \(a(t) \) is a rate function for \(v \text{.}\)
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, and then the constant acceleration function is \(a(t) = g \text{.}\) 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 9.4.7 (just as in Example 9.4.10) 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{align*} v(t) \amp = v(0) + \ccmint{0}{t}{a(t)}{u} \\ \amp = 0 + \ccmint{0}{t}{g}{u} \\ \amp = g t \text{.} \end{align*}

Subsection 10.4.2 Linear rate functions

Next we consider a linear rate of accumulation.

Example 10.4.4. Linear velocity.

In Example 10.4.1, 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 9.3.7.
Graph of a velocity function with bounded oriented area as accumulation of displacement.
Figure 10.4.5. Accumulation of displacement as net oriented area under a velocity graph.
The triangle in Figure 10.4.5 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 10.4.4 with Pattern 10.4.2 to deal with all linear rate functions, since Property 3 of Pattern 9.3.11 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 10.4.7. Parabolic motion.

In Example 5.1.2 and Example 10.3.10, we justified the height-versus-time formula we used for an object thrown upwards against gravity simply by appealing to knowledge “from physics.” We are know in a position to justify this formula ourselves.
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 10.4.2 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 10.4.2, 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.
If we apply this formula to the situation of Example 5.1.2 and Example 10.3.10, where we had \(g = 10 \) and \(v_0 = 30 \text{,}\) we obtain
\begin{equation*} s(t) = -5 t^2 + 30 t \text{,} \end{equation*}
which agrees with the \(h(t) \) function used in those examples.
Now let’s answer the same questions as in Example 10.3.10 in a more general setting.
  • 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 10.3.11) 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 10.3.11), 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 10.4.8. A position and velocity graph for parabolic motion due to gravity.

Subsection 10.4.3 Polynomial rate functions

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 9.3.11, 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{equation} \ccmint{0}{t}{u^m}{u} = \frac{t^{m + 1}}{m + 1} \text{.}\tag{✶} \end{equation}

Example 10.4.9. 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 \(m \) except \(m = -1 \) (see below). For powers of \(t \) with a positive exponent, the formula in (✶) can be applied directly. However, 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 10.4.11.

  • 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 9.3.11 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 10.4.10 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 10.4.9.

Example 10.4.13. Using Pattern 10.4.12.

Suppose we want to compute
\begin{equation*} \ccmint{2}{5}{ 4 - 5 t^2 + 9 t^3 }{t} \text{.} \end{equation*}
Pattern 10.4.12 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 9.3.14 to re-base at \(t = 0 \text{,}\) and then apply Pattern 10.4.12:
\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 10.4.14. Using Pattern 10.4.10.

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 9.3.14 to re-base the integral at \(t = 1 \text{,}\) and then we’ll be able to apply Formula 2 from Pattern 10.4.10 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*}

Section 10.5 Differentiation versus integration for power functions

Let’s compare our differentiation and integration formulas for power functions side-by-side (for positive exponents, at least).
\begin{align*} \ddt{(t^m)} \amp = m t^{m - 1} \amp \ccmint{0}{t}{u^m}{u} \amp = \frac{t^{m + 1}}{m + 1} \end{align*}
We might describe these formulas in words as follows.
derivative of a power function
multiply by the (old) exponent, then reduce the exponent by one
integral of a power function
increase the exponent by one, then divide by the new exponent
These certainly seem to be reverse of each other; let’s see what happens if we integrate a derivative (on the left below) or if we differentiate an integral (on the right below), in both cases starting with function \(t^m \text{.}\)
\begin{align*} \amp \ccmint{0}{t}{ \dd{u^m}{u} }{u} \amp \amp \qquad \amp \amp \opddt \, \ccmint{0}{t}{u^m}{u}\\ \amp = \ccmint{0}{t}{ m u^{m - 1} }{u} \amp \amp \amp \amp = \opddt \left( \frac{t^{m + 1}}{m + 1} \right)\\ \amp = m \ccmint{0}{t}{ u^{m - 1} }{u} \amp \amp \amp \amp = \frac{1}{m + 1} \cdot \ddt{t^{m + 1}}\\ \amp = m \frac{t^{(m - 1) + 1}}{(m - 1) + 1} \amp \amp \amp \amp = \frac{1}{\bcancel{m + 1}} \cdot \bcancel{(m + 1)} t^{(m + 1) - 1}\\ \amp = \bcancel{m} \frac{t^m}{\bcancel{m}} \amp \amp \amp \amp = t^m\\ \amp = t^m \end{align*}
Both cases returned us to the starting point of \(t^m \text{.}\) Is this a coincidence? We will return to this question after studying some other fundamental types of functions.

Checkpoint 10.5.1. Differentiation versus integration for power functions with negative exponent.

(a)

Compute both of the following using Formula 2 from Pattern 10.4.10.
\begin{equation*} \ccmint{1}{t}{ \dd{u^m}{u} }{u} \qquad\qquad \opddt \, \ccmint{1}{t}{u^m}{u} \end{equation*}
PrevTopNext