3.6: The second derivative and higher order derivatives.

You may read or hear statements such as “the rate of population growth is decreasing”, or “the rate of inflation is increasing”, or the velocity of the particle is increasing.” In each case the underlying quantity is a rate and its rate of change is important.

Definition 3.6.1 The second derivative

The second derivative of a function, \(P\), is the derivative of the derivative of \(P\), or the derivative of \(P ^{\prime}\).

The second derivative of P may be denoted by

\[P^{\prime \prime}, \quad P^{\prime \prime}(t), \quad \frac{d^{2} P}{d t^{2}}, \quad \ddot{P}, \quad \ddot{P}(t), \quad P^{(2)}, \quad P^{(2)}(t), \quad \text { or } \quad D_{t}^{2} P(t)\]

Geometry of the first and second derivatives. That a function, \(P\) is increasing on an interval \([a, b]\) means that

\[\text { if } s \text { and } t \text { are in }[a, b] \text { and } \quad s<t \quad \text { then } \quad P(s)<P(t)\]

It should be fairly intuitive that if the first derivative of a function, \(P\), is positive throughout [\(a, b\)], then \(P\) is increasing on [\(a, b\)]. Both graphs in Figure 3.6.1 have positive first derivatives and are increasing. The graphs also illustrate the geometry of the second derivative. In Figure 3.6.1A \(P ^{\prime}\) is increasing (\(P^{\prime}(s)<P^{\prime}(t)\)). \(P\) has a positive second derivative (\(P^{\prime \prime}>0\)) and the graph of \(P\) is concave upward. In Figure 3.6.1B, \(P ^{\prime}\) is decreasing (\(P^{\prime}(s)>P^{\prime}(t)\)). \(P\) has a negative second derivative (\(P^{\prime \prime}<0\)) and the graph of \(P\) is concave downward. We will expand on this interpretation in Chapters 8 and 12.

Figure \(\PageIndex{1}\): A. Graph of a function with a positive second derivative; it is concave up. B. Graph of a function with a negative second derivative; it is concave down.

The higher order derivatives are a natural extension of the transition from the derivative to the second derivative. The third derivative is the derivative of the second derivative; the fourth derivative is the derivative of the third derivative; and the process continues. In this language, the derivative of \(P\) is called the first derivative of \(P\) (and sometimes \(P\) itself is called the zero-order derivative of \(P\)).

Definition 3.6.2 Inductive definition of higher order derivatives

The derivative of a function, \(P\), is the order 1 derivative of \(P\). For \(n\) an integer greater than 1, the order \(n\) derivative of \(P\) is the derivative of the order \(n − 1\) derivative of \(P\).

The third order derivative of P may be denoted by

\[P^{\prime \prime \prime}, \quad P^{\prime \prime \prime}(t), \quad \frac{d^{3} P}{d t^{3}}, \quad P^{(3)}, \quad P^{(3)}(t), \quad \text { or } \quad D_{t}^{3} P(t)\]

For \(n > 3\) the n^th order derivative of \(P\) may be denoted by

\[P^{(n)}, \quad P^{(n)}(t), \quad \frac{d^{n} P}{d t^{n}}, \quad \text { or } \quad D_{t}^{n} P(t)\]

If \(P(t) = \mu t + \beta\) is a linear function, then \(P ^{\prime} (t) = \mu\), is a constant function, and \(P^{\prime \prime}(t)=[\mu]^{\prime}=0\) is the zero function. A similar pattern occurs with quadratic polynomials. Suppose \(P(t) = a t^{2} + b t + c\) is a quadratic polynomial. Then

\[\begin{array}{rlr}
P^{\prime}(t) & =\left[a t^{2}+b t+c\right]^{\prime} & \\
& =2 a t+b & P^{\prime}(t) \text { is a linear function. } \\
P^{\prime \prime}(t) & =[2 a t+b]^{\prime} & \\
& =2 a & P^{\prime \prime}(t) \text { is a constant function. } \\
P^{\prime \prime \prime}(t) & =[2 a]^{\prime} & \\
& =0 & P^{\prime \prime \prime}(t) \text { is the zero function. }
\end{array} \label{3.35}\]

Explore 3.6.1 Suppose \(P(t) = a + b t + c t^{2} + d t^{3}\) is a cubic polynomial. Show that \(P^{\prime}\) is a quadratic polynomial, \(P^{\prime \prime}\) is a linear function, \(P^{\prime \prime \prime}\) is a constant function, and \(P ^{(4)}\) is the zero function.

If s(t) is the position of a particle along an axis at time \(t\), then \(s ^{\prime} (t)\) is the velocity of the particle and the rate of change of velocity, \(s ^{\prime \prime} (t)\), is called the acceleration of the particle. Sometimes when \(s ^{\prime \prime} (t)\) is negative the word deceleration is used to describe the motion of the particle. The word acceleration is used to describe second derivatives in other contexts. An accelerating economy is one in which the rate of increase of the gross national product is increasing.

Example 3.6.1 Shown in Figure 3.6.2A is the graph of the logistic function, \(L(t)\), first shown in Figure 3.3.1. Some tangents are drawn on the graph of \(L\). The slope at \(b\) is greater than the slope at \(a\); the slope, \(L ^{\prime} (t)\), is increasing on the interval to the left of the point marked, \(I\) = Inflection Point.

Figure \(\PageIndex{2}\): A. The graph of a logistic equation and an inflection point I. B. The same graph with tangent segments. The slope of the segment at a is less than the slope at b. The slope at c is greater than the slope at d.

Explore 3.6.2 Do this.

1. The slopes of \(L\) at \(b\) and \(d\) are approximately 1.5 and 0.75. Estimate the slopes at \(a\) and \(c\). Confirm that the slope at a is less than the slope at \(b\) and that the slope at \(c\) is greater than the slope at \(d\).
2. Let \(t_I\) be the time of the inflection point \(I\). Argue that \(L ^{\prime \prime} (t) > 0\) on \([0, t_{I} ]\).
3. Argue that \(L ^{\prime \prime} (t) < 0\) for \(t > t_I\).
4. Argue that \(L ^{\prime \prime} (t_{I}) = 0\).
5. On what interval is the graph of \(L\) concave downward?

The tangent at the inflection point is shown in Figure 3.6.2B, and it is interesting that the tangent crosses the curve at \(I\). The slope of that tangent \(\doteq\) 1.733, and is the largest of the slopes of all of the tangents. The maximum population growth rate occurs at \(t_I\) and is approximately 1733 individuals per year.

Explore 3.6.3 Danger: Obnubilation Zone. You have to think about this. Suppose the population represented by the logistic curve in Figure 3.6.2 is a natural population such as deer, fish, geese or shrimp, and suppose you are responsible for setting the size of harvest each year. What is your strategy?

Argue with a friend about this. You should observe that the population size at the inflection point, \(I\), is 5 which is one-half the maximum supportable population of 10. You should discuss the fact that variable weather, disease and other factors may disrupt the ideal of logistic growth. You should discuss how much harvest you could have if you maintained the population at points \(a, b, c\) or \(d\) in Figure 3.6.2A.

 

3.6.1 Falling objects.

We describe the position, \(s(t)\), of an object falling in the earth’s gravity field near the Earth’s surface \(t\) seconds after release. We assume that the velocity of the object at time of release is \(v_0\) and the height of the object above the earth (or some reference point) at time of release is \(s_0\).

Gravity near the Earth’s surface is constant, equal to \(g = -980 cm/sec^2\). We write the

Mathematical Model 3.6.1 Free falling object. The acceleration of a free falling object near the Earth’s surface is the acceleration of gravity, \(g\).

Because \(s ^{\prime prime} (t)\) is the acceleration of the falling object, we write

\[s^{\prime \prime}(t)=g\]

Now, \(s ^{\prime \prime}\) is a constant and is the derivative of \(s ^{\prime}\). The derivative of a linear function, \(P(t) = at + b\), is a constant (\(P ^{\prime} (t) = a\)). We invoke some advertising logic⁹ and guess that \(s ^{\prime} (t)\) is a linear function (proved in Chapter 10).

\[s ^{\prime} (t) = g t + b \quad \textbf{ Bolt out of Chapter 10. }\]

Because \(s ^{\prime} (0) = v_0\) is assumed known, and \(s ^{\prime} (0) = g 0 + b = b\), we write

\[s ^{\prime} (t) = g t + s_0\]

Using equally compelling logic, because the derivative of a quadratic function, \(P(t) = a t^{2} + b t + c\), is a linear function (\(P ^{\prime} (t) = 2a t + b\)), and because \(s ^{\prime}\) is a linear function and \(s(0) = s_0\), we write

\[s(t) = \frac{g}{2} t^{2} + v_{0} t + s_{0} \label{3.36}\]

Example 3.6.2 Students measured height vs time of a falling bean bag using a Texas Instruments Calculator Based Laboratory Motion Detector, and the data are shown in Figure 3.6.3A. Average velocities were computed between data points and plotted against the midpoints of the data intervals in Figure 3.6.3B. Mid-time is \((\text{Time}_{i+1} + \text{Time}_{i})/2\) and Ave. Vel. is \((\text{Height}_{i+1} - \text{Height}_{i})/(\text{Time}_{i+1} - \text{Time}_{i})\).

Figure \(\PageIndex{3}\): Graph of height vs time and velocity vs time of a falling bean bag.

An equation of the line fit by least squares to the graph of Average Velocity vs Midpoint of time interval is

\[v_{\text {ave }}=-849 t_{\text {mid }}+126 \quad \mathrm{~cm} / \mathrm{s}\]

If we assume a continuous model based on this data, we have

\[s^{\prime}(t)=-849 t+126, \quad s(t)=\frac{-849}{2} t^{2}+126 t+s_{0} \quad \mathrm{~cm}\]

From Figure 3.6.3, the height of the first point is about 240. We write

\[s(t)=\frac{-849}{2} t^{2}+126 t+240 \quad \mathrm{~cm}\]

The graph of \(s\) along with the original data is shown in Figure 3.6.3C. The match is good. Instead of \(g ~ -980 ~ \mathrm{cm/s}^2\) that applies to objects falling in a vacuum we have acceleration of the bean bag in air to be \(-849 ~ \mathrm{cm/sec}^2\).

 

Exercises for Section 3.6, The second derivative and higher order derivatives.

Exercise 3.6.1 Compute \(P ^{\prime} , P ^{\prime \prime}\) and \(P ^{\prime \prime \prime}\) for the following functions.

1. \(P(t) = 17\)
2. \(P(t) = t\)
3. \(P(t) = t ^2\)
4. \(P(t) = t ^3\)
5. \(P(t) = t ^{1/2}\)
6. \(P(t) = t ^{−1}\)
7. \(P(t) = t ^8\)
8. \(P(t) = t ^{125}\)
9. \(P(t) = t ^{5/2}\)

Exercise 3.6.2 Find the acceleration of a particle at time \(t\) whose position, \(P(t)\), on an axis is described by

1. \(P(t) = 15\)
2. \(P(t) = 5 t + 7\)
3. \(P(t) = −4.9 t^{2} + 22 t + 5\)
4. \(P(t) = t − \frac{t^{3}}{6} + \frac{t^{5}}{120}\)

Exercise 3.6.3 Compute \(P ^{\prime} , P ^{\prime \prime}\) and \(P ^{\prime \prime \prime}\) and \(P ^{(4)}\) for \(P(t) = a + b t + c t^{2} + d t^{3}\).

Exercise 3.6.4 For each figure in Exercise Figure 3.6.4, state whether:

1. \(P\) is increasing or decreasing?
2. \(P ^{\prime}\) is positive or negative?
3. \(P ^{\prime}\) is increasing or decreasing?
4. \(P ^{\prime \prime} (a)\) positive or negative?
5. The graph of \(P\) is concave up or concave down

Figure for Exercise 3.6.4 Graphs for exercise 3.6.4

Exercise 3.6.5 The function, \(P\), graphed in Figure Ex. 3.6.5 has a local minimum at the point \((a, P(a))\).

1. What is \(P ^{\prime} (a)\)?
2. For \(t < a, P ^{\prime} (t)\) is (positive or negative)?
3. For \(a < t, P ^{\prime} (t)\) is (positive or negative)?
4. \(P ^{\prime} (t)\) is (increasing or decreasing)?
5. \(P ^{\prime \prime (a)\) is (positive or negative)?

Figure for Exercise 3.6.5 Graph of a function \(P\) with a minimum at \((a, P(a))\). See Exercise 3.6.5.

Exercise 3.6.6 The function, \(P\), graphed in Figure Ex. 3.6.6 has a local maximum at the point \((a, P(a))\).

1. What is \(P ^{\prime} (a)\)?
2. For \(t < a, P ^{\prime} (t)\) is (positive or negative)?
3. For \(a < t, P ^{\prime} (t)\) is (positive or negative)?
4. \(P ^{\prime} (t)\) is (increasing or decreasing)?
5. \(P ^{\prime \prime (a)\) is (positive or negative)?

Figure for Exercise 3.6.6 Graph of a function \(P\) with a maximum at \((a, P(a))\). See Exercise 3.6.6.

Exercise 3.6.7 Show that \(A(t)\) of Equation 3.4.3,

\[A(t)=\left(\frac{K}{2} t+2\right)^{2} \quad t \geq 0\]

satisfies Equation 3.4.2,

\[A(0)=4 \quad A^{\prime}(t)=K \sqrt{A(t)} \quad t \geq 0\]

You will need to compute \(A ^{\prime} (t)\) and to do so expand

\[A(t)=\left(\frac{K}{2} t+2\right)^{2} \quad \text { to } \quad A(t)=\frac{K^{2}}{4} t^{2}+K t+4\]

and show that

\[A^{\prime}(t)=K\left(\frac{K}{2} t+2\right)=K \sqrt{A(t)}\]

Exercise 3.6.8 Show that for \(s(t) = \frac{g}{2} t^{2} + v_{0}t + \gamma, s ^{\prime} (t) = gt + v_0\).

Exercise 3.6.9 Evaluate \(\gamma\) if

1. \(s(t) = 5t^{2} + \gamma\) and \(s(0) = 15\)
2. \(s(t) = -8t^{2} + 12t + \gamma\) and \(s(0) = 11\)
3. \(s(t) = -8t^{2} + 12t + \gamma\) and \(s(1) = 11\)
4. \(s(t) = - \frac{849}{2} t^{2} + 126t + \gamma\) and \(s(0.232) = 245.9\)

Exercise 3.6.10 Show that

1. \(P(t) = 5t + 3\) satisfies \(P(0) = 3\) and \(P ^{\prime} (t) = 5\)
2. \(P(t) = 8t + 2\) satisfies \(P(0) = 2\) and \(P ^{\prime}(t) = 8\)
3. \(P(t) = t^{2} + 3t + 7\) satisfies \(P(0) = 7\) and \(P ^{\prime}(t) = 2t + 3\)
4. \(P(t) = −2t^{2} + 5t + 8\) satisfies \(P(0) = 8\) and \(P ^{\prime}(t) = -4t + 5\)
5. \(P(t) = (3t + 4)^{2}\) satisfies \(P(0) = 16\) and \(P ^{\prime}(t) = 6 \sqrt{P(t)}\)
6. \(P(t) = (5t + 1)^{2}\) satisfies \(P(0) = 1\) and \(P ^{\prime}(t) = 10 \sqrt{P(t)}\)
7. \(P(t) = (1 − 2t)^{−1}\) satisfies \(P(0) = 1\) and \(P ^{\prime}(t) = 2 ( P(t) )^2\)
8. \(P(t) = \frac{5}{1 − 15t}\) satisfies \(P(0) = 5\) and \(P ^{\prime}(t) = 3 ( P(t) )^2\)
9. \(P(t) = (6t + 9)^{1/2}\) satisfies \(P(0) = 3\) and \(P ^{\prime}(t) = 3/P(t)\)
10. \(P(t) = (4t + 4)^{1/2}\) satisfies \(P(0) = 2\) and \(P ^{\prime}(t) = 2/P(t)\)
11. \(P(t) = (4t + 4)^{3/2}\) satisfies \(P(0) = 8\) and \(P ^{\prime}(t) = 6 \sqrt[3] {P(t)}\)

For parts g - k, use the Definition of Derivative 3.2.2 to compute \(P^{\prime}\).

Exercise 3.6.11 Add the equations,

\[\begin{array}\
H_{2}-H_{1} &=\frac{-849}{2}\left(t_{2}^{2}-t_{1}^{2}\right)+126\left(t_{2}-t_{1}\right) \\
H_{3}-H_{2} &=\frac{-849}{2}\left(t_{3}^{2}-t_{2}^{2}\right)+126\left(t_{3}-t_{2}\right) \\
\vdots & \vdots \\
H_{n}-H_{n-1} &=\frac{-849}{2}\left(t_{n}^{2}-t_{n-1}^{2}\right)+126\left(t_{n}-t_{n-1}\right) .
\end{array}\]

to obtain

\[H_{n}-H_{1}=\frac{-849}{2}\left(t_{n}^{2}-t_{1}^{2}\right)+126\left(t_{n}-t_{1}\right)\]

Exercise 3.6.12 In a chemical reaction of the form

\[2 A + B \rightarrow A_{2}B,\]

where the reaction does not involve intermediate compounds, the reaction rate is proportional to \([A]^{2} [B]\) where \([A]\) and \([B]\) denote, respectively, the concentrations of the components \(A\) and \(B\). Let \(a\) and \(b\) denote \([A]\) and \([B]\), respectively, and assume that \([B]\) is much greater than \([A]\) so that (\(b(t) \gg a(t)\). The rate at which a changes may be written

\[a^{\prime}(t)=-k(a(t))^{2} b(t)=-K(a(t))^{2}\]

We have assumed that \(b(t)\) is (almost) constant because \([A]\) is the limiting concentration of the reaction.

Let

\[a(t)=\frac{a_{0}}{1-a_{0} K t}\]

Show that \(a(0) = a_0\).

Use the Definition of Derivative 3.3.3,

\[a^{\prime}(t)=\lim _{b \rightarrow t} \frac{a(b)-a(t)}{b-t}\]

to compute \(a ^{\prime} (t)\). Then compute \(( a(t) )^2\) and show that

\[a ^{\prime} (t) = −K ( a(t) )^2\]

Exercise 3.6.13 Show that for any quadratic function, \(Q(t) = a + bt + ct^2\) (\(a, b\) and \(c\) are constants), and any interval, \([u, v]\), the average rate of change of \(Q\) on \([u, v]\) is equal to the rate of change of \(Q\) at the midpoint, \((u + v)/2\), of \([u, v]\).

 

⁹ Advertising Logic: A tall, muscular, rugged man drives a Dodge Ram. If you buy a Dodge Ram, you will be a tall muscular, rugged man.