Mathematics 265 Introduction to Calculus I

Study Guide :: Unit 4

Differentiation

Objectives

When you have completed this unit, you should be able to

1. interpret the derivative as a rate of change.
2. interpret the derivative as a slope of a tangent line.
3. differentiate algebraic and trigonometric functions.
4. apply differentials to estimate numbers and relative errors.
5. apply the derivative to solve problems in the natural and social sciences.
6. apply implicit differentiation to solve related rate problems.

In this unit, we address the question of how, in a function $f(x)$, the dependent variable $f$ changes with respect to the independent variable $x$. That is, for a specific value of $x$, we want to know if $f$ is increasing, deceasing or neither, and if $f(x)$ does change, we want to know how quickly or how slowly the change is occurring. The answers to these questions are intrinsically related to the geometric concept of slope.

The Average Rate of Change and the Slope of the Secant Line

If $d$ is the distance traveled by a moving object (in metres) at time $t$ (in seconds), then $d$ depends on $t$—in terms of functions, $d\left(t\right)$. If we want to know how $d$ is increasing or decreasing with respect to an interval of time, we are referring to the “average velocity” of the object. The average velocity is defined as follows:

${\displaystyle \text{average velocity}=\frac{\text{distance traveled}}{\text{time elapsed}}\text{.}}$

If the interval is from the time $t_0$ to the time $t_1$, then the time elapsed is $t_1-t_0$, and the distance traveled is $d\left(t_1\right)-d\left(t_0\right)$; hence,

${\displaystyle \text{average velocity}=\frac{d\left(t_1\right)-d\left(t_0\right)}{t_1-t_0}\text{with units of m/s}\text{.}}$

Example 4.1. If $d\left(2\right)=7$ (the distance of the object at $2$ seconds is $7$ metres), and $d\left(5\right)=12$, then the time elapsed between $5$ and $2$ is $5-2=3$ seconds, and the distance traveled during this period of time is $d\left(5\right)-d\left(2\right)=12-7=5$ metres. The average velocity is

${\displaystyle \text{average velocity}=\frac{d\left(5\right)-d\left(2\right)}{5-3}=\frac{5}{3}\text{m/s}}$

On average, from $2$ to $5$ seconds, the distance is increasing at a rate of $5∕3$ m/s, that is, the average velocity of the object, from $2$ to $5$ seconds, is $5∕3$ m/s.

Example 4.2. If $P$ is the population (in millions) at time $t$ (in minutes), then the average rate of growth in a period of time is given by the population over the time elapsed. Hence, if $P\left(12\right)=120$ and $P\left(26\right)=324$, then

${\displaystyle \text{average rate of growth }=\frac{P\left(26\right)-P\left(12\right)}{26-12}=\frac{324-120}{14}=\frac{102}{7}\text{.}}$

From $12$ to $26$ minutes, the population is growing on average, at a rate of $102/7$ million/min.

Example 4.3. If $C$ is the cost (in thousands of dollars) of extracting $x$ tonnes of gold, then $C$ depends on $x$ and we evaluate $C\left(x\right)$. If $C\left(20\right)=50$ and $C\left(50\right)=96\text{,}$ then the quotient

${\displaystyle \frac{C\left(50\right)-C\left(20\right)}{50-20}=\frac{23}{15}}$

is the average cost of gold extraction from $20$ to $50$ tonnes. Hence, when we are extracting between $20$ and $50$ tonnes of gold, the total cost is increasing on average, at a rate of $23/15$ thousand dollars/tonne.

Example 4.4. If $S$ is the number of people who have been ill for a time $t$ (in months), and $S\left(28\right)=568$ and $S\left(36\right)=236\text{,}$ then the quotient

${\displaystyle \frac{S\left(36\right)-S\left(28\right)}{36-28}=\frac{236-568}{8}=-\frac{83}{2}}$

indicates that from $28$ to $36$ months, the number of ill people decreases, on average, at a rate of $83/2$ people/month.

In general,

Let us consider the geometric interpretation of the average rate of change.

Figure 4.1. Average rate of change

In Figure 4.1, above, we see that the points $\left(a,f\left(a\right)\right)$ and $\left(b,f\left(b\right)\right)$ are on the graph of $f$, and the line that passes through these points, by definition, has a slope given by the quotient

${\displaystyle \frac{\text{rise}}{\text{run}}=\frac{f\left(b\right)-f\left(a\right)}{b-a}\text{.}}$

Comparing this equation with Definition 4.1, we conclude that, if $f(x)$ is a continuous curve on an interval $\left[a,b\right]$, then the quotient

${\displaystyle \frac{f\left(b\right)-f\left(a\right)}{b-a}}$

has two different interpretations:

1. it is the slope of the secant line through the points $\left(a,f\left(a\right)\right)$ and $\left(b,f\left(b\right)\right)$.
2. it is the average rate of change of $f$ on $\left[a,b\right]\text{.}$

The rate of change, a physical concept, is equal to the slope of a line, a geometrical concept. It is because of this relationship between the physical and the geometrical that we constantly refer to the graphs of functions in calculus.

Example 4.5. In Example 4.1, the quotient

${\displaystyle \frac{d\left(5\right)-d\left(2\right)}{5-3}=\frac{5}{2}}$

is the slope of the line passing through the points $\left(5,d\left(5\right)\right)$ and $\left(2,d\left(2\right)\right)$. The slope is positive, so the distance is increasing.

Example 4.6. In Example 4.4, the quotient

${\displaystyle \frac{S\left(36\right)-S\left(28\right)}{36-28}=-\frac{83}{2}}$

indicates that the slope of the secant line passing through $\left(28,S\left(28\right)\right)$ and $\left(36,S\left(36\right)\right)$ is $-83∕2$; hence, the average number of people who are ill is decreasing.