Mathematics 265 Introduction to Calculus I

Study Guide :: Unit 4

Differentiation

The Instantaneous Rate of Change and the Slope of the Tangent Line

If average velocity indicates how quickly or slowly an object moves in a period of time, instantaneous velocity indicates how quickly or slowly the object moves at a precise instant.

If $d\left(t\right)$ is, again, the distance (in metres) traveled by a moving object at time $t$ (in seconds), then the average velocity on the interval $\left[t,a\right]$—or $\left[a,t\right]$—corresponds to the quotient

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

By “the instantaneous velocity at time $a$,” we mean the velocity at precisely the time $a\text{.}$ Intuitively, this value is obtained by taking the average velocity on intervals $\left[a,t\right]$—or $\left[t,a\right]$—for $t$ very close to $a\text{.}$ Hence, the instantaneous velocity at time $a$ is defined as the limit of the average velocity as $t$ approaches $a\text{.}$ That is, the velocity of the object at time $a$ is the limit

${\displaystyle \underset{t\to a}{\lim }{\displaystyle \frac{f\left(t\right)-f\left(a\right)}{t-a}}\text{.}}$

Example 4.7. The displacement (in centimetres) of a particle at time $t$ (in seconds) is given by the function

${\displaystyle s\left(t\right)=t^2-{\displaystyle \frac{t}{2}}+1\text{.}}$

The average velocity on the interval $\left[1,t\right]$ is

${\displaystyle {\displaystyle \frac{s\left(t\right)-s\left(1\right)}{t-1}}={\displaystyle \frac{\left(t^2-{\displaystyle \frac{t}{2}}+1\right)-\left(1-{\displaystyle \frac{1}{2}}+1\right)}{t-1}}={\displaystyle \frac{t^2-{\displaystyle \frac{t}{2}}-{\displaystyle \frac{1}{2}}}{t-1}}\text{.}}$

The average velocity on the interval $\left[0,1\right]$ is

${\displaystyle {\displaystyle \frac{s\left(1\right)-s\left(0\right)}{1-0}}={\displaystyle \frac{{\displaystyle \frac{-1}{2}}}{-1}}={\displaystyle \frac{1}{2}}}$

and the average velocity on the interval $\left[1,2\right]$ is

${\displaystyle {\displaystyle \frac{s\left(2\right)-s\left(1\right)}{2-1}}={\displaystyle \frac{{\displaystyle \frac{5}{2}}}{1}}={\displaystyle \frac{5}{2}}\text{.}}$

The limit

${\displaystyle \begin{array}{ll}\hfill \underset{t\to 1}{\lim }{\displaystyle \frac{s\left(t\right)-s\left(1\right)}{t-1}}& =\underset{t\to 1}{\lim }{\displaystyle \frac{t^2={\displaystyle \frac{t}{2}}-{\displaystyle \frac{1}{2}}}{t-1}}\hfill \\ \hfill & =\underset{t\to 1}{\lim }{\displaystyle \frac{\left(t-1\right)\left(t+{\displaystyle \frac{1}{2}}\right)}{t-1}}\hfill \\ \hfill & =\underset{t\to 1}{\lim }t+{\displaystyle \frac{1}{2}}={\displaystyle \frac{3}{2}}\hfill \end{array}}$

is the velocity of the particle at $1$ second.

That is to say, at precisely $1$ second, the velocity of the particle is $3∕2$ cm/s, or the distance traveled by the particle at $1$ second is increasing at a rate of $3∕2$ cm/s.

Example 4.8. The temperature $T$ (in $\text{°}\text{C}$) at time $t$ (in hours) is given by $T\left(t\right)=5-2t-t^3\text{.}$ The average temperature on the interval $\left[t,1\right]$ is the quotient

${\displaystyle \begin{array}{ll}\hfill {\displaystyle \frac{T\left(1\right)-T\left(t\right)}{1-t}}& ={\displaystyle \frac{\left(5-2-1\right)-\left(5-2t-t^3\right)}{1-t}}\hfill \\ \hfill & ={\displaystyle \frac{-3+2t+t^3}{1-t}}\text{.}\hfill \end{array}}$

Hence, the average temperature on the interval $\left[0,1\right]$ is $-3\text{.}$ On average, during the first hour, the temperature is decreasing at a rate of $3\text{°}\text{C/h.}$ The limit

${\displaystyle \underset{t\to 1}{\lim }{\displaystyle \frac{T\left(t\right)-T\left(1\right)}{t-1}}=\underset{t\to 1}{lim}{\displaystyle \frac{3-2t-t^3}{t-1}}=\underset{t\to 1}{lim}-t^2-t-3=-5}$

indicates that, at precisely $1$ hour, the temperature is decreasing at a rate of $5\text{°}\text{C/h}\text{.}$

In general, then,

Figure 4.2, below, shows the geometric interpretation of this limit.

Figure 4.2. Instantaneous rate of change

The quotient

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

is the slope of the secant line through the points $\left(a,f\left(a\right)\right)$ and $\left(x,f\left(x\right)\right)\text{,}$ as shown on Figure 4.2, above.

For each $x$ near $a$ (in Figure 4.2, from the right), we consider the point $\left(x,f\left(x\right)\right)$ and the corresponding secant line through $\left(a,f\left(a\right)\right)$ and $\left(x,f\left(x\right)\right)\text{.}$ If we move $x$ towards $a\text{,}$ we see that the secant line tends toward the line tangent to the curve at the point $\left(a,f\left(a\right)\right)$ (see Figure 1 on page 103 of the textbook). The same is true for $x$ near $a$ from the left. Make a similar sketch and convince yourself that this is indeed the case.

Hence, the slope of the secant lines tend to the slope of the tangent line at $\left(a,f\left(a\right)\right)\text{.}$

We know that the limit

${\displaystyle \underset{x\to a}{\lim }{\displaystyle \frac{f\left(x\right)-f\left(a\right)}{x-a}}}$

has two interpretations:

1. it is the instantaneous rate of change at (precisely) $a\text{.}$
2. it is the slope of the tangent to the curve at the point $\left(a,f\left(a\right)\right)\text{.}$

Example 4.9. Consider the graph of the function $f\left(x\right)=4-x^2$ shown in Figure 4.3, below.

Figure 4.3. Function $f\left(x\right)=4-x^2$

We can see by the inclination of the tangent line that the slope must be negative, and indeed the slope at the point $\left(1,f\left(1\right)\right)=\left(1,3\right)$ is given by the limit

${\displaystyle \underset{x\to 1}{\lim }{\displaystyle \frac{f\left(x\right)-f\left(1\right)}{x-1}}=\underset{x\to 1}{lim}{\displaystyle \frac{4-x^2-3}{x-1}}=\underset{x\to 1}{lim}{\displaystyle \frac{1-x^2}{x-1}}=\underset{x\to 1}{lim}-\left(x+1\right)=-2}$

If $f\left(x\right)=4-x^2$ is the altitude at time $x$ (in seconds) of an object falling freely from a height of $4$ m, then at precisely $1$ second, the altitude is decreasing at a rate of $2$ m/s.

We also see from the graph that the tangent line at $x=0$ is horizontal; hence, the slope must be 0. Indeed,

${\displaystyle \underset{x\to 0}{\lim }{\displaystyle \frac{f\left(x\right)-f\left(0\right)}{x-0}}=\underset{x\to 0}{lim}{\displaystyle \frac{4-x^2-4}{x}}=\underset{x\to 0}{lim}-x=0\text{.}}$

At $x=0$ (the initial position), the height of the object is neither increasing nor decreasing—the object is stationary.

The limit in Definition 4.2 has a special name and notation: it is called the “derivative of the function $f$ at $a\text{,}$ ” and we write it $f^{\prime }\left(a\right)\text{.}$

Example 4.10. If $C$ is the cost of producing $x$ tonnes of wheat, then the equation $C^{\prime }\left(10\right)=12$ has two interpretations. First, the “real world” interpretation: when the production is $10$ tonnes, the cost is increasing at a rate of $\$12\text{.}00$ dollars/tonne. The geometric interpretation is that the slope of the tangent line to the curve $C\left(x\right)$ at the point $\left(10,C\left(10\right)\right)$ is $12\text{.}$

Example 4.11. See Exercise 30 on page 119 of the textbook.

The function $f$ represents the quantity (in pounds) of coffee sold at a price $p$ dollars per pound. The derivative function $f^{\prime }\left(p\right)$ is the rate at which the quantity of coffee increases or decreases for a certain price $p\text{.}$ When the price of coffee sold is $\$8\text{.}00\text{,}$ the quantity of coffee increases or decreases at a rate of $f^{\prime }\left(8\right)$ pounds/dollar. We estimate that the higher the value of $p$, the less coffee is sold. For a price of $\$8\text{.}00$ per pound, we believe that the quantity of coffee sold decreases; hence, $f^{\prime }\left(8\right)$ is negative.

Example 4.12. If the derivative function of a function $s\left(t\right)$ that represents the motion of a particle measured in metres at $t$ seconds is $s^{\prime }\left(t\right)=3t^2+5$, then $s^{\prime }\left(t\right)$ is the velocity of the particle at time $t$ and its units are metres per second. Since

${\displaystyle \underset{t\to 5}{\lim }{\displaystyle \frac{s^{\prime }\left(t\right)-s^{\prime }\left(5\right)}{t-5}}=\underset{t\to 5}{lim}{\displaystyle \frac{3t^2-75}{t-5}}=\underset{t\to 5}{lim}3\left(t+5\right)=30\text{,}}$

we find that at time $5$ seconds, the velocity of the particle is increasing at a rate of $30$ (m/s)/s. That is, the acceleration of the particle is $30$ m/s ${}^2\text{.}$

On the other hand, the slope of the tangent line at the point $\left(5,s^{\prime }\left(5\right)\right)=\left(5,80\right)$ is $30$.

Example 4.13.

1. Refer to Example 4.7. If ${\displaystyle s\left(t\right)=t^2-{\displaystyle \frac{t}{2}}+1\text{,}}$ then ${\displaystyle s^{\prime }\left(1\right)={\displaystyle \frac{3}{2}}\text{.}}$
2. Refer to Example 4.8. If $T\left(t\right)=5-2t-t^3\text{,}$ then $T^{\prime }\left(1\right)=-5$.
3. Refer to Example 4.9. If $f\left(x\right)=4-x^2\text{,}$ then $f^{\prime }\left(0\right)=0$ and $f^{\prime }\left(1\right)=-2\text{.}$

We know that a limit may or may not exist. The limit in Definition 4.3 is no exception; we will use the geometric interpretation of this limit to find out when the limit itself does not exist.

For the limit in Definition 4.3 to exist, we need a tangent line to the curve at the point $\left(a,f\left(a\right)\right)$. If the curve does not have such a tangent line, then the limit cannot exist. The tangent line is the line closest to the curve at the point $\left(a,f\left(a\right)\right)$; that is, the tangent line lies on the point—we do not have tangent lines when the function is not defined at $a\text{.}$

In Figure 4.4, below, it is clear that there cannot be a tangent line at the point $\left(a,f\left(a\right)\right)$ in either graph. But breaks do not constitute the only case where a curve is not defined at a point.

Figure 4.4. Curves that are not defined at the point $\left(a,f\left(a\right)\right)$

For example, it may be that, while the function is continuous, it has a kink at the point $\left(a,f\left(a\right)\right)$ that makes a tangent line impossible, examples of this case are shown in Figure 4.5.

Figure 4.5. Curves showing a kink at the point $\left(a,f\left(a\right)\right)$

The slope of vertical lines is undefined; hence, if the tangent line at the point $\left(a,f\left(a\right)\right)$ is vertical, the limit in Definition 4.3 is also undefined. This case is shown in Figure 4.6.

Figure 4.6. Curve with a vertical tangent at the point $\left(a,f\left(a\right)\right)$

Summarizing: If the limit in Definition 4.3 does not exist, it is because

• the function is not continuous at $a$, or
• the function is continuous, but there is no clearly defined tangent line at the point $\left(a,f\left(a\right)\right)$, or
• the function is continuous and there is a tangent line at the point $\left(a,f\left(a\right)\right)$, but that tangent line is vertical.

Conversely if the limit in Definition 4.3 exists, it is because the curve around the point $\left(a,f\left(a\right)\right)$ is “smooth” enough to have a tangent line.

From the graphs of the basic functions in Table 2.1, we see that the absolute value function $g\left(x\right)=\left|x\right|$ is not differentiable at $0$, the root function is not differentiable at $0$, the tangent and secant functions are not differentiable at $\pi ∕2$, and the sine and cosine functions are differentiable everywhere.

In the next examples, we evaluate the limit using Definition 4.3 for a fixed number $a$. As you will see, for relatively simple functions, this limit is not difficult to evaluate. In the following examples, you must identify the algebraic manipulations and results we apply to evaluate each limit.

Example 4.14. Let $f\left(x\right)=\sqrt{x}$

${\displaystyle \underset{x\to a}{\lim }{\displaystyle \frac{f\left(x\right)-f\left(a\right)}{x-a}}=\underset{x\to a}{lim}{\displaystyle \frac{\sqrt{x}-\sqrt{a}}{x-a}}=\underset{x\to a}{lim}{\displaystyle \frac{1}{\sqrt{x}+\sqrt{a}}}={\displaystyle \frac{1}{2\sqrt{a}}}\text{.}}$

So, for example, $f^{\prime }\left(3\right)={\displaystyle \frac{1}{2\sqrt{3}}}$ is the slope of the tangent line at $\left(3,f\left(3\right)\right)=\left(3,\sqrt{3}\right)$.

Since ${\displaystyle f^{\prime }\left(0\right)}$ is undefined, the function is not differentiable at ${\displaystyle 0}\text{,}$ as we observed earlier.

Example 4.15. Let ${\displaystyle f\left(x\right)={\displaystyle \frac{1}{x+3}}}$

${\displaystyle \begin{array}{ll}\hfill \underset{x\to a}{\lim }{\displaystyle \frac{f\left(x\right)-f\left(a\right)}{x-a}}& =\underset{x\to a}{\lim }{\displaystyle \frac{{\displaystyle \frac{1}{x+3}}-{\displaystyle \frac{1}{a+3}}}{x-a}}\hfill \\ \hfill & =\underset{x\to a}{\lim }{\displaystyle \frac{a-x}{\left(x-a\right)\left(x+3\right)\left(a+3\right)}}\hfill \\ \hfill & =\underset{x\to a}{\lim }-{\displaystyle \frac{1}{\left(x+3\right)\left(a+3\right)}}\hfill \\ \hfill & =-{\displaystyle \frac{1}{{\left(a+3\right)}^2}}\text{.}\hfill \end{array}}$

For example,

${\displaystyle f^{\prime }\left(5\right)=-{\displaystyle \frac{1}{8^2}}=-{\displaystyle \frac{1}{64}}}$

and the slope of the tangent line at the point ${\displaystyle \left(5,f\left(5\right)\right)=\left(2\text{,}{\displaystyle \frac{1}{8}}\right)}$ is $-1∕64\text{.}$

We see that $f^{\prime }\left(-3\right)$ is undefined; hence, the function is not differentiable at $-3$, because the function $f$ is not continuous at $-3\text{.}$

Example 4.16. Let $f\left(x\right)=x^3\text{.}$

${\displaystyle \underset{x\to a}{\lim }{\displaystyle \frac{f\left(x\right)-f\left(a\right)}{x-a}}=\underset{x\to a}{lim}{\displaystyle \frac{x^3-a^3}{x-a}}=\underset{x\to a}{lim}{x}^2+xa+a^2=3a^2\text{.}}$

So, for example, the slope of the tangent line at ${\displaystyle \left({\displaystyle \frac{1}{2}}\text{,}{\displaystyle \frac{1}{8}}\right)}$ is

${\displaystyle f^{\prime }\left({\displaystyle \frac{1}{2}}\right)=3{\left({\displaystyle \frac{1}{2}}\right)}^2={\displaystyle \frac{3}{4}}\text{.}}$

The function is differentiable everywhere. [Why?]

These examples show that if we have a general expression for the derivative, then we can find the slope of the tangent line or the instantaneous rate of change for any number where $f^{\prime }\left(a\right)$ is defined, without having to evaluate a limit at each particular number.

There is an equivalent form for the limit in Definition 4.3. If we set $x=h+a$, then $h=x-a$, and as $x\to a$, we observe that $h\to 0$, and

${\displaystyle f^{\prime }\left(a\right)=\underset{x\to a}{lim}{\displaystyle \frac{f\left(x\right)-f\left(a\right)}{x-a}}=\underset{h\to 0}{lim}{\displaystyle \frac{f\left(a+h\right)-f\left(a\right)}{h}}\text{.}}$

We can use either form to find the derivative at $a$, depending on which is more convenient for the given function.