### Subsection2.3.1Instantaneous Rate of Change

In fact we have already (secretly) used a derivative to compute an instantaneous rate of change in Section 1.2. For your convenience we'll review that computation here, in Example 2.3.1, and then generalise it.

You drop a ball from a tall building. After $t$ seconds the ball has fallen a distance of $s(t)=4.9 t^2$ metres. What is the velocity of the ball one second after it is dropped?

• In the time interval from $t=1$ to $t=1+h$ the ball travels a distance
\begin{gather*} s(1+h)-s(1)=4.9 (1+h)^2 - 4.9 (1)^2 =4.9\big[2h+h^2\big] \end{gather*}
• So the average velocity over this time interval is
\begin{align*} &\text{average velocity from $t=1$ to $t=1+h$}\\ &=\frac{\text{distance travelled from $t=1$ to $t=1+h$}} {\text{length of time from $t=1$ to $t=1+h$}}\\ &=\frac{s(1+h)-s(1)}{h}\\ &=\frac{4.9\big[2h+h^2\big]}{h}\\ &=4.9[2+h] \end{align*}
• The instantaneous velocity at time $t=1$ is then defined to be the limit
\begin{align*} &\text{instantaneous velocity at time } t=1\\ &\hskip0.5in=\lim_{h\rightarrow 0} \big[\text{average velocity from $t=1$ to $t=1+h$}\big]\\ &\hskip0.5in=\lim_{h\rightarrow 0}\frac{s(1+h)-s(1)}{h} = s'(1)\\ &\hskip0.5in= \lim_{h\rightarrow 0} 4.9[2+h]\\ &\hskip0.5in= 9.8\text{m/sec} \end{align*}
• We conclude that the instantaneous velocity at time $t=1\text{,}$ which is the instantaneous rate of change of distance per unit time at time $t=1\text{,}$ is the derivative $s'(1)=9.8\text{m/sec}\text{.}$

Now suppose, more generally, that you are taking a walk and that as you walk, you are continuously measuring some quantity, like temperature, and that the measurement at time $t$ is $f(t)\text{.}$ Then the

\begin{align*} &\text{average rate of change of $f(t)$ from $t=a$ to $t=a+h$}\\ &\hskip0.5in=\frac{\text{change in $f(t)$ from $t=a$ to $t=a+h$}} {\text{length of time from $t=a$ to $t=a+h$}}\\ &\hskip0.5in=\frac{f(a+h)-f(a)}{h} \end{align*}

so the

\begin{align*} &\text{instantaneous rate of change of $f(t)$ at $t=a$}\\ &\hskip0.5in=\lim_{h\rightarrow 0} \big[\text{average rate of change of $f(t)$ from $t=a$ to $t=a+h$}\big]\\ &\hskip0.5in=\lim_{h\rightarrow 0}\frac{f(a+h)-f(a)}{h}\\ &\hskip0.5in= f'(a) \end{align*}

In particular, if you are walking along the $x$–axis and your $x$–coordinate at time $t$ is $x(t)\text{,}$ then $x'(a)$ is the instantaneous rate of change (per unit time) of your $x$–coordinate at time $t=a\text{,}$ which is your velocity at time $a\text{.}$ If $v(t)$ is your velocity at time $t\text{,}$ then $v'(a)$ is the instantaneous rate of change of your velocity at time $a\text{.}$ This is called your acceleration at time $a\text{.}$

You might expect that if the instantaneous rate of change of a function at time $c$ is strictly positive, then, in some sense, the function is increasing at $t=c\text{.}$ You would be right. Indeed, if $f'(c)\gt 0\text{,}$ then, by definition, the limit of $\frac{f(t)-f(c)}{t-c}$ as $t$ approaches $c$ is strictly bigger than zero. So

• for all $t\gt c$ that are sufficiently close 2 This is typical mathematician speak — it allows us to be completely correct, without being terribly precise. In this context, sufficiently close means The following need not be true for all $t$ bigger than $c\text{,}$ but there must exist some $b\gt c$ so that the following is true for all $c\lt t\lt b$. Typically we do not know what $b$ is. And typically it does not matter what the exact value of $b$ is. All that matters is that $b$ exists and is strictly bigger than $c\text{.}$ to $c$
\begin{align*} \frac{f(t)-f(c)}{t-c}\gt 0 \amp \implies f(t)-f(c) \gt 0\qquad (\text{since }t-c\gt 0) \\ \amp \implies f(t)\gt f(c) \end{align*}
• for all $t\lt c$ that are sufficiently close to $c$
\begin{align*} \frac{f(t)-f(c)}{t-c}\gt 0 \amp \implies f(t)-f(c) \lt 0\qquad (\text{since }t\lt c) \\ \amp \implies f(t)\lt f(c) \end{align*}

Consequently we say that “$f(t)$ is increasing at $t=c$”. If we wish to emphasise that the inequalities above are the strict inequalities $\gt$ and $\lt\text{,}$ as opposed to $\ge$ and $\le\text{,}$ we will say that “$f(t)$ is strictly increasing at $t=c$”.