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Subsection 2.3.1 Instantaneous 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 \(t=c\)”.

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