# 2.3: Interpretations of the Derivative

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In the previous sections we defined the derivative as the slope of a tangent line, using a particular limit. This allows us to compute “the slope of a curve” ^{1} and provides us with one interpretation of the derivative. However, the main importance of derivatives does not come from this application. Instead, (arguably) it comes from the interpretation of the derivative as the instantaneous rate of change of a quantity.

## 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 \(a\text{.}\)

## Slope

Suppose that \(y=f(x)\) is the equation of a curve in the \(xy\)–plane. That is, \(f(x)\) is the \(y\)–coordinate of the point on the curve whose \(x\)–coordinate is \(x\text{.}\) Then, as we have already seen,

\begin{gather*} \big[\text{the slope of the secant through $\big(a,f(a)\big)$ and $\big(a+h,f(a+h)\big)$}\big] =\frac{f(a+h)-f(a)}{h} \end{gather*}

This is shown in Figure 2.3.2 below.

In order to create the tangent line (as we have done a few times now) we squeeze \(h \to 0\text{.}\) As we do this, the secant through \(\big(a,f(a)\big)\) and \(\big(a+h,f(a+h)\big)\) approaches ^{2} the tangent line to \(y=f(x)\) at \(x=a\text{.}\) Since the secant becomes the tangent line in this limit, the slope of the secant becomes the slope of the tangent and

\begin{align*} \big[\text{the slope of the tangent line to $\;y=f(x)$ at $x=a$}\big] &=\lim_{h\rightarrow 0}\frac{f(a+h)-f(a)}{h}\\ &=f'(a). \end{align*}

Let us go a little further and work out a general formula for the equation of the tangent line to \(y=f(x)\) at \(x=a\text{.}\) We know that the tangent line

- has slope \(f'(a)\) and
- passes through the point \(\big(a, f(a)\big)\text{.}\)

There are a couple of different ways to construct the equation of the tangent line from this information. One is to observe, as in Figure 2.3.3, that if \((x,y)\) is any other point on the tangent line then the line segment from \(\big(a,f(a)\big)\) to \((x,y)\) is part of the tangent line and so also has slope \(f'(a)\text{.}\) That is,

\begin{gather*} \frac{y- f(a)}{x-a} =\big[\text{the slope of the tangent line}\big] =f'(a) \end{gather*}

Cross multiplying gives us the equation of the tangent line:

\begin{gather*} y-f(a) = f'(a)\,(x-a)\qquad\text{or}\qquad y=f(a)+f'(a)\,(x-a) \end{gather*}

A second way to derive the same equation of the same tangent line is to recall that the general equation for a line, with finite slope, is \(y=mx+b\text{,}\) where \(m\) is the slope and \(b\) is the \(y\)-intercept. We already know the slope — so \(m=f'(a)\text{.}\) To work out \(b\) we use the other piece of information — \((a,f(a))\) is on the line. So \((x,y)=(a,f(a))\) must solve \(y=f'(a)\,x+b\text{.}\) That is,

\begin{align*} f(a) &= f'(a) \cdot a + b & \text{and so } && b&=f(a)- af'(a) \end{align*}

Hence our equation is, once again,

\begin{align*} y &= f'(a) \cdot x + \left(f(a)-af'(a) \right) && \text{or, after rearranging a little,}\\ y &= f(a) + f'(a) \, (x-a) \end{align*}

This is a very useful formula, so perhaps we should make it a theorem.

The tangent line to the curve \(y=f(x)\) at \(x=a\) is given by the equation

\begin{align*} y &= f(a) + f'(a) \, (x-a) \end{align*}

provided the derivative \(f'(a)\) exists.

The caveat at the end of the above theorem is necessary — there are certainly cases in which the derivative does not exist and so we do need to be careful.

Find the tangent line to the curve \(y=\sqrt{x}\) at \(x=4\text{.}\)

Rather than redoing everything from scratch, we can, and for efficiency, should, use Theorem 2.3.4. To write this up properly, we must ensure that we tell the reader what we are doing. So something like the following:

- By Theorem 2.3.4, the tangent line to the curve \(y=f(x)\) at \(x=a\) is given by
\begin{align*} y &= f(a) + f'(a) (x-a) \end{align*}

provided \(f'(a)\) exists. - In Example 2.2.9, we found that, for any \(a \gt 0\text{,}\) the derivative of \(\sqrt{x}\) at \(x=a\) is
\begin{align*} f'(a) &= \frac{1}{2\sqrt{a}} \end{align*}

- In the current example, \(a=4\) and we have
\begin{align*} f(a)&=f(4)=\sqrt{x}\big|_{x=4}=\sqrt{4}=2\\ \text{and}\qquad f'(a)&=f'(4)=\frac{1}{2\sqrt{a}}\Big|_{a=4}=\frac{1}{2\sqrt{4}}=\frac{1}{4} \end{align*}

- So the equation of the tangent line to \(y=\sqrt{x}\) at \(x=4\) is
\begin{gather*} y= 2+\frac{1}{4}\,\big(x-4\big)\qquad\text{or}\qquad y=\frac{x}{4}+1 \end{gather*}

We don't have to write it up using dot-points as above; we have used them here to help delineate each step in the process of computing the tangent line.

## Exercises

Stage 2

Suppose \(h(t)\) gives the height at time \(t\) of the water at a dam, where the units of \(t\) are hours and the units of \(h\) are meters.

- What is the physical interpretation of the slope of the secant line through the points \((0,h(0))\) and \((24,h(24))\text{?}\)
- What is the physical interpretation of the slope of the tangent line to the curve \(y=h(t)\) at the point \((0,h(0))\text{?}\)

Suppose \(p(t)\) is a function that gives the profit generated by selling \(t\) widgets. What is the practical interpretation of \(p'(t)\text{?}\)

\(T(d)\) gives the temperature of water at a particular location \(d\) metres below the surface. What is the physical interpretation of \(T'(d)\text{?}\) Would you expect the magnitude of \(T'(d)\) to be larger when \(d\) is near 0, or when \(d\) is very large?

\(C(w)\) gives the calories in \(w\) grams of a particular dish. What does \(C'(w)\) describe?

The velocity of a moving object at time \(t\) is given by \(v(t)\text{.}\) What is \(v'(t)\text{?}\)

The function \(T(j)\) gives the temperature in degrees Celsius of a cup of water after \(j\) joules of heat have been added. What is \(T'(j)\text{?}\)

A population of bacteria, left for a fixed amount of time at temperature \(T\text{,}\) grows to \(P(T)\) individuals. Interpret \(P'(T)\text{.}\)

Stage 3

You hammer a small nail into a wooden wagon wheel. \(R(t)\) gives the number of rotations the nail has undergone \(t\) seconds after the wagon started to roll. Give an equation for how quickly the nail is rotating, measured in degrees per second.

A population of bacteria, left for a fixed amount of time at temperature \(T\text{,}\) grows to \(P(T)\) individuals. There is one ideal temperature where the bacteria population grows largest, and the closer the sample is to that temperature, the larger the population is (unless the temperature is so extreme that it causes all the bacteria to die by freezing or boiling). How will \(P'(T)\) tell you whether you are colder or hotter than the ideal temperature?