Graphs of change · The Lumenary

MotivationH

A story in a graph

Keza lives near Huye in Rwanda. One Saturday she cycles to the market, $6$ kilometres away, and back. Her phone records how far she is from home, minute by minute. Here is the graph.

Keza's distance from home during her trip to market. Segment A: riding out; B: stopped to fix a loose chain; C: riding on; D: at the market; E: riding home.

Without being told anything else, we can read off a lot. When was she stopped? When was she riding fastest? How far did she ride altogether? And here is a harder one: the graph starts and ends at zero. Does that mean that, on average, nothing happened? By the end of this lesson you will be able to answer all of these precisely.

IntuitionH

Reading the shape of a graph

A graph of change, drawn in the coordinate plane, shows how one quantity depends on another — very often, how something changes over time. Read it from left to right, like a sentence:

Going up (segments A and C): the distance from home is increasing — Keza is riding away.
Flat (B and D): the distance is not changing — she has stopped.
Going down (E): the distance is decreasing — she is coming back.

The steepness tells us how fast the quantity changes. Segment E is the steepest: she covers $6$ km in $20$ minutes on the way home, faster than on the way out. (Perhaps the road home is downhill — but the graph does not show hills!)

DefinitionsH

Rates of change

Definition— average rate of change

If a quantity has the value $y(a)$ at $x = a$ and $y(b)$ at $x = b$ , with $a \neq b$ , its average rate of change between $a$ and $b$ is

\frac{y(b) - y(a)}{b - a} = \frac{\text{change in } y}{\text{change in } x}.

On the graph, it is the slope of the straight segment joining the two points (see linear relationships). Its units are "units of $y$ per unit of $x$ ": kilometres per minute, degrees per hour, litres per second. A rate is a ratio of two changes (see ratios and rates).

Definition— increasing, decreasing, constant

On an interval of $x$ -values, a graph is

increasing if $y$ gets larger whenever $x$ gets larger: $x_1 < x_2$ implies $y(x_1) < y(x_2)$ ;
decreasing if $y$ gets smaller whenever $x$ gets larger: $x_1 < x_2$ implies $y(x_1) > y(x_2)$ ;
constant if $y$ has the same value at every $x$ in the interval.

Definition— average speed

For a journey, the average speed is the total distance travelled divided by the total time taken.

If a graph shows the distance from a starting point along a route, the slope of a segment is the average rate of change of that distance. As long as the traveller does not turn round during the segment, the size of the slope is the average speed, and its sign says whether they are moving away ( $+$ ) or coming back ( $-$ ). If the graph shows the total distance travelled, it can never go down, and its slopes are exactly the average speeds.

Worked examplesH

Worked examples

Example— Rates on Keza's journey

segment	time	change in distance	average rate	in km/h
A	$0$ – $20$ min	$+4$ km	$\tfrac{4}{20} = 0.2$ km/min	$12$
B	$20$ – $25$ min	$0$ km	$0$ km/min	$0$
C	$25$ – $35$ min	$+2$ km	$\tfrac{2}{10} = 0.2$ km/min	$12$
D	$35$ – $55$ min	$0$ km	$0$ km/min	$0$
E	$55$ – $75$ min	$-6$ km	$\tfrac{-6}{20} = -0.3$ km/min	$-18$

(To turn km per minute into km per hour, multiply by $60$ .) A and C have the same slope, so she rode at the same average speed on both; E is steepest, $18$ km/h, and the minus sign records that she was heading home.

Example— The whole trip: zero change, but not nothing

Over the whole trip, from $0$ to $75$ minutes, the distance from home changes by $0 - 0 = 0$ , so the average rate of change is $0$ km/h. But Keza rode $4 + 2 + 6 = 12$ km in $75$ minutes $= 1.25$ hours: her average speed was $\tfrac{12}{1.25} = 9.6$ km/h. A zero average rate of change does not mean nothing happened — only that she ended where she started.

Example— A table: a hot day in Delhi

Air temperatures on a hot day in Delhi (illustrative numbers):

time	6 a.m.	9 a.m.	noon	3 p.m.	6 p.m.	9 p.m.
temperature (°C)	$29$	$33$	$38$	$40$	$37$	$33$

From 6 a.m. to noon the average rate of change is $\tfrac{38 - 29}{6} = 1.5$ °C per hour. From noon to 6 p.m. it is $\tfrac{37 - 38}{6} = -\tfrac{1}{6}$ °C per hour: nearly zero, even though the temperature rose to $40$ °C and fell again in between. From 3 p.m. to 9 p.m. it is $\tfrac{33 - 40}{6} \approx -1.17$ °C per hour. The table only tells us the values at six moments; what happened between them is a guess.

Example— Sketching from a description

"The bath fills steadily for 8 minutes. Then the tap is turned off, and Tomás gets in — the water level jumps up. He lies still for 10 minutes, gets out, and pulls the plug; the bath empties steadily in 4 minutes."

A sketch of water depth against time: a straight rising segment for 8 minutes; a short, very steep rise when Tomás gets in; a flat segment for 10 minutes; a short, very steep drop when he gets out (back to the depth before he got in); then a straight falling segment down to $0$ , steeper than the filling segment because emptying takes $4$ minutes rather than $8$ . The heights come from the story's quantities; the steepness from how fast things happen.

ProofH

What rates of change can tell us

Let $y$ be a quantity that depends on $x$ , and write $R(a, b)$ for its average rate of change between $a$ and $b$ .

Theorem— change = rate × time

For $a \neq b$ : $\;y(b) - y(a) = R(a, b) \cdot (b - a)$ . On a distance–time graph: distance covered $=$ average speed $\times$ time taken.

Proof

Multiply the definition $R(a, b) = \dfrac{y(b) - y(a)}{b - a}$ by $b - a$ .

∎

Theorem— the sign of the rate

A graph is increasing on an interval if and only if every average rate of change $R(x_1, x_2)$ with $x_1 < x_2$ in the interval is positive. Likewise, it is constant on the interval if and only if all these average rates are $0$ .

Proof

Take $x_1 < x_2$ . The run $x_2 - x_1$ is positive, so the quotient $\dfrac{y(x_2) - y(x_1)}{x_2 - x_1}$ is positive exactly when the rise $y(x_2) - y(x_1)$ is positive, that is, exactly when $y(x_1) < y(x_2)$ . "Increasing" says this for every such pair, and so does "all rates are positive". In the same way, the quotient is $0$ exactly when $y(x_1) = y(x_2)$ , so all the rates are $0$ exactly when $y$ never changes.

∎

So a horizontal segment really does mean "not changing", and a straight segment that slopes upwards means steady increase. (Decreasing works the same way, with negative rates.)

Next: how do the rates over two pieces of an interval combine? Not by simply averaging them.

Theorem— combining intervals

For $a < b < c$ :

R(a, c) = \frac{(b - a)\,R(a, b) + (c - b)\,R(b, c)}{c - a}.

The rate over the whole interval is the average of the two rates, weighted by the lengths of the pieces. In particular, $R(a, c)$ lies between $R(a, b)$ and $R(b, c)$ .

Proof

The total change is the sum of the changes on the two pieces, and by "change = rate × time":

y(c) - y(a) = \big(y(b) - y(a)\big) + \big(y(c) - y(b)\big) = (b - a)R(a, b) + (c - b)R(b, c).

Dividing by $c - a$ gives the formula. Now suppose $R(a, b) \leq R(b, c)$ (otherwise swap the roles). Replacing $R(b, c)$ by the smaller number $R(a, b)$ makes the right-hand side no larger, so $(c - a)R(a, c) \geq (b - a)R(a, b) + (c - b)R(a, b) = (c - a)R(a, b)$ . Dividing by $c - a > 0$ gives $R(a, c) \geq R(a, b)$ . Replacing $R(a, b)$ by the larger $R(b, c)$ shows $R(a, c) \leq R(b, c)$ in the same way.

∎

For Keza's first $35$ minutes: $R(0, 35) = \dfrac{20 \cdot 12 + 5 \cdot 0 + 10 \cdot 12}{35} = \dfrac{72}{7} \approx 10.3$ km/h — between $0$ and $12$ , as it must be, but not their plain average.

Theorem— same average, different journeys

Two quantities can have the same average rate of change over an interval but behave differently within it.

Proof

One example is enough. Ana runs $100$ m at a steady $5$ m/s, so she is $5t$ metres along after $t$ seconds. Ben runs at $7$ m/s for $10$ seconds and then, tiring, at $3$ m/s. Both finish $100$ m in $20$ s, so both average $5$ m/s. But after $10$ s Ana has run $50$ m and Ben $70$ m.

∎

Theorem— round trips

If you travel a distance $d$ at speed $u$ and return at speed $v$ (with $d, u, v > 0$ ), your average speed is

\frac{2d}{\frac{d}{u} + \frac{d}{v}} = \frac{2uv}{u + v},

which is never more than $\tfrac{u + v}{2}$ , and equals it only when $u = v$ .

Proof

The total distance is $2d$ , and the times for the two legs are $\tfrac{d}{u}$ and $\tfrac{d}{v}$ . Multiplying the top and bottom of $\tfrac{2d}{d/u + d/v}$ by $\tfrac{uv}{d}$ gives $\tfrac{2uv}{v + u}$ . For the comparison,

\frac{u + v}{2} - \frac{2uv}{u + v} = \frac{(u + v)^2 - 4uv}{2(u + v)} = \frac{(u - v)^2}{2(u + v)},

which is never negative (a square is never negative) and is $0$ only when $u = v$ .

∎

The classic puzzle: drive $60$ km at $30$ km/h ( $2$ hours) and back at $60$ km/h ( $1$ hour). The average speed is $\tfrac{120}{3} = 40$ km/h — not $45$ . You spend more time at the slower speed, so it counts for more.

ExplanationH

From average to instantaneous

A stone dropped from a cliff falls faster and faster. In the idealised model with no air resistance, it has fallen $d(t) = 4.9\,t^2$ metres after $t$ seconds. The graph is curved, so its average rate of change depends on the interval we choose. What is the stone's speed at the moment $t = 2$ ?

Take a short interval from $2$ to $2 + h$ seconds (with $h \neq 0$ ). The average speed over it is

\frac{4.9(2 + h)^2 - 4.9 \cdot 2^2}{h} = \frac{4.9\,(4h + h^2)}{h} = 4.9\,(4 + h) = 19.6 + 4.9\,h \;\text{ m/s}.

As $h$ gets smaller and smaller — from either side — this gets as close as we like to $19.6$ m/s. That number is the stone's instantaneous speed at $t = 2$ : what a speedometer would read. On the graph, the segments become the tangent line. Making this idea precise is the starting point of calculus (see introductory calculus).

Average speed over a shrinking interval

The straight line's slope is the average speed between P and Q. Shrink h.

d = 4.9t²
segment through P and Q

aa = 2hh = 1

Trace along xx = 1

d = 4.9t²: y = 4.9
segment through P and Q: y = −4.9

Formal mathematicsF

Rates of change in Lean

In Lean, a changing quantity is a function f : Rat → Rat, and the average rate of change is a definition — the slope of the segment between two points of the graph.

avgRateLean 4Formally verified

/-- The average rate of change of `f` between `a` and `b`: change in the quantity divided by
the change in time, `(f b − f a) / (b − a)`. It is the slope of the segment of the graph from
`(a, f a)` to `(b, f b)`. Meaningful only for `a ≠ b`; the theorems below assume that. -/
def avgRate (f : Rat → Rat) (a b : Rat) : Rat := (f b - f a) / (b - a)