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Where is change fastest?

Change and motion · level 6, lesson 6 of 6

A question to start

A new playground game spreads through a school of 1000 students. On day 0, 15 students play it. Each day, players teach it to friends. Here is how many students know the game:

Day 0 1 2 3 4 5 6 7 8 9 10 11 12
Students who know 15 30 59 111 200 333 500 667 800 889 941 970 985

At first it spreads slowly (few players to teach it), and at the end it spreads slowly too (almost everyone already knows). Somewhere in the middle it spreads fastest. Where exactly, and why there?

And a second question. A ball is thrown up and comes back down. At the very top, how fast is its height changing?

The idea

Finding the fastest change in a table

The number of new students each day is the change from one day to the next:

Day to day 0-1 1-2 2-3 3-4 4-5 5-6 6-7 7-8 8-9 9-10 10-11 11-12
New students 15 29 52 89 133 167 167 133 89 52 29 15

The daily growth rises, peaks at 167 on the days either side of day 6, then falls in a mirror image. The game spreads fastest at day 6, when exactly 500 students, half the school, know it. The central estimate at day 6 is \((667 - 333) \div 2 = 167\) students per day (and a careful calculation gives about 173 per day at that instant).

The S-shaped curve

steepest: day 6, 500 know it slow startslow finish 024681012 day 02505007501000

A curve like this is called an S-curve (or a logistic curve). It turns up whenever growth starts out exponential but runs out of room: a game or a rumour in a school, a new kind of phone spreading through a country, yeast in a jar of sugar, a sunflower approaching its full height.

Before day 6 the curve gets steeper and steeper; after day 6 it gets flatter. The point where it switches from "getting steeper" to "getting flatter" is where the rate of change is greatest. It has a name: the point of inflection.

Why halfway?

How many new students learn the game on a day depends on two things:

  • how many students already know it (they do the teaching), and
  • how many don't know it yet (only they can learn).

If few know it, there are few teachers. If almost everyone knows it, there are few left to learn. In this model the rate is proportional to the product of the two:

\[\text{rate} \approx 0.000693 \times (\text{know}) \times (\text{don't know}).\]

The two numbers always add up to 1000. When is their product biggest?

Know 100 200 300 400 500 600 700
Don't know 900 800 700 600 500 400 300
Product 90 000 160 000 210 000 240 000 250 000 240 000 210 000

The product is biggest when the two are equal, 500 and 500. Here is why, with a picture: think of the two numbers as the sides of a rectangle with a fixed perimeter. Moving from a square to a rectangle that is \(d\) longer one way and \(d\) shorter the other changes the area from \(500 \times 500\) to \((500 + d)(500 - d) = 250\,000 - d^2\). The \(d^2\) is always taken away, so the square wins. (Do you recognise \((a + b)(a - b) = a^2 - b^2\) from the picture proof in level 5?)

So the game spreads fastest when half the school knows it, and the fastest rate is about \(0.000693 \times 500 \times 500 \approx 173\) students per day. At the start, while "don't know" is close to 1000, the rule is nearly "rate \(= 0.693 \times\) know": the exponential growth of the last lesson.

Highest and lowest points: the rate is zero

Now the ball. Its height is \(H(t) = 20t - 5t^2\) metres. In lesson 2 you found its upward speed is 10 m/s at \(t = 1\) and \(-10\) m/s at \(t = 3\). The same algebra at any time \(t\) gives the average over \([t, t + h]\) as \(20 - 10t - 5h\), which tends to

\[\text{rate of change of height} = 20 - 10t.\]
+10 m/s0 m/s−10 m/s 01234 time (seconds) 01020 m
Time (s) 0 1 2 3 4
Height (m) 0 15 20 15 0
Rate of change of height (m/s) 20 10 0 \(-10\) \(-20\)

The rate is zero at \(t = 2\), exactly at the top, where the height is 20 m. The tangent there is flat.

Why must it be zero at the top? Just before the top the ball is still rising, so the rate is positive. Just after, it is falling, so the rate is negative. The rate changes smoothly (here it is \(20 - 10t\), a straight line), and to go from positive to negative without jumping it has to pass through zero. It does that exactly at the top. If the rate at some point were positive, the curve would still be going up there, so that point could not be the highest; if it were negative, the curve would have been higher just before. Only a rate of zero is left.

The same is true at a lowest point of a smooth curve: going down before, up after, zero in between. So:

At the highest or lowest point of a smooth curve, the rate of change is zero and the tangent is flat.

This gives a powerful way to find the biggest or smallest value of something: look for where its rate of change is zero. That idea, called optimization, is one of the most useful things calculus does (much more in level 8).

Two warnings

  • Flat does not always mean top or bottom. The curve \(y = x^3\) has rate 0 at \(x = 0\) (the average over \([0, h]\) is \(h^3 \div h = h^2\), which tends to 0). But the curve keeps going up on both sides; it just levels off for an instant. So after finding a zero rate, check what happens on each side.
  • A corner can be a top with no tangent at all. A ball bouncing on the floor has its lowest points at sharp corners, where the rate jumps from negative to positive without passing through zero. Zero rates find the smooth tops and bottoms; corners and endpoints must be checked separately.

Worked examples

1. The highest point of a throw. A ball's height is \(H(t) = 30t - 5t^2\) m. When is it highest, and how high?

The same algebra as before gives the rate \(30 - 10t\). It is zero at \(t = 3\) s. Height: \(H(3) = 90 - 45 = 45\) m. Check the sides: at \(t = 2\) the rate is \(+10\) (rising), at \(t = 4\) it is \(-10\) (falling). So 45 m is the top.

2. The biggest field. A farmer has 40 m of fence for a rectangular pen. If one side is \(x\) m, the other is \(20 - x\), and the area is \(A = x(20 - x) = 20x - x^2\). What is the biggest area?

The average rate over \([x, x + h]\) is \(\dfrac{20h - 2xh - h^2}{h} = 20 - 2x - h\), which tends to \(20 - 2x\). That is zero at \(x = 10\): a 10 m by 10 m square, area 100 m². (A 9 by 11 pen gives only 99 m².) It is the same "equal sides win" fact as the spreading game.

3. A daily temperature. Temperatures on a summer day were 18 °C at 8:00, 24 at 11:00, 28 at 14:00, 27 at 15:00 and 23 at 18:00. Around when was it hottest, and what was the rate there?

The temperature was still rising fast between 11:00 and 14:00 and was already falling between 14:00 and 15:00, so the top came around 14:00. At the hottest moment, the rate of change of temperature was zero: it had stopped rising and not yet started to fall.

4. Steepest point from data. Using the sunflower of level 5 (8, 20, 56, 120, 164 cm on days 0, 10, 20, 30, 40), which ten days had the fastest growth?

Changes: 12, 36, 64, 44 cm. The fastest was from day 20 to day 30 (6.4 cm per day). Like the game, a sunflower's growth is an S-curve: slow, then fast, then slow as it reaches full height. Its point of inflection is somewhere in that stretch.

Try it

  1. From the game table, between which two days did the most students learn it? How many did on day 2 to 3?
  2. Show that \(450 \times 550\) is less than \(500 \times 500\), using \((500 + 50)(500 - 50)\).
  3. A ball's height is \(H(t) = 15t - 5t^2\) m. Its rate of change of height is \(15 - 10t\). When is it highest, and how high?
  4. A stone thrown upwards on Mars has height \(H(t) = 12t - 1.855t^2\) m, and its rate of change of height is \(12 - 3.71t\). When is it highest, and how high (to 1 decimal place)?
  5. 60 m of fence makes a rectangle. Find the biggest area. What shape is it?
  6. Find the rate of change of \(y = x^2 - 6x + 10\) by the shrinking-interval method, and the lowest point of its graph.
  7. The rate of change of \(y = x^3\) at \(x = 0\) is 0. Is \((0, 0)\) a highest point, a lowest point, or neither? Explain.
  8. For the spreading game, use rate \(\approx 0.000693 \times\) know \(\times\) don't know to estimate the rate when 200 students know it. Compare with the table.
Answers
  1. Days 5-6 and 6-7 (167 each). From day 2 to 3, 52 students learned it.
  2. \((500 + 50)(500 - 50) = 500^2 - 50^2 = 250\,000 - 2500 = 247\,500\), which is less than 250 000.
  3. \(15 - 10t = 0\) at \(t = 1.5\) s. Height: \(22.5 - 11.25 = 11.25\) m.
  4. \(12 - 3.71t = 0\) at \(t = 12 \div 3.71 \approx 3.23\) s. Height: \(12 \times 3.2345 - 1.855 \times 3.2345^2 \approx 38.81 - 19.41 \approx 19.4\) m. (On Earth the same throw would peak at only about 7.3 m.)
  5. Sides \(x\) and \(30 - x\); area \(30x - x^2\); rate \(30 - 2x = 0\) at \(x = 15\). A 15 m by 15 m square, 225 m².
  6. Average over \([x, x + h]\): \(\dfrac{(x + h)^2 - 6(x + h) + 10 - (x^2 - 6x + 10)}{h} = \dfrac{2xh + h^2 - 6h}{h} = 2x - 6 + h \to 2x - 6\). Zero at \(x = 3\), where \(y = 9 - 18 + 10 = 1\). Lowest point \((3, 1)\) (falling before, rising after).
  7. Neither. Just left of 0, \(x^3\) is negative; just right, it is positive. The curve goes up through \((0, 0)\), only levelling off for an instant.
  8. \(0.000693 \times 200 \times 800 \approx 111\) students per day. The table's central estimate at day 4 (where 200 know it) is \((333 - 111) \div 2 = 111\). They agree.

Common mistake

Mixing up "biggest value" and "biggest rate". At the top of the ball's flight the height is biggest and the rate is zero. On the S-curve, the rate is biggest at the inflection point, where the amount is only halfway. Always ask: the biggest what?

Believing every flat point is a top or a bottom. \(y = x^3\) is flat at 0 and is neither. Check the sign of the rate on both sides.

Where it leads