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Shrinking the interval

Change and motion · level 6, lesson 2 of 6

A question to start

A stone is dropped, and falls \(d = 4.9t^2\) metres in \(t\) seconds. In the last lesson, its average speeds over shorter and shorter stretches starting at \(t = 2\) were 24.5, then 22.05, then 20.09 m/s. They were heading somewhere.

Where exactly? Will they ever get there? And here is the surprising part: with a little algebra we can find the exact answer without working out a single extra average, and without ever dividing by zero.

The idea

A name for the small stretch

Let us call the length of the stretch of time \(h\) (any small positive number, like 0.1 or 0.001). The stretch goes from \(t = 2\) to \(t = 2 + h\). Then

\[\text{average speed} = \frac{d(2 + h) - d(2)}{h}.\]

Here \(d(2 + h)\) means "the distance fallen at time \(2 + h\)", and \(d(2) = 4.9 \times 4 = 19.6\) m.

The word for a stretch of numbers from one value to another is an interval. The interval from 2 to 2.1 is written \([2, 2.1]\).

Squeezing from both sides

Now make \(h\) smaller and smaller. And for good measure, also look backwards, at intervals that end at 2: \([2 - h, 2]\).

\(h\) (s) Average over \([2, 2 + h]\) (m/s) Average over \([2 - h, 2]\) (m/s)
1 24.5 14.7
0.5 22.05 17.15
0.1 20.09 19.11
0.01 19.649 19.551
0.001 19.6049 19.5951
0.0001 19.60049 19.59951

The averages after 2 are all a bit more than 19.6; the averages before 2 are all a bit less. The stone is speeding up, so the later stretch is always faster. As \(h\) shrinks, both columns crowd in on the same number from opposite sides.

141618202224 19.6 after 2: from the rightbefore 2: from the left

Do the averages ever equal 19.6? Not for any \(h\) that is bigger than zero: the "after" averages are always a little above and the "before" ones a little below. But they get as close as we like. Want them within 0.001 of 19.6? Take \(h = 0.0001\). Within a millionth? Take \(h\) smaller still. No other number has this property: any other number, say 19.59, is eventually left behind by both columns.

The algebra: finding the exact answer

Tables suggest the answer; algebra proves it. We need \(d(2 + h) = 4.9(2 + h)^2\). What is \((2 + h)^2\)? A picture helps: a square of side \(2 + h\) splits into a 2-by-2 square, two strips 2 by \(h\), and a tiny \(h\)-by-\(h\) square.

2 × 2 = 42h2hh² 2h 2h

So \((2 + h)^2 = 4 + 2h + 2h + h^2 = 4 + 4h + h^2\). Then

\[d(2 + h) - d(2) = 4.9(4 + 4h + h^2) - 4.9 \times 4 = 4.9(4h + h^2).\]

Divide by \(h\). Since \(h\) is not zero (just small), this is allowed, and each term loses one \(h\):

\[\frac{4.9(4h + h^2)}{h} = 4.9(4 + h) = 19.6 + 4.9h.\]

That is a formula for every average in the table at once. Check: \(h = 0.1\) gives \(19.6 + 0.49 = 20.09\). ✓ And \(h = 0.01\) gives \(19.6 + 0.049 = 19.649\). ✓

Now the question "where are the averages heading?" is easy. As \(h\) shrinks towards 0, the extra bit \(4.9h\) shrinks towards 0 too, so the averages head to exactly 19.6. (For the "before" column, the same algebra with \(-h\) gives \(19.6 - 4.9h\), which heads to the same place.)

The stone's speed at the instant \(t = 2\) is exactly 19.6 m/s.

The word "limit"

The number that the averages approach as \(h\) gets closer and closer to 0 is called the limit of the averages. Mathematicians write

\[\lim_{h \to 0} \,(19.6 + 4.9h) = 19.6,\]

read "the limit, as \(h\) tends to 0, of 19.6 plus 4.9\(h\), is 19.6". The arrow \(\to\) means "gets closer and closer to". The key point: we never put \(h = 0\) into the fraction \(\dfrac{d(2+h) - d(2)}{h}\), which would be \(0 \div 0\). We first simplify it to \(19.6 + 4.9h\), a formula with no division by \(h\), and only then let \(h\) head to 0. Limits are studied properly in level 7, and their exact definition (how to make "as close as we like" airtight) is in Infinity.

Every instant at once

Nothing was special about \(t = 2\). At any time \(t\), the same algebra (with \(t\) in place of 2) gives

\[\frac{4.9(t + h)^2 - 4.9t^2}{h} = \frac{4.9(2th + h^2)}{h} = 9.8t + 4.9h \;\longrightarrow\; 9.8t.\]

So the falling stone's speed at time \(t\) is \(9.8t\) m/s:

Time (s) 0 1 2 3 4
Speed at that instant (m/s) 0 9.8 19.6 29.4 39.2

It gains 9.8 m/s every second, exactly the acceleration of gravity from level 5. Two completely different routes, one by thinking about acceleration and one by shrinking intervals, agree. On Mars, where \(d \approx 1.855t^2\), the same steps give a speed of \(3.71t\) m/s.

Worked examples

1. The stone at 3 s. Find the exact speed at \(t = 3\).

  1. \((3 + h)^2 = 9 + 6h + h^2\).
  2. \(d(3 + h) - d(3) = 4.9(9 + 6h + h^2) - 44.1 = 4.9(6h + h^2)\).
  3. Divide by \(h\): \(4.9(6 + h) = 29.4 + 4.9h\).
  4. As \(h \to 0\), this tends to 29.4 m/s. (Check with a table: \(h = 0.1\) gives 29.89, \(h = 0.01\) gives 29.449.)

2. A thrown ball. A ball's height is \(H(t) = 20t - 5t^2\) m. Find its upward speed at \(t = 1\).

  1. \(H(1) = 15\). \(H(1 + h) = 20 + 20h - 5(1 + 2h + h^2) = 15 + 10h - 5h^2\).
  2. Change: \(10h - 5h^2\). Divide by \(h\): \(10 - 5h\).
  3. As \(h \to 0\): 10 m/s upwards. (The table: \(h = 0.1\) gives 9.5, \(h = 0.01\) gives 9.95.)

3. A growing square. A square's side \(s\) grows. How fast does its area \(A = s^2\) grow, per cm of side, at the moment when \(s = 5\) cm?

\(\dfrac{(5 + h)^2 - 25}{h} = \dfrac{10h + h^2}{h} = 10 + h \to 10\). The area grows at 10 cm² per cm of side. The picture says why: growing a 5 cm square by a thin \(h\) adds two strips of \(5 \times h\), plus a crumb \(h^2\) that hardly matters.

4. A cube. For \(f(x) = x^3\), find the rate of change at \(x = 2\).

\((2 + h)^3 = 8 + 12h + 6h^2 + h^3\) (multiply \((4 + 4h + h^2)\) by \((2 + h)\)). So \(\dfrac{f(2 + h) - f(2)}{h} = 12 + 6h + h^2 \to\) 12. (Table: \(h = 0.1\) gives 12.61, \(h = 0.01\) gives 12.0601.)

Try it

  1. Use \(19.6 + 4.9h\) to find the average speed of the stone over \([2, 2.2]\), and check it the long way.
  2. How small must \(h\) be for the average over \([2, 2 + h]\) to be within 0.01 m/s of 19.6?
  3. Find the stone's exact speed at \(t = 1\) by the algebra method. Does it fit \(9.8t\)?
  4. On Mars, \(d = 1.855t^2\). Find the average speed over \([2, 2 + h]\) as a formula, and the speed at \(t = 2\).
  5. For the ball \(H(t) = 20t - 5t^2\), find its upward speed at \(t = 3\). What does the sign tell you?
  6. For \(f(x) = x^2\), find the rate of change at \(x = 4\) and at \(x = -1\).
  7. For \(f(x) = 3x + 1\), find \(\dfrac{f(2 + h) - f(2)}{h}\). Why does \(h\) disappear completely?
  8. For \(f(x) = x^3\), show that the average rate of change over \([1, 1 + h]\) is \(3 + 3h + h^2\), and find the rate at \(x = 1\).
Answers
  1. \(19.6 + 4.9 \times 0.2 = 20.58\) m/s. Long way: \(d(2.2) = 4.9 \times 4.84 = 23.716\); \((23.716 - 19.6) \div 0.2 = 4.116 \div 0.2 = 20.58\). ✓
  2. Need \(4.9h < 0.01\), so \(h < 0.01 \div 4.9 \approx 0.00204\) s. Any \(h\) smaller than about 0.002 s works.
  3. \(\dfrac{4.9(1 + h)^2 - 4.9}{h} = \dfrac{4.9(2h + h^2)}{h} = 9.8 + 4.9h \to 9.8\) m/s. Yes: \(9.8 \times 1 = 9.8\).
  4. \(\dfrac{1.855(4 + 4h + h^2) - 1.855 \times 4}{h} = 1.855(4 + h) = 7.42 + 1.855h\). The speed at \(t = 2\) is 7.42 m/s (\(= 3.71 \times 2\)).
  5. \(H(3) = 60 - 45 = 15\). \(H(3 + h) = 60 + 20h - 5(9 + 6h + h^2) = 15 - 10h - 5h^2\). Change ÷ \(h\) \(= -10 - 5h \to -10\) m/s. Negative: the ball is coming down at 10 m/s.
  6. At 4: \(\dfrac{(4 + h)^2 - 16}{h} = 8 + h \to 8\). At \(-1\): \(\dfrac{(-1 + h)^2 - 1}{h} = \dfrac{-2h + h^2}{h} = -2 + h \to -2\).
  7. \(\dfrac{3(2 + h) + 1 - 7}{h} = \dfrac{3h}{h} = 3\). A straight line has the same slope everywhere, so every average is already exactly 3: there is nothing left to shrink.
  8. \((1 + h)^3 = 1 + 3h + 3h^2 + h^3\), so the change is \(3h + 3h^2 + h^3\); divided by \(h\): \(3 + 3h + h^2\). As \(h \to 0\) it tends to 3.

Common mistake

Putting \(h = 0\) before simplifying. \(\dfrac{4.9(2 + h)^2 - 19.6}{h}\) with \(h = 0\) is \(\dfrac{0}{0}\), which tells you nothing. Always simplify first, until no \(h\) is left on the bottom; then let \(h\) head to 0.

Thinking the averages must reach the limit. For every real stretch (\(h > 0\)) the average is not 19.6. The limit is the number they approach, not a value they land on. That is exactly what makes it a new idea.

Expanding \((2 + h)^2\) as \(4 + h^2\). The square picture shows two strips of \(2h\) that must not be forgotten: \((2 + h)^2 = 4 + 4h + h^2\).

Where it leads