Demonstration 1 of 4
Rate times duration, block by block
What does a block below the baseline do to the running total?
Each block contributes height times width, keeping its sign. The right panel adds those contributions to the starting 10, one block at a time.
\[6 + (-1) = 5\ \text{units}\]
\[10 + 5 = 15\ \text{units}\]
The rate is in units per hour, the width in hours, so each rectangle's height times width is a change in units. The starting amount is 10 units.
Predict first. Make the second rate -3 instead of -1. Where does the running total end?
Choose an example
Constructed example: the chapter's +3 and -1 blocks starting from 10 units, plus other second rates.
Calculated values
- First change
- 3 x 2 = 6 units
- Second change
- (-1) x 1 = -1 unit
- Net change
- 6 + (-1) = 5 units
- Ending amount
- 10 + 5 = 15 units
First block: 3 units per hour for 2 hours, 3 x 2 = 6 units. Second block: -1 unit per hour for 1 hour, (-1) x 1 = -1. Net change 6 + (-1) = 5 units, and from a start of 10 the ending amount is 10 + 5 = 15 units. The block below the baseline pulls the running total down.
Use the idea
Before adding a rate to an amount, multiply it by a duration. A rate and an amount have different units and cannot be added directly.
Where the conclusion applies
The rate is declared constant on each block, which is what makes each rectangle exact. If the rate varied inside a block, one rectangle would only be an estimate.
Check your understanding: If the second rate were -2 units per hour for 1 hour, what would the ending amount be?
Chapter 31 source: section "Constant-rate intervals give exact rectangles". Demonstration C31-D01.
Demonstration 2 of 4
Net change or total magnitude?
The same two pieces, +6 and -1, give 5 or 7. Which one answers your question?
For net change the second arrow points back. For total magnitude it points forward, because the sign has been dropped on purpose.
\[6 - 1 = 5\]
\[6 + 1 = 7\]
+6 is the contribution above the baseline. The second piece lies below it. The bars | | mean magnitude: the size without the sign.
Predict first. Keep the question on net change and make the second piece -6. What is the net change?
Choose an example
Constructed example: the chapter's +6 and -1 pieces, plus larger negative pieces.
Calculated values
- Question
- net change (signed)
- Pieces
- +6 and -1
- Net change
- 6 - 1 = 5 units
- Total magnitude
- 6 + 1 = 7 units
- Answer to this question
- 5 units
6 - 1 = 5 units. Signs kept: the second piece subtracts. Both numbers are correct answers to different questions.
Use the idea
Decide first whether you want where things ended up (net) or how much happened in all (magnitude). Then add with or without signs.
Where the conclusion applies
Two pieces with known signed contributions. Large pieces of opposite sign can nearly cancel, so a small net change can hide a lot of activity.
Check your understanding: With pieces +6 and -3, what are the net change and the total magnitude?
Chapter 31 source: section "Below the baseline means negative contribution". Demonstration C31-D02.
Demonstration 3 of 4
Where you sample matters
For the rising rate r(t) = t, why do left and right rectangles disagree?
Each rectangle's height is the rate at its sample time. On a rising line, left samples sit low and right samples sit high; more rectangles shrink the gap.
\[r(t) = t\]
\[\text{sum of }r(t_i^*)\,\Delta t_i.\]
r(t) is the rate in units per hour at time t, from 0 to 2 hours. The starred t in piece i is the sample time chosen there, and the Delta t beside it is that piece's width in hours.
Predict first. With 2 rectangles, which rule gives 1, which gives 2, and which gives 3?
Choose an example
Constructed example: the chapter's r(t) = t on 0 to 2 hours, with 2 and 4 rectangles.
Calculated values
- Sample rule
- midpoint
- Rectangles
- 2, each 1 hour wide
- Sample heights
- 0.5, 1.5
- Estimate
- 2 units
- Triangle under the line
- 2 x 2 / 2 = 2 units
Estimate = 0.5 + 1.5 = 2 units, which is equal to the triangle area 2 x 2 / 2 = 2 under the line. Each rectangle is a sample height times its width. For a straight line the midpoint rectangles happen to match exactly.
Use the idea
When you estimate a total from samples, say both the partition and the sampling rule. Different rules give different estimates.
Where the conclusion applies
A rising straight-line rate. The left-low, right-high pattern needs an increasing rate; a falling or wavy rate can order the estimates differently.
Check your understanding: With 4 rectangles and right endpoints, what are the heights and the estimate?
Chapter 31 source: section "Changing rates need representative rectangles". Demonstration C31-D03.
Demonstration 4 of 4
Unequal widths, each in its own place
When the pieces have different widths, what must each rate be multiplied by?
Each rectangle is labelled rate x width. The widths differ, so there is no single width to factor out; every rate keeps its own.
\[r(t)=2t\]
\[2.5(0.5)+4.5(1.5)=1.25+6.75=8\]
r(t) = 2t is the rate in units per hour from 1 to 3 hours. The book partition 1, 1.5, 3 gives widths 0.5 and 1.5 hours.
Predict first. Switch the book partition to left endpoints. Will the sum be above or below 8?
Choose an example
Constructed example: the chapter's worked example on 1 to 3 hours, plus an even partition.
Calculated values
- Partition
- 1, 1.5, 3
- Widths (hours)
- 0.5 and 1.5
- Sample rates
- 2.5 and 4.5
- Sum
- 8 units
- Exact accumulation
- average height 4 x 2 hours = 8 units
Midpoint sum: 2.5(0.5) + 4.5(1.5) = 1.25 + 6.75 = 8 units. It matches the exact 8, because each midpoint rectangle equals its trapezoid for a straight line.
Use the idea
Data often comes at uneven spacing. Multiply each reading by its own interval before adding.
Where the conclusion applies
A straight-line rate, which is why the midpoint sum is exact here. For a curved rate the same recipe is only an estimate.
Check your understanding: On the even partition 1, 2, 3 with right endpoints, what is the sum?
Chapter 31 source: section "Worked example: unequal widths still belong in the sum". Demonstration C31-D04.