Demonstration 1 of 4
An average rate needs two times
Between 2 seconds and another time, how fast was the position changing on average?
The two points on the curve fix a secant line, and its slope is the average rate. Everything that happened in between is squeezed into that one number.
\[[f(b) - f(a)]/(b - a)\]
\[\text{slope} = \text{vertical change} / \text{horizontal change}\]
a and b are two input times, here a = 2 seconds. f(b) - f(a) is the change in position in meters and b - a the change in time in seconds, so the quotient is in meters per second.
Predict first. Set the second time to 2 as well. What happens to the average rate?
Choose an example
Constructed example: the chapter's position model with positions 4 meters at 2 seconds and 9 meters at 3 seconds.
Calculated values
- Interval
- from 2 to 3 seconds
- Change in position
- 9 - 4 = 5 m
- Change in time
- 3 - 2 = 1 s
- Average rate
- 5 meters per second
[s(3) - s(2)]/(3 - 2) = (9 - 4)/1 = 5/1 = 5 meters per second. This one number summarizes the whole interval.
Use the idea
When someone reports an average rate, ask for the two endpoints. A different interval can give a different rate for the same motion.
Where the conclusion applies
The constructed model s = t^2 with fixed units. Reversing both changes keeps the rate; reversing only one flips its sign. Equal endpoints leave a zero denominator and no rate.
Check your understanding: What is the average rate from t = 2 to t = 4?
Chapter 28 source: section "Average rate uses two inputs". Demonstration C28-D01.
Demonstration 2 of 4
Shrink the interval without making it zero
What do the average rates near t = 2 do as the interval gets shorter from either side?
For every nonzero h the quotient simplifies to 4 + h, so the rates lie on a straight line with a hole at h = 0. From both sides they head toward 4.
\[[s(2 + h) - s(2)]/h\]
\[= h(4 + h)/h\]
\[= 4 + h\]
h is the interval length in seconds and is never 0. Positive h looks later than t = 2; negative h looks earlier. The quotient is the average rate over that interval.
Predict first. Switch h from 0.1 to -0.1. Will the rate land above 4 or below it?
Choose an example
Constructed example: the chapter's values h = 0.5, 0.1 and -0.1, plus h = -0.5.
Calculated values
- Second time 2 + h
- 2 + 0.1 = 2.1 seconds
- s(2 + h)
- 2.1^2 = 4.41
- Difference quotient
- (4.41 - 4)/0.1 = 4.1
- Simplified rule
- 4 + 0.1 = 4.1
- Rate approached as h shrinks
- 4 meters per second
With h = 0.1: [s(2.1) - s(2)]/0.1 = (4.41 - 4)/0.1 = 0.41/0.1 = 4.1. The simplified rule agrees: 4 + 0.1 = 4.1. The cancellation h(4 + h)/h = 4 + h is allowed because h is not 0.
Use the idea
When a calculation needs a rate at an instant, keep the interval nonzero, simplify, and only then ask what the result approaches.
Where the conclusion applies
The model s = t^2 near t = 2. Cancelling h is legal only because h is not 0; substituting h = 0 into the original quotient gives 0/0, which is not a rate.
Check your understanding: Using 4 + h, what is the average rate from t = 2 to t = 2.01?
Chapter 28 source: section "Shrink the interval without making it zero". Demonstration C28-D02.
Demonstration 3 of 4
Nearby secants approach a direction
As the second point slides toward (2, 4), what happens to the secant and to the tangent estimate?
The base point (2, 4) stays fixed while the second point moves in. The dashed secant turns toward the dotted line of slope 4, the limiting direction.
\[s = t^2\]
\[4(0.1) = 0.4\]
The secant joins (2, 4) to (2 + h, (2 + h)^2). Its slope is 4 + h. The tangent estimate of the position change is 4 times h; the discrepancy is how far that estimate is off.
Predict first. Shrink h from 0.1 to 0.01. How big is the discrepancy now?
Choose an example
Constructed example: the chapter's table of intervals from 2 to 3, 2.5, 2.1 and 2.01.
Calculated values
- Interval length h
- 0.1 s
- Secant slope 4 + h
- 4 + 0.1 = 4.1 m/s
- Exact change
- 2.1^2 - 4 = 0.41 m
- Tangent estimate 4h
- 4(0.1) = 0.4 m
- Discrepancy
- 0.41 - 0.4 = 0.01 m
Over h = 0.1 s, the exact change is 2.1^2 - 4 = 0.41 m and the tangent estimate is 4(0.1) = 0.4 m, a discrepancy of 0.01 m. The secant slope 4.1 sits 0.1 above 4; shrink h and both the slope gap and the discrepancy shrink.
Use the idea
Near a point where a rate exists, change is roughly the rate times the small step. The word roughly is honest only for small steps.
Where the conclusion applies
The model s = t^2 near t = 2. The local estimate 4h is good over small h, not over every interval: at h = 1 it is off by a whole meter.
Check your understanding: At h = 0.5, what are the exact change, the tangent estimate, and the discrepancy?
Chapter 28 source: section "Nearby secants approach a direction". Demonstration C28-D03.
Demonstration 4 of 4
A corner has no single rate
At the corner of |x|, do the slopes from the left and from the right agree?
On the right the graph is the line y = x, on the left the line y = -x. Every right-hand quotient is 1 and every left-hand quotient is -1, however small h gets.
\[[g(h) - g(0)]/h = |h|/h = 1\]
\[|h|/h = -1\]
g(x) = |x| is the distance of x from 0. h is a nonzero step from 0; |h|/h is the slope of the secant from (0, 0) to (h, |h|).
Predict first. Switch h from 0.01 to -0.01. Does the secant slope move a little or flip completely?
Choose an example
Constructed example: the chapter's absolute value function at 0, with steps chosen for the reader.
Calculated values
- Step h
- 0.01
- g(h) - g(0)
- |0.01| - 0 = 0.01
- Quotient |h|/h
- 0.01/0.01 = 1
- Right-hand quotients
- 1 for every positive h
- Left-hand quotients
- -1 for every negative h
- g'(0)
- undefined (the sides approach 1 and -1)
With h = 0.01: [g(0.01) - g(0)]/0.01 = 0.01/0.01 = 1. Every right-hand quotient gives 1, however small h is. The other side gives -1. The sides disagree, so g'(0) does not exist, even though |x| is continuous at 0.
Use the idea
When a graph has a sharp corner, check both one-sided rates before naming a single rate.
Where the conclusion applies
Only the example g(x) = |x| at 0. The function is continuous there; it is the derivative that fails, because the two sides approach different numbers.
Check your understanding: For g(x) = |x|, what is the quotient [g(h) - g(0)]/h at h = -0.2?
Chapter 28 source: section "Derivatives can fail to exist". Demonstration C28-D04.