Recognize the limits of an estimate
Slopes that come close but are never exact
The chalk dots show a table of values of a function f near x = 2, and the orange dot is the reading (2, 8).
Press the buttons to move the second reading closer to x = 2. The slope of the chalk line falls from 19 to 15.25, 12.61 and 12.0601, and the line turns toward the faint orange tangent.
The slopes come closer and closer to 12, so the derivative at 2 is about 12.
A table has only finitely many readings and cannot show what happens between them, so it suggests a value but does not prove one.
An estimate and its error
The slope of the chalk line through A and B is an estimate, and the slope of the orange tangent at A is the derivative.
The red bracket shows how far apart the two lines are at B, so it measures the error of the estimate.
Drag B toward A, and the error becomes smaller. Bend the curve more, and the error becomes larger.
The error depends on the distance between the readings and on how much the curve bends between them. A table does not show the bend.
Notation
Write an estimate with the word about and with its units.
\[f'(a) \approx \cy{m}\]The left slope uses the reading before \(a\), and the right slope uses the reading after it. If the graph bends one way between them, the tangent slope lies between the two.
\[f'(a) \ \text{is between} \ \cy{m_{\text{left}}} \ \text{and} \ \cy{m_{\text{right}}}\]State your assumption: \(f\) does not change sharply between the readings. Use the estimate only near \(a\).
An exact derivative needs a formula for \(f\), or the limit of the slopes as the gap between readings goes to 0.
\[f'(a) = \lim_{h\to 0}\dfrac{f(a+h)-f(a)}{h}\]
When it fails
A mass on a spring is read once each second. Every reading is 0 cm, so every slope between readings is 0, but the mass rises and falls in between.
A reading at 0.5 seconds is 1 cm, so the slope from 0.5 to 1 seconds is −2 cm/s. Readings must be close together to show the fastest change.
Example 1: How many digits can you report?
For the function in the opening, the slope over a gap of −0.1 from x = 2 is 11.41, and the slope over a gap of 0.1 is 12.61. Report the derivative at 2 from these two slopes.
If the curve bends one way near 2, as in the figure, the tangent slope lies between the two slopes.
\[11.41 < f'(2) < 12.61\]The two slopes differ by 1.2, so the data fix the derivative only to about one unit. Their average, 12.01, has more digits than the data support.
Report the derivative as about 12.
\[f'(2) \approx \cy{12}\]
Example 2: Predicting too far
A drone is 4 m from its base at 2 s, 9 m at 3 s and 16 m at 4 s. Use its speed at 3 s to predict its distance at 13 s. A later reading shows 169 m.
The centered estimate of the speed at 3 s uses the readings at 2 s and 4 s.
\[\cy{\dfrac{16-4}{4-2} = 6}\]A tempting wrong move: The drone moves at 6 m/s for 10 more seconds. It covers 60 m, so it is at 69 m at 13 s.
\[\cr{9 + 6 \cdot 10 = 69}\]The reading at 13 s is 169 m, not 69 m, because the speed does not stay at 6 m/s. The orange tangent follows the curve only near 3 s, and the red bracket shows an error of 100 m.
Use the estimate for a small step. For one more second, the prediction is 9 + 6 = 15 m. The reading at 4 s is 16 m, so the error is 1 m.
Practice
A sensor reads a pressure of 102 kPa at 8 s, 110 kPa at 12 s and 114 kPa at 16 s. Give a range for the rate at 12 s, give the centered estimate, and state one assumption.
The lines before and after 12 s give the ends of the range, and the centered line spans 8 s.
Solution
The left slope and the right slope give the ends of the range.
\[\begin{aligned} \dfrac{110-102}{4} &= 2 \\ \dfrac{114-110}{4} &= 1 \end{aligned}\]The rate at 12 s is between 1 and 2 kPa per second, and the centered estimate is 1.5 kPa per second.
\[\cy{\dfrac{114-102}{8} = 1.5}\]This assumes that the pressure does not change sharply between the readings.
Questions
Questions 1 and 2 use the position of a mass on a spring. Question 3 needs a calculation.
The dots at 0 s and 1 s are both 0 cm, so the line through them is flat. What does the curve show about the slope at 0 s?
- The curve rises steeply at 0 s, so the flat line is a poor estimate.
- The slope is 0, because both readings are 0 cm.
- The slope cannot be estimated with any line.
Answer
The curve rises steeply at 0 s, so the flat line is a poor estimate.
A reading at 0.5 s is 1 cm. Which line gives the better estimate of the slope at 0 s?
- The line through the dots at 0 s and 0.5 s.
- The line through the dots at 0 s and 1 s.
- Both lines are equally good, because both dots are on the curve.
Answer
The line through the dots at 0 s and 0.5 s.
A table gives \(h(2) = 10\), \(h(3) = 16\) and \(h(4) = 20\). Which statement does the table support?
- \(h'(3)\) is between 4 and 6, and about 5 if \(h\) changes smoothly.
- \(h'(3)\) is exactly 5.
- \(h'(3)\) is exactly 6.
- \(h'(3)\) is 4, because the table ends at 4.
Answer
\(h'(3)\) is between 4 and 6, and about 5 if \(h\) changes smoothly.
Summary
You report a derivative from data with the word about, its units, its assumption and, with readings on both sides, a range.
This is needed because a table cannot show how the curve bends between its readings.
An exact derivative needs a formula for the function or the limit of the slopes.
This is one of four skills in topic 2.3 of Which & Why: AP Calculus.