The chalk line joins the dots at 5 and 20 minutes. How does its slope compare with the slope of the tangent?

It is less steep than the tangent.

It is steeper than the tangent.

It has the same slope as the tangent.

The chalk line joins the dots at 5 and 10 minutes, and it slopes down. What does this say about the tea?

The temperature falls, so the estimate of the rate is negative.

The temperature rises, because the run is positive.

Topic 2.3

Estimate a derivative from nearby data

Builds on checking whether the defining limit of a derivative exists. 6 minutes.

A table of readings and no formula

The chalk dots show a table of the temperature of a cup of tea at five times. The curve through them shows how the tea cools in between. You want to know how fast the tea cools at the orange dot, the reading of 68 °C at 5 minutes.

Press a button to choose a second reading. The slope of the orange line through the two dots is the rise divided by the run. The red leg is the rise, the change in temperature, red because the temperature falls. The faint leg is the run, the minutes between the readings.

The reading at 20 minutes gives a slope of −2 °C per minute. The readings at 10, 2 and 0 minutes give −2.8, −4 and −4.4.

The slopes differ because the curve bends between the dots. Press Show the tangent. It touches the curve at the orange dot, and its slope is the derivative. On each side, the reading closer to 5 minutes gives a line closer to the tangent.

Two points on a curve

Point \(A\) is the orange point, and point \(B\) is a second point on the same curve. The red leg is the rise, red because the curve falls, and the faint leg is the run.

Drag \(B\) to either side of \(A\). The orange line through the two points turns, but its slope is always the change in output divided by the change in input.

If \(B\) is close to \(A\), the line is close to the faint orange tangent at \(A\), so its slope estimates the derivative at \(A\).

Notation

Choose the input \(a\) where you want the rate, and a reading at a nearby input \(b\).
The rise is the change in output. The run is the gap between the two inputs, not the number of table rows between them.
\[\begin{aligned} \text{rise} &= f(b)-f(a) \\ \text{run} &= b-a \end{aligned}\]
The derivative at \(a\) is about equal to the rise divided by the run.
\[f'(a) \approx \cy{\dfrac{f(b)-f(a)}{b-a}}\]
A negative answer means that \(f\) falls. The unit is the output unit per input unit.

When it fails

If the readings are far apart, the curve bends between them, so the line through the two points turns away from the tangent.

A table cannot show how the curve bends between its readings, so the estimate is good only if the readings are close together.

Example 1: How fast does the tea cool at 2 minutes?

The table shows 80 °C at 2 minutes and 68 °C at 5 minutes. Estimate how fast the tea cools at 2 minutes.

The rise is the change in temperature from the orange dot at 2 minutes to the dot at 5 minutes. It is red because the temperature falls.
\[\cr{68-80=-12}\]
The run is the gap between the readings, 3 minutes. It is not 1 minute, although the rows are neighbors.
\[\cq{5-2=3}\]
Divide the rise by the run. The slope is −4, so the tea cools at about 4 °C per minute at 2 minutes.
\[\cy{\dfrac{-12}{3} = -4}\]

Example 2: How fast does a tank fill?

A sensor reads 200 liters in a tank at 4 minutes and 380 liters at 10 minutes. Estimate how fast the tank fills at 4 minutes.

The rise is the change in volume, green because the volume grows.
\[\cg{380-200=180}\]
A tempting wrong move: The two readings come one after the other, so divide by 1.
\[\cr{\dfrac{180}{1} = 180}\]
A rate of 180 liters per minute adds 180 liters in each minute, but the tank gains 180 liters in 6 minutes. The red line with slope 180 climbs far above the dots.
The run is the 6 minutes between the readings, so the tank fills at about 30 liters per minute at 4 minutes.
\[\cy{\dfrac{180}{6} = 30}\]

Practice

A sapling is measured at weeks 0, 1, 3 and 6, and its heights are 20, 26, 35 and 41 cm. Estimate how fast it grows at week 1, using the readings at weeks 1 and 3.

Move the second reading to week 3. The rise is the change in height, and the run is the number of weeks between the two dots.

Solution

The rise is the change in height, green because the sapling grows. The run is the gap between the weeks.
\[\cg{35-26 = 9}, \qquad \cq{3-1 = 2}\]
The sapling grows at about 4.5 cm per week at week 1.
\[\cy{\dfrac{9}{2} = 4.5}\]

Questions

Questions 1 and 2 use the tea readings. The orange line is the tangent at 5 minutes. Question 3 needs a calculation.

The chalk line joins the dots at 5 and 20 minutes. How does its slope compare with the slope of the tangent?
- It is less steep than the tangent.
- It is steeper than the tangent.
- It has the same slope as the tangent.
Answer
It is less steep than the tangent.
The chalk line joins the dots at 5 and 10 minutes, and it slopes down. What does this say about the tea?
- The temperature falls, so the estimate of the rate is negative.
- The temperature rises, because the run is positive.
- The rate is 54 °C per minute.
Answer
The temperature falls, so the estimate of the rate is negative.
A cyclist is 40 m from the start at 10 seconds and 64 m from the start at 14 seconds. Estimate the speed at 10 seconds.
- 6 m/s
- 24 m/s
- 1.6 m/s
- about 0.17 m/s
Answer
6 m/s

Summary

You divide the change in output between two close readings by the real gap between their inputs.

This works because the line through two close points on a smooth curve has almost the slope of the tangent.

You use it when a table or a set of measurements gives data but no formula.

This is one of four skills in topic 2.3 of Which & Why: AP Calculus.