Topic 2.3

Choose a left, right, or centered estimate

Builds on estimating a derivative from two nearby readings. 7 minutes.

Which reading to use at 2 km

A table gives the height of a hiking trail every kilometer, and the chalk dots show the readings. The orange dot is the reading at 2 km, and the faint orange line touches the curve there.

Its slope is the slope of the trail at 2 km. You estimate it from the reading before 2 km, the reading after it, or both.

Press the first button. The line joins the dots at 1 km and 2 km and rises 28 m over 1 km. The left estimate is therefore 28 meters per kilometer, and the rise is green because it is positive.

The second button joins the dots at 2 km and 3 km, so the right estimate is 22. The third button joins the dots at 1 km and 3 km. Its run is 2 km, so the centered estimate is 50 divided by 2, which is 25.

The trail is less steep after 2 km than before it, so the left estimate is the largest and the right the smallest.

Three lines around one point

The orange point \(A\) is on the curve at the input you want. \(L\) and \(R\) are the readings before and after \(A\), at equal gaps.

The lines through \(L\) and \(A\), through \(A\) and \(R\), and through \(L\) and \(R\) give the left, right and centered estimates.

The centered line rises as much as the other two together, over twice the run. Its slope is therefore the average of their slopes, so it lies between them.

Make the gap smaller. The three lines turn toward the faint orange tangent at A.

Notation

  1. Let \(h\) be the gap between neighboring readings. The left estimate uses the reading before \(\cy{a}\), at \(a-h\).

    \[\dfrac{f(a)-f(a-h)}{h}\]
  2. The right estimate uses the reading after \(a\), at \(a+h\).

    \[\dfrac{f(a+h)-f(a)}{h}\]
  3. The centered estimate uses both. Its run is \(2h\), the gap between the outer readings.

    \[\cy{\dfrac{f(a+h)-f(a-h)}{2h}}\]
  4. If the question names a side, use only the readings on that side. If it names none, use the centered estimate.

When it fails

At the last reading there is no reading after it, so only the left estimate exists. At the first reading, only the right estimate exists.

A centered estimate needs a reading on each side, so it does not exist at either end.

Example 1: The speed of a car at 1 hour

A car is 30 km from its start at 0.5 h, 70 km at 1 h and 120 km at 1.5 h. Estimate its speed at 1 h, using readings on both sides.

  1. The readings at 0.5 h and 1.5 h lie on each side of 1 h, so use the centered estimate.

  2. The rise is the change in distance between these outer readings, green because it is positive.

    \[\cg{120-30 = 90}\]
  3. The run is 1 hour, so the speed is about 90 km/h.

    \[\cy{\dfrac{90}{1} = 90}\]
  4. As a check, the left and right estimates are 80 km/h and 100 km/h, and their average is 90.

Example 2: A medicine in the blood

A medicine has a concentration of 7 mg/L at 2 h, 5 mg/L at 4 h and 4 mg/L at 6 h. Estimate the rate of change at 4 h, using readings on both sides.

  1. The rise between the outer readings is the change in concentration, red because it is negative.

    \[\cr{4-7 = -3}\]
  2. A tempting wrong move: The readings are 2 hours apart, so divide by 2.

    \[\cr{\dfrac{-3}{2} = -1.5}\]

    The line joins the outer readings, which are 4 hours apart. The red line of slope −1.5 is steeper than both faint lines, but the centered slope lies between their slopes, −1 and −0.5.

  3. The run is 4 hours, so the rate is about −0.75 mg/L per hour. The negative sign means the concentration falls.

    \[\cy{\dfrac{-3}{4} = -0.75}\]

Practice

The depth of snow is 2 cm at 0 h, 8 cm at 2 h, 12 cm at 4 h and 14 cm at 6 h. Find the three estimates of the rate at 4 h.

The centered line joins the dots at 2 h and 6 h, so its run is 4 hours.

Solution
  1. The left estimate uses the readings at 2 h and 4 h, and the right estimate uses those at 4 h and 6 h.

    \[\dfrac{12-8}{2} = 2, \qquad \dfrac{14-12}{2} = 1\]
  2. The centered estimate uses the readings at 2 h and 6 h. It is 1.5 cm per hour, the average of 2 and 1.

    \[\cy{\dfrac{14-8}{4} = 1.5}\]

Questions

Questions 1 and 2 use the trail readings. Question 3 needs a calculation.

  1. The orange dot is at 2 km. Which of the three lines has the largest slope?

    • The left line, through the dots at 1 km and 2 km.
    • The right line, through the dots at 2 km and 3 km.
    • The centered line, through the dots at 1 km and 3 km.
    Answer

    The left line, through the dots at 1 km and 2 km.

  2. The orange dot is at 4 km, the last reading. Which estimates can you find from the table?

    • The left estimate only.
    • The left and centered estimates.
    • The right estimate only.
    • All three estimates.
    Answer

    The left estimate only.

  3. A table gives \(g(8) = 31\), \(g(10) = 40\) and \(g(12) = 46\). Find the centered estimate of \(g'(10)\).

    • 3.75
    • 7.5
    • 4.5
    • 3
    Answer

    3.75

Summary

You use the readings on the side that the question names, or both sides for a centered estimate.

Each estimate is the slope of a line through two readings, and with equal gaps the centered slope is the average of the other two.

A centered estimate needs a reading on each side, so at the end of a table you use a one-sided estimate.

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