Wemazi

Topic 4.6

Use supplied values or a table

Builds on predicting whether a tangent-line estimate is high or low. 7 minutes.

Tea cooling in a table

The dots show a table of the temperature \(T\) of a cup of tea, in degrees Celsius, at five times. There is no formula for \(T\).

To use the tangent line at \(t = 10\), you need the value \(T(10) = 65\), which is the orange dot, and the slope \(T'(10)\). The table gives the value but not the slope.

Press the first button to draw the line through the dots at 8 and 12, the nearest times on each side of 10. A line through two points like this is a secant, and its slope is \(\dfrac{61 - 72}{12 - 8} = -2.75\) degrees each minute.

Press the second button to draw the line with this slope through the orange dot. At \(t = 10.5\) the line gives \(65 + (-2.75)(0.5) = 63.625\) degrees.

A value and a slope

The approximation needs two numbers: the value at a and the slope at a. A formula is one way to find them, but not the only way.

A problem can give the slope directly. Press the first button to see the line with the given slope.

If the problem gives only values, divide the change in the output by the change in the input between two points close to a. Press the second button to see this case.

Notation

  1. Find the value at the center a, where the line touches. The problem gives it, or you read it from the table.

    \[\cy{f(a)} = T(10) = 65\]

  2. Find the slope. If the problem gives the slope, use it. If not, divide the change in the values by the change in the inputs between the nearest data points on each side.

    \[\cy{f'(a)} \approx \frac{T(12) - T(8)}{12 - 8} = -2.75\]

  3. Write the line, where the offset \(t - 10\) is the distance from the center to the target. Then substitute the target.

    \[T(t) \approx T(10) + T'(10)\,(t - 10)\]

    The approximation is useful only when the target is near the center.

When it fails

The slope is only as good as the data near \(a\). If you use the far dots at 6 and 14 in place of 8 and 12, the slope becomes \(\dfrac{59 - 84}{14 - 6} = -3.125\).

This is the average slope over a long stretch, not the slope at 10. In the picture, the new line is steeper, so it falls further below the dots at 12 and 14.

Example 1: Values given

A function \(g\) has \(g(5) = 12\) and \(g'(5) = -3\). Use the tangent line at \(a = 5\) to estimate \(g(5.2)\).

  1. The problem gives the value and the slope, so no formula for g is needed.

    \[g(5) = 12, \qquad g'(5) = -3\]

  2. The offset is \(5.2 - 5 = 0.2\). The slope is negative, so the change is negative and the red segment points down.

    \[\Delta g \approx (-3)(0.2) = -0.6\]

  3. Add the change to the value.

    \[g(5.2) \approx 12 - 0.6 = \cy{11.4}\]

Example 2: A table of speeds

The table gives the speed \(v\) of a cyclist, in m/s, at times \(t\) in seconds: \(v(0) = 0\), \(v(2) = 6\), \(v(4) = 9\), \(v(6) = 10\), \(v(8) = 10\). Estimate \(v(4.5)\) with a tangent line at \(t = 4\).

  1. The value at the center is 9. The slope comes from the nearest dots on each side, at 2 and 6, and the speed changes by 4 between them.

    \[v(6) - v(2) = 10 - 6 = 4\]

  2. A tempting wrong move: Use 4 as the slope.

    \[v(4.5) \approx 9 + 4(0.5) = 11\]

    The speed rises only from 9 to 10 between \(t = 4\) and \(t = 6\), so 11 at \(t = 4.5\) is too high. In the picture, the red line climbs above the dots at 6 and 8.

  3. A slope is a rate, so divide the change in speed by the change in time.

    \[v'(4) \approx \frac{10 - 6}{6 - 2} = 1 \ \text{m/s}^2\]

  4. The offset is 4.5 − 4 = 0.5.

    \[v(4.5) \approx 9 + 1(0.5) = \cy{9.5}\ \text{m/s}\]

Practice

A function \(h\) has \(h(20) = 40\) and \(h(20.01) = 40.07\). Estimate \(h(20.3)\) with a tangent line at \(a = 20\).

The two dots are very close together, so their rise over run is close to the slope at 20.

Solution

  1. Find the slope from the two nearby dots. The step between them is 0.01.

    \[h'(20) \approx \frac{40.07 - 40}{0.01} = 7\]

  2. The offset is 20.3 − 20 = 0.3.

    \[\Delta h \approx 7(0.3) = 2.1\]

  3. Add the change to the value.

    \[h(20.3) \approx 40 + 2.1 = 42.1\]

Questions

The dots show the height \(H\) of a plant, in centimeters, at weeks 2, 4, 6, 8 and 10. The heights are 4, 9, 13, 16 and 18.

  1. Which value is the best estimate of the growth rate \(H'(6)\)?

    • 1.75 cm per week
    • About 2.17 cm per week
    • 7 cm per week
    Answer

    1.75 cm per week

  2. Use that slope to estimate \(H(6.5)\).

    • 13.875 cm
    • 13.5 cm
    • 14.75 cm
    Answer

    13.875 cm

  3. A function \(f\) has \(f(10) = -2\) and \(f'(10) = 0.6\). Estimate \(f(9.7)\).

    • \(-2.18\)
    • \(-1.82\)
    • \(-2.6\)
    Answer

    \(-2.18\)

Summary

You find the value and the slope at a from the given numbers, and you add the slope times the offset to the value.

This works because the tangent line needs only those two numbers, and a slope can come from rise over run between nearby data.

You use it when a problem gives a table or a few values and no formula.

This is one of four skills in topic 4.6 of Which & Why: AP Calculus.