Wemazi

Topic 2.1

Compare the two kinds of rate

Builds on interpreting the sign and units of a rate. 7 minutes.

A sled on a hill

A sled slides down a hill. After \(t\) seconds it has gone \(3t^2\) meters. The chalk line joins the points of the graph at 1 second and 5 seconds. The sled goes from 3 meters to 75 meters in 4 seconds, so the chalk line has slope \(72 \div 4 = 18\) meters per second.

This slope is the average rate, which describes the whole 4 seconds.

Drag the orange point. The orange line is the tangent, which touches the curve and points in its direction. Its slope is the instantaneous rate, the rate at that one moment. At 1 second the sled moves at 6 meters per second, and at 5 seconds at 30.

Go to 3 seconds. The tangent is parallel to the chalk line, so the instantaneous rate is 18, the same as the average.

Two rates on one curve

The chalk line through the points at \(a\) and \(b\) is called a secant. Its slope, the rise divided by the run, is the average rate over the interval from \(a\) to \(b\).

The orange tangent touches the curve at \(c\), and its slope is the instantaneous rate at \(c\).

Move c. The tangent turns, but the secant does not move. The two slopes are equal only where the lines are parallel.

Move \(c\) to \(a\). Then move \(b\) toward \(a\). The secant turns onto the tangent, so the average rate over \([a, b]\) approaches the instantaneous rate at \(a\).

Notation

  1. The average rate over \([a, b]\) is the slope of the secant, and it describes an interval.

    \[\cy{\frac{f(b)-f(a)}{b-a}}\]

  2. The instantaneous rate at \(c\) is the slope of the tangent, and it describes one input.

    \[\cy{f'(c)}=\lim_{h\to 0}\frac{f(c+h)-f(c)}{h}\]

  3. As \(b\) approaches \(a\), the average rate over \([a, b]\) approaches \(f'(a)\). If the rate is constant, as on a line, the two rates are equal everywhere.

  4. To find an input where the two rates match, set \(f'(c)\) equal to the average rate. Then solve for \(c\). For the sled, \(6c = 18\) gives \(c = 3\).

    \[f'(c)=\frac{f(b)-f(a)}{b-a}\]

When it fails

At the start of the interval the sled moves at 6 meters per second, and at the end it moves at 30. The average rate of 18 is the rate at neither end.

An average rate does not tell you the rate at a given input. The two rates match only at inputs where the tangent is parallel to the secant.

Example 1: where the rates match

Let \(f(x) = x^2 - 6x\). Find the average rate of change on \([0, 4]\). Then find the input \(c\) in \((0, 4)\) where \(f'(c)\) equals it.

  1. Find the average rate. The values are \(f(0) = 0\) and \(f(4) = 16 - 24 = -8\).

    \[\cy{\frac{-8-0}{4-0}=-2}\]

  2. Find the rate at a general input. The average from \(x\) to \(x + h\) simplifies to \(2x - 6 + h\), and it approaches \(2x - 6\).

    \[\cy{f'(x)=2x-6}\]

  3. Set \(f'(c)\) equal to \(-2\) and solve. The tangent at \(c = 2\) is parallel to the secant.

    \[2c-6=-2 \quad\Rightarrow\quad \cy{c=2}\]

Example 2: a reciprocal

Let \(g(x) = \dfrac1x\). Find the average rate of change on \([1, 4]\). Then find the input \(c\) in \((1, 4)\) where \(g'(c)\) equals it.

  1. Find the average rate.

    \[\cy{\frac{\frac14-1}{4-1}=-\frac14}\]

  2. Find the rate at a general input. The average from \(x\) to \(x + h\) simplifies to \(-\dfrac{1}{x(x+h)}\), and it approaches \(-\dfrac1{x^2}\).

    \[\cy{g'(x)=-\frac{1}{x^2}}\]

  3. A tempting wrong move: The rates match in the middle of the interval, at \(c = 2.5\).

    \[\cr{\begin{aligned} g'(2.5) &= -\frac{1}{2.5^2} \\ &= -0.16 \neq -0.25 \end{aligned}}\]

    At 2.5 the tangent has slope −0.16, so it is flatter than the chalk line of slope −0.25. The middle works for a parabola, as in Example 1, but this curve is not a parabola.

  4. Solve \(g'(c) = -\frac14\). This gives \(c^2 = 4\), so \(c = 2\) or \(c = -2\), and only \(c = 2\) lies in \((1, 4)\).

    \[-\frac{1}{c^2}=-\frac14 \quad\Rightarrow\quad \cy{c=2}\]

Practice

Let \(f(x) = x^2 + 4x\). Find the average rate of change on \([0, 6]\) and the instantaneous rate \(f'(6)\). Which one is larger?

Find the slope of the chalk secant first. Then find the slope of the tangent at \(x = 6\) and compare the two lines.

Solution

  1. Find the average rate. The values are \(f(0) = 0\) and \(f(6) = 36 + 24 = 60\).

    \[\cy{\frac{60-0}{6-0}=10}\]

  2. Find the rate at 6. The average from 6 to \(6 + h\) simplifies to \(16 + h\).

    \[\cy{f'(6)=16}\]

  3. The instantaneous rate at 6 is larger. The curve bends upward, so the tangent at the right end is steeper than the secant.

Questions

Questions 1 and 2 use the graph of \(f(x) = \tfrac12 x^2\). Question 3 needs a calculation.

  1. The chalk secant joins the points at \(x = 2\) and \(x = 6\). What is the average rate of change of \(f\) over \([2, 6]\)?

    • 4
    • 16
    • 3
    Answer

    4

  2. The orange tangent can sit at \(x = 2\), \(x = 4\) or \(x = 6\). At which input does the instantaneous rate equal the average rate over \([2, 6]\)?

    • \(x = 4\)
    • \(x = 2\)
    • \(x = 6\)
    • No input, because an average is not a rate at an input
    Answer

    \(x = 4\)

  3. For \(f(x) = x^2 + 2x\), the average rate over \([1, 7]\) is 10. At which input \(c\) is \(f'(c) = 10\)?

    • \(c = 4\)
    • \(c = -1 + \sqrt{11}\)
    • \(c = 10\)
    Answer

    \(c = 4\)

Summary

The secant's slope is the average rate over an interval, and the tangent's slope is the instantaneous rate at one input.

The rates are equal where the tangent is parallel to the secant, and you find those inputs by setting \(f'(c)\) equal to the average rate.

A question about a whole interval asks for the average rate, and a question about one moment asks for the instantaneous rate.

This is one of four skills in topic 2.1 of Which & Why: AP Calculus.