Wemazi

Topic 8.4

Identify the upper and lower curves

Builds on finding the horizontal limits of the region. 6 minutes.

Which curve is on top

The curve \(y = \sqrt{x}\) and the line \(y = \tfrac{x}{2}\) meet at \(x = 0\) and at \(x = 4\), and they enclose the shaded region.

The orange segment is a vertical strip, with one end on the curve and the other on the line. Drag it across the region.

At \(x = 1\) the curve is 1 high and the line is \(\tfrac{1}{2}\) high. The curve is higher, so it is the upper curve and the line is the lower curve.

The length of the strip is the upper height minus the lower height. At \(x = 1\) it is \(1 - \tfrac{1}{2} = \tfrac{1}{2}\), and it shrinks to 0 at \(x = 0\) and at \(x = 4\).

Any two curves

The curves \(f\) and \(g\) enclose a region. At each x-value between the limits \(a\) and \(b\), one curve is higher than the other.

The upper curve has the larger height there, and the lower curve the smaller. Swap the curves: \(g\) is now higher, so the tag upper is next to \(g\).

The strip length is the upper height minus the lower height, so it is positive everywhere inside the region.

Notation

  1. Choose one x-value between the limits, called the test value \(\cy{x_0}\). Do not use a limit, because the two curves have the same height there.

    \[a < \cy{x_0} < b\]

  2. Find the height of each curve at the test value. The curve with the larger height is the upper curve.

    \[\cy{f(x_0)} > \cy{g(x_0)}\]

    The curves must be continuous and must not meet between \(a\) and \(b\). Then one curve cannot pass the other, so it stays on top.

  3. Here \(f\) is the upper curve, so the strip length is \(f(x) - g(x)\). It is positive, because \(f\) is higher.

    \[\text{length} = \cy{f(x) - g(x)}\]

When it fails

Past the right limit \(b\), the curve \(g\) is higher than \(f\). The expression \(f(x) - g(x)\) is negative there, so the strip and the region turn red.

If the curves meet inside the interval, the order can change there, so test each piece between meeting points on its own.

Example 1: a hyperbola and a line

Find the upper curve, the lower curve and the strip length for the region enclosed by \(y = \dfrac{4}{x}\) and \(y = 5 - x\).

  1. The hyperbola \(y = \frac{4}{x}\) and the line meet where \(\frac{4}{x} = 5 - x\). Multiply both sides by \(x\) and move every term to one side. Then factor.

    \[x^2 - 5x + 4 = (x - 1)(x - 4) = 0\]

  2. The solutions are \(x = 1\) and \(x = 4\). Choose the test value \(x = 2\), which lies between them. There the hyperbola is 2 high and the line is 3 high.

    \[\frac{4}{2} = 2 < 3 = 5 - 2\]

  3. The line is higher, so it is the upper curve, and the hyperbola is the lower curve. Subtract the lower height from the upper height.

    \[\cy{(5 - x) - \frac{4}{x}}\]

  4. As a check, at \(x = 2\) this gives 1, which is \(3 - 2\).

    \[(5 - 2) - \tfrac{4}{2} = 1\]

Example 2: a cubic and a line

Find the upper curve and the strip length for the region between \(y = x^3\) and \(y = x\), for \(0 \le x \le 1.5\).

  1. The curves meet where \(x^3 = x\). Factor the equation. The solution \(x = -1\) lies outside the interval, so the curves meet at \(x = 0\) and \(x = 1\).

    \[x^3 - x = x(x - 1)(x + 1) = 0\]

  2. A tempting wrong move: Test the right end \(x = 1.5\). The cubic is 3.375 high and the line is 1.5 high, so the cubic is the upper curve everywhere.

    \[x^3 - x\]

    At \(x = 0.5\) the cubic is 0.125 high and the line is 0.5 high, so \(x^3 - x = -0.375\) is negative. The strip and the region from 0 to 1 are red, because the line is above the cubic there.

  3. One curve can pass the other only where they meet, at \(x = 1\). This splits the interval into two pieces, and each needs its own test value. At \(x = 0.5\) the line is higher, and at \(x = 1.5\) the cubic is higher.

    \[\begin{aligned} \cy{x - x^3} &\text{ on } [0, 1] \\ \cy{x^3 - x} &\text{ on } [1, 1.5] \end{aligned}\]

Practice

Find the upper curve and the strip length for the region between \(y = 9 - x^2\) and \(y = 3 - x\), for \(0 \le x \le 3\).

Drag the strip to x = 1 and read the two heights. Which is larger? Does the order change before x = 3?

Solution

  1. Test \(x = 1\). The parabola is 8 high and the line is 2 high, so the parabola is the upper curve.

    \[9 - 1^2 = 8 > 2 = 3 - 1\]

  2. The curves meet where the difference \(6 + x - x^2\) is 0, at \(x = -2\) and \(x = 3\). Neither lies inside \(0 < x < 3\), so the order does not change.

    \[\begin{aligned} &(9 - x^2) - (3 - x) \\ &\quad = 6 + x - x^2 \\ &\quad = -(x - 3)(x + 2) \end{aligned}\]

  3. The strip length is this difference, upper minus lower.

    \[\cy{6 + x - x^2}\]

Questions

The graph shows the parabola \(y = x^2 - 4x + 5\) and the line \(y = x + 1\).

  1. What is the length of the vertical strip at \(x = 3\)?

    • 2
    • \(-2\)
    • 4
    • 6
    Answer

    2

  2. For which x-values is the line the upper curve?

    • \(1 < x < 4\)
    • \(x < 1\) or \(x > 4\)
    • All x
    Answer

    \(1 < x < 4\)

  3. For \(0 \le x \le \ln 2\), which expression is the length of the strip between \(y = e^x\) and \(y = 2\)?

    • \(2 - e^x\)
    • \(e^x - 2\)
    • \(e^x + 2\)
    Answer

    \(2 - e^x\)

Summary

You compare the heights of the curves at a test value between the limits, and you subtract the lower height from the upper height.

This works because the strip runs from the lower curve to the upper curve, so its length is the difference of the two heights.

You use it to find the expression that you integrate for the area between curves.

This is one of four skills in topic 8.4 of Which & Why: AP Calculus.