Wemazi

Topic 7.4

Determine increasing and decreasing behavior

Builds on following a solution from its initial point. 6 minutes.

A path that turns

The field shows \(\dfrac{dy}{dx} = (x - 1)y\). The orange dot marks the starting point \((0, 1)\).

Press "Trace the solution". The path is green where it rises, because its slope is positive there, and red where it falls, because its slope is negative there.

For \(x < 1\), the factor \(x - 1\) is negative and \(y\) is positive, because the path stays above the x-axis. A negative number times a positive number is negative, so the slope is negative and the path falls.

For \(x > 1\), both factors are positive, so the slope is positive and the path rises. At \(x = 1\) the slope is 0, and the path turns.

Sign sets direction

A solution rises where its slope is positive, and it falls where its slope is negative.

Drag the orange start. The path is green where the segments rise and red where they fall.

The strip along the bottom of the plot repeats these colors along the x-axis.

The color changes where the slope is 0, which is on a dashed orange line. If the sign changes there, the path turns.

Notation

  1. The slope along a solution is the right side of the equation, evaluated at the points of the solution.

    \[\cy{\frac{dy}{dx}} = f(x, y)\]

  2. If the slope is positive, the solution increases, so that part of the path is green. If the slope is negative, the solution decreases, so that part is red.

    \[\begin{aligned} \cg{f > 0} &\Rightarrow y \text{ increases} \\ \cr{f < 0} &\Rightarrow y \text{ decreases} \end{aligned}\]

  3. To find the sign, factor the right side and find the sign of each factor along the path. A product is positive when its factors have the same sign.

    \[f(x, y) = (x - 1)\,y\]

    The sign of the slope can change only where a factor is 0, on a dashed line.

When it fails

The second path starts at \((0, -1)\). At each \(x\) the factor \(x - 1\) is the same as on the first path, but \(y\) is negative, so the product has the opposite sign. The second path rises where the first falls.

The sign depends on \(y\) as well as on \(x\), so use the \(y\)-values of the solution that you follow.

Example 1: Two turning points

A solution of \(\dfrac{dy}{dx} = x^2 - 1\) has \(y(0) = 0\). On which intervals does it increase and decrease?

  1. Factor the right side. The slope is 0 at \(x = -1\) and at \(x = 1\).

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

  2. The strip along the bottom shows the sign of the slope in each interval. For \(x < -1\), both factors are negative, so the slope is positive. For \(-1 < x < 1\), the factors have different signs, so the slope is negative. For \(x > 1\), both factors are positive, so the slope is positive.

  3. The solution increases for \(x < -1\), decreases for \(-1 < x < 1\), and increases for \(x > 1\). At the start, in the middle interval, the slope is \(0^2 - 1 = -1\) and the path falls.

Example 2: A path below the axis

A solution of \(\dfrac{dy}{dx} = (x - 1)y\) has \(y(3) = -1\). On which intervals does it increase and decrease?

  1. At the start, the slope is negative.

    \[(3 - 1)(-1) = -2\]

  2. A tempting wrong move: The factor \(x - 1\) is positive for \(x > 1\), so the solution increases there.

    \[x - 1 > 0 \Rightarrow \text{increases}\]

    This uses one factor only. At \((3, -1)\) the factor \(y\) is \(-1\), so the slope is \(-2\) and the orange tangent falls.

  3. Use both factors. The x-axis, \(y = 0\), is itself a solution, because the right side is 0 there. Only one path passes through each point, so this path cannot cross the axis and \(y < 0\) along the whole path.

  4. For \(x > 1\), a positive number times a negative number is negative, so the solution decreases. For \(x < 1\), both factors are negative, so the solution increases.

Practice

A solution of \(\dfrac{dy}{dx} = (x + 2)(y - 1)\) has \(y(1) = 0\). On which intervals does it increase and decrease?

The line \(y = 1\) is a solution, so the path stays on one side of it. Which side? That sets the sign of \(y - 1\).

Solution

  1. The line \(y = 1\) is a solution, because the right side is 0 there. The path starts at \(y = 0\), below it, and cannot cross it, so \(y - 1 < 0\) along the whole path.

  2. For \(x > -2\), the factor \(x + 2\) is positive. A positive number times a negative number is negative, so the solution decreases.

  3. For \(x < -2\), both factors are negative. The product is positive, so the solution increases.

Questions

The figure shows the slope field of \(\dfrac{dy}{dx} = \dfrac{y - 2}{2}\), the dashed line \(y = 2\), and four faint solutions.

  1. A solution passes through the orange point \((0, 3)\). What does it do as \(x\) increases?

    • It increases.
    • It decreases.
    • It stays at \(y = 3\).
    Answer

    It increases.

  2. A solution passes through \((0, 1)\). Which statement is correct?

    • It decreases for all \(x\).
    • It decreases at first and then increases.
    • It increases toward the dashed line.
    Answer

    It decreases for all \(x\).

  3. A solution of \(\dfrac{dy}{dx} = (x - 4)(y + 2)\) has \(y(0) = -5\). Where does it increase?

    • For \(x < 4\)
    • For \(x > 4\)
    • For all \(x\)
    Answer

    For \(x < 4\)

Summary

You find the sign of the right side along the solution: positive means that the solution increases, and negative means that it decreases.

This works because the right side of the equation is the slope of the solution at each of its points.

You use it when a problem asks where a solution rises, falls or turns, and gives the equation or the field.

This is one of five skills in topic 7.4 of Which & Why: AP Calculus.