Locate zero-slope sets
Where the segments are flat
The field shows \(\dfrac{dy}{dx} = x(y - 2)\). A flat segment has slope 0, and the orange dashed lines pass through every flat segment.
On the vertical line, \(x = 0\), so the slope is \(0 \cdot (y - 2) = 0\).
On the horizontal line, \(y = 2\), so the slope is \(x \cdot 0 = 0\). Drag the orange point onto either line, and the readout shows 0.
Press the button. The solution from \((-2, 2)\) follows the horizontal line, because every segment on that line is flat.
Touching and staying
For any equation, the zero-slope set is every point where the slope is 0, and it can be a line or a curve. In this field it is the dashed orange wave and the dashed horizontal line.
Drag the start point. The curve through it is the solution that starts there, and it follows the segments of the field.
Each highest or lowest point of the solution lies on the dashed wave, because the solution is flat there. Where a solution meets the wave, it crosses and leaves it, so it is flat for only an instant.
Press the button. A solution that starts on the horizontal dashed line stays on it, because every segment on that line is flat.
Notation
The zero-slope set is every point where the right side of the equation, \(f(x, y)\), is 0.
\[\cy{f(x, y)} = 0\]A product is 0 when one factor is 0, so each factor gives one part of the set.
\[x(y - 2) = 0 \;\Rightarrow\; \cy{x = 0} \text{ or } \cy{y = 2}\]A constant solution is a solution with the same \(y\)-value everywhere. Its graph is a horizontal line, and its slope is 0 at every point.
\[y = c\]In words, the line \(y = c\) is a solution when every point on it has slope 0.
\[f(x, c) = 0 \text{ for every } x\]A solution cannot follow a vertical line, because a function has only one \(y\)-value for each \(x\). A curved set is not a solution either, because a curve is not flat at every point.
When it fails
The point \((0, 1)\) is on the vertical dashed line, so the segment there is flat.
The solution through it rises to the left of the line and falls to the right, so it is flat at that one point only. A flat point does not make a flat solution, so the line \(x = 0\) is not a solution.
Example 1: a curved set
Find the zero-slope set of \(\dfrac{dy}{dx} = y - x^2\). Decide whether it is a solution.
Set the right side equal to 0 and solve for \(y\).
\[y - x^2 = 0 \;\Rightarrow\; \cy{y = x^2}\]The zero-slope set is the parabola \(y = x^2\). The segment at \((2, 4)\) is flat, because \(4 - 4 = 0\), so the solution through that point has a horizontal tangent there.
If the parabola were a solution, its slope would be 0 at every point. But the slope of \(y = x^2\) is \(2x\), which is 4 at \(x = 2\), so the parabola is not a solution.
\[\dfrac{d}{dx}\,x^2 = 2x\]
Example 2: a fraction
Find the zero-slope set of \(\dfrac{dy}{dx} = \dfrac{y - 3}{x + 1}\).
Set the right side equal to 0.
\[\dfrac{y - 3}{x + 1} = 0\]A tempting wrong move: Treat the fraction like a product, so each part gives one line of the set.
\[y - 3 = 0 \text{ and } x + 1 = 0\]On the line \(x = -1\) the equation divides by 0, so the slope does not exist. The field has no segments there, and a missing segment is not a flat one.
A fraction is 0 only when its top is 0 and its bottom is not 0. The zero-slope set is the line \(y = 3\), without the point where \(x = -1\).
\[y = 3, \quad x \ne -1\]On this line the right side is 0 for every \(x\) except \(-1\), so \(y = 3\) is a constant solution on any interval without \(-1\).
Practice
Find the zero-slope set of \(\dfrac{dy}{dx} = 1 - xy\). Decide whether it is a solution.
Drag the point to \((1, 1)\), where its segment is flat because \(1 - 1 \cdot 1 = 0\). Where else is \(1 - xy\) equal to 0?
Solution
Set the right side equal to 0.
\[1 - xy = 0 \;\Rightarrow\; \cy{y = \dfrac{1}{x}}, \quad x \ne 0\]The zero-slope set is the curve \(y = \tfrac1x\). It is not horizontal, so it is not a constant solution.
The curve \(y = \tfrac1x\) has slope \(-\tfrac{1}{x^2}\), which is \(-1\) at \((1, 1)\) and never 0, so the curve is not a solution.
\[\dfrac{d}{dx}\,\dfrac{1}{x} = -\dfrac{1}{x^2}\]
Questions
The figure shows the slope field of \(\dfrac{dy}{dx} = x - y\) and four of its solutions.
Which line contains all of the flat segments?
- The line \(y = x\)
- The \(x\)-axis
- The \(y\)-axis
- The line \(y = -x\)
Answer
The line \(y = x\)
Every segment on the line \(y = x\) is flat. Is the line \(y = x\) a solution?
- No. A curve along the line would have slope 1, but the segments on the line have slope 0.
- Yes. Every segment on the line is flat.
- No. No solution touches the line.
Answer
No. A curve along the line would have slope 1, but the segments on the line have slope 0.
The equation \(\dfrac{dy}{dx} = (x - 4)y\) has flat segments on \(y = 0\) and on \(x = 4\). Which statement is correct?
- The line \(y = 0\) is a constant solution, and the line \(x = 4\) is not.
- The line \(x = 4\) is a constant solution, and the line \(y = 0\) is not.
- Both lines are constant solutions.
- Only the line \(x = 4\) has flat segments.
Answer
The line \(y = 0\) is a constant solution, and the line \(x = 4\) is not.
Summary
You set the right side of the equation equal to 0 and solve, which gives the points where the segments are flat.
A solution stays on that set only along a horizontal line, where the slope is 0 at every point.
You use it when a problem asks where the slope is zero or whether a horizontal line is a constant solution.
This is one of four skills in topic 7.3 of Which & Why: AP Calculus.