Match a slope segment to a point

A segment that turns

The field shows \(\dfrac{dy}{dx} = \dfrac{y - x}{2}\). The orange point sits on the row \(y = 2\), and the orange segment through it is the field segment there. The two faint dashed lines through the point have slope 1 and slope \(-1\).

At \(x = -2\), the slope is \(\dfrac{2 + 2}{2} = 2\), so the orange segment rises and is steeper than the faint rising line.

Drag the point to the right along its row. At \(x = 0\) the slope is 1, so the segment lies on the faint rising line. At \(x = 2\) the slope is 0, so the segment is flat. At \(x = 4\) the slope is \(-1\), so the segment falls and lies on the faint falling line.

Sign and size

The orientation of a segment is the direction in which it leans, and its steepness is how sharply it leans. The slope m sets both.

Move the slider. If m is positive, the segment rises to the right, and if m is negative, it falls.

The size of m, which ignores its sign, sets the steepness. If the size is less than 1, the segment is flatter than the faint dashed lines. If it is more than 1, the segment is steeper.

If m is 0, the segment lies on the faint horizontal line. If the equation divides by 0 at the point, it gives no slope, so no segment fits. The figure then shows a red dashed vertical line.

Notation

1. Use the differential equation to calculate the slope \(\cy{m}\) at the point \(\cy{(a, b)}\).

   \[\cy{m} = f(a, b)\]

2. Read the sign of \(m\). If \(m > 0\), the segment rises to the right, and if \(m < 0\), it falls. If \(m = 0\), it is flat.

3. Compare the size of \(m\), which is \(|m|\), with 1. The faint dashed lines have slope 1 and slope \(-1\), so they show a size of 1.

   \[|m| < 1 \text{ flatter}, \quad |m| > 1 \text{ steeper}\]

4. In words, the sign of \(m\) fixes whether the segment at the point rises or falls, and the size of \(m\) fixes how steeply.

   The slope changes gradually from one point to the next, so the field segments near your choice must lean nearly the same way.

When it fails

The equations \(\dfrac{y - x}{2}\) and \(y - x\) give the same sign at every point. At \((-1, 3)\) the first gives 2 and the second gives 4.

In both fields the segment rises, but in the second it is steeper. The sign alone cannot tell the two fields apart, so find the size of the slope too.

Example 1: a steep rise

For \(\dfrac{dy}{dx} = \dfrac{y - x}{2}\), which segment belongs at \((-2, 2)\)?

1. Calculate the slope at the point.

   \[\dfrac{2 - (-2)}{2} = \cy{2}\]

2. The slope is positive, so the segment rises to the right.

3. The size of the slope is 2, which is more than 1, so the segment is steeper than the faint lines. Choose the steep rising segment.

   As a check, the slope field appears. Its segment at the point lies along the orange one.

Example 2: a steep fall

For \(\dfrac{dy}{dx} = \dfrac{y - x}{2}\), which segment belongs at \((5, 2)\)?

1. Calculate the slope at the point.

   \[\dfrac{2 - 5}{2} = \cy{-\tfrac{3}{2}}\]

2. A tempting wrong move: The slope is negative, so any falling segment fits. Choose the gentle one.

   \[m < 0 \;\Rightarrow\; \text{any fall fits}\]

   A gentle fall has a slope between \(-1\) and 0, but \(-\tfrac32\) is below \(-1\). The red segment does not line up with the field segments around the point, which fall more steeply.

3. The slope is negative, so the segment falls. Its size \(\tfrac32\) is more than 1, so it falls more steeply than the faint line. Choose the steep falling segment.

   \[\left|-\tfrac{3}{2}\right| > 1\]

Practice

For \(\dfrac{dy}{dx} = \dfrac{y - x}{2}\), which segment belongs at \((5, 4)\)?

Drag the point to \((5, 4)\). The sign of the slope there gives the direction of the segment, and its size, compared with 1, gives the steepness.

Questions

The figure shows the slope field of \(\dfrac{dy}{dx} = x - \dfrac{y}{2}\). One segment is missing at each open circle, and the faint dashed lines through each circle have slope 1 and \(-1\).

1. Which segment belongs at the open circle at \((-3, 1)\)?

   • A steep fall
   • A steep rise
   • A gentle fall
   • A flat segment
   Answer

   A steep fall

2. Which segment belongs at the open circle at \((2, 3)\)?

   • A gentle rise
   • A steep rise
   • A gentle fall
   • A flat segment
   Answer

   A gentle rise

3. For \(\dfrac{dy}{dx} = x^2 - 2y\), which segment belongs at \((1, 4)\)?

   • A steep fall
   • A steep rise
   • A gentle fall
   Answer

   A steep fall

Summary

You calculate the slope at the point, read its sign for the orientation of the segment, and compare its size with 1 for the steepness.

This works because a segment has two features, the direction it leans and how sharply it leans, and the slope fixes both.

You use it when a problem gives a field and asks which segment, rising or falling, steep or gentle, belongs at a point.

This is one of four skills in topic 7.3 of Which & Why: AP Calculus.