Calculate slopes at supplied points

Slopes from an equation

The equation \(\dfrac{dy}{dx} = 2x - y\) gives the slope of a solution at each point. The figure is a slope field: each short segment sits at a grid point and tilts to the slope that the equation gives there.

Drag the orange point, or use the sliders. The orange segment shows the slope at that point, and the readouts show the numbers that give it.

At \((1, 3)\), substitute \(x = 1\) and \(y = 3\) into \(2x - y\). The slope is \(2 - 3 = -1\), so the segment falls 1 unit for each unit that it moves to the right.

Press "Move to (2, 1)". There the slope is \(4 - 1 = 3\), so the segment rises steeply.

Any equation

The right side of the equation, written \(f(x, y)\), can be any rule in \(x\) and \(y\). It takes the \(x\)-value and the \(y\)-value of a point and gives the slope there.

Drag the orange point. The orange segment turns, because the value of the rule changes as the point moves.

Press "Show the curve through the point". The curve is a solution, and it touches the orange segment and has the same slope there.

Only the point sets the slope, so every solution through the same point has the same slope at that point.

Notation

1. The equation writes the slope as a rule in \(x\) and \(y\).

   \[\cq{\dfrac{dy}{dx} = f(x, y)}\]

2. To find the slope at the point \(\cy{(a, b)}\), substitute \(a\) for \(x\) and \(b\) for \(y\). The order matters, because the first number is \(x\).

   \[\cy{\text{slope at } (a, b)} = f(a, b)\]

   In words, the slope at a point equals the value of the right side at that point.

3. A positive result gives a segment that rises to the right, and a negative result gives a segment that falls. A result of 0 gives a flat segment.

When it fails

Press the first button. The new rule divides by \(y - 1\), so on the red dashed line \(y = 1\) the right side divides by 0, which is not a number. The field has no segment on this line.

Press the second button. The segment and the slope label disappear, because a point on this line has no slope. Before you substitute, ask whether the point keeps every denominator nonzero.

Example 1: a point with positive coordinates

Find the slope of \(\dfrac{dy}{dx} = 2x - y\) at the point \((3, 2)\).

1. The point is \((3, 2)\), so \(x = 3\) and \(y = 2\).

2. Substitute these values into the right side.

   \[2(3) - 2 = \cy{4}\]

3. The slope is positive, so the segment rises 4 units for each unit that it moves to the right.

Example 2: a negative coordinate

Find the slope of \(\dfrac{dy}{dx} = 2x - y\) at the point \((-1, -3)\).

1. The point is \((-1, -3)\), so \(x = -1\) and \(y = -3\).

2. A tempting wrong move: Write the \(y\)-value as 3 and subtract it.

   \[2(-1) - 3 = -5\]

   This gives a steep fall, but the field shows a segment that rises at this point. The value of \(y\) is \(-3\), and subtracting \(-3\) adds 3.

3. Put \(y = -3\) in parentheses. Subtracting \(-3\) is the same as adding 3.

   \[2(-1) - (-3) = -2 + 3 = \cy{1}\]

4. The slope is 1, so the segment rises 1 unit for each unit that it moves to the right.

Practice

Find the slope of \(\dfrac{dy}{dx} = 2x - y\) at the point \((-2, 1)\).

Drag the point to \((-2, 1)\). The segment there rises or falls, and your slope must have the same sign.

Questions

The figure shows the slope field of \(\dfrac{dy}{dx} = x + 2y\). Three segments are orange.

1. What is the slope of the orange segment at \((1, 1)\)?

   • 3
   • 1
   • \(-3\)
   Answer

   3

2. Which orange segment has slope \(-2\)?

   • The segment at \((0, -1)\)
   • The segment at \((1, 1)\)
   • The segment at \((-2, 1)\)
   Answer

   The segment at \((0, -1)\)

3. For \(\dfrac{dy}{dx} = 3x - y^2\), what is the slope at the point \((2, -3)\)?

   • \(-3\)
   • 9
   • 15
   Answer

   \(-3\)

Summary

You substitute the x-value and the y-value of the point into the right side of the equation, and the result is the slope.

This works because the equation gives the slope of every solution through the point, so the segment is a piece of the tangent line.

You use it when a problem gives a point and asks for a slope, a direction, or the start of a sketch.

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