Determine concavity when supported

How the slope changes

The field shows \(\dfrac{dy}{dx} = y(6 - y)\), and the orange dot marks the start \((0, 1)\). Move the slider. An orange segment moves along the path and shows its slope.

At the start the slope is 5. As the point moves to the right, the slope grows to 9 at \(y = 3\). When the slope increases, the path bends upward, which is concave up, and the path is green.

After \(y = 3\), the slope decreases from 9 toward 0, so the path bends downward. This is concave down, and the path is red.

The bend changes at \(y = 3\), where the slope is largest.

The bend and the slope

Move the point along the curve. The orange segment is the tangent, and it turns as the point moves.

If the slope increases as \(x\) increases, the curve is concave up. This part is green, and the tangent lies below it.

If the slope decreases as \(x\) increases, the curve is concave down. This part is red, and the tangent lies above it.

The bend depends on how the slope changes, not on the size of the slope.

Notation

1. The second derivative \(y''\) is the rate of change of the slope, and its sign gives the concavity.

   \[\begin{aligned} \cg{y'' > 0}&: \text{concave up} \\ \cr{y'' < 0}&: \text{concave down} \end{aligned}\]

2. The equation gives \(y'\). To find \(y''\), differentiate both sides with respect to \(x\). Because \(y\) is a function of \(x\), the derivative of a \(y\)-term has the factor \(y'\).

   \[y' = 6y - y^2 \;\Rightarrow\; y'' = 6y' - 2y\,y'\]

3. Factor out \(y'\). Then replace it with the right side of the equation, so that \(y''\) depends on \(x\) and \(y\) only.

   \[\begin{aligned} y'' &= (6 - 2y)\,y' \\ &= (6 - 2y)\,y(6 - y) \end{aligned}\]

4. Put the \(y\) of a point on the path into this formula. The result is positive for \(0 < y < 3\) and negative for \(3 < y < 6\), so the path is concave up below \(y = 3\) and concave down above it.

   An inflection point is a point where the concavity changes. At \(y = 3\) the sign of \(y''\) changes, so the path has an inflection point there. A zero of \(y''\) alone is not enough.

When it fails

Press the button to change the equation to \(\dfrac{dy}{dx} = x + y\). Then \(y'' = 1 + y' = 1 + x + y\), which is 0 on the dashed line \(y = -x - 1\).

The line is itself a solution, so no other solution crosses it, and \(y''\) keeps one sign along each solution. Above the line \(y'' > 0\), so the green solutions are concave up. Below it \(y'' < 0\), so the red solutions are concave down, and none has an inflection point.

Example 1: A linear equation

A solution of \(\dfrac{dy}{dx} = 2x + y\) passes through \((1, -1)\). Is it concave up or concave down there?

1. Differentiate both sides with respect to \(x\). The derivative of \(y\) is \(y'\).

   \[y'' = 2 + y'\]

2. Replace \(y'\) with \(2x + y\).

   \[y'' = 2 + 2x + y\]

3. At \((1, -1)\), \(y''\) is positive, so the solution is concave up there. The slope grows from 1 to about 2.9 at \(x = 1.5\).

   \[\cg{y'' = 2 + 2(1) + (-1) = 3}\]

Example 2: A flat point

A solution of \(\dfrac{dy}{dx} = x - y^2\) passes through \((1, 1)\). Is it concave up or concave down there?

1. At \((1, 1)\), the slope is 0.

   \[y' = 1 - 1^2 = 0\]

2. A tempting wrong move: Differentiate \(x - y^2\) as \(1 - 2y\).

   \[y'' = 1 - 2(1) = -1\]

   This gives concave down, a hill. But the field segment at \((0.5, 1)\) falls, with slope \(-0.5\), and the one at \((1.5, 1)\) rises, with slope 0.5. The slope increases from left to right, so the path is a valley.

3. The variable \(y\) depends on \(x\), so the derivative of \(y^2\) is \(2y\,y'\). Do not leave out \(y'\).

   \[y'' = 1 - 2y\,y'\]

4. Put in \(y = 1\) and \(y' = 0\). The result is positive, so the solution is concave up, with a low point at \((1, 1)\).

   \[\cg{y'' = 1 - 2(1)(0) = 1}\]

Practice

A solution of \(\dfrac{dy}{dx} = xy\) passes through \((-2, -1)\). Is it concave up or concave down there?

Slide the tangent to the right of \((-2, -1)\) and see whether its slope grows or shrinks. Then differentiate \(xy\) with the product rule, where \(y\) depends on \(x\).

Questions

The figure shows the slope field of \(\dfrac{dy}{dx} = x(2 - y)\), the dashed line \(y = 2\), and the orange start at the origin.

1. Is the solution concave up or concave down at \(x = 2\)?

   • Concave down
   • Concave up
   • Neither, because the path is straight there
   Answer

   Concave down

2. At which \(x\) is the slope of the solution greatest?

   • \(x = 1\)
   • \(x = 0\)
   • \(x = 2\)
   • \(x = 3\)
   Answer

   \(x = 1\)

3. A solution of \(\dfrac{dy}{dx} = 3x - 2y\) passes through \((2, 1)\). What does \(y''\) show there?

   • \(y'' = -5\), so the solution is concave down.
   • \(y'' = 3\), so the solution is concave up.
   • \(y'' = 4\), so the solution is concave up.
   • It cannot be found without solving the equation.
   Answer

   \(y'' = -5\), so the solution is concave down.

Summary

You differentiate the equation with respect to \(x\), replace \(y'\) with the right side, and find the sign of \(y''\) at the point.

This works because \(y''\) is the rate of change of the slope, and a slope that increases bends the path upward.

You use it when a problem asks for the concavity at a point, or for where solutions bend up or down.

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