Interpret the result at a valid point

The sign of the second derivative

A point moves with \(x = 3t\) and \(y = \frac{t^4}{4} - 2t^2\). The orange line is the tangent at the orange point. The strip under the graph is green where \(\frac{d^2y}{dx^2}\) is positive and red where it is negative.

At \(t = 0\), \(\frac{d^2y}{dx^2} = -\frac49\), which is negative. The curve lies below its tangent line, so it is concave down, which means it bends downward.

At \(t = 2\), \(\frac{d^2y}{dx^2} = \frac89\), which is positive. The curve lies above its tangent line, so it is concave up, which means it bends upward.

Between these values, at \(t = \frac{2}{\sqrt3}\), the second derivative is 0, the strip changes color and the bend changes direction.

Which side of the tangent line

The scene is the same without numbers, and a label beside the orange point says whether \(\frac{d^2y}{dx^2}\) is positive or negative.

Move t. When the second derivative is positive, the strip is green and the curve lies above its tangent line. When it is negative, the strip is red and the curve lies below it.

This sign describes the curve as a graph of \(y\) against \(x\), so it does not depend on the direction in which the point travels.

Notation

1. Find \(\frac{d^2y}{dx^2}\) as a function of \(t\), as in the last skill.

2. Check that \(\frac{dx}{dt}\) is not 0 at your value of \(t\), because the formula divides by \(\frac{dx}{dt}\).

3. If \(\frac{d^2y}{dx^2}\) is positive there, the curve is concave up. If it is negative, the curve is concave down.

   \[\begin{aligned} \cg{\dfrac{d^2y}{dx^2} > 0}&:\ \text{concave up} \\ \cr{\dfrac{d^2y}{dx^2} < 0}&:\ \text{concave down} \end{aligned}\]

4. Read the sign of the whole quotient \(\frac{dm/dt}{dx/dt}\), where \(m = \frac{dy}{dx}\), because a negative \(\frac{dx}{dt}\) reverses the sign of \(\frac{dm}{dt}\).

5. If \(\frac{d^2y}{dx^2} = 0\), the test does not decide. Compare the sign on both sides of that value of \(t\).

   If the sign changes, the curve has an inflection point there.

When it fails

On the curve \(x = t^2\), \(y = t\), the second derivative is \(-\frac{1}{4t^3}\). It is negative for \(t > 0\) and positive for \(t < 0\).

Move \(t\) to 0. The formula has no value there, because \(\frac{dx}{dt} = 0\), and the tangent is vertical. The formula describes only the branches on either side of this point.

Example 1: A point that moves right

For \(x = t^2\) and \(y = t^3 + 2t\), the second derivative is \(\dfrac{d^2y}{dx^2} = \dfrac{3t^2 - 2}{4t^3}\). Is the curve concave up or concave down at \(t = 2\)?

1. Find the point and check the rate \(\frac{dx}{dt}\). At \(t = 2\), the point is \((4, 12)\) and \(\frac{dx}{dt} = 2t = 4\), which is not 0, so the formula applies.

   \[(x,\,y) = (4,\,12),\qquad \dfrac{dx}{dt} = 4\]

2. Substitute \(t = 2\) in the second derivative.

   \[\cy{\dfrac{d^2y}{dx^2}} = \dfrac{3\cdot4 - 2}{4\cdot8} = \dfrac{10}{32} = \dfrac{5}{16}\]

3. The value is positive, so the curve is concave up at \((4, 12)\). It lies above its tangent line.

   \[\cg{\dfrac{5}{16} > 0}\]

Example 2: A point that moves left

For the same curve, \(x = t^2\) and \(y = t^3 + 2t\), the derivative of the slope is \(\dfrac{d}{dt}\!\left(\dfrac{dy}{dx}\right) = \dfrac{3t^2 - 2}{2t^2}\). Is the curve concave up or concave down at \(t = -1\)?

1. At \(t = -1\), the point is \((1, -3)\). The rate \(\frac{dx}{dt} = 2t = -2\) is not 0, and it is negative, so the point moves to the left.

   \[\dfrac{dx}{dt} = \cr{-2}\]

2. A tempting wrong move: The derivative of the slope with respect to \(t\) is \(\frac12\), which is positive. The slope is increasing, so the curve is concave up.

   \[\begin{aligned} \dfrac{3 - 2}{2} &= \dfrac12 > 0 \\ &\Rightarrow\ \cg{\text{concave up}}\ ? \end{aligned}\]

   Because the point moves to the left as \(t\) increases, a slope that rises with \(t\) falls as \(x\) increases. The curve lies below its tangent line at \((1, -3)\), while the red curve that bends up lies above it.

3. Concavity depends on the sign of \(\frac{d^2y}{dx^2}\), so divide by \(\frac{dx}{dt} = -2\).

   \[\cy{\dfrac{d^2y}{dx^2}} = \dfrac{1/2}{-2} = -\dfrac14\]

4. The value is negative, so the curve is concave down at \((1, -3)\). It lies below its tangent line.

   \[\cr{-\dfrac14 < 0}\]

Practice

For \(x = t - t^3\) and \(y = t^2\), find where the curve is concave up and where it is concave down. Say which values of \(t\) need a separate look.

The numerator of the second derivative is positive for every \(t\), so only the sign of \(1 - 3t^2\) matters. The label beside the point shows the sign.

Questions

The graph shows the curve \(x = 1 - 2t\), \(y = 3t - t^3\). As \(t\) increases, the point moves to the left. The orange line is the tangent at the orange point. The strip is green where \(\frac{d^2y}{dx^2}\) is positive and red where it is negative.

1. At the orange point, the peak of the curve, is the curve concave up or concave down?

   • Concave down
   • Concave up
   • Neither, because the tangent is horizontal
   Answer

   Concave down

2. The orange point moves to \(t = 0\), where the strip changes color. What is \(\frac{d^2y}{dx^2}\) there?

   • 0, and the concavity changes
   • Positive
   • It has no value
   Answer

   0, and the concavity changes

3. For \(x = e^{-t}\) and \(y = t^2\), the second derivative is \(\frac{d^2y}{dx^2} = 2(1 + t)e^{2t}\), and \(\frac{d}{dt}\!\left(\frac{dy}{dx}\right) = -2(1 + t)e^{t}\). Is the curve concave up or concave down at \(t = -2\)?

   • Concave down
   • Concave up
   • The formula has no value
   Answer

   Concave down

Summary

Where \(\frac{dx}{dt}\) is not 0, a positive \(\frac{d^2y}{dx^2}\) means the curve is concave up, and a negative value means concave down.

This works because the sign tells whether the slope rises or falls as \(x\) increases, whichever way the point travels.

On any parametric curve, where \(\frac{dx}{dt}\) or \(\frac{d^2y}{dx^2}\) is 0, compare the sign on both sides of that value of \(t\).

This is one of four skills in topic 9.2 of Which & Why: AP Calculus.