Determine the accumulation's concavity

Water that arrives faster, then slower

A tank fills with water. The upper graph shows the rate \(f(t) = 6t - t^2\) in liters per minute. The lower graph shows the amount of water \(A(t)\), which equals the shaded area. The orange line is tangent to it, and its slope is the rate.

At \(t = 1\) and at \(t = 5\), the rate has the same height, \(f = 5\), so the orange line has the same slope at both. Drag \(t\) to each. The lower graph is green where it bends up and red where it bends down.

At \(t = 1\), the rate is rising. Each minute the water arrives faster, so the slope of the lower graph increases and the graph bends upward.

At \(t = 5\), the rate is falling. Each minute the water arrives more slowly, so the slope decreases and the graph bends downward.

The rate is highest at \(t = 3\), so there the lower graph changes from bending upward to bending downward.

The trend of the rate decides the bend

The orange line on the lower graph has the slope \(f(x)\), the height of the upper curve.

Where the upper curve rises, drag \(x\) to the right. The slope of the orange line increases, so the lower graph bends upward. This is called concave up.

Where the upper curve falls, the slope decreases, so the lower graph bends downward. This is called concave down.

Drag \(x\) to the stretch where the curve is above the axis but falling. The lower graph rises there, but it bends down. The height of \(f\) tells whether the lower graph rises, and its trend tells how it bends.

Notation

1. The accumulation \(F(x) = \int_a^x f(t)\,dt\) has the slope \(f\). The derivative of the slope is the second derivative, so \(F''\) is the derivative of \(f\).

   \[F'(x) = f(x), \qquad F''(x) = \cy{f'(x)}\]

   This needs \(f\) to be differentiable.

2. If \(f\) is increasing, then \(f'\) is positive, so \(F\) is concave up.

   \[\cg{f'(x) > 0} \;\Rightarrow\; F \text{ concave up}\]

3. If \(f\) is decreasing, then \(f'\) is negative, so \(F\) is concave down.

   \[\cr{f'(x) < 0} \;\Rightarrow\; F \text{ concave down}\]

4. An inflection point of \(F\) is a point where the concavity changes, so \(f'\) changes sign there and \(f\) has a relative maximum or minimum.

5. To find the intervals, solve \(f'(x) = 0\). Then test the sign of \(f'\) on each side of each zero.

Example 1: a rate that is negative and rising

\(F(x) = \displaystyle\int_0^x (t^3 - 3t)\,dt\). Find the intervals of concavity and the inflection points of \(F\).

1. The second derivative of \(F\) is the derivative of the integrand.

   \[\begin{aligned} F''(x) &= 3x^2 - 3 \\ &= 3(x - 1)(x + 1) \end{aligned}\]

2. Its zeros, \(x = -1\) and \(x = 1\), cut the \(x\)-axis into three parts. Test one point in each.

   \[\begin{aligned} F''(-2) &= 9 \\ F''(0) &= -3 \\ F''(2) &= 9 \end{aligned}\]

3. \(F\) is concave up on \((-\infty, -1)\) and on \((1, \infty)\), and concave down on \((-1, 1)\). The concavity changes at \(x = -1\) and \(x = 1\), so these are inflection points.

   At \(x = 1.5\), the integrand is \(-1.125\), so \(F\) is falling. It is still concave up, because the integrand is rising.

Example 2: a rate that is positive

\(F(x) = \displaystyle\int_0^x t e^{-t}\,dt\) for \(x > 0\). Find the intervals where \(F\) is concave up and concave down.

1. The second derivative of \(F\) is the derivative of the integrand, by the product rule.

   \[\begin{aligned} F''(x) &= \frac{d}{dx}\left(x e^{-x}\right) \\ &= e^{-x} - x e^{-x} \\ &= (1 - x)e^{-x} \end{aligned}\]

2. A tempting wrong move: The integrand \(t e^{-t}\) is positive for \(t > 0\), so \(F\) is concave up for all \(x > 0\).

   \[f(x) > 0 \;\Rightarrow\; F \text{ concave up}\]

   After \(x = 1\), the lower graph rises more and more slowly. Its slope decreases, so it bends downward and is red. A positive rate makes \(F\) rise, but it does not make \(F\) bend up.

3. The factor \(e^{-x}\) is positive for every \(x\), so the sign of \(F''\) is the sign of \(1 - x\).

   \[\begin{aligned} F''(x) &> 0 \text{ for } 0 < x < 1 \\ F''(x) &< 0 \text{ for } x > 1 \end{aligned}\]

4. \(F\) is concave up on \((0, 1)\) and concave down on \((1, \infty)\), with an inflection point at \(x = 1\), the peak of the integrand.

Practice

\(R(x) = \displaystyle\int_1^x \left(t + \frac{4}{t}\right)dt\) for \(x > 0\). Find where \(R\) is concave up and concave down, and find any inflection point.

The integrand falls and then rises. Find where its derivative is zero, and test a point on each side.

Questions

The upper graph shows a rate \(f\) on 0 to 8, made of straight pieces through \((0, 2)\), \((2, -2)\), \((5, 4)\) and \((8, 0)\). The amount is \(F(x) = \int_0^x f(t)\,dt\).

1. On which interval is \(F\) concave up?

   • From 2 to 5
   • From 0 to 2
   • From 3 to 8
   Answer

   From 2 to 5

2. At which value of \(x\) does \(F\) have an inflection point?

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

   \(x = 2\)

3. \(K(x) = \displaystyle\int_1^x (t^3 - 6t^2)\,dt\) for \(x > 0\). Where does \(K\) have an inflection point?

   • \(x = 4\)
   • \(x = 6\)
   • There is none.
   Answer

   \(x = 4\)

Summary

The accumulation is concave up where the rate rises and concave down where it falls, with an inflection point where the trend changes.

This works because the second derivative of the accumulation is the derivative of the rate.

You use it when a question asks how the accumulated amount bends and gives a graph or a formula for the rate.

This is one of five skills in topic 6.5 of Which & Why: AP Calculus.