The second derivative, concavity and acceleration

Warm-up

Three quick questions from earlier lessons.

Discover

The violet curve is $y = x^3$ . The blue line is its tangent at $x = a$ . Slide $a$ from $-2$ to $2$ .

The cubic y = x³ and its tangent at x = a. For a < 0 the curve bends down and lies below the tangent; for a > 0 it bends up and lies above it. The bending changes at x = 0.

For $a > 0$ , the curve lies above its tangent lines and the slopes $3a^2$ grow. How does the graph bend there?

It bends up (concave up)
It bends down (concave down)

Key idea

The derivative of the derivative tells how the slope changes.

f''(x)

is the derivative of

f'(x)

. If

f'' > 0

,

f'

increases and

f

is concave up; if

f'' < 0

,

f'

decreases and

f

is concave down. A point where the concavity changes is an inflection point. For motion: acceleration

a(t) = v'(t) = s''(t)

.

Worked example

$v(t) = t^2 + 4t$ m/s. Find the acceleration at $t = 3$ .

a(t) = v'(t) = 2t + 4

From the derivative-function lesson: the derivative of t² + 4t.

a(3) = 10

m/s²

Substitute t = 3.

Your turn, guided

Your turn: same $v(t)$ , at $t = 1$ .

a(1) = 6

m/s²

2 · 1 + 4.

Practice ★

On $(2, 8)$ , the graph of $f$ is increasing and concave down. Tick all the true statements about that interval.
1. $f'(x) > 0$
2. $f''(x) < 0$
3. $f'$ is decreasing: the graph rises more and more slowly
4. $f''(x) > 0$ , because $f$ is increasing
5. $f'(x) < 0$ , because the graph bends down

A particle's velocity is $v(t) = t^2 + 4t$ m/s. Its acceleration is $a(t) = v'(t)$ . Find $a(6)$ , in m/s².
A particle's velocity is $v(t) = t^2 + 1t$ m/s. Its acceleration is $a(t) = v'(t)$ . Find $a(6)$ , in m/s².

Watch out!

Practice ★★

Here is the graph of the derivative $y = f'(x)$ . Tap the point of this graph at the $x$ -value where $f$ has an inflection point.
Here is the graph of the derivative $y = f'(x)$ . Tap the point of this graph at the $x$ -value where $f$ has an inflection point.

Exit ticket

Three questions to check you have got it.

A particle's velocity is $v(t) = t^2 + 4t$ m/s. Its acceleration is $a(t) = v'(t)$ . Find $a(6)$ , in m/s².
A particle's velocity is $v(t) = t^2 + 1t$ m/s. Its acceleration is $a(t) = v'(t)$ . Find $a(6)$ , in m/s².
A particle's velocity is $v(t) = t^2 + 6t$ m/s. Its acceleration is $a(t) = v'(t)$ . Find $a(6)$ , in m/s².

Summary card

The second derivative

$f'' > 0$ : slopes grow, concave up. $f'' < 0$ : slopes shrink, concave down.
Inflection point: the concavity changes; there $f'$ has a maximum or minimum.
Acceleration is the derivative of velocity.

Watch out: $f'$ says rising or falling; $f''$ says how it bends.

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