The rate of change and the derivative at a point

A small pool is filled from a tap. At first it holds 20 litres, and every minute 5 litres flow in.
$\displaystyle {V}={5t+20}$ – the amount of water in litres after $\displaystyle t$ minutes.

• Every minute the amount grows by the same number, 5 litres: the rate of change is constant – 5 litres a minute.
• The graph of a function with a constant rate of change is a line, and the rate of change is its slope.

A car sets off and speeds up. $\displaystyle {s(t)}={t^2}$ – the distance in metres it covers in the first $\displaystyle t$ seconds.

• In the first second the car covers one metre, in the second second 3 metres, in the third 5, in the fourth 7.
• Every second it covers more: the rate of change varies – the speed grows.
• The graph is a parabola, not a line: it has no single slope.

The average rate of change between two points is the change in $\displaystyle y$ divided by the change in $\displaystyle x$:

The example below works out the car’s average speed between $\displaystyle {t}={1}$ and $\displaystyle {t}={3}$:

• The car covered 8 metres in 2 seconds: on average 4 metres a second.
• On the board: the dashed line passes through $\displaystyle A(1,1)$ and $\displaystyle B(3,9)$. The segment $\displaystyle AB$ is a chord of the graph.

The average rate of change is the slope of the chord: the slope of the dashed line is 4.

The average speed does not say what the speed was exactly at the moment $\displaystyle {t}={1}$. To know it, bring the second point closer to the first.
The second point: $\displaystyle {t}={1+h}$, where $\displaystyle h$ is a small number other than 0.

• Dividing by $\displaystyle h$ is allowed, because $\displaystyle h\neq0$.
• As $\displaystyle h$ gets closer to 0, the average speed gets closer to 2.
• So at the moment $\displaystyle {t}={1}$ the car’s speed is 2 metres a second: this is the instantaneous rate of change.

The instantaneous rate of change of a function at a point $\displaystyle x_0$ is called the derivative of $\displaystyle f$ at $\displaystyle x_0$, written $\displaystyle f'(x_0)$:

• $\displaystyle \lim\limits_{h\to0}$ is read “the limit as $\displaystyle h$ tends to 0”: the number the expression gets closer to as $\displaystyle h$ gets closer to 0.
• On the board: $\displaystyle {f(x)}={x^2}$, the line of the chord from $\displaystyle A(1,1)$ to $\displaystyle (2,4)$ (the dashed one, slope 3), and a solid line the chords get closer to.

• As $\displaystyle h$ gets smaller, the second point comes closer to $\displaystyle A$, and the chord turns about $\displaystyle A$.
• The line the chords get closer to is the tangent to the graph at $\displaystyle A$: it touches the graph at $\displaystyle A$.
• The derivative at a point is the slope of the tangent: here $\displaystyle {f'(1)}={2}$, and the tangent is $\displaystyle {y}={2x-1}$.

$\displaystyle {f(x)}={x^2}$.
The example below finds $\displaystyle f'(3)$ by the definition.

• As $\displaystyle h$ gets closer to 0, $\displaystyle 6+h$ gets closer to 6.
• So $\displaystyle {f'(3)}={6}$: the slope of the tangent to the graph at the point $\displaystyle (3,9)$ is 6.

$\displaystyle {f(x)}={x^2+1}$.

Work out the average rate of change of $\displaystyle f$ between $\displaystyle {x}={1}$ and $\displaystyle {x}={4}$.

Work out the average rate of change of $\displaystyle f$ between $\displaystyle {x}={1}$ and $\displaystyle {x}={2}$.

The two chords start at the same point, $\displaystyle (1,2)$.
Which of them is steeper? What does that tell you about the graph of $\displaystyle f$ to the right of $\displaystyle {x}={1}$?

$\displaystyle {f(x)}={x^2}$.

Simplify the average rate of change between $\displaystyle {x}={2}$ and $\displaystyle {x}={2+h}$.

What does your expression get closer to as $\displaystyle h$ gets closer to 0?
Write $\displaystyle f'(2)$ as a number.

$\displaystyle {f(x)}={x^3}$.
A hint: $\displaystyle {(1+h)^3}={(1+h)^2(1+h)}$.

Simplify the average rate of change between $\displaystyle {x}={1}$ and $\displaystyle {x}={1+h}$.

Write $\displaystyle f'(1)$ as a number.

$\displaystyle {f(x)}={x^2-4x}$.
$\displaystyle {f(1)}={-3}$.

Simplify the average rate of change between $\displaystyle {x}={1}$ and $\displaystyle {x}={1+h}$.

Write $\displaystyle f'(1)$ as a number.

The derivative you found is negative.
What does that say about the tangent at the point $\displaystyle (1,-3)$, and about the graph of $\displaystyle f$ around the point?

A ball is thrown upwards.
$\displaystyle {s(t)}={20t-5t^2}$ – the ball’s height in metres after $\displaystyle t$ seconds.

Work out the ball’s average speed in the first two seconds.

Simplify the average speed between $\displaystyle {t}={1}$ and $\displaystyle {t}={1+h}$.

What is the ball’s speed exactly at the moment $\displaystyle {t}={1}$?

At the moment $\displaystyle {t}={1}$, is the ball going up or down?
How does the sign of $\displaystyle s'(1)$ show it?