Rates of change and the derivative: VCE Math Methods Unit 1 Year 11

What this dot point is asking

VCAA wants you to introduce calculus through the concept of a rate of change: average rates between two points (gradient of a chord), instantaneous rates at a point (gradient of a tangent), and the derivative formula via the power rule for polynomials. The dot point is the introduction to Unit 3's full differentiation work.

Average rate of change

The average rate of change of $y = f(x)$ between $x = a$ and $x = b$ is:

This is the gradient of the chord joining the points $(a, f(a))$ and $(b, f(b))$.

Interpretation: how much the function changes on average per unit of $x$ between the two points.

Instantaneous rate of change: the gradient at a point

The instantaneous rate of change at $x = a$ is the limit of the average rates as the second point approaches $a$:

Geometrically, this is the gradient of the tangent to the curve at $x = a$.

The derivative function $f'(x)$ gives the gradient at any value of $x$.

The power rule

For $f(x) = x^n$, the derivative is:

This rule extends to polynomial sums via linearity:

For Unit 1, the power rule applies for non-negative integer $n$; Unit 3 extends to all rational $n$.

Worked examples

$f(x) = x^4$: $f'(x) = 4 x^3$.

$f(x) = 3 x^5 - 2 x^2 + 7$: $f'(x) = 15 x^4 - 4 x + 0 = 15 x^4 - 4 x$.

The derivative of a constant is zero (constants do not change).

Tangent and normal

Tangent. The line touching the curve at one point with the same gradient as the curve.

For $y = f(x)$ at $x = a$: tangent has gradient $f'(a)$. Equation: $y - f(a) = f'(a)(x - a)$.

Normal. Perpendicular to the tangent at the same point.

Gradient of normal $= -1 / f'(a)$ (negative reciprocal of tangent gradient).

Worked example

$y = x^2 - 4 x + 3$ at $x = 1$. $f(1) = 1 - 4 + 3 = 0$.

$f'(x) = 2 x - 4$. $f'(1) = -2$.

Tangent: $y - 0 = -2(x - 1)$, i.e. $y = -2 x + 2$.

Normal: gradient $1/2$. $y - 0 = (1/2)(x - 1)$, i.e. $y = (x - 1)/2$.

Increasing and decreasing functions

A function is increasing on an interval if $f'(x) > 0$ throughout, decreasing if $f'(x) < 0$.

At a stationary point, $f'(x) = 0$. The point is a local maximum if $f'$ changes from positive to negative there, local minimum if $f'$ changes from negative to positive, or a stationary inflection if the sign does not change.

Worked example: stationary points

$f(x) = x^3 - 3 x + 2$. $f'(x) = 3 x^2 - 3 = 3(x - 1)(x + 1)$.

$f'(x) = 0$ at $x = 1$ and $x = -1$.

Sign of $f'$:

• $x < -1$: positive (e.g. $x = -2$, $f' = 9 > 0$). Function increasing.
• $-1 < x < 1$: negative. Function decreasing.
• $x > 1$: positive. Function increasing.

So at $x = -1$ the function changes from increasing to decreasing: local maximum. $f(-1) = -1 + 3 + 2 = 4$.

At $x = 1$ from decreasing to increasing: local minimum. $f(1) = 1 - 3 + 2 = 0$.

Examples in context

Example 1. Average rate from a chord. A ball's height (metres) is $h(t) = 20t - 5t^2$ for $t$ in seconds. The average rate of change between $t = 1$ and $t = 3$ is $\frac{h(3) - h(1)}{3 - 1} = \frac{(60 - 45) - (20 - 5)}{2} = \frac{15 - 15}{2} = 0$ m/s. The ball is at the same height at both times, so the average velocity over the interval is zero.

Example 2. Instantaneous rate via the power rule. For the same $h(t) = 20t - 5t^2$, the derivative is $h'(t) = 20 - 10t$. At $t = 1$, $h'(1) = 10$ m/s (rising); at $t = 3$, $h'(3) = -10$ m/s (falling). The instantaneous velocity is zero at $t = 2$, the top of the flight.

Try this

Q1. Find the gradient of the chord of $y = x^2$ between $x = 2$ and $x = 5$. [2 marks]

• Cue. $\frac{25 - 4}{5 - 2} = \frac{21}{3} = 7$.

Q2. For $f(x) = x^3 - 6x$, find $f'(x)$ and the gradient of the tangent at $x = 2$. [3 marks]

• Cue. $f'(x) = 3x^2 - 6$; $f'(2) = 12 - 6 = 6$.

Q3. A car's position is $s(t) = t^3 - 9t^2 + 24t$ metres. (a) Find $s'(t)$. (b) Find the times when the velocity is zero. [2+2 marks]

• Cue. (a) $s'(t) = 3t^2 - 18t + 24 = 3(t - 2)(t - 4)$. (b) $t = 2$ and $t = 4$ seconds.