Year 12 Methods (Unit 3 & 4) Further differentiation and applications

Rates of change

Q: What is the average rate of change?

The average rate of change of a function over an interval from a to b is the change in the function divided by the change in the input: (f(b) minus f(a)) divided by (b minus a). Geometrically it is the gradient of the chord joining the two points on the graph.

Q: What is the instantaneous rate of change?

The instantaneous rate of change at a point is the value of the derivative at that point. It is the gradient of the tangent line to the graph at that single point, and it tells you how fast the quantity is changing at that exact instant.

Q: How do you find an instantaneous rate of change?

Differentiate the function to get the derivative, then substitute the required value. For example if V equals 2000 times e to the power 0.08t, then dV by dt equals 160 times e to the power 0.08t, and at t equals 10 the rate is about 356 dollars per year.

Q: What does the sign of a rate of change tell you?

A positive rate means the quantity is increasing, a negative rate means it is decreasing, and a zero rate means it is momentarily stationary. The size of the rate tells you how quickly the change is happening.

Q: How is the average rate different from the instantaneous rate?

The average rate is the gradient of the chord over a whole interval; the instantaneous rate is the gradient of the tangent at one point. They are usually different unless the graph is a straight line.

Q: Why is the derivative called a rate of change?

Because the derivative measures how the output responds to a small change in the input. For a quantity varying with time, the derivative with respect to time gives the rate at which the quantity is changing at each instant.

20 practice questions 0 video lessons Theory + worked examples

In Year 12 Mathematical Methods (Queensland, QCAA), a rate of change measures how one quantity changes as another changes. The average rate of change over an interval is the gradient of the chord, $\dfrac{f(b)-f(a)}{b-a}$; the instantaneous rate of change at a point is the derivative there, and equals the gradient of the tangent. You read the sign of a rate to say whether a quantity is increasing or decreasing, and give its units in context — applied to power, exponential, logarithmic and trigonometric models.

A rate of change tells you how fast one quantity changes with respect to another. There are two kinds.

The average rate of change of $f$ over the interval $[a,b]$ is

$\dfrac{f(b)-f(a)}{b-a}$,

the gradient of the chord (the straight line joining $(a,f(a))$ and $(b,f(b))$). It summarises the change across the whole interval.

The instantaneous rate of change at $x=a$ is the derivative $f'(a)$. It is the gradient of the tangent to the curve at that single point, and it describes how fast the quantity is changing at that exact instant. For a quantity $Q$ that varies with time $t$, the instantaneous rate is $\dfrac{dQ}{dt}$; when $Q$ is a displacement, this rate is the velocity.

The sign of a rate carries meaning: positive means increasing, negative means decreasing, and zero means momentarily stationary. Always state the units — the quantity per unit of the variable (for example L/min, people/year or $^{\circ}$C/min).

Key idea. Average rate \(=\) gradient of the chord \(=\dfrac{f(b)-f(a)}{b-a}\). Instantaneous rate \(=\) derivative \(=f'(a)=\) gradient of the tangent. Sign \(+\Rightarrow\) increasing, \(-\Rightarrow\) decreasing.

Average rate over $[a,b]$: the gradient of the chord joining $A$ and $B$

Instantaneous rate at $P$: the gradient of the tangent touching the curve at $P$

Average rate of change over $[a,b]$ (gradient of the chord):

\[\text{average rate}=\dfrac{f(b)-f(a)}{b-a}\]

\frac{f (b) - f (a)}{b - a}

Instantaneous rate of change at $x=a$ (the derivative, the gradient of the tangent):

\[\text{instantaneous rate}=f'(a)=\left.\dfrac{dy}{dx}\right|_{x=a}\]

f^{'} (a)

The standard derivatives used in rate models (QCAA Units 3 & 4):

\[\dfrac{d}{dx}\,x^{n}=nx^{n-1},\quad \dfrac{d}{dx}\,e^{kx}=k\,e^{kx},\quad \dfrac{d}{dx}\,\ln(ax+b)=\dfrac{a}{ax+b},\quad \dfrac{d}{dx}\,\sin(kx)=k\cos(kx)\]

\frac{d}{d x} e^{k x} = k e^{k x}

Interpretation. A rate has a sign and a unit. If \(\dfrac{dQ}{dt}>0\) the quantity \(Q\) is increasing; if \(\dfrac{dQ}{dt}<0\) it is decreasing. The unit is \(Q\)-units per \(t\)-unit (e.g. litres per minute).

How to work with a rate of change

Average or instantaneous? "Over an interval" or "on average" $\Rightarrow$ average rate; "at this instant" or "when $t=\ldots$" $\Rightarrow$ instantaneous rate.
Average rate. Evaluate $f$ at both endpoints, then compute $\dfrac{f(b)-f(a)}{b-a}$ — the gradient of the chord.
Instantaneous rate. Differentiate to get $f'(x)$ (or $\dfrac{dQ}{dt}$), then substitute the required value. This is the gradient of the tangent at that point.
Interpret. State the sign (increasing/decreasing), the size, and the units in context.

From a graph. If you are given a graph rather than a rule, the instantaneous rate is the gradient of the drawn tangent — read two clear points on the tangent and use \(\dfrac{\text{rise}}{\text{run}}\).

Example 1 — Average rate

Find the average rate of change of $f(x)=x^{2}$ over $[1,3]$.

Solution

Evaluate $f$ at each endpoint, then take the gradient of the chord.

$f(1)$	$=$	$1$
$f(3)$	$=$	$9$
$\dfrac{f(3)-f(1)}{3-1}$	$=$	$\dfrac{9-1}{2}=4$

\frac{9 - 1}{2} = 4

Example 2 — Instantaneous rate

For the same $f(x)=x^{2}$, find the instantaneous rate of change at $x=3$.

Solution

Differentiate, then substitute $x=3$.

$f'(x)$	$=$	$2x$
$f'(3)$	$=$	$2(3)=6$

The instantaneous rate $6$ differs from the average rate $4$: the curve is steeper at $x=3$ than on average across $[1,3]$.

f^{'} (3) = 6

Example 3 — Exponential model

An investment is $V=2000e^{0.08t}$ dollars after $t$ years. Find the rate of growth at $t=10$.

Solution

Differentiate (the chain rule brings the $0.08$ down), then substitute $t=10$.

$\dfrac{dV}{dt}$	$=$	$160e^{0.08t}$
$\left.\dfrac{dV}{dt}\right\|_{t=10}$	$=$	$160e^{0.8}\approx 356$

The rate is about $\$356$ per year, and it is positive, so the value is increasing.

160 e^{0.8} \approx 356

Example 4 — Sign of a rate

A coffee cools as $T=25+60e^{-0.1t}$ degrees Celsius after $t$ minutes. Find and interpret the rate at $t=5$.

Solution

Differentiate, then substitute $t=5$.

$\dfrac{dT}{dt}$	$=$	$-6e^{-0.1t}$
$\left.\dfrac{dT}{dt}\right\|_{t=5}$	$=$	$-6e^{-0.5}\approx -3.6$

The rate is $-3.6\,^{\circ}$C/min. It is negative, so the coffee is cooling at that instant.

- 6 e^{- 0.5} \approx - 3.6

Common pitfalls

Average is not instantaneous. The average rate over an interval (the chord gradient) is usually different from the instantaneous rate at a point (the tangent gradient). Read the question: "over" or "average" $\Rightarrow$ chord; "at $t=\ldots$" $\Rightarrow$ derivative.

Do not forget to differentiate. An instantaneous rate is $f'(a)$, not $f(a)$. Substituting into $f$ itself gives the value of the quantity, not its rate of change.

Keep the sign and the units. A negative rate means the quantity is decreasing — do not drop the minus sign. Always give the units of the rate (quantity per unit of the variable).

Frequently asked questions

What is the average rate of change?

The change in the function over the change in the input, $\dfrac{f(b)-f(a)}{b-a}$ — the gradient of the chord joining the two points on the graph.

What is the instantaneous rate of change?

The value of the derivative at a point, $f'(a)$. It equals the gradient of the tangent to the curve there and describes how fast the quantity changes at that instant.

How do you find an instantaneous rate of change?

Differentiate the function, then substitute the required value. E.g. $V=2000e^{0.08t}\Rightarrow \dfrac{dV}{dt}=160e^{0.08t}$, which is about $\$356$/year at $t=10$.

What does the sign of a rate of change tell you?

Positive means the quantity is increasing, negative means decreasing, and zero means momentarily stationary. The size shows how quickly it is changing.

How is the average rate different from the instantaneous rate?

The average rate is the chord gradient over an interval; the instantaneous rate is the tangent gradient at one point. They agree only when the graph is a straight line.

Why is the derivative called a rate of change?

Because it measures how the output responds to a small change in the input. For a quantity varying with time, the time derivative gives its rate of change at each instant, such as a velocity from a displacement.

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\(f(1)\)	\(=\)	\(1\)
\(f(3)\)	\(=\)	\(9\)
\(\dfrac{f(3)-f(1)}{3-1}\)	\(=\)	\(\dfrac{9-1}{2}=4\)

\(f'(x)\)	\(=\)	\(2x\)
\(f'(3)\)	\(=\)	\(2(3)=6\)

\(\dfrac{dV}{dt}\)	\(=\)	\(160e^{0.08t}\)
\(\left.\dfrac{dV}{dt}\right\|_{t=10}\)	\(=\)	\(160e^{0.8}\approx 356\)

\(\dfrac{dT}{dt}\)	\(=\)	\(-6e^{-0.1t}\)
\(\left.\dfrac{dT}{dt}\right\|_{t=5}\)	\(=\)	\(-6e^{-0.5}\approx -3.6\)