Open Math›Year 11›Mathematics Advanced›Introduction to differentiation›Rates of change

Rates of change

Find average rates of change using the secant gradient formula and estimate instantaneous rates of change by evaluating average rates over successively smaller intervals.

Worked examples

Finding and interpreting average rate of change

Straightforward

Problem

Find the average rate of change of

f(x) = x^2 + 3

from

x = 1

x = 4

. Interpret your answer geometrically.

Evaluate the function at both endpoints.

f(1) = 1^2 + 3 = 4

f(4) = 4^2 + 3 = 19

Apply the average rate of change formula with

a = 1

and

b = 4

\frac{f(b) - f(a)}{b - a} = \frac{f(4) - f(1)}{4 - 1} = \frac{19 - 4}{3} = \frac{15}{3} = 5

State the geometric interpretation.

The average rate of change of

5

is the gradient of the secant line joining

(1,\, 4)

and

(4,\, 19)

on the graph of

f

Answer

The average rate of change is

5

. Geometrically, this is the gradient of the secant through

(1, 4)

and

(4, 19)

Estimating instantaneous rate of change using smaller intervals

Moderate

Problem

For

f(x) = x^2

, estimate the instantaneous rate of change at

x = 3

by calculating the average rate of change over

[3,\, 3+h]

for

h = 1

h = 0.1

, and

h = 0.01

. Hence state the value the instantaneous rate of change appears to approach.

Find the average rate of change for

h = 1

using

[3, 4]

f(3) = 9, \quad f(4) = 16

\frac{f(4) - f(3)}{4 - 3} = \frac{16 - 9}{1} = 7

Find the average rate of change for

h = 0.1

using

[3, 3.1]

f(3.1) = (3.1)^2 = 9.61

\frac{f(3.1) - f(3)}{3.1 - 3} = \frac{9.61 - 9}{0.1} = \frac{0.61}{0.1} = 6.1

Find the average rate of change for

h = 0.01

using

[3, 3.01]

f(3.01) = (3.01)^2 = 9.0601

\frac{f(3.01) - f(3)}{3.01 - 3} = \frac{9.0601 - 9}{0.01} = \frac{0.0601}{0.01} = 6.01

Identify the value that the average rates are approaching as

h \to 0

h = 1 \implies 7, \quad h = 0.1 \implies 6.1, \quad h = 0.01 \implies 6.01

The values are approaching

6

Answer

The instantaneous rate of change of

f(x) = x^2

x = 3

is approximately

6

. (The exact value, found by differentiation, is

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

Average and instantaneous velocity in context

Challenging

Problem

A particle moves along a straight line. Its position in metres is

s(t) = -t^2 + 6t

at time

t

seconds. Find the average velocity from

t = 1

t = 5

, and estimate the instantaneous velocity at

t = 2

using the interval

[2,\, 2.01]

. Explain the physical meaning of each result.

Find

s(1)

and

s(5)

s(1) = -(1)^2 + 6(1) = -1 + 6 = 5 \text{ m}

s(5) = -(5)^2 + 6(5) = -25 + 30 = 5 \text{ m}

Calculate the average velocity from

t = 1

t = 5

\text{Average velocity} = \frac{s(5) - s(1)}{5 - 1} = \frac{5 - 5}{4} = \frac{0}{4} = 0 \text{ m/s}

Interpret the average velocity. A result of

0

m/s does not mean the particle was stationary.

The particle was at

5

m at

t = 1

and also at

5

m at

t = 5

. The net displacement over the interval was zero, so the average velocity is

0

m/s. The particle moved during the interval but returned to the same position.

Estimate the instantaneous velocity at

t = 2

using the interval

[2, 2.01]

. First calculate

s(2)

and

s(2.01)

s(2) = -(2)^2 + 6(2) = -4 + 12 = 8 \text{ m}

s(2.01) = -(2.01)^2 + 6(2.01) = -4.0401 + 12.06 = 8.0199 \text{ m}

Divide the change in position by the time interval to estimate the instantaneous velocity.

\text{Instantaneous velocity} \approx \frac{s(2.01) - s(2)}{0.01} = \frac{8.0199 - 8}{0.01} = \frac{0.0199}{0.01} \approx 1.99 \text{ m/s}

This is very close to

2

m/s. (The derivative

s'(t) = -2t + 6

gives

s'(2) = 2

m/s exactly.)

Answer

Average velocity from

t = 1

t = 5

0

m/s (the particle returned to its position at

t = 1

, so net displacement is zero). Instantaneous velocity at

t = 2

is approximately

1.99

m/s, which approaches

2

m/s as the interval shrinks.

Practise

Q1·Straightforward

Find the average rate of change of

f(x) = x^2

from

x = 1

x = 4

Explanation

f(1) = 1^2 = 1

and

f(4) = 4^2 = 16

\frac{f(4) - f(1)}{4 - 1} = \frac{16 - 1}{3} = \frac{15}{3} = 5

Q2·Straightforward

Find the average rate of change of

f(x) = 2x + 3

from

x = 0

x = 5

Explanation

f(0) = 3

and

f(5) = 10 + 3 = 13

\frac{f(5) - f(0)}{5 - 0} = \frac{13 - 3}{5} = \frac{10}{5} = 2

This equals the gradient of the line, as expected for a linear function.

Q3·Straightforward

Find the average rate of change of

g(x) = x^2 + 4x

from

x = 0

x = 3

Explanation

g(0) = 0 + 0 = 0

and

g(3) = 9 + 12 = 21

\frac{g(3) - g(0)}{3 - 0} = \frac{21 - 0}{3} = \frac{21}{3} = 7

Q4·Straightforward

The average rate of change of a function

f

from

x = a

x = b

can be interpreted geometrically as the gradient of which line?

The tangent to the graph of $f$ at $x = a$The tangent to the graph of $f$ at $x = b$The secant through $(a,\, f(a))$ and $(b,\, f(b))$The normal to the graph of $f$ at $x = a$

Explanation

The average rate of change

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

is the rise over run between the two points

(a, f(a))

and

(b, f(b))

. This is exactly the gradient of the secant line through those two points. The tangent only touches the curve at a single point and represents the instantaneous rate of change.

Q5·Moderate

For

f(x) = x^2

, find the average rate of change over the interval

[2,\, 3]

Explanation

f(2) = 4

and

f(3) = 9

\frac{f(3) - f(2)}{3 - 2} = \frac{9 - 4}{1} = 5

Q6·Moderate

For

f(x) = x^2

, find the average rate of change over the interval

[2,\, 2.1]

. Give your answer correct to one decimal place.

Explanation

f(2) = 4

and

f(2.1) = (2.1)^2 = 4.41

\frac{f(2.1) - f(2)}{2.1 - 2} = \frac{4.41 - 4}{0.1} = \frac{0.41}{0.1} = 4.1

Q7·Moderate

For

f(x) = x^2

, find the average rate of change over the interval

[2,\, 2.01]

. Give your answer correct to two decimal places.

Explanation

f(2) = 4

and

f(2.01) = (2.01)^2 = 4.0401

\frac{f(2.01) - f(2)}{2.01 - 2} = \frac{4.0401 - 4}{0.01} = \frac{0.0401}{0.01} = 4.01

Q8·Moderate

The average rates of change of

f(x) = x^2

near

x = 2

are:

5

(over

[2, 3]

4.1

(over

[2, 2.1]

4.01

(over

[2, 2.01]

). As the interval width approaches zero, what value does the average rate of change appear to approach?

$2$$4$$5$$8$

Explanation

The values

5 \to 4.1 \to 4.01

are converging to

4

h \to 0

. This limiting value is the instantaneous rate of change (the derivative) of

f(x) = x^2

x = 2

. Since

f'(x) = 2x

, we have

f'(2) = 4

, confirming the pattern.

Q9·Challenging

A particle moves along a straight line. Its position in metres after

t

seconds is

s(t) = 3t^2 + 2t

. Find the particle's average velocity (in m/s) from

t = 1

t = 4

Explanation

s(1) = 3(1)^2 + 2(1) = 3 + 2 = 5

s(4) = 3(4)^2 + 2(4) = 48 + 8 = 56

\text{Average velocity} = \frac{s(4) - s(1)}{4 - 1} = \frac{56 - 5}{3} = \frac{51}{3} = 17 \text{ m/s}

This is the gradient of the secant on the position-time graph from

t = 1

t = 4

Q10·Challenging

For a particle with position

s(t) = 3t^2 + 2t

metres, estimate the instantaneous velocity (in m/s) at

t = 2

by finding the average velocity over the interval

[2,\, 2.01]

. Give your answer correct to two decimal places.

Explanation

s(2) = 3(4) + 2(2) = 12 + 4 = 16

s(2.01) = 3(2.01)^2 + 2(2.01) = 3(4.0401) + 4.02 = 12.1203 + 4.02 = 16.1403

\text{Average velocity} = \frac{s(2.01) - s(2)}{0.01} = \frac{16.1403 - 16}{0.01} = \frac{0.1403}{0.01} = 14.03 \text{ m/s}

The derivative

s'(t) = 6t + 2

gives

s'(2) = 14

m/s, confirming this estimate.

Q11·Challenging

On a position-time graph, the instantaneous velocity of a particle at a given moment corresponds to the gradient of which of the following?

The secant joining the start and end of the journeyThe tangent to the position-time graph at that momentThe secant joining the start of the journey to that momentA horizontal line through the position at that moment

Explanation

As the time interval

[t, t+h]

shrinks (

h \to 0

), the secant through

(t, s(t))

and

(t+h, s(t+h))

rotates towards the tangent at the point

(t, s(t))

. In the limit, the gradient of the secant becomes the gradient of the tangent, which is the instantaneous velocity.

Q12·Challenging

A ball is thrown upward and lands back at the same height after 4 seconds. Its height is modelled by

h(t) = 20t - 5t^2

. The average rate of change of height from

t = 0

t = 4

0

m/s. Which statement correctly interprets this result?

The ball was stationary for the entire 4 seconds.The ball's instantaneous velocity was $0$ m/s at every point during the flight.The ball returned to its initial height, so the net change in position was zero.The ball moved at a constant speed of $0$ m/s throughout.

Explanation

h(0) = 0

and

h(4) = 80 - 80 = 0

\frac{h(4) - h(0)}{4 - 0} = \frac{0 - 0}{4} = 0 \text{ m/s}

This does not mean the ball was stationary. It means the net displacement over the interval was zero — the ball went up and came back to the same height. The instantaneous velocity was positive during the upward phase, zero at the peak, and negative during the descent. The average rate of change only captures the overall start-to-finish change.