Instantaneous rates and limiting secants | Year 11

01 / LEARN

The idea, step by step

Shortening an interval

An instantaneous rate is the tangent slope at one input. Hold one point fixed and move the other closer from either side. The secant gradients may approach one number.

A small interval gives an approximation, not automatically an exact answer. Compare several interval sizes. At a corner the limiting slopes from left and right may disagree.

Difference quotient and limit

Change the input from a to a+h, with h nonzero. Form the quotient, simplify, then inspect its value as h tends to zero. Polynomial expansion often removes the factor of h that initially appears in the denominator.

Do not substitute h=0 into the unsimplified quotient: 0/0 is not a tangent gradient. The limiting argument explains the exact rate approached by nearby secants.

f^{'} (a) = lim_{h \to 0} \frac{f (a + h) - f (a)}{h} .

A common trap

Calling one small-interval calculation an exact derivative.

Instead: A secant is an approximation; justify the exact derivative through a limit or an established rule.

02 / WORKED EXAMPLES

See the thinking

Example 1

Worked example 1

Find the instantaneous gradient of $f (x) = x^{2}$ at 3 from a difference quotient.

$f (3 + h) - f (3) = 6 h + h^{2}$ .
For $h \neq 0$ , divide by h to get $6 + h$ .
As $h \to 0$ , the quotient tends to 6.

Example 2

Worked example 2

For $s (t) = t^{3}$ , estimate the rate at 1 using $h = 0.1, 0.01$ , then find its limit.

$[s (1 + h) - s (1)] / h = 3 + 3 h + h^{2}$ for $h \neq 0$ .
The two secant estimates are 3.31 and 3.0301.
As $h \to 0$ , both $3 h$ and $h^{2}$ tend to 0. The exact limiting rate is 3; negative h gives the same limit.

Instantaneous rate 3 output units per time unit.

Check your understanding

A simplified difference quotient is $8 + 2 h$ . What is its limit as h tends to 0?

Show answer

The term $2 h$ tends to 0.

03 / WORKSHEET

Your turn

6 questions · 18 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Use a difference quotient for $f (x) = x^{2}$ at 1.
3 marks
Use a difference quotient for $f (x) = 3 x^{2}$ at 2.
3 marks
Find the gradient of $f (x) = x^{3}$ at 1 using its difference quotient.
3 marks
For $f (x) = x^{2}$ at 2, calculate secant gradients for $h = 1, 0.1, 0.01$ and state the limiting value.
3 marks
For $f (x) = ∣ x ∣$ at 0, compare the left and right difference quotients. Does one derivative exist?
3 marks
$s (t) = 5 t^{2}$ metres. Use a limiting quotient to find velocity at 3 seconds.
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Use a difference quotient for $f (x) = x^{2}$ at 1.
3 marks
$[{(1 + h)}^{2} - 1] / h = 2 + h$ for $h \neq 0$ .
The limit is 2.
2
Use a difference quotient for $f (x) = 3 x^{2}$ at 2.
3 marks
$[3 {(2 + h)}^{2} - 12] / h = 12 + 3 h$ .
Its limit is 12.
12
Find the gradient of $f (x) = x^{3}$ at 1 using its difference quotient.
3 marks
$[{(1 + h)}^{3} - 1] / h = 3 + 3 h + h^{2}$ .
The limit is 3.
3
For $f (x) = x^{2}$ at 2, calculate secant gradients for $h = 1, 0.1, 0.01$ and state the limiting value.
3 marks
The quotient is $4 + h$ .
Substitute the three h-values and then let h approach 0.
5, 4.1, 4.01; limit 4
For $f (x) = ∣ x ∣$ at 0, compare the left and right difference quotients. Does one derivative exist?
3 marks
$∣ h ∣ / h$ equals 1 for positive h and −1 for negative h.
The two limiting slopes disagree.
Left slope −1, right slope 1; no derivative.
$s (t) = 5 t^{2}$ metres. Use a limiting quotient to find velocity at 3 seconds.
3 marks
$[5 {(3 + h)}^{2} - 45] / h = 30 + 5 h$ .
The limit is 30 metres per second.
30 m/s

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 Mathematics topic 6: rates of change, derivatives and applications. Introductory polynomial calculus with explicit modelling and interpretation.

Official curriculum source ↗

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