Rates and motion applications | Year 12

01 / LEARN

Understand the idea.

Position, velocity and acceleration

For straight-line motion, position $s (t)$ gives velocity $v = s^{'}$ and acceleration $a = v^{'}$ . Speed is $∣ v ∣$ : a negative velocity indicates direction, while speed remains nonnegative.

A particle is instantaneously at rest when $v = 0$ , but reverses only if velocity changes sign. It speeds up when velocity and acceleration have the same sign, and slows down when their signs differ, away from rest instants.

Rates involving a changing measurement

When a geometric measurement changes over time, its dependent quantities do too. Differentiate the relationship with respect to time: from $A = π r^{2}$ , the chain rule gives $\frac{d A}{d t} = 2 π r \frac{d r}{d t}$ .

Substitute the radius and its rate after differentiating. Substituting a fixed numerical radius first would erase its change. Check dimensions: an area rate uses square units per time and a volume rate uses cubic units per time.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Rest times and reversals

A particle has $s (t) = t^{3} - 6 t^{2} + 9 t$ m for $0 \leq t \leq 4$ s. Find rest times and direction changes.

Worked solution

1
Differentiate and factor velocity.
$v (t) = 3 t^{2} - 12 t + 9 = 3 (t - 1) (t - 3)$
Velocity is the derivative of position with respect to time.
2
Find the rest times in the model interval.
$v (t) = 0 t = 1 or t = 3, 0 \leq t \leq 4$
Both roots are allowed times and give zero instantaneous velocity.
3
Determine velocity signs throughout the interval.
$\begin{aligned} v (t) & > 0 & for 0 \leq t < 1 \\ v (t) & < 0 & for 1 < t < 3 \\ v (t) & > 0 & for 3 < t \leq 4 \end{aligned}$
The two linear factors have the same sign outside their roots and opposite signs between them.
4
Identify both direction changes.
$t = 1 s and t = 3 s$
Velocity changes positive to negative at 1 second and negative to positive at 3 seconds, so both rest instants are reversals.

Answer

It reverses direction at both $t = 1$ s and $t = 3$ s.

EXAMPLE 02

A changing circular area

A circle has radius increasing at 0.4 m/s. Find its area rate when $r = 5$ m.

Worked solution

1
Write the changing-area relationship.
$A (t) = π {[r (t)]}^{2}, \frac{d r}{d t} = 0.4 m s^{- 1}$
The radius depends on time, so area is a composite function of time.
2
Differentiate before substituting the radius.
$\frac{d A}{d t} = 2 π r \frac{d r}{d t}$
The chain rule accounts for the radius change.
3
Substitute the current radius and rate.
$\frac{d A}{d t} = 2 π (5) (0.4) = 4 π m^{2} s^{- 1}$
The positive result means area is increasing; metres times metres per second gives square metres per second.

Answer

$4 π m^{2} /s$

03 / YOUR TURN

Try it yourself.

Check your understanding

If $v = - 3$ m/s and $a = - 2$ m/s squared, is speed increasing or decreasing?

Show answer & reasoning

Increasing.

Here $v = - 3$ and $a = - 2$ have the same sign. The velocity is moving further below zero, so $∣ v ∣$ increases and the particle speeds up.

Keep practising

6 worksheet questions · 18 marks

Open worksheet Worked solutions

Read the questions online

For $s (t) = 2 t^{2} - 5 t + 1$ metres, with $t$ in seconds, find velocity and acceleration.
2 marks
For $s (t) = t^{3} - 3 t$ metres on $t \geq 0$ seconds, find the rest time and whether direction changes.
3 marks
For $s (t) = t^{3} - 6 t^{2} + 9 t$ metres on $0 \leq t \leq 4$ seconds, find displacement and total distance.
4 marks
For velocity $v (t) = t^{2} - 4 t + 3$ metres per second, with $t$ in seconds, find the speed and its instantaneous rate of change at $t = 2$ seconds.
3 marks
A cube has side length increasing at 0.2 cm/s. Find the volume rate when the side is 3 cm.
3 marks
A cost model is $C (q) = 0.02 q^{2} + 5 q + 100$ dollars. Find and interpret marginal cost at 50 units.
3 marks

Read the worked solutions

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

For $s (t) = 2 t^{2} - 5 t + 1$ metres, with $t$ in seconds, find velocity and acceleration.
2 marks
Differentiate position once for velocity.
Differentiate again for acceleration.
$v = 4 t - 5$ m/s; $a = 4$ m/s squared.
For $s (t) = t^{3} - 3 t$ metres on $t \geq 0$ seconds, find the rest time and whether direction changes.
3 marks
The velocity is $3 (t - 1) (t + 1)$ .
On the physical domain it changes sign at 1.
$t = 1$ s; direction changes from negative to positive.
For $s (t) = t^{3} - 6 t^{2} + 9 t$ metres on $0 \leq t \leq 4$ seconds, find displacement and total distance.
4 marks
The reversal times are 1 and 3.
Positions at $0, 1, 3, 4$ are $0, 4, 0, 4$ .
Add absolute position changes for distance.
Displacement 4 m; distance 12 m.
For velocity $v (t) = t^{2} - 4 t + 3$ metres per second, with $t$ in seconds, find the speed and its instantaneous rate of change at $t = 2$ seconds.
3 marks
The velocity at 2 is $- 1$ .
Acceleration is $2 t - 4$ , which is zero at 2.
Speed 1 m/s; acceleration is zero at that instant, so the instantaneous rate of change of speed is zero.
A cube has side length increasing at 0.2 cm/s. Find the volume rate when the side is 3 cm.
3 marks
Use $V = x^{3}$ and differentiate with respect to time.
Then $\frac{d V}{d t} = 3 x^{2} \frac{d x}{d t} = 3 (9) (0.2)$ .
$5.4 {cm}^{3} /s$
A cost model is $C (q) = 0.02 q^{2} + 5 q + 100$ dollars. Find and interpret marginal cost at 50 units.
3 marks
Differentiate with respect to quantity: $C^{'} (q) = 0.04 q + 5$ .
At 50, this is 7; it is a local rate, not the exact finite cost of the next unit.
$7$ per additional unit approximately.

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Curriculum reference & lesson scope

SACE Stage 2 Mathematical Methods

Further differentiation and applications from the current SACE Methods outline. Logarithmic derivatives have their own chapter; integration follows later. Trigonometric calculus uses radians.

Official curriculum source ↗

In this lesson

Understand the idea.

Position, velocity and acceleration

Rates involving a changing measurement

See the method in action.

Rest times and reversals

Differentiate and factor velocity.

Find the rest times in the model interval.

Determine velocity signs throughout the interval.

Identify both direction changes.

A changing circular area

Write the changing-area relationship.

Differentiate before substituting the radius.

Substitute the current radius and rate.

Try it yourself.

Check your understanding

Keep practising