01 · Volume rate
V(t) = 40 + 12t − t² litres for 0 ≤ t ≤ 4 minutes. Find the filling rate at t = 3.
Hint
Differentiate V with respect to t.
Worked solution
V′(t) = 12 − 2t
V′(3) = 6 litres/min.
The volume is increasing at that instant.
Understand · explore · practise
Rates of change
Interpret derivatives with units and distinguish average from instantaneous rates. Apply differentiation to volume, displacement, velocity, acceleration and physical graphs.
Before you startDifferentiation rules and interpreting graphs
01 / What does dy/dx measure?
Name both the changing quantity and its input.
If V is a volume in litres and t is time in minutes, dV/dt is a rate in litres per minute. If A is area in cm² and r is radius in cm, dA/dr measures area change per centimetre of radius.
Units of derivative = units of output / units of input
A positive derivative means the output is increasing with the named input. A negative derivative means it is decreasing. The notation tells you which relationship is being measured; dV/dr is not the same quantity as dV/dt.
02 / Average and instantaneous rates
A whole interval and a single instant answer different questions.
Suppose water volume is modelled by V(t) = 120 + 18t − t² litres for 0 ≤ t ≤ 8 minutes.
Average rate from t = 2 to t = 6:
[V(6) − V(2)]/(6 − 2)
= (192 − 152)/4 = 10 litres/min
The instantaneous rate is V′(t) = 18 − 2t. It is 14 litres/min at t = 2 and 6 litres/min at t = 6. The average need not equal either endpoint rate.
Move the clock yourself to inspect the value and its rate at the same instant. You can also switch to a displacement model and compare position with velocity.
Move the clock; compare value and rateMove at your pace
At t = 4 minutes, the water volume is 176 litres and its instantaneous rate is 10 litres per minute. The model is restricted to 0–8 minutes. Both charts show the same instant; the vertical quantities and units differ.
Watch a value increase while its rate decreases
Pause, replay or seek freely. The notes explain the same idea and stay in view.
03 / A decreasing rate can still be positive
Differentiate twice to describe how the rate changes.
For the water model, V′(t) = 18 − 2t stays positive throughout 0 ≤ t ≤ 8. The volume is still increasing. However V″(t) = −2 litres/min², so the filling rate is falling by 2 litres/min each minute.
The formula is a model for the stated interval. Extending it far beyond that interval would eventually predict falling and then negative water volume. A correct derivative does not justify an unrealistic extrapolation.
04 / Rates with respect to size
A geometric sensitivity need not be a time rate.
For a sphere of radius r cm, V = (4π/3)r³ cm³ and A = 4πr² cm².
dV/dr = 4πr²
dA/dr = 8πr
At r = 3 cm, dV/dr = 36π cm³ per cm and dA/dr = 24π cm² per cm. Writing the units as “per cm of radius” makes the meaning clear, even though dimensions can be simplified algebraically.
When radius depends on time
If r = 2t cm with t in seconds, substitute first: V(t) = (32π/3)t³. Then dV/dt = 32πt² cm³/s. At t = 1 this is 32π cm³/s; it is not dV/dr at r = 2, which is 16π cm³ per cm.
05 / Displacement, velocity and acceleration
Differentiate with respect to time.
For displacement s(t) = t³ − 6t² + 9t metres on 0 ≤ t ≤ 4 seconds:
Velocity v(t) = s′(t) = 3t² − 12t + 9 m/s
Acceleration a(t) = s″(t) = 6t − 12 m/s²
The velocity is zero at t = 1 and t = 3. It is positive before 1, negative between 1 and 3, and positive after 3 within the model interval. Speed is |v|, so a negative velocity does not mean a negative speed.
At t = 2, velocity is −3 m/s and acceleration is zero. Zero acceleration at an instant does not mean zero velocity. At a turning instant, velocity can be zero while acceleration is non-zero.
06 / Read a rate from a physical graph
The horizontal coordinate stays time; the vertical units change.
The upper schematic shows a damped oscillation in displacement, measured in centimetres. Its rate graph uses centimetres per second. At the marked turning times t = 1, 2 and 3 s, the displacement has a horizontal tangent and the velocity is zero.
Where the displacement rises, velocity is positive; where it falls, velocity is negative. Crossing the equilibrium position does not imply zero velocity. A qualitative sketch locates signs and turning times without determining every exact speed.
07 / Your turn
Include the derivative’s units and what its sign means.
Differentiate with respect to the variable named in each question.
02 · Average rate
For the same model, find the average rate from t = 0 to t = 4.
Hint
Use the difference in volumes divided by 4.
Worked solution
V(0) = 40, V(4) = 72
Average = (72 − 40)/4 = 8 litres/min.
03 · A circumference
For circumference C = 2πr, find dC/dr and interpret it.
Hint
π is a constant.
Worked solution
dC/dr = 2π.
Circumference increases by 2π length units per unit increase of radius.
04 · Sphere area
A = 4πr² cm². Find dA/dr at r = 5 cm.
Hint
Differentiate the square.
Worked solution
dA/dr = 8πr = 40π cm² per cm of radius.
05 · A reciprocal rate
r(t) = 18/t cm for t > 0 seconds. Find dr/dt at t = 3.
Hint
Use 18t⁻¹.
Worked solution
dr/dt = −18t⁻²
At t = 3: −2 cm/s.
The radius is decreasing.
06 · Velocity and acceleration
s(t) = 2t³ − 3t² + t metres. Find velocity and acceleration at t = 2 seconds.
Hint
Differentiate once and twice.
Worked solution
v(t) = 6t² − 6t + 1 ⇒ v(2) = 13 m/s
a(t) = 12t − 6 ⇒ a(2) = 18 m/s².
07 · Speed from signed velocity
A particle has velocity −5 m/s. State its speed and explain the minus sign.
Hint
Speed is a magnitude.
Worked solution
Its speed is 5 m/s. The minus sign says it is moving in the negative coordinate direction.
08 · A fractional displacement
For s(t) = (t² + 4)/√t metres, t > 0 seconds, find the acceleration.
Hint
Rewrite as t3/2 + 4t−1/2 and differentiate twice.
Worked solution
v(t) = (3/2)t1/2 − 2t−3/2
a(t) = (3/4)t−1/2 + 3t−5/2 m/s².
09 · Graph interpretation
A differentiable displacement graph rises to a maximum at t = 2 s, falls to a minimum at t = 5 s, then rises. These are its only horizontal tangents. Describe the velocity signs.
Hint
Read rising and falling rather than whether the displacement is above zero.
Worked solution
Velocity is zero at both turning times, changes from positive to negative at 2 s, and from negative to positive at 5 s. It is negative between the turns.
10 · A model limit
A volume model has a negative second derivative but a positive first derivative. Is its volume decreasing?
Hint
Distinguish volume from filling rate.
Worked solution
No. The volume is increasing because its first derivative is positive. Its rate of increase is decreasing because its second derivative is negative.
08 / Recap
A rate is a derivative with a meaning and units.
- Name the output and the input variable.
- Average rates use an interval; instantaneous rates use a derivative.
- The second derivative describes how the rate changes.
- Displacement, velocity and acceleration are successive derivatives in time.
- Speed is the magnitude of velocity.
- Keep model domains and physical limitations explicit.
Section 1 of 8 · What does dy/dx measure?