Rates of Change
Understand how derivatives describe instantaneous rates of change, including velocity, acceleration, and practical applications.
Understanding Rates of Change
A rate of change measures how quickly one quantity changes with respect to another. The derivative dy/dx gives the instantaneous rate of change of y with respect to x at any given point.
Average Rate of Change
The change in y divided by the change in x between two points: (y2 - y1) / (x2 - x1). This is the gradient of the secant line.
Instantaneous Rate of Change
The derivative dy/dx at a specific point. This is the gradient of the tangent line and tells us the exact rate at that instant.
Velocity and Acceleration
The most common application of rates of change in Year 11 is motion. If x(t) represents the position (displacement) of an object at time t:
Velocity = dx/dt
The first derivative of displacement. Positive means moving forward; negative means moving backward.
Acceleration = dv/dt = d2x/dt2
The second derivative of displacement (or first derivative of velocity). Positive means speeding up (in the positive direction); negative means decelerating.
Key insight: An object is at rest when v(t) = 0, and changes direction when v(t) changes sign.
Practical Applications
Rates of change appear in many real-world contexts beyond motion:
Population Growth
dP/dt represents how fast a population is growing at time t.
Volume Flow
dV/dt is the rate at which volume changes, e.g. water filling a tank.
Temperature Change
dT/dt measures how fast temperature changes over time (cooling/heating).
Cost and Revenue
dC/dx is the marginal cost -- the cost of producing one additional unit.
Key Vocabulary
Instantaneous Rate
The rate of change at a specific instant, found by evaluating the derivative at a particular value of the independent variable.
Velocity
The rate of change of displacement with respect to time: v = ds/dt. It includes direction (positive or negative).
Acceleration
The rate of change of velocity with respect to time: a = dv/dt = d2s/dt2.
Displacement
The position of an object relative to a reference point, often written as s(t) or x(t) as a function of time.
Worked Examples
A particle's displacement is s = 3t2 - 12t + 5 metres. Find its velocity at t = 4 seconds.
Step 1: Find v(t) = ds/dt = 6t - 12.
Step 2: Substitute t = 4: v(4) = 6(4) - 12 = 24 - 12 = 12.
Answer: The velocity at t = 4 is 12 m/s.
When is the particle at rest if s = t3 - 6t2 + 9t?
Step 1: v(t) = ds/dt = 3t2 - 12t + 9.
Step 2: Set v(t) = 0: 3t2 - 12t + 9 = 0, so t2 - 4t + 3 = 0.
Step 3: Factorise: (t - 1)(t - 3) = 0, so t = 1 or t = 3.
Answer: The particle is at rest at t = 1 s and t = 3 s.
Water flows into a tank so that the volume is V = 50t - 2t2 litres after t minutes. Find the rate of flow at t = 10.
Step 1: dV/dt = 50 - 4t.
Step 2: At t = 10: dV/dt = 50 - 4(10) = 50 - 40 = 10.
Answer: The rate of flow at t = 10 is 10 litres per minute.
Knowledge Check
Select the correct answer for each question. Click "Check Answer" to see if you are right.
Question 1
If s = 5t2 + 3t, what is the velocity v(t)?
Question 2
A particle moves with displacement s = t3 - 3t. At what time is the particle at rest?
Question 3
If the velocity is v(t) = 6t - 2, what is the acceleration?
Question 4
A population is modelled by P(t) = 200 + 50t - t2. What is the rate of population change at t = 10?
Question 5
What does a negative velocity indicate about the motion of a particle?
Key Concepts Summary
- ● The derivative gives the instantaneous rate of change of one quantity with respect to another.
- ● Velocity = ds/dt (first derivative of displacement); Acceleration = dv/dt (second derivative).
- ● A particle is at rest when v(t) = 0, and changes direction when velocity changes sign.
- ● Rates of change apply to many fields: physics, biology, economics, and engineering.
- ● Always include correct units in your answer (e.g. m/s, L/min, people/year).