Rates of Change Revision Notes for VCE SSCE Mathematical Methods

Rates of Change

Introduction

Differentiation is a powerful tool that goes beyond finding gradients of curves. It allows us to analyse how quantities change in many practical situations. Whether it's the speed of a falling object, the rate at which a balloon deflates, or how quickly coffee cools down, derivatives help us understand and measure these changes.

The key insight is that derivatives represent rates of change - they tell us how fast one quantity changes with respect to another. This extends the geometric interpretation of derivatives as gradients to solve real-world problems involving motion, growth, decay, and many other dynamic processes.

Average rate of change

The average rate of change tells us how much a function changes, on average, over a specific interval. It's like calculating the average speed of a car journey by dividing the total distance by the total time.

For a function with rule $f(x)$ , the average rate of change over the interval $x \in [a, b]$ is:

\text{Average rate of change} = \frac{f(b) - f(a)}{b - a}

This formula calculates the change in function values divided by the change in x values.

infoNote

Geometric Interpretation

The average rate of change represents the gradient of the straight line connecting the points $(a, f(a))$ and $(b, f(b))$ on the curve. This line is called a secant line, and its gradient gives us the average slope over that interval.

Instantaneous rate of change

The instantaneous rate of change measures how fast a function is changing at one specific point, rather than over an interval. This is where derivatives become essential.

For a function with rule $f(x)$ , the instantaneous rate of change at $x = a$ is given by the derivative:

f'(a)

The derivative $\frac{dy}{dx}$ gives the instantaneous rate of change of y with respect to x.

chatImportant

The Power of the Derivative

While the average rate of change tells us about overall change over an interval, the instantaneous rate of change tells us the exact rate at a specific moment. This is like the difference between your average speed on a journey versus your speed at one particular instant shown on your speedometer.

Interpreting the sign of the derivative

The sign of the derivative tells us about the behaviour of the function:

If $\frac{dy}{dx} > 0$ , then $y$ is increasing as $x$ increases
If $\frac{dy}{dx} < 0$ , then $y$ is decreasing as $x$ increases

Think of it this way: a positive derivative means you're going uphill, while a negative derivative means you're going downhill.

Worked example: Polynomial function

lightbulbExample

Worked Example: Finding Different Rates of Change

For the function with rule $f(x) = x^2 + 2x$ , we can find different types of rates of change.

Part (a): Finding the average rate of change for $x \in [2, 3]$

\text{Average rate of change} = \frac{f(3) - f(2)}{3 - 2}

First, calculate $f(3) = 3^2 + 2(3) = 9 + 6 = 15$

Then, calculate $f(2) = 2^2 + 2(2) = 4 + 4 = 8$

\text{Average rate of change} = \frac{15 - 8}{1} = 7

Part (b): Finding the average rate of change for the interval $[2, 2 + h]$

\text{Average rate of change} = \frac{f(2 + h) - f(2)}{2 + h - 2}

Calculate $f(2 + h) = (2 + h)^2 + 2(2 + h) = 4 + 4h + h^2 + 4 + 2h = 8 + 6h + h^2$

= \frac{(8 + 6h + h^2) - 8}{h}

= \frac{6h + h^2}{h}

= 6 + h

Part (c): Finding the instantaneous rate of change at $x = 2$

The derivative is $f'(x) = 2x + 2$

When $x = 2$ : $f'(2) = 2(2) + 2 = 6$

Connection: Notice that this confirms our result from part (b): as $h$ approaches $0$ in the expression $6 + h$ , we get 6. This shows how the instantaneous rate of change is the limit of average rates of change as the interval shrinks to a point.

Worked example: Balloon volume

lightbulbExample

Worked Example: Practical Application with Volume

A balloon develops a microscopic leak and gradually decreases in volume. Its volume, $V$ cm³, at time $t$ seconds is given by:

V = 600 - 10t - \frac{1}{100}t^2, \quad t \geq 0

Part (a): Finding the rate of change of volume

To find how fast the volume is changing, we differentiate with respect to time:

\frac{dV}{dt} = -10 - \frac{t}{50}

Part (b): After 10 seconds

When $t = 10$ :

\frac{dV}{dt} = -10 - \frac{10}{50} = -10 - \frac{1}{5} = -10\frac{1}{5}

The volume is decreasing at a rate of $10\frac{1}{5}$ cm³ per second.

Part (c): After 20 seconds

When $t = 20$ :

\frac{dV}{dt} = -10 - \frac{20}{50} = -10 - \frac{2}{5} = -10\frac{2}{5}

The volume is decreasing at a rate of $10\frac{2}{5}$ cm³ per second.

Observation: Notice that the rate of decrease is getting faster as time passes (the negative value is becoming larger in magnitude). This makes physical sense - as the balloon deflates, the pressure difference might cause faster leaking.

Part (d): Validity of the model

The model only makes sense when $V \geq 0$ . To find when the volume reaches zero:

600 - 10t - \frac{1}{100}t^2 = 0

Multiply through by 100:

60000 - 1000t - t^2 = 0

t^2 + 1000t - 60000 = 0

Using the quadratic formula or completing the square:

t = 100(\sqrt{31} - 5) \quad \text{or} \quad t = -100(\sqrt{31} + 5)

Since time cannot be negative, the model is suitable for $0 \leq t \leq 100(\sqrt{31} - 5)$ seconds (approximately $0 \leq t \leq 57$ seconds).

chatImportant

Always Check Model Validity

When working with real-world applications, always check whether your mathematical model makes physical sense. In this case, the volume cannot be negative, so we need to find when $V = 0$ to determine the valid time range for our model.

Worked example: Cooling liquid

lightbulbExample

Worked Example: Temperature Change with Exponential Decay

A pot of liquid is removed from the stove when it reaches 80°C and placed on a kitchen bench where the temperature is 20°C. The temperature of the liquid, $T$ °C, at time $t$ minutes is given by:

T = 20 + 60e^{-0.3t}

Part (a): Finding the rate of change in terms of $T$

First, rearrange the equation to express the exponential term:

T = 20 + 60e^{-0.3t}

T - 20 = 60e^{-0.3t}

e^{-0.3t} = \frac{T - 20}{60}

Now differentiate with respect to time:

\frac{dT}{dt} = -18e^{-0.3t}

Substitute the expression for $e^{-0.3t}$ :

\frac{dT}{dt} = -18 \times \frac{T - 20}{60}

= -3 \times \frac{T - 20}{10}

= 0.3(20 - T)

Part (b): When $T = 80$ °C

\frac{dT}{dt} = 0.3(20 - 80) = 0.3(-60) = -18

The liquid is cooling at a rate of 18°C per minute.

Part (c): When $T = 30$ °C

\frac{dT}{dt} = 0.3(20 - 30) = 0.3(-10) = -3

The liquid is cooling at a rate of 3°C per minute.

Physical Interpretation: Notice that the liquid cools much faster when it's hotter (greater temperature difference from room temperature), and the cooling rate decreases as the liquid approaches room temperature. This follows Newton's Law of Cooling, where the rate of cooling is proportional to the temperature difference.

infoNote

Understanding the Formula $\frac{dT}{dt} = 0.3(20 - T)$

This elegant form shows that the cooling rate depends on $(20 - T)$ , which is the temperature difference between the room (20°C) and the liquid ( $T$ °C). When the liquid is much hotter than the room, this difference is large (negative), so cooling is rapid. As the liquid approaches room temperature, the difference approaches zero, and cooling slows down.

bookmarkSummary

Key Points to Remember:

Average rate of change over $[a, b]$ is calculated using $\frac{f(b) - f(a)}{b - a}$ and represents the overall change over an interval
Instantaneous rate of change at $x = a$ is given by the derivative $f'(a)$ and measures the exact rate at that specific point
A positive derivative indicates the function is increasing, whilst a negative derivative indicates the function is decreasing
The derivative $\frac{dy}{dx}$ tells us how fast y is changing with respect to x
Real-world applications often use derivatives to model rates such as volume change, temperature change, or velocity
Always check the validity of your model by considering physical constraints (like volume cannot be negative, time cannot be negative)
The sign and magnitude of the derivative give us valuable information about the behaviour of the function

Rates of Change (VCE SSCE Mathematical Methods): Revision Notes

Rates of Change

Introduction

Average rate of change

Instantaneous rate of change

Interpreting the sign of the derivative

Worked example: Polynomial function

Worked example: Balloon volume

Worked example: Cooling liquid

Explore VCE SSCE Mathematical Methods Model Answers by Topics

Tangents and Normals

Rates of Change

Stationary Points

Absolute Maximum and Minimum Values

Maximum and Minimum Problems

Families of Functions

Applications to Motion in a Straight Line

Newton’s Method for Finding Solutions to Equations

Explore VCE SSCE Mathematical Methods Quizzes by Topics

Tangents and Normals

Rates of Change

Stationary Points

Absolute Maximum and Minimum Values

Maximum and Minimum Problems

Families of Functions

Applications to Motion in a Straight Line

Newton’s Method for Finding Solutions to Equations

Explore VCE SSCE Mathematical Methods Flashcards by Topics

Tangents and Normals

Rates of Change

Stationary Points

Absolute Maximum and Minimum Values

Maximum and Minimum Problems

Families of Functions

Applications to Motion in a Straight Line

Newton’s Method for Finding Solutions to Equations

Join 100,000+ SSCE students studying Revision Notes with us.