Introduction to Calculus: Rates of Change

What calculus is really about

Calculus is the mathematics of change. Its first big idea is the rate of change: how fast one quantity changes compared to another. How fast is a car speeding up? How quickly is a population growing? How steep is a hill at one exact spot? Each is a rate of change, and calculus gives us a precise way to measure it.

You already know one rate of change well: the gradient of a straight line. If you have not met it recently, review straight-line graphs and gradients before going on. Calculus extends that same idea from straight lines to curves.

Average rate of change

On a straight line the gradient is the same everywhere. On a curve the steepness keeps changing, so we start with the average rate of change between two points.

The average rate of change of y between x₁ and x₂ is:

average rate = (y₂ − y₁) / (x₂ − x₁)

This is exactly the gradient of the chord — the straight line joining the two points on the curve.

Worked example 1: For the curve y = x², find the average rate of change between x = 2 and x = 4.

1. At x = 2: y = 2² = 4, giving the point (2, 4).
2. At x = 4: y = 4² = 16, giving the point (4, 16).
3. average rate = (16 − 4) / (4 − 2) = 12 / 2 = 6.

So between x = 2 and x = 4 the curve rises, on average, 6 units of y per unit of x.

Worked example 2: A ball's height is h = 20t − 5t² metres after t seconds. Find its average speed between t = 1 and t = 3.

1. At t = 1: h = 20(1) − 5(1)² = 20 − 5 = 15 m.
2. At t = 3: h = 20(3) − 5(3)² = 60 − 45 = 15 m.
3. average rate = (15 − 15) / (3 − 1) = 0 / 2 = 0 m/s.

The average velocity is 0 — the ball ended at the same height it started, having gone up and come back down. Average rate of change can hide what happens in between.

Instantaneous rate of change

The average rate tells us nothing about a single instant. To find the rate at one exact point, we use the gradient of the tangent — the straight line that just touches the curve there.

The clever idea: take a chord from our point to a second point close by, then slide the second point closer and closer. As the gap shrinks, the chord swings round until it becomes the tangent. The gradient it approaches is the instantaneous rate of change.

Second point x	Chord gradient on y = x² from (3, 9)
4	(16 − 9)/(4 − 3) = 7
3.5	(12.25 − 9)/(3.5 − 3) = 6.5
3.1	(9.61 − 9)/(3.1 − 3) = 6.1
3.01	(9.0601 − 9)/(3.01 − 3) = 6.01

The chord gradients are closing in on 6. That target value is what we call a limit, and it is the gradient of the tangent at x = 3.

Finding the gradient with algebra

Numbers only suggest the answer is 6. Algebra proves it. Let the second point be a small step h to the right of x.

Worked example 3: Find the gradient of y = x² at a general point x.

1. Two points: (x, x²) and (x + h, (x + h)²).
2. Chord gradient = ((x + h)² − x²) / ((x + h) − x).
3. Expand the top: (x² + 2xh + h² − x²) / h = (2xh + h²) / h.
4. Factor and cancel h: = (h(2x + h)) / h = 2x + h.
5. Now let h → 0: the gradient approaches 2x.

So the gradient of y = x² at any x is 2x. At x = 3 that is 2 × 3 = 6 — matching the table exactly, with proof.

Why cancel before letting h → 0? If you set h = 0 too early you get 0/0, which is meaningless. Simplifying first removes the division by h, so the limit becomes safe to take. This is the heart of every calculus calculation.

Worked example 4: Find the gradient of y = x² + 3x at x = 5, using the limit method.

1. Points: (x, x² + 3x) and (x + h, (x + h)² + 3(x + h)).
2. Top: (x + h)² + 3(x + h) − (x² + 3x) = x² + 2xh + h² + 3x + 3h − x² − 3x = 2xh + h² + 3h.
3. Gradient = (2xh + h² + 3h)/h = 2x + h + 3.
4. Let h → 0: gradient = 2x + 3.
5. At x = 5: 2(5) + 3 = 13.

A picture of the three ideas

• A chord joins two points → its gradient is the average rate of change.
• A tangent touches one point → its gradient is the instantaneous rate of change.
• Letting the chord shrink to the tangent is the limit.

Curves and their tangents connect to other graph work — see functions and graphs to revise plotting and shapes of curves.

Where rates of change are used

Every "per" is a rate of change. Speed is the rate of change of distance with time; acceleration is the rate of change of speed. In biology, rates describe how fast a population or an infection grows. In economics, marginal cost is the rate of change of total cost as you make one more item. Engineers use rates to control how fast a tank fills or a circuit charges. Calculus is the toolkit for all of them — and it starts with the simple gradient you already know.

Practice activity

1. Find the average rate of change of y = x² between x = 1 and x = 5.
2. A car's distance is d = 4t² metres. Find its average speed between t = 2 and t = 5 seconds.
3. Using the limit method, find the gradient of y = x² at x = 7.
4. Find the gradient of y = x² − 4x at a general point x (use h → 0).

Answers:

1. (25 − 1)/(5 − 1) = 24/4 = 6.
2. At t = 2, d = 16; at t = 5, d = 100. (100 − 16)/(5 − 2) = 84/3 = 28 m/s.
3. Gradient of x² is 2x, so at x = 7 it is 14.
4. Top = 2xh + h² − 4h, divide by h → 2x + h − 4, let h → 0 → 2x − 4.

Summary

Calculus measures change. The average rate of change between two points is the gradient of the chord: (y₂ − y₁)/(x₂ − x₁). The instantaneous rate of change at one point is the gradient of the tangent, found by letting a nearby point approach it — a limit. Algebraically, simplify the chord gradient ((x+h)... − ...)/h, cancel the h, then let h → 0. For y = x² this gives a gradient of 2x everywhere. These rates of change describe speed, growth, cost and slope throughout science and the real world.