Describe the rate of change of a quantity, distinguishing the average rate of change over an interval (the gradient of the chord joining two points on the curve) from the instantaneous rate of change at a point (the gradient of the tangent there), estimate an instantaneous rate of change, read and interpret rates from a graph, and recognise that the gradient of the tangent at each point defines a new function, the derivative

What this dot point is asking

A rate of change measures how fast one quantity changes as another changes: how fast a tank fills, how fast a hot drink cools, how fast a car covers ground. You already know one rate intimately, the gradient of a straight line, rise over run. The whole of calculus grows out of one question: what does "rate of change" mean when the graph is a curve, where the steepness keeps changing from point to point? This dot point answers it in the two ways the exam tests.

The average rate of change over an interval is the steepness of the straight line joining the two endpoints of that interval on the curve, the chord. It is the single rate that, if held constant across the interval, would produce the same overall change. The instantaneous rate of change at one point is the steepness of the curve at that exact point, captured by the tangent there, the straight line that just grazes the curve. Average rate looks across an interval; instantaneous rate looks at a single instant.

This page is the gateway to differentiation. It builds the intuition (chord, tangent, estimating a gradient, reading rates off a graph, and the derivative as a new function) without yet using the formal limit or any differentiation rules, which come on the next pages. Everything here is geometry and gradients.

The answer

Average rate of change is the gradient of a chord

A chord (also called a secant) is the straight line joining two points on a curve. If a quantity has value $y_1$ at $x_1$ and value $y_2$ at $x_2$, the average rate of change of $y$ with respect to $x$ over that interval is the gradient of the chord:

This is exactly the gradient formula you already know, applied to the two endpoints. The units come straight from the quantities: litres per minute for a filling tank, degrees per minute for cooling, metres per second for distance over time. The average rate answers "over this whole stretch, how much did $y$ change for each unit of $x$, on average?" and a single chord captures it no matter how the curve wiggles in between.

Instantaneous rate of change is the gradient of a tangent

The instantaneous rate of change is the rate at one precise moment, not averaged over a stretch. On a graph it is the steepness of the curve at a single point, and the line that captures that steepness is the tangent: the straight line that touches the curve at that point and heads off in the exact direction the curve is going. Think of the beam from a car's headlights as it rounds a bend, always pointing along the curve at the instant.

A chord needs two points; a tangent lives at one. So the instantaneous rate is the gradient of the tangent at that point. The diagram below shows both on one curve, here a cooling temperature against time: the dashed chord through $A$ and $B$ gives the average rate of cooling between those two times, and the solid tangent at $P$ gives the instantaneous rate of cooling at that single instant. The chord cuts the curve at two points; the tangent touches it at one.

The sign and size of the instantaneous rate read off the tangent directly. A tangent that slopes up to the right means the quantity is increasing (positive rate); one that slopes down means it is decreasing (negative rate); a horizontal tangent means the rate is momentarily zero, neither rising nor falling. A steeper tangent means a faster rate. In the cooling curve above the tangent slopes downward, so the rate is negative: the drink is losing heat, and because the curve flattens as it approaches room temperature, the tangents get less steep and the cooling slows.

Estimating an instantaneous rate: shrink the chord onto the tangent

Drawing the perfect tangent by eye is hard, and reading its gradient off a graph is rough. There is a cleaner numerical idea hiding in the picture: a chord between two nearby points is almost the tangent. As you bring the two ends of a chord closer together around a point, the chord swings into line with the tangent at that point, so its gradient gets closer and closer to the instantaneous rate. This is the seed of the formal limit you meet on the next page; here we just use it to estimate.

The best estimate for the instantaneous rate at $x = a$ comes from a short symmetric chord, from $x = a - h$ to $x = a + h$, with $h$ small. Centring the chord on the point makes the over-shoot on one side cancel the under-shoot on the other, so a symmetric chord is a sharper estimate than a one-sided one of the same width. The four panels below show this for $y = x^2$, estimating the instantaneous rate at $x = 3$. The chord shrinks from half-width $h = 2$, to $h = 1$, to $h = 0.5$, and finally the two ends merge at the single point $P$ where the line is the tangent.

Stage 1, a wide symmetric chord. Take the chord from $x = 1$ to $x = 5$ (half-width $h = 2$, centred on $x = 3$). Its gradient is $\dfrac{5^2 - 1^2}{5 - 1} = \dfrac{24}{4} = 6$. Even this wide chord already gives $6$.

Stage 2, halve the half-width. Now take the chord from $x = 2$ to $x = 4$ (half-width $h = 1$). Its gradient is $\dfrac{4^2 - 2^2}{4 - 2} = \dfrac{12}{2} = 6$. The chord is hugging the curve more tightly and still reads $6$.

Stage 3, shrink it again. The chord from $x = 2.5$ to $x = 3.5$ (half-width $h = 0.5$) has gradient $\dfrac{3.5^2 - 2.5^2}{3.5 - 2.5} = \dfrac{6}{1} = 6$. The two points are nearly on top of each other and the reading holds at $6$.

Stage 4, the chord becomes the tangent. When the half-width shrinks to nothing the two points merge into the single point $P$ at $x = 3$, and the chord has become the tangent there. Its gradient is the instantaneous rate of change, $6$. The chord readings were heading to this all along.

For the parabola $y = x^2$ the symmetric chord gives exactly $6$ at every width, which is why all four readings agree. That is a happy feature of symmetric chords on a parabola, not a general guarantee: on most curves the estimate is very close but not exact, and it improves as $h$ shrinks. For a cooling curve, for instance, a symmetric chord of half-width $1$ minute typically lands within a fraction of a percent of the true instantaneous rate, which is plenty for an estimate.

Reading and interpreting a rate from a graph

Many exam questions give a graph, not a formula, and ask you to read a rate off it. Two skills cover almost all of them.

For an average rate, find the two relevant points on the curve, read their coordinates, and compute the gradient of the chord between them. For an instantaneous rate, find the point, draw (or imagine) the tangent there, and read its gradient as rise over run by counting grid squares. Always attach the units, which come from the axes: on a distance-time graph the gradient is a speed; on a volume-time graph it is a flow rate; on a temperature-time graph it is a rate of heating or cooling.

The distance-time graph below makes the headline case concrete. The curve gives distance $s$ against time $t$, and the tangent at the point $P$ (at $t = 4$ s) has been drawn. Its gradient, rise in metres over run in seconds, is the instantaneous speed at that moment. A chord across two times would instead give the average speed between them.

A subtlety worth holding onto: the average speed over an interval and the instantaneous speed at a point inside it are usually different numbers, because the steepness of the curve changes across the interval. They coincide only when the graph over that stretch is a straight line, that is, when the rate is constant. On a curving distance-time graph the average over the whole trip can be slower than a top-speed instant in the middle, exactly as a car's trip-average is slower than its fastest moment.

The derivative: collecting the gradient at every point

Here is the idea the rest of the calculus course is built on. At every point of a curve there is a tangent, and that tangent has a gradient, the instantaneous rate of change there. If you read off that gradient at point after point, you get a new function: input an $x$ value, output the gradient of the curve at that $x$. This new function is the derivative of the original, written $f'(x)$ (read "$f$ dashed of $x$").

So $f'(x)$ is not a single number; it is a function in its own right, with its own table of values and its own graph, manufactured by recording the gradient at each point of $y = f(x)$. The single instantaneous rate at one point $x = a$ is just one value of it, $f'(a)$. This is the payoff of the whole topic: instead of redrawing a tangent every time you want a rate, you can find the derivative once and read any gradient straight off it. (How to compute $f'(x)$ exactly, by the first-principles limit and then by rules, is the next two pages; here we only need the idea of what it is.)

You can sketch the broad shape of the derivative just by reading the sign of the gradient along the curve, with no algebra at all. Where the curve rises, the tangent slopes up, so $f'(x)$ is positive; where it falls, the tangent slopes down, so $f'(x)$ is negative; at a peak or a trough, the tangent is horizontal, so $f'(x) = 0$. The figure below marks these regions on a curve that rises, turns, falls, turns again and rises: positive, zero, negative, zero, positive.

The two points where the gradient is zero are the stationary points, where the curve is momentarily neither increasing nor decreasing. Year 11 only needs you to recognise that the instantaneous rate (the gradient of the tangent) is zero there; the full machinery of turning points, maxima, minima and concavity is Year 12 work. For now, "the rate of change is zero" and "the tangent is horizontal" are two ways of saying the same thing.

How exam questions ask about rates of change

The wording tells you whether a chord or a tangent is wanted:

• "Find the average rate of change of ... over the interval ..." or "... between $t = \ldots$ and $t = \ldots$" is a chord gradient: take the two endpoint values and compute $\dfrac{y_2 - y_1}{x_2 - x_1}$.
• "Average speed / average velocity" is the average rate of change of distance with time: total change in distance over total change in time. "Speed at the instant ..." or "velocity when $t = \ldots$" is the instantaneous rate, the tangent gradient.
• "Estimate the instantaneous rate of change at ..." wants a short-chord estimate: state the small interval you used and compute its gradient, ideally a symmetric chord centred on the point.
• "Find the gradient of the tangent at ..." or "the rate of change at the point ..." is the instantaneous rate: read the tangent's gradient (or compute a tight chord estimate from a graph).
• "From the graph, when is the quantity increasing / decreasing / stationary?" asks for the sign of the gradient: increasing where the tangent slopes up, decreasing where it slopes down, stationary where the tangent is horizontal.
• "When is the rate of change greatest?" points to the steepest part of the curve (the steepest tangent), not the highest point of the curve.
• "Sketch the rate of change against time" or "describe the derivative" wants the new function: positive where the original rises, zero at its stationary points, negative where it falls.
• Always attach units from the axes: a distance-time gradient is a speed (m/s), a volume-time gradient is a flow rate (L/min), a temperature-time gradient is degrees per minute.