1.1 Change at an instant

Your car’s odometer measures distance; your speedometer measures speed. Over a two-hour trip of 120 km your average speed is $\frac{120}{2}=60$ km/h — but at any given moment the needle might read 0 (at a stop light) or 100 (on the highway). The speedometer reports an instantaneous rate: how fast the distance is changing at that instant. Calculus is the mathematics that makes “at that instant” precise, and it does so for every quantity that changes, not just distance.

This lesson assumes a little precalculus. You should be comfortable with function notation — evaluating $f(x)$ at a specific input, including expressions like $f(a+h)$, where you substitute the entire quantity $a+h$ in place of $x$. You should know the slope of a line as rise over run, $\dfrac{\Delta y}{\Delta x}$, and the average rate of change of a function over an interval as the slope of the line between two points on its graph. Finally, you need the algebra that simplifies a difference quotient: expanding $(a+h)^2=a^2+2ah+h^2$, combining like terms, and cancelling a common factor of $h$ from the top and bottom. The idea of a limit is introduced here, so you do not need it in advance — but the smoother your algebra, the more clearly the new calculus idea will stand out.

Start with what you already know. For a function $f$, the average rate of change on the interval from $x=a$ to $x=a+h$ is the change in output divided by the change in input: $\frac{f(a+h)-f(a)}{(a+h)-a}=\frac{f(a+h)-f(a)}{h}.$ Geometrically this is the slope of the secant line through the two points $\big(a,\,f(a)\big)$ and $\big(a+h,\,f(a+h)\big)$ on the graph. The letter $h$ is the width of the interval — the “run” between the two points.

Now shrink the interval. Make $h$ smaller and smaller, so the second point $Q$ slides down the curve toward the first point $P$. Each choice of $h$ gives a secant line with its own slope. As $Q$ approaches $P$, those secant lines pivot and settle toward one special line — the tangent line at $P$. The instantaneous rate of change at $x=a$ is the slope of that tangent line, and it is the value the average rates approach as $h\to 0$.

Putting the idea in symbols gives the definition that launches the entire course. The instantaneous rate of change of $f$ at $x=a$ — also called the derivative of $f$ at $a$, written $f'(a)$ — is $f'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h},$ provided this limit exists. The expression $\frac{f(a+h)-f(a)}{h}$ is the difference quotient: the slope of a secant. The limit turns a family of secant slopes into a single tangent slope.

Read the figure as a process, not a picture. Each dashed secant is a genuine average rate over a real interval you could measure. The tangent is the limit of that process. This is the pattern of every idea in Unit 1: take something you can compute on an interval, then ask what it approaches as the interval collapses to a point.

Notice three things as you slide the handle: the offset $h$ shrinks toward $0$, the secant line rotates until it lies along the tangent, and the “secant slope” readout closes in on $6$ while “secant − tangent” shrinks to $0$. You can also flip $Q$ to the left side: the secant slopes approach $6$ from that side too. That agreement from both sides is the numerical signature of a genuine instantaneous rate — and $6$ is the value the next worked example proves exactly.

Let $f(x)=x^2$ and find the instantaneous rate at $x=3$. First form the difference quotient at $a=3$: $\frac{f(3+h)-f(3)}{h}=\frac{(3+h)^2-3^2}{h}=\frac{9+6h+h^2-9}{h}=\frac{6h+h^2}{h}.$ For every $h\neq 0$ we may cancel the common factor of $h$: $\frac{6h+h^2}{h}=6+h.$ This says the average rate over $[3,\,3+h]$ is exactly $6+h$ — a genuinely useful fact on its own. Now take the limit: as $h\to 0$, $6+h\to 6$. Therefore $f'(3)=6$. The tangent to $y=x^2$ at $x=3$ has slope $6$.

The same limit describes velocity. If $s(t)$ is the position of an object at time $t$, then the average velocity over $[t,\,t+h]$ is $\frac{s(t+h)-s(t)}{h}$, and the instantaneous velocity is $s'(t)=\lim_{h\to0}\frac{s(t+h)-s(t)}{h}$. A speedometer is a physical limit machine. The very same structure will later give us acceleration (the rate of change of velocity), marginal cost in economics, reaction rate in chemistry, and current in a circuit — anywhere a quantity changes, its instantaneous rate is this limit.

When the difference quotient is hard to simplify — or when the function is given only as measured data — you estimate the instantaneous rate by computing average rates over shrinking intervals and watching the trend. For $f(x)=x^2$ near $x=3$ we found the average rate over $[3,\,3+h]$ is exactly $6+h$. Tabulating shrinking $h$: at $h=0.1$ it is $6.1$; at $h=0.01$, $6.01$; at $h=0.001$, $6.001$. The values march toward $6$. Approaching from the left, $h=-0.1$ gives $5.9$ and $h=-0.01$ gives $5.99$ — also heading to $6$. Both sides converging to the same number is the numerical signature of a well-defined instantaneous rate, and it is exactly how you handle a function known only through a table of points (a recurring AP skill developed in 1.4).

A rate always carries units: output units per input unit. For a position $s(t)$ in metres and time $t$ in seconds, $s'(t)$ is in metres per second — a velocity. For a cost $C(q)$ in dollars and a quantity $q$ in items, $C'(q)$ is dollars per item — a marginal cost. Attaching and interpreting units is not decoration: on the AP exam an instantaneous-rate answer stated without correct units, or without an interpretation in the context of the problem, loses credit even when the number is right.

Once you have the rate, you can write the tangent line itself. At $x=a$ the tangent passes through $\big(a,\,f(a)\big)$ with slope $f'(a)$, so its equation is $y-f(a)=f'(a)\,(x-a).$ For $f(x)=x^2$ at $a=3$: $f(3)=9$ and $f'(3)=6$, giving $y-9=6(x-3)$, i.e. $y=6x-9$. Near $x=3$ this line is the best straight-line approximation to the curve — the seed of the linearization technique you will use in Unit 4 to approximate hard values quickly.

Try each of these with the answer hidden first, then reveal to check your reasoning — the goal is to run the difference quotient yourself and take the limit cleanly.

The instantaneous rate lives in four representations at once, and fluency means moving between them. Symbolically it is the limit of the difference quotient, $f'(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}$. Graphically it is the slope of the tangent line — the limit of secant slopes as $Q\to P$ in Figure 1.1. Numerically it is the value a table of average rates converges to, as with $6+h\to 6$ for $x^2$ at $x=3$. Verbally, in context, it is a rate carrying units — a velocity in metres per second, a marginal cost in dollars per item, a flow in litres per minute. On a well-behaved function all four agree, and a strong habit is to compute in one representation and sanity-check in another: an algebraic slope of $6$ should look like a tangent rising steeply to the right, and a table near $x=3$ should be closing in on $6$. When two representations disagree, you have found a mistake to fix.

Answer these to confirm the core ideas before moving on. Each gives feedback; keep trying until you get it.