Average and Instantaneous Rates of Change

📈 Average & Instantaneous Rates of Change

Part 1 of 7 — Average Rate of Change as a Secant Slope

Topics in This Part

🔑 Key Concept: The average rate of change (AROC) of a function over an interval is just the slope of the line connecting the two endpoints of the graph on that interval. If you can find a slope, you can find an average rate of change.

The Average Rate of Change Formula

For a function $f$ on the interval from $x = a$ to $x = b$:

$\text{AROC} = \frac{f(b) - f(a)}{b - a} = \frac{\Delta y}{\Delta x}$

This is "rise over run" — the change in output divided by the change in input.

Worked Example: $f(x) = x^2$ on $[1, 3]$

$\text{AROC} = \frac{f(3) - f(1)}{3 - 1} = \frac{9 - 1}{3 - 1} = \frac{8}{2} = 4$

So between $x = 1$ and $x = 3$, the function rises by $4$ units of $y$ for every $1$ unit of $x$, on average.

💡 The word average matters. Between $x=1$ and $x=3$ the curve isn't a straight line, so its steepness changes — but $4$ is the single slope of the straight secant line joining the two endpoints.

Secant Line = Average Rate

A secant line passes through two points on a curve. The slope of that secant line is the average rate of change.

So these three phrases all mean the same number:

• "average rate of change of $f$ on $[a, b]$"
• "slope of the secant line through $(a, f(a))$ and $(b, f(b))$"
• "$\dfrac{\Delta y}{\Delta x}$ over the interval"

⚠️ Order matters in pairs, not overall. You may compute $\dfrac{f(b)-f(a)}{b-a}$ or $\dfrac{f(a)-f(b)}{a-b}$ — both give the same answer. But you may not mix them: $\dfrac{f(b)-f(a)}{a-b}$ flips the sign and is wrong.

Concept Check 🎯

Compute the Average Rate 🧮

Find the average rate of change on the given interval. Fractions or decimals are fine.

1) $f(x) = x^2$ on $[2, 5]$ 2) $f(x) = x^2$ on $[-1, 4]$ 3) $h(x) = 3x - 7$ on $[10, 20]$ (it's linear!)

What You've Got

You can now turn any "average rate of change" question into a slope computation:

$\boxed{\;\text{AROC} = \frac{f(b) - f(a)}{b - a}\;}$

In Part 2 we attach units and meaning to that number, so an average rate of change becomes a real-world statement like "the car gained $15$ miles per hour each second."

📈 Average & Instantaneous Rates of Change

Part 2 of 7 — Units, Signs & Interpretation

🔑 The Idea: A rate of change isn't just a number — it carries units (output units per input unit) and a sign (increasing vs. decreasing). Reading both is the heart of AP Precalculus rate questions.

Units: Output per Input

The units of a rate of change are always:

$\frac{\text{units of } y}{\text{units of } x}$

Example

A tank holds $V(t)$ liters of water after $t$ minutes. If the average rate of change of $V$ on $[0, 4]$ is $-6$, then the tank loses water at an average of $6$ liters per minute over those $4$ minutes.

💡 To interpret a rate in a sentence, fill in: "$y$ changes by about [rate] [$y$-units] for each [$x$-unit]."

What the Sign Tells You

⚠️ Zero average rate does NOT mean "no movement." On $[-2, 2]$ for $f(x)=x^2$, AROC $=\frac{4-4}{2-(-2)}=0$ — yet the function dips down and comes back up. Average rate only compares the two endpoints.

Interpret the Rate 🔽

A balloon's height is $H(t)$ meters after $t$ seconds. Its average rate of change on $[0, 10]$ is $+4$.

Concept Check 🎯

📈 Average & Instantaneous Rates of Change

Part 3 of 7 — Rate at a Point from Shrinking Intervals

🔑 The Big Leap: The rate of change at a point (often called the instantaneous rate of change) describes how fast the output would change if the input changed right at that point. You estimate it with average rates of change over smaller and smaller intervals containing the point.

From Average to "At a Point"

An average rate of change needs two points. A rate at a point is about one point, so you sneak up on it with intervals that contain the point and keep getting narrower.

To estimate the rate of change of $f$ at $x = a$, compute the average rate of change on intervals like:

As the intervals shrink, the average rates settle on one value. That value is your estimate of the rate of change at $x = a$.

💡 Calculus later makes this idea exact. AP Precalculus asks you to estimate it from formulas, tables, and graphs, and to compare rates at different points.

Worked Example: $f(x) = x^2$ at $x = 2$

Shrinking From the Right

Shrinking From the Left

The right-hand values sit just above $4$ and the left-hand values just below it. The centered interval $[1.9, 2.1]$ gives $\dfrac{4.41 - 3.61}{0.2} = 4$.

So the rate of change of $x^2$ at $x = 2$ is about $4$: near $x = 2$, the output grows about $4$ units per unit of input.

Concept Check 🎯

Shrinking Intervals 🧮

Let $f(x) = x^2 + x$.

1) Average rate on $[3, 4]$ 2) Average rate on $[3, 3.1]$ 3) Your estimate of the rate of change at $x = 3$ (a whole number)

📈 Average & Instantaneous Rates of Change

Part 4 of 7 — Estimating the Rate at a Point from a Table

🔑 The Idea: On the AP exam you often get a table of values, not a formula. You can't shrink the interval past the data you have, so you estimate the rate of change at a point using the average rate over the smallest interval available around the point.

Three Ways to Estimate the Rate at $x = a$

💡 The symmetric (centered) difference uses one point on each side and is usually the best estimate, because it averages the slopes coming in and going out.

Worked Example

A function $P(t)$ (a population, in thousands) is sampled:

Estimate the rate of change of $P$ at $t = 4$.

Use the symmetric difference with the neighbors $t=2$ and $t=6$:

$\text{rate at } t = 4 \approx \frac{P(6) - P(2)}{6 - 2} = \frac{30 - 10}{4} = \frac{20}{4} = 5$

So the population is growing at about $5$ thousand people per year at $t = 4$.

⚠️ Use the points closest to $t=4$ that bracket it ($t=2$ and $t=6$). Don't reach all the way out to $t=8$ — that would average over too wide an interval and lose accuracy.

Estimate from the Table 🧮

Using the same table:

1) Symmetric-difference estimate of the rate at $t = 6$ (use $t=4$ and $t=8$) 2) Forward-difference estimate of the rate at $t = 2$ (use $t=2$ and $t=4$) 3) Backward-difference estimate of the rate at $t = 8$ (use $t=6$ and $t=8$)

Pick the Right Method 🔽

For each situation, choose the best difference method to estimate the rate of change at $x = a$.

📈 Average & Instantaneous Rates of Change

Part 5 of 7 — Comparing Rates at Two Points from a Graph

🔑 The Idea: On a graph, the rate of change at a point shows up as steepness there. To estimate it, read two points on the graph just to the left and right of the point and compute the slope between them. To compare rates at two points, do the same at each point.

Reading a Rate off a Graph

To estimate the rate of change at $x = a$ from a graph:

1. Read two points on the graph close to $a$, one on each side, ideally where the curve crosses gridlines.
2. Compute the slope $\dfrac{\Delta y}{\Delta x}$ between them. That is the average rate on a small interval containing $a$.
3. Repeat at the second point if you are comparing, using an interval of the same width.

Example

A graph of $f$ passes through these points:

$\text{rate at } x = 3 \approx \frac{4.9 - 3.1}{3.5 - 2.5} = 1.8 \qquad \text{rate at } x = 6 \approx \frac{7.4 - 7.0}{6.5 - 5.5} = 0.4$

Both rates are positive, so $f$ is increasing at both points, but it rises much faster near $x = 3$: about $4.5$ times as fast.

Steepness Tells You the Shape

⚠️ At a peak, average rates just to the left are positive and just to the right are negative, so a small interval centered on the peak gives a rate near $0$. That is why the rate of change is $0$ at the top of a thrown ball's path.

Concept Check 🎯

Compare Rates on a Graph 🔽

On a single curve, the graph is steep and rising near point $A$, shallow and rising near point $B$, and it peaks at point $C$.

📈 Average & Instantaneous Rates of Change

Part 6 of 7 — When the Rate Itself Is Changing

🔑 The Idea: A rate of change can increase or decrease as you move along the curve. AP Precalculus asks you to recognize when a function is changing at an increasing rate vs. a decreasing rate — and to connect that to the curve's shape.

Increasing Rate vs. Decreasing Rate

Look at how the average rates over successive equal intervals behave:

Example

Successive average rates: $\frac{2-1}{1}=1$, then $\frac{4-2}{1}=2$, then $\frac{8-4}{1}=4$. They grow ($1, 2, 4$), so $f$ is increasing and doing so at an increasing rate — the graph curves upward.

💡 "Increasing at a decreasing rate" still means the function is going up — just less and less steeply. Think of a car still speeding up, but easing off the gas.

Read the Behavior 🔽

A function's average rates of change over consecutive equal intervals are $12,\ 8,\ 5,\ 3$ — all positive but shrinking.

Concept Check 🎯

Track the Changing Rate 🧮

Compute the average rate of change on each consecutive interval.

1) AROC on $[0, 2]$ 2) AROC on $[2, 4]$ 3) AROC on $[4, 6]$

📈 Average & Instantaneous Rates of Change

Part 7 of 7 — Mixed Applications & Mastery Check

You can now (1) compute average rates of change as secant slopes, (2) interpret their units and signs, (3) estimate a rate at a point from shrinking intervals, (4) estimate it from tables and (5) compare rates on graphs, and (6) describe how a rate of change itself changes. Time to put it all together.

Quick Reference

⚠️ Don't confuse the two rates. Average rate uses a fixed interval (two points); a rate at a point is estimated from small intervals around one point. And a zero average rate only means the endpoints match — not that nothing happened in between.

Mixed Practice 🎯

Applied Computation 🧮

The position of a particle is $s(t) = t^2$ meters at time $t$ seconds.

1) Average velocity on $[1, 3]$ (in m/s) 2) Estimate the velocity at $t = 1$ using $[0.9, 1.1]$ 3) Estimate the velocity at $t = 5$ using $[4.9, 5.1]$

Exit Quiz ✅

Answer all three to finish the lesson.