Lesson 1: What Is Calculus? Rates of Change

(a) Opening Question

A drone rises straight up. Its height (in meters) after t seconds is given by s(t) = t² + 1.

1. How high is the drone at t = 2? At t = 4?
2. What is the drone's average speed over the time interval from t = 2 to t = 4 seconds?
3. Here is the hard one — and the question that calculus was invented to answer: how fast is the drone moving at the exact instant t = 2? Not over an interval, but at a single frozen moment.

Try parts 1 and 2 with arithmetic you already know (speed = distance ÷ time). For part 3, just write down your best guess and a sentence about why it's hard. You don't have the tool yet — building that tool is the whole point of this course. Hold onto your guess; we'll return to it.

(b) Core Concepts

Two ancient problems, one new idea

Calculus grew out of two questions that stumped mathematicians for two thousand years.

• The tangent problem: Given a curve, how do we find the slope of the line that just touches it at a single point? Slope means "rate of change," so this is really the question how fast is something changing at one instant?
• The area problem: How do we find the exact area of a region with curved boundaries?

These look unrelated, but the Fundamental Theorem of Calculus (Unit 6) reveals they are two sides of one coin. The single idea that cracks both is the limit: the value a quantity approaches as we zoom in closer and closer. This lesson is about the tangent problem, and it starts with something you already understand — slope.

Average rate of change = slope of a secant line

For a function f, the average rate of change on the interval [a, b] is the change in output divided by the change in input:

average rate of change = [f(b) − f(a)] / (b − a)

Geometrically, this is exactly the slope of the secant line — the straight line through the two points (a, f(a)) and (b, f(b)) on the graph. A secant line crosses the curve at two points. Average rate of change answers: "On average, how fast did the output change per unit of input across this whole interval?"

Worked computation. Let f(x) = 3x − x². Find the average rate of change on [0, 2].

f(0) = 3(0) − 0² = 0
f(2) = 3(2) − 2² = 6 − 4 = 2

avg rate of change = [f(2) − f(0)] / (2 − 0) = (2 − 0) / (2 − 0) = 1

So over [0, 2], the function rises by 1 unit of output per unit of input on average. The secant line through (0, 0) and (2, 2) has slope 1. Notice the word average: in between, the function might be climbing faster or slower than 1 — the secant slope smooths all of that into a single number.

Instantaneous rate of change = slope of a tangent line

The instantaneous rate of change of f at a point x = a is the slope of the tangent line — the line that touches the curve at just that one point and matches its direction there. This answers a sharper question: "How fast is the output changing at the exact instant x = a?"

This is the quantity our drone problem demanded and arithmetic couldn't give. Average rate of change needs two points. A tangent line touches at only one point — so the formula [f(b) − f(a)] / (b − a) collapses into the meaningless 0/0 if we naively set b = a. We need a smarter approach.

The difference quotient and the shrinking-secant idea

Here is the central insight of the lesson. Pick the point a where we want the instantaneous rate. Pick a second point a little distance h away, at a + h. The secant line through (a, f(a)) and (a + h, f(a + h)) has slope:

[f(a + h) − f(a)] / h

This expression is called the difference quotient. It is just the average-rate-of-change formula written with the interval width h instead of b − a.

Now shrink h toward 0. As the second point slides toward the first, the secant line pivots and gets closer and closer to the tangent line. The secant slopes close in on the tangent slope. The instantaneous rate of change is the limit of these average rates as the interval shrinks:

instantaneous rate at a  =  limit of  [f(a + h) − f(a)] / h  as h → 0

We write lim_{h→0} for "the limit as h approaches 0." We can't plug in h = 0 (that gives 0/0), but we can watch where the values head. That target value is the answer.

• Smooth upward-opening parabola through (0,0), (1,1), (2,4)
• Point P marked at (1, 1)
• Secant line 1 through P and (2, 4): slope 3 (steeper, drawn lighter)
• Secant line 2 through P and (1.5, 2.25): slope 2.5 (drawn lighter)
• Secant line 3 through P and (1.1, 1.21): slope 2.1 (drawn lighter, closer to tangent)
• Tangent line at P: slope 2, drawn bold/solid, just touching the curve at P
• Caption idea: as the second point slides down toward P, the secant slopes 3 → 2.5 → 2.1 → ... close in on the tangent slope 2]

Watching the limit numerically

Let's see the limit for f(x) = x² at x = 1. The difference quotient is [(1 + h)² − 1²] / h. Compute it for shrinking h:

The secant slopes march toward 2. So the instantaneous rate of change of x² at x = 1 — the tangent slope — appears to be 2. (In Unit 2 you'll prove it's exactly 2.) For now, the takeaway is the picture: a limit is the value a quantity homes in on, and the tangent slope is the limit of secant slopes.

Notice the pattern. No single secant slope in the table equals 2. We never reach h = 0. The limit isn't about arriving — it's about where you're headed. That distinction is the soul of calculus, and the next four lessons make it precise.

(c) Worked Examples

Example 1 — Average rate of change [NO CALC]

Problem. Find the average rate of change of g(x) = x² + 1 on the interval [1, 4].

Strategy. Average rate of change is the secant slope [g(b) − g(a)] / (b − a). Evaluate g at both endpoints, then divide.

Solution.

g(1) = 1² + 1 = 2
g(4) = 4² + 1 = 17

avg rate = [g(4) − g(1)] / (4 − 1) = (17 − 2) / 3 = 15 / 3 = 5

Justification. The average rate of change of g on [1, 4] is 5. This is the slope of the secant line through (1, 2) and (4, 17); on average the output increases 5 units per unit increase in x across this interval.

Example 2 — Average velocity vs. estimating instantaneous velocity [NO CALC]

Problem. A ball is thrown upward; its height in feet after t seconds is h(t) = −16t² + 64t. (a) Find the average velocity on [0, 2]. (b) Estimate the instantaneous velocity at t = 1 using the difference quotient with h = 0.01.

Strategy. Average velocity is the secant slope of the position function. For the instantaneous estimate, compute the difference quotient [h(1 + Δt) − h(1)] / Δt with a small Δt.

Solution.

(a)

h(0) = −16(0)² + 64(0) = 0
h(2) = −16(2)² + 64(2) = −64 + 128 = 64

avg velocity = [h(2) − h(0)] / (2 − 0) = (64 − 0) / 2 = 32 ft/sec

(b) With Δt = 0.01, evaluate h(1.01):

h(1.01) = −16(1.01)² + 64(1.01) = −16(1.0201) + 64.64 = −16.3216 + 64.64 = 48.3184
h(1)    = −16(1)² + 64(1) = −16 + 64 = 48

difference quotient = (48.3184 − 48) / 0.01 = 0.3184 / 0.01 = 31.84 ft/sec

Justification. The average velocity on [0, 2] is 32 ft/sec (a secant slope). The instantaneous velocity at t = 1 is approximately 31.84 ft/sec (a tangent slope, estimated by a narrow secant). The true value is 32 ft/sec; our estimate is close and would tighten as Δt → 0. Note these answer different questions — one summarizes a whole interval, the other captures a single instant.

Example 3 — Estimating an instantaneous rate from a table [CALC]

Problem. A bacteria population P(t), in thousands, is measured every hour:

Estimate the instantaneous rate of growth at t = 3 hours. Include units.

Strategy. We can't take a true limit from a table, but the best estimate of the tangent slope at t = 3 uses the symmetric interval centered on t = 3 — that is, [2, 4]. A secant straddling the point hugs the tangent better than a one-sided secant.

Solution.

estimate = [P(4) − P(2)] / (4 − 2) = (48 − 30) / 2 = 18 / 2 = 9 thousand bacteria per hour

Justification. Using the symmetric interval [2, 4] around t = 3, the average rate of change is 9 thousand bacteria per hour, which best estimates the instantaneous rate at t = 3. (For comparison, the forward secant on [3, 4] gives 10 and the backward secant on [2, 3] gives 8; the centered estimate, 9, sits between them and is more accurate.)

Example 4 — Secant slopes approaching a tangent slope [CALC]

Problem. For f(x) = x³, compute the difference quotient at x = 2 for h = 0.1 and h = 0.01, and state what the instantaneous rate of change at x = 2 appears to be.

Strategy. Compute [f(2 + h) − f(2)] / h for each h. A graphing calculator speeds the arithmetic; watch where the results converge.

Solution.

f(2) = 2³ = 8

h = 0.1:   [f(2.1) − f(2)] / 0.1 = (9.261 − 8) / 0.1 = 1.261 / 0.1 = 12.61
h = 0.01:  [f(2.01) − f(2)] / 0.01 = (8.120601 − 8) / 0.01 = 0.120601 / 0.01 = 12.0601

Calculator check.

TI-84: MATH → 8:nDeriv( → nDeriv(X³, X, 2) → 12
Means the instantaneous rate of change of f(x) = x³ at x = 2 is 12.

Justification. As h shrinks from 0.1 to 0.01, the secant slopes drop from 12.61 to 12.0601, closing in on 12. The instantaneous rate of change of f(x) = x³ at x = 2 appears to be 12 — the limit of the secant slopes as h → 0, which is the slope of the tangent line.

(d) Common Mistakes

Confusing average rate with instantaneous rate.

What students do: They report a secant slope over an interval when asked for the rate "at" a single instant. Why it's wrong: Average rate needs two points and summarizes an interval; instantaneous rate lives at one point. Fix: Read carefully. "On [a, b]" or "from a to b" → average (secant). "At x = a" or "at the instant" → instantaneous (tangent).

Flipping the difference quotient.

What students do: They write (b − a) / [f(b) − f(a)] — change in x over change in y. Why it's wrong: Slope is rise over run, output-change over input-change. Fix: Output (the f values) goes on top; input goes on the bottom. Always.

Plugging h = 0 into the difference quotient.

What students do: They set h = 0 immediately and get 0/0, then conclude the answer is 0 or undefined. Why it's wrong: 0/0 is indeterminate — it tells you nothing. Fix: You take a limit — watch where the values head as h shrinks, without ever setting h = 0.

Forgetting units (and the rate language) in context.

What students do: They answer "32" instead of "32 ft/sec." Why it's wrong: On the AP exam, a rate in context requires units, or you lose the point. Fix: Output units per input units — feet per second, thousands of bacteria per hour, dollars per item.

Using a one-sided table estimate when a symmetric one is available.

What students do: They estimate the rate at t = 3 using [3, 4] only. Why it's wrong: A centered (symmetric) secant approximates the tangent more accurately. Fix: When the table allows it, center the interval on the point — use [2, 4] for an estimate at t = 3.

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## (e) Practice Problems

The average rate of change of f on [a, b] is given by which expression?

- (A) [f(b) − f(a)] / (b − a)

- (B) (b − a) / [f(b) − f(a)]

- (C) [f(b) + f(a)] / (b + a)

- (D) f(b) − f(a)

Find the average rate of change of f(x) = x² on [2, 5].

- (A) 3

- (B) 7

- (C) 21

- (D) 10

Find the average rate of change of f(x) = √x on [1, 4].

- (A) 1

- (B) 1/3

- (C) 3

- (D) 2/3

A car's position is s(t) = t² + 1 meters at time t seconds. What is its average velocity on [2, 4]?

- (A) 4 m/s

- (B) 5 m/s

- (C) 6 m/s

- (D) 12 m/s

A line that intersects a curve at two points is called a:

- (A) tangent line

- (B) secant line

- (C) normal line

- (D) limit line

For f(x) = x², the difference quotient [(1 + h)² − 1] / h is computed for shrinking h: 2.5, 2.1, 2.01, 2.001. The instantaneous rate of change of f at x = 1 is best estimated as:

- (A) 0

- (B) 1

- (C) 2

- (D) undefined

A quantity Q(t) is sampled below. Estimate the instantaneous rate of change at t = 2 using the most accurate interval available.

| t | 0 | 1 | 2 | 3 | 4 |

|---|---|---|---|---|---|

| Q(t) | 5 | 9 | 11 | 17 | 27 |

- (A) 2

- (B) 4

- (C) 6

- (D) 8

The slope of the tangent line to a curve at a point represents the:

- (A) average rate of change over an interval

- (B) instantaneous rate of change at that point

- (C) area under the curve

- (D) total change across the domain

As h → 0 in the difference quotient [f(a + h) − f(a)] / h, the secant line through (a, f(a)) and (a + h, f(a + h)) approaches:

- (A) a horizontal line

- (B) the tangent line at x = a

- (C) the x-axis

- (D) a vertical line

Short answer. For g(x) = 3x − x², find the average rate of change on [1, 3]. Show the secant-slope computation.

Short answer. For f(x) = x³, fill in the difference quotient [f(1 + h) − f(1)] / h for h = 0.1 and h = 0.01, then state your best estimate of the instantaneous rate of change at x = 1.

Justification. A student computes the average rate of change of a function on [0, 4] to be 7, and claims "therefore the instantaneous rate of change at x = 2 is also 7." Is the student's reasoning valid? Explain.

Interpretation. The temperature T(t) of a cup of coffee, in °F, is recorded:

| t (min) | 0 | 5 | 10 | 15 | 20 |

|---|---|---|---|---|---|

| T(t) (°F) | 200 | 170 | 150 | 138 | 132 |

(a) Find the average rate of change of temperature on [0, 20], with units. (b) Estimate the instantaneous rate of change at t = 10 minutes, with units. (c) Interpret the sign of your answer to (b) in context.

Justification. Explain why we cannot find the instantaneous rate of change at x = a by simply substituting b = a into [f(b) − f(a)] / (b − a). What idea do we use instead?

A ball's height is h(t) = −16t² + 64t feet. Find the average velocity on [0, 2] and state whether the ball is, on average, rising or falling over this interval.

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